Mechanics - User contributions [en]

D7. Examples of 3D dynamics

2026-03-18T18:26:05Z

Eantem: /* ✏️ EXAMPLE D7.4: rotating ball */

<div class="noautonum">__TOC__</div>
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<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\textrm{E}}(\textrm{J}) = {\vvec}_{\textrm{E}}(\textrm{G})+\vec{\Omega}_{\textrm{E}}^\text{ring} \times \vec{\textrm{GJ}} = (\odot \textrm{L} \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>

:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math> \begin{Bmatrix}{\vec{\Hs}_{\text{RTO}}(\Os)}\end{Bmatrix}_{\text{B}} = \left(\frac{1}{2}\ms\Rs^2\mat{2}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\vector{-\ms\Rs^2\dot\varphi_0}{0}{m((\Rs^2/2)+\Ls^2)\dot\psi}
</math>

:<math>\begin{Bmatrix}{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)}\end{Bmatrix}_{\text{B}} = \vector{0}{0}{m((\Rs^2/2)+\Ls^2)\ddot\psi} + \vector{0}{0}{\dot\psi}\times\vector{-\ms\Rs^2\dot\varphi_0}{0}{m((\Rs^2/2)+\Ls^2)\dot\psi} = \vector{0}{-\ms\Rs^2\dot\psi\dot\varphi_0}{m((\Rs^2/2)+\Ls^2)\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\
\vector{\mu\Ns\Rs+\Ms_1}{\text{mgL}-\text{NL}}{\mu\text{NL}} = \vector{-\ms\Rs^2\dot\psi\dot\varphi_0}{-\ms\Rs^2\dot\psi\dot\varphi_0}{m((\Rs^2/2)+\Ls^2)\ddot\psi}</math>

:The second component yields <math>\Ns = \ms\gs+\ms\frac{\Rs^2}{\Ls}\dot\psi\dot\varphi_0</math>. Substitution of that value into the third component yields the equation of motion:

:<center><math>\boxed{\ddot\psi = \frac{2\mu}{\Rs^2+2\Ls^2}(\gs\Ls+\Rs^2\dot\psi\dot\varphi_0)}</math></center>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{2\mu gL}{\Rs^2+2\Ls^2}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\text{R}/\text{L})\dot\varphi_0</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = [\otimes (R \dot{\varphi_0}- \text{L}\dot{\psi})] = 0 ⇒ \dot{\psi} = \frac{\Rs}{\Ls}\dot{\varphi_0} </math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint force appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\Rs=\mu\ms\Rs^2\left(\frac{\gs}{\Rs} + \frac{\Rs}{\Ls}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:In option 2, the equation of movement and the motor torque cannot be obtained from just one theorem. For that reason, it will be disregarded.

====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-03-18T18:25:45Z

Eantem: /* ✏️ EXAMPLE D7.4: rotating ball */

D7. Examples of 3D dynamics

2026-03-18T18:24:23Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
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\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
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</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\textrm{E}}(\textrm{J}) = {\vvec}_{\textrm{E}}(\textrm{G})+\vec{\Omega}_{\textrm{E}}^\text{ring} \times \vec{\textrm{GJ}} = (\odot \textrm{L} \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>

:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math> \begin{Bmatrix}{\vec{\Hs}_{\text{RTO}}(\Os)}\end{Bmatrix}_{\text{B}} = \left(\frac{1}{2}\ms\Rs^2\mat{2}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\vector{-\ms\Rs^2\dot\varphi_0}{0}{m((\Rs^2/2)+\Ls^2)\dot\psi}
</math>

:<math>\begin{Bmatrix}{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)}\end{Bmatrix}_{\text{B}} = \vector{0}{0}{m((\Rs^2/2)+\Ls^2)\ddot\psi} + \vector{0}{0}{\dot\psi}\times\vector{-\ms\Rs^2\dot\varphi_0}{0}{m((\Rs^2/2)+\Ls^2)\dot\psi} = \vector{0}{-\ms\Rs^2\dot\psi\dot\varphi_0}{m((\Rs^2/2)+\Ls^2)\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\
\vector{\mu\Ns\Rs+\Ms_1}{\text{mgL}-\text{NL}}{\mu\text{NL}} = \vector{-\ms\Rs^2\dot\psi\dot\varphi_0}{-\ms\Rs^2\dot\psi\dot\varphi_0}{m((\Rs^2/2)+\Ls^2)\ddot\psi}</math>

:The second component yields <math>\Ns = \ms\gs+\ms\frac{\Rs^2}{\Ls}\dot\psi\dot\varphi_0</math>. Substitution of that value into the third component yields the equation of motion:

:<center><math>\boxed{\ddot\psi = \frac{2\mu}{\Rs^2+2\Ls^2}(\gs\Ls+\Rs^2\dot\psi\dot\varphi_0)}</math></center>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{2\mu gL}{\Rs^2+2\Ls^2}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\text{R}/\text{L})\dot\varphi_0</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = [\otimes (R \dot{\varphi_0}- \text{L}\dot{\psi})] = 0 ⇒ \dot{\psi} = \frac{\Rs}{\Ls}\dot{\varphi_0} </math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint force appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\Rs=\mu\ms\Rs^2\left(\frac{\gs}{\Rs} + \frac{\Rs}{\Ls}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:In option 2, the equation of movement and the motor torque cannot be obtained from just one theorem. For that reason, it will be disregarded.

====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
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D7. Examples of 3D dynamics

2026-03-13T15:38:24Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
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\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\textrm{E}}(\textrm{J}) = {\vvec}_{\textrm{E}}(\textrm{G})+\vec{\Omega}_{\textrm{E}}^\text{ring} \times \vec{\textrm{GJ}} = (\odot \textrm{L} \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
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''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-03-13T15:37:38Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
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\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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\newcommand{\vec}[1]{\overline{#1}}
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</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\textrm{E}}(\textrm{J}) = {\vvec}_{\textrm{E}}(\textrm{G})+\vec{\Omega}_{\textrm{E}}^\text{ring} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-03-13T15:37:22Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
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</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\textrm{E}}(\textrm{J}) = {\vvec}_{\textrm{E}}(\textrm{G})+\vec{\Omega}_{\Es}^\text{ring} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-03-13T15:36:56Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
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\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
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\newcommand{\ts}{\textrm{t}}
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\newcommand{\xs}{\textsf{x}}
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\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
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\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
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\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
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\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
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\begin{Bmatrix}
{#1}\\
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\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
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\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\textrm{E}}(\textrm{J}) = {\vvec}_{\textrm{E}}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{ring} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-03-13T15:36:20Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
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\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\mat}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
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\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
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\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
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\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
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\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\textrm{E}}(J) = {\vvec}_{\textrm{E}}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{ring} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-03-13T15:35:35Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\mat}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
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\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Es}(J) = {\vvec}_{\Es}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{ring} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-03-13T15:33:57Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\mat}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{ring} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

C2. Movement of a mechanical system

2026-03-03T19:53:20Z

Eantem: /* 3. Find the velocity and the acceleration of point G of the plate relative to the ground. . */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\boldsymbol\alpha}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\is}{\textrm{i}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textrm{P}}
\newcommand{\Es}{\textrm{E}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\OPvec}{\vec{\Os\Ps}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\dth}{\dot\theta}
\newcommand{\ddth}{\ddot\theta}
\newcommand{\sth}{\sin{\theta}}
\newcommand{\cth}{\cos{\theta}}
\newcommand{\spsi}{\sin{\psi}}
\newcommand{\cpsi}{\cos{\psi}}
\definecolor{blau}{RGB}{39, 127, 255}
\definecolor{verd}{RGB}{9, 131, 9}
</math>
==C2.1 Velocity of a particle==

The '''velocity of a particle <math>\Qs</math>''' (or a point that belongs to a rigid body) '''relative to a reference frame R''', <math>\vvec_{\Rs}(\Qs)</math>, is the rate of change of its position vector with time. Mathematically, it is the time derivative of a position vector (relative to R). The time derivative of two different position vectors (<math>\overline{\Or\Qs}</math>, <math>\overline{\Os'_\Rs\Qs}</math> ) yield the same velocity because points <math>\Os_\Rs</math> and <math>\Os'_\Rs</math> are mutually fixed and fixed to the reference frame, hence <math>\overline{\Os_\Rs\Os'_\Rs}</math> is constant in R:

<center>
<math>
\vvec_\Rs(\Qs) = \dert{\vec{\Os_{\Rs}\Qs}}{R} =
\dert{\vec{\Os_{\Rs}\Os_{\Rs}'}}{R} + \dert{\vec{\Os_{\Rs}'\Qs}}{R} =
\dert{\vec{\Os_{\Rs}'\Qs}}{R}
</math>
</center>

One must bear in mind that the time derivative of a vector depends on the reference frame where it is being calculated. For that reason, there is a subscript R in the preceding equations which reminds of that dependency.

The [[Vector calculus #V.2 Operations between vectors with geometric representation|'''time derivative of a vector relative to a reference frame R ''']] assesses the evolution of the characteristics of that vector (direction and value) between two close time instants, separated by a time differential. Hence, the velocity <math>\vvec_\Rs(\Qs)</math> is nonzero whenever the value of the position vector, or its direction, or both change.

====✏️ EXAMPLE C2-1.1: rotating platform====
---------

::The platform (RP) rotates about an axis perpendicular to the ground (R). The movement of a point <math>\Qs</math> on the platform periphery depends on whether it is observed from the ground or from the platform.

[[File:C2-Ex1-1-1-neut.png|thumb|center|350px|link=]]

::The center of the platform (<math>\Os</math>) is fixed to both reference frames. Hence, <math>\vec{\Os\Qs}</math> is a position vector for point <math>\Qs</math> both in R and RP. It is evident that <math>\vvec_\Rs(\Qs)\neq \vec{0}</math> i <math>\vvec_{\Rs\Ps}(\Qs)= \vec{0}</math>, though the vector whose time derivative is being calculated is the same.

::As <math>\abs{\OQvec}</math> is the platform radius r, its value is constant. Hence, the time derivative of <math>\abs{\OQvec}</math> can only be associated with a change of direction.

::To assess the change of orientation of <math>\abs{\OQvec}</math> relative to the ground or to the platform, we have to define an angle between a straight line fixed in the reference frame (“departure” line) and vector <math>\OQvec</math> (“arrival” line). For the sake of clarity, we have represented the “departure” line as the direction of the arm of an observer located in the reference frame (thus not moving relative to it).

:[[File:C2-Ex1-1-2-neut.png|thumb|center|450px|link=]]

<center>
<math>\psi(t)\neq\psi(t+dt) \implies \OQvec</math> changes its direction relative to R <math>\implies \textcolor{blau}{\vvec_\Rs(}\Qs\textcolor{blau}{) \neq \vec{0}}</math>
</center>

::As seen in [[Vector calculus#V.1 Geometric representation of a vector|'''section V.1''']], <math>\textcolor{blau}{\vvec_\Rs(}\Qs\textcolor{blau}{)}</math> is perpendicular to <math>\OQvec</math>, and its value is that of <math>\OQvec\:(\textrm{r})</math> times the rate of change of orientation of <math>\OQvec</math> relative to R <math>(\dot{\psi})</math>:

[[File:C2-Ex1-1-3-neut.png|thumb|center|450px|link=]]

::Velocity of <math>\Qs</math> relative to the platform (RP):

[[File:C2-Ex1-1-4-neut.png|thumb|center|450px|link=]]

<center>
<math>\psi(t)=\psi(t+dt) \implies \OQvec</math> does not change its direction relative to RP <math>\textcolor{verd}{\implies \vvec_\Rs(}\Qs\textcolor{verd}{) = \vec{0}}</math>
</center>

<div>
=====Analytical calculation ➕=====

::The two logical vector bases for the calculation are:

:::* Basis B (1,2,3) fixed in R (thus moving in RP): <math>\velang{B}{R}=\vec{0},\velang{B}{RP}= \vec{\dot{\psi}}</math>
:::* Basis B' (1',2',3') fixed in RP (thus moving in R): <math>\velang{B'}{RP}=\vec{0},\velang{B'}{R} = -\vec{\dot{\psi}}</math>

[[File:C2-Ex1-1-5-neut.png|thumb|200px|right|link=]]
::Projection of the position vector <math>\OQvec</math> on both bases:

<center>
<math>\braq{\OQvec}{B}=\vector{r\cpsi}{r\spsi}{0}, \: \: \braq{\OQvec}{B'}=\vector{r}{0}{0}</math>
</center>

::Velocity of <math>\Qs</math> relative to R:

<center>
<math>
\braq{\vvec_\Rs(\Qs)}{B} = \braq{\dert{\OQvec}{R}}{B}=\frac{\ds}{\ds\ts}\braq{\OQvec}{B} = \vector{-r\dot \psi \spsi}{r\dot{\psi} \cpsi}{0}
</math>

<math>
\braq{\vvec_\Rs(\Qs)}{B'}=\braq{\dert{\OQvec}{R}}{B'} = \frac{\ds}{\ds\ts}\braq{\OQvec}{B'}+\braq{\velang{B'}{R}\times \OQvec}{B'}=\braq{\velang{B'}{R}\times \OQvec}{B'}=\vector{0}{0}{\dot\psi} \times \vector{r}{0}{0}= \vector{0}{r\dot\psi}{0}
</math>
</center>

::Velocity of <math>\Qs</math> relative to RP:
<center>
<math>
\braq{\vvec_{\Rs\Ps}(\Qs)}{B} =\braq{\dert{\OQvec}{RP}}{B}= \frac{\ds}{\ds\ts}\braq{\OQvec}{B}+\braq{\velang{B}{RP}\times \OQvec}{B}=\vector{-r\dot\psi \spsi}{r\dot\psi \cpsi}{0}+ \vector{0}{0}{-\dot\psi}\times\vector{r\cpsi}{r\spsi}{0}= \vector{0}{0}{0}
</math>
<math>
\braq{\vvec_{\Rs\Ps}(\Qs)}{B'} =\braq{\dert{\OQvec}{RP}}{B'}= \frac{\ds}{\ds\ts}\braq{\OQvec}{B'}+\braq{\velang{B'}{RP}\times \OQvec}{B'}=\frac{\ds}{\ds\ts}\braq{\OQvec}{B'} = \vector{0}{0}{0}
</math>
</center>
</div>

====✏️ EXAMPLE C2-1.2: Euler pendulum====
------------

::The endpoint <math>\Qs</math> of [[C1. Configuration of a mechanical system #✏️ EXAMPLE C1-5.4: Euler pendulum|'''Euler pendulum''']] describes a circular motion relative to the block. The corresponding velocity <math>\vel{Q}{BL} = \dert{\vecbf{CQ}}{BL}</math> can be obtained in a similar way as that used in the previous example.

[[File:C2-Ex1-2-1-neut.png|thumb|center|300px|link=]]

::The angle <math>\psi</math> orientates the bar both relative to the block and the ground, as its origin (vertical line) has a constant orientation in both reference frames
::The velocity of <math>\Qs</math> relative to the ground can be obtained as the time derivative of vector <math>\vec{\Or\Qs} (=\vec{\Or\Cbf}+\vecbf{CQ})</math> relative to the ground:

<center>
<math>
\vel{Q}{R} = \dert{\vec{\Or\Qs}}{R} = \dert{\vec{\Or\Cbf}}{R}+ \dert{\vec{\Cbf\Qs}}{R}
</math>
</center>

::Vector <math>\vec{\Or\Cbf}</math> has a constant direction in R but a variable value. Hence, its time derivative is parallel to <math>\vec{\Or\Cbf}</math> with value <math>\dot x</math>. Vector <math>\vec{\Cbf\Qs}</math>, however, has a constant value (L) but variable direction. Consequently, its time derivative is perpendicular to <math>\vec{\Cbf\Qs}</math>, and its value is that of<math>\vec{\Cbf\Qs}</math> times the rate of change of orientation of <math>\vec{\Cbf\Qs}</math> relative to R (<math>\dot\psi</math>):

[[File:C2-Ex1-2-2-neut-new.png|thumb|center|300px|link=]]

::The <math>\vel{Q}{R}</math> direction is not any of the directions associated with the system (it is not vertical, not horizontal, not parallel to the bar, not perpendicular to the bar). For that reason, it is better to represent it as the addition of the terms <math>\dot x</math> and <math>L\dot\psi</math>, whose directions do correspond to one of those singular directions.

::The first term of the expression <math>\vel{Q}{R} = \dert{\vec{\Or\Cbf}}{R}+\dert{\vecbf{CQ}}{R}</math> corresponds to the velocity of <math>\Cs</math> relative to the ground <math>\left(\vel{Q}{R} = \dert{\vec{\Or\Cbf}}{R} \right)</math>, whereas the second one has no physical interpretation: point <math>\Cs</math> is not fixed in R, thus it is not a position vector in that reference frame.

<div>
=====Analytical calculation ➕=====
::The two logical vector bases for the calculation are:

[[File:C2-Ex1-2-3-neut.png|thumb|right|200px|link=]]

:::* Basis B (1,2,3) fixed relative to R and BL <math>\Omegavec_\Rs^\Bs=\vec{0},\Omegavec_{\Bs\Ls}^\Bs = \vec{0}</math>
:::* Basis B' (1',2',3') fixed relative to the bar, thus moving in R and BL: <math>\velang{P}{B'}=\vec{0}</math>, <math>\velang{RL}{B'} = -\vec{\dot{\psi}}</math>

::Projection of the position vector <math>\OQvec</math> in both bases:

::<math>\braq{\OQvec}{B} = \vector{x+\Ls \spsi}{-\Ls \cpsi}{0}\:\:\:\:\:\:\:\:\:\: \braq{\OQvec}{B'} = \vector{\Ls+ x \spsi}{x\cpsi}{0}</math>

::Velocity of <math>\Qs</math> relative to R:
<center>
<math>\braq{\vel{Q}{R}}{B} = \braq{\dert{\OQvec}{R}}{B} = \frac{\ds}{\ds\ts}\braq{\OQvec}{B} = \vector{\dot x+\Ls\dot\psi \cpsi}{\Ls\dot\psi \spsi}{0}</math>
<math>\braq{\vel{Q}{R}}{B'} = \frac{\ds}{\ds\ts}\braq{\OQvec}{B'} + \braq{\velang{B'}{R} \times \OQvec}{B'} = \vector{\dot x \spsi+ x\dot\psi \cpsi}{\dot x \cpsi - x \dot\psi sin \psi}{0} + \vector{0}{0}{\dot\psi}\times \vector{\Ls+ x \spsi}{ x \cpsi}{0}=\vector{\dot x sin \psi}{\dot x \cpsi + \Ls\dot\psi}{0}
</math>
</center>
::If we want to calculate the velocity of <math>\Qs</math> relative to BL, the position vector to be differentiated is <math>\vecbf{CQ}</math>:
<center><math>
\braq{\vecbf{CQ}}{B} = \vector{\Ls sin \psi}{-\Ls cos \psi}{0}, \:\:\:\:\:\:\:\:\:\: \braq{\vecbf{CQ}}{B'}=\vector{\Ls}{0}{0}
</math></center>
</div>

---------
---------

==C2.2 Acceleration of a particle==

The '''acceleration of a particle ''' <math>\Qs</math> (or of a point belonging to a rigid body) '''relative to a reference frame R''', <math>\acc{Q}{R}</math>, is the rate of change of its velocity with time:
<center>
<math>\acc{Q}{R} = \dert{\vel{Q}{R}}{R}</math>
</center>

====✏️ EXAMPLE C2-2.1: rotating platform====
------


::In the circular motion of point <math>\Qs</math> of the [[C2. Movement of a mechanical system #✏️ EXAMPLE C2-1.1: rotating platform|'''platform relative to the ground ''']], the acceleration <math>\acc{Q}{R}</math> comes both from the change of value and the change of orientation of <math>\vel{Q}{R}</math>. As <math>\vel{Q}{R}</math> is always perpendicular to <math>\OQvec</math>, its rate of change of orientation is <math>\dot\psi</math>, the same as that of <math>\OQvec</math> :
[[File:C2-Ex2-1-eng.png|thumb|center|450px|link=]]

::The <math>\acc{Q}{R}</math> direction is not any of the directions associated to the system (not the radial direction, not that perpendicular to the radius). For that reason, it is better to represent it as the addition of the two terms <math>\rs\ddot\psi</math> and <math>r\dot\psi^2</math> , whose directions do correspond to one of those singular directions.

<div>
=====Analytical calculation ➕=====
::The vector bases B and B’ are the same as in [[C2. Movement of a mechanical system #✏️ EXAMPLE C2-1.1: rotating platform|'''example C2-1.1''']].
<center><math>
\braq{\acc{Q}{R}}{B} = \braq{\dert{\vel{Q}{R}}{R}}{B} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{R}}{B} = \vector{-\rs \ddot\psi \spsi - \rs \dot\psi^2\cpsi}{\rs\ddot\psi \cpsi - \rs \dot\psi^2\spsi}{0}
</math>
<math>
\braq{\acc{Q}{R}}{B'} = \braq{\dert{\vel{Q}{R}}{R}}{B'} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{R}}{B'} + \braq{\velang{B'}{R} \times \vel{Q}{R}}{B} = \vector{0}{\rs\ddot\psi}{0} + \vector{0}{0}{\dot\psi} \times \vector{0}{\rs\dot\psi}{0} = \vector{-\rs\dot\psi^2}{\rs\ddot\psi}{0}
</math>
</center>
</div>

====✏️ EXAMPLE C2-2.2: Euler pendulum====
------

::The calculation of the acceleration of <math>\Qs</math> relative to the ground (R) is laborious because the velocity <math>\vel{Q}{R}</math> comes from the addition of two terms:
::<math>\vel{Q}{R} = \dert{\vec{\Os_\Rs\Cbf}}{R} + \dert{\vecbf{CQ}}{R}</math>.

:::* <math>\dert{\vec{\Os_\Rs\Cbf}}{R}</math>: constant direction (horizontal), variable value <math>(\dot x)</math>. Thus, its time derivative <math>\ddert{\vec{\Os_\Rs\Cbf}}{R}</math> is horizontal with value <math>\ddot x</math>.
:::* <math>\dert{\vecbf{CQ}}{R}</math>: direction perpendicular to the bar, thus variable; variable value <math>\Ls\dot\psi</math>. Thus, its time derivative <math>\ddert{\vecbf{CQ}}{R}</math> has a component perpendicular to <math>\dert{\vecbf{CQ}}{R}</math> (parallel to the bar) with value <math>\Ls\dot\psi\cdot\dot\psi</math> , and another one parallel to <math>\dert{\vecbf{CQ}}{R}</math> (perpendicular to the bar) with value <math>\Ls\ddot\psi</math>.

[[File:C2-Ex2-2-neut.png|thumb|center|300px|link=]]
<div>
=====Analytical calculation ➕=====
::The vector bases B and B’ are the same as in [[C2. Movement of a mechanical system #✏️ EXAMPLE C2-1.2: Euler pendulum|'''example C2-1.2''']].

[[File:C2-Ex1-2-3-neut.png|thumb|right|200px|link=]]

::Acceleration of <math>\Qs</math> relative to BL:
<center>
<math>\braq{\acc{Q}{BL}}{B} = \braq{\dert{\vel{Q}{BL}}{BL}}{B} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{BL}}{B} = \vector{\Ls\ddot\psi \cpsi-\Ls \dot\psi^2 \spsi}{\Ls \ddot\psi \spsi +\Ls \dot\psi^2 \cpsi}{0}</math>

<math>\braq{\acc{Q}{BL}}{B'} = \braq{\dert{\vel{Q}{BL}}{BL}}{B'} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{BL}}{B'}+ \braq{\velang{B'}{BL}\times \OQvec}{B} = \vector{0}{\Ls\ddot\psi}{0} + \vector{0}{0}{\dot\psi}\times \vector{0}{\Ls\dot\psi}{0} = \vector{-\Ls\dot\psi^2}{\Ls\ddot\psi}{0}</math>
</center>
::Acceleration of <math>\Qs</math> relative to R:
<center>
<math>
\braq{\acc{Q}{R}}{B}=\braq{\dert{\vel{Q}{R}}{R}}{B} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{R}}{B}=\vector{\ddot x+\Ls\ddot\psi \cpsi-\Ls\dot\psi^2\spsi}{\Ls\ddot\psi \spsi+\Ls\dot\psi^2\cpsi}{0}
</math>
<math>
\braq{\acc{Q}{R}}{B'}=\braq{\dert{\vel{Q}{R}}{R}}{B'} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{R}}{B'}+\braq{\velang{B'}{R}\times \OQvec}{B}=\vector{\ddot x \spsi+\dot x\dot\psi \cpsi}{\ddot x \cpsi-\dot x\dot\psi \spsi +\Ls\ddot\psi}{0}+\vector{0}{0}{\dot\psi}\times\vector{\dot x \spsi}{\dot x \cpsi+\Ls\dot\psi}{0}=\vector{\ddot x \spsi - \Ls \dot\psi^2}{\ddot x \cpsi+\Ls\ddot\psi}{0}
</math>
</center>
</div>

-----------
-----------

==C2.3 Intrinsic directions. Intrinsic components of the acceleration==
A simple drawing shows that the velocity of a point <math>\Qs</math> relative to a reference frame R is always tangent to the trajectory it describes in R ('''Figure C2.1'''). Its direction is the '''tangential direction'''.
[[File:C2-1-eng.png|thumb|center|375px|link=|]]
<center>'''Figure C2.1''' The velocity vector is always tangent to the trajectory</center>

In a general case, the velocity <math>\vel{Q}{R}</math> changes both its value and its direction. Hence, the acceleration <math>\acc{Q}{R}</math> has two components, one associated with the change of value (parallel to <math>\vel{Q}{R}</math>) and another one associated with the change of direction (orthogonal to <math>\vel{Q}{R}</math>). Those components are the '''intrinsic components of the acceleration''', and they are called '''tangential component''' <math>\accs{Q}{R}</math> and '''normal component''' <math>\accn{Q}{R}</math>, respectively:

<center>
<math>\acc{Q}{R}=\accs{Q}{R}+\accn{Q}{R}</math>
</center>

In a [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-2.1: rotating platform|'''circular motion''']], the tangential component is perpendicular to the radius, and the normal one is parallel to the radius and pointing to the center of the trajectory ('''Figure C2.2'''):
[[File:C2-2-neut.png|thumb|center|275px|link=]]
<center>'''Figure C2.2''' Intrinsic components of the acceleration in a circular motion</center>
That result may be used locally for any other movement. Indeed, as the calculation of the velocity of a point <math>\Qs</math> with respect to a reference frame R (<math>\vel{Q}{R}</math>) calls for two consecutive position vectors (or, what is the same, two consecutive points of the trajectory), that of the acceleration <math>\acc{Q}{R}</math> calls for three:
<center>
<math>\acc{Q}{R}=\dert{\vel{Q}{R}}{R}\simeq\frac{\vvec_\Rs(\textbf{Q},\textrm{t+dt})-\vvec_\Rs(\textbf{Q},\textrm t)}{\Delta t(\rightarrow0)}\equiv\frac{\Delta\vvec_\Rs(\textbf{Q},\textrm t)}{\Delta t(\rightarrow0)}</math>
</center>

The calculation of vector <math>\Delta\vvec_\Rs(\textbf{Q},\textrm t)</math> calls for three consecutive points of the trajectory (two for each velocity, where the last point to calculate <math>\vvec_\Rs(\textbf{Q},\textrm t)</math> and the first point to calculate <math>\vvec_\Rs(\textbf{Q},\textrm{t+dt})</math> are the same). These three points define a plane ('''osculating plane'''), and there is just one circle containing the three of them. That is: any trajectory may be approximated locally by a circle ('''osculating circle'''). The center and the radius of that circle are the '''center of curvature''' and the '''radius of curvature''' of the trajectory of Q relative R (<math>\textrm{CC}_\textrm{R}(\textbf{Q})</math> and <math>\Re_\textrm{R}(\textbf Q)</math> respectively). The results obtained for the [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-2.1: rotating platform|'''circular motion''']] may be used locally to calculate <math>\Re_\textrm{R}(\textbf Q)</math> ('''Figure C2.3''').

[[File:C2-3-eng.png|thumb|center|500px|link=]]

<center>'''Figure C2.3''' local geometry of the trajectory of a particle Q relative to a reference frame R</center>
Both the radius of curvature and the position of the center of curvature change along the trajectory in general. In rectilinear spans, as there is no change in the velocity direction, the normal component of the acceleration is zero, and the radius of curvature becomes infinite.

The '''tangential unit vector ''' <math>\vecbf{s}</math> (<math>\vecbf{s}=\velo{R}/|\velo{R}|=\accso{R}/|\accso{R}|</math>) and the '''normal unit vector ''' <math>\vecbf{n}</math> (<math>\vecbf{n}=\accno{R}/|\accno{R}|</math>) may be completed with a third unit vector <math>\vecbf{b}</math> orthogonal to the other two ('''binormal unit vector''', <math>\vecbf{b}\equiv\vecbf{s}\times\vecbf{n}</math>), and constitute the '''intrinsic basis''' or '''Frenet basis''' for the motion of <math>\Qs</math> in the reference frame R.

====✏️ EXAMPLE C2-3.1: Euler pendulum====
---------

::In the circular motion of the endpoint <math>\Qs</math> of the [[C2. Movement of a mechanical system #✏️ EXAMPLE C2-1.2: Euler pendulum|'''bar relative to the block''']], the two intrinsic components of the acceleration <math>\acc{Q}{BL}</math> are nonzero. Their values and directions are those of the [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-2.1: rotating platform|'''circular motion''']]:
:::* tangential acceleration <math> \accs{Q}{BL}</math>: parallel to <math>\vel{Q}{BL}</math> with value L<math>\ddot\psi</math>.
:::* normal acceleration <math>\accn{Q}{BL}</math> : perpendicular to <math>\vel{Q}{BL}</math> with value L<math>\dot\psi^2</math>.
[[File:C2-Ex3-1-1-neut.png|thumb|center|300px|link=]]
::Though it is evident that the radius of curvature of the trajectory of <math>\Qs</math> relative to BL is L (it is a circular motion), it can also be obtained as <math>\frac{\vecbf{v}_{\textrm{BL}}^2(\Qs)}{|\accn{Q}{BL}|}=\frac{(\Ls\dot\psi)^2}{\Ls\dot\psi^2}=\Ls</math>.

::The acceleration <math>\acc{Q}{R}</math> has been described in [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-2.2: Euler pendulum|'''example C2-2.2''']] as the addition of three terms (the two horizontal ones corresponding to <math>\acc{Q}{BL}</math> plus a permanently horizontal one with value <math>\ddot x</math>). Identifying in that case which is the tangential component (parallel to <math>\vel{Q}{R}</math>) and which is the normal one (orthogonal to <math>\vel{Q}{R}</math>) is not straightforward, as the <math>\vel{Q}{R}</math> direction is not that of a singular direction of the problem [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-1.2: Euler pendulum|'''example C2-1.2''']].
::That identification is straightforward in two particular configurations where the <math>\vel{Q}{R}</math> direction (which is the tangential direction) is horizontal:
[[File:C2-Ex3-1-2-eng.png|thumb|center|400px|link=]]
::The radius of curvature of the pendulum endpoint relative to the ground for the <math>\psi=0</math> configuration is:
[[File:C2-Ex3-1-3-eng.png|thumb|center|400px|link=]]
::The center of curvature is always above <math>\Qs</math> because the normal acceleration points in that direction.
::Particular cases:

[[File:C2-Ex3-1-5-neut.png|thumb|center|400px|link=]]
::The dotted circular lines correspond to the approximation of the trajectory in the neighbourhood of the <math>\psi=0</math> configuration for those two particular cases.

::Though it is a laborious, it is possible to calculate <math>\re{Q}{R}</math> for a general configuration if we remember that only the parallel components participate in the scalar product <math>\vel{Q}{R}\cdot\acc{Q}{R}</math> (and so <math>\accs{Q}{R}</math>), and that only the orthogonal components participate in the cross product <math>\vel{Q}{R}\times\acc{Q}{R}</math>, (and so <math>\accn{Q}{R}</math>) [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-3.1: Euler pendulum|('''example C2-3.1''']] analytical). The result is:

<center><math>\re{Q}{R}=\frac{\textbf{v}_{\Rs}^2(\Qs)}{|\accn{Q}{R}|}=\frac{\left[\dot x^2+\left(\Ls\dot\psi\right)^2+2\Ls\dot x\dot\psi \cpsi\right]^{3/2}}{\left|\Ls(\ddot x\dot\psi-\dot x\ddot\psi)\spsi-\Ls\dot\psi^2(\Ls\dot\psi+\dot x \cpsi)\right|}</math></center>

::When the calculated expressions are complicated (as the previous one), it is advisable to check that it works in simple situations to avoid easily detectable errors. For example:

:::*If <math>\dot x=0</math> permanently (that is, <math>\ddot x=0</math>), the trajectory of <math>\Qs</math> relative to R is circular with radius L:
:::<math>\re{Q}{R}\big]_{\dot x=0, \ddot x=0}=\frac{\left(\Ls^2\dot\psi^2\right)^{3/2}}{\Ls\dot\psi^2\Ls\dot\psi}=\Ls</math>

:::*If <math>\dot\psi=0</math> permanently (<math>\ddot\psi=0</math>), the trajectory of <math>\Qs</math> relative to R is rectilinear, and the radius of curvature has to be infinite:

:::<math>\re{Q}{R}\big]_{\dot\psi=0, \ddot\psi=0}=\frac{(\dot x^2)^{3/2}}{0}\rightarrow\infty</math>
<div>
=====Analytical calculation ➕=====
::The vector bases B and B’ are the same as in [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-2.1: rotating platform|'''example C2-2.1''']].

::Acceleration of <math>\Qs</math> relative to BL:
<center>
<math>\braq{\acc{Q}{BL}}{B} = \braq{\dert{\vel{Q}{BL}}{BL}}{B} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{BL}}{B} = \vector{\Ls\ddot\psi \cpsi-\Ls \dot\psi^2 \spsi}{\Ls \ddot\psi \spsi +\Ls \dot\psi^2 \cpsi}{0}</math>
<math>\braq{\acc{Q}{BL}}{B'} = \braq{\dert{\vel{Q}{BL}}{BL}}{B'} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{BL}}{B'}+ \braq{\velang{B'}{BL}\times \OQvec}{B} = \vector{0}{\Ls\ddot\psi}{0} + \vector{0}{0}{\dot\psi}\times \vector{0}{\Ls\dot\psi}{0} = \vector{-\Ls\dot\psi^2}{\Ls\ddot\psi}{0}</math>
</center>
::Acceleration of <math>\Qs</math> relative to R:
<center>
<math>
\braq{\acc{Q}{R}}{B}=\braq{\dert{\vel{Q}{R}}{R}}{B} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{R}}{B}=\vector{\ddot x+\Ls\ddot\psi \cpsi-\Ls\dot\psi^2\spsi}{\Ls\ddot\psi \spsi+\Ls\dot\psi^2\cpsi}{0}
</math>
<math>
\braq{\acc{Q}{R}}{B'}=\braq{\dert{\vel{Q}{R}}{R}}{B'} = \frac{\ds}{\ds\ts}\braq{\vel{Q}{R}}{B'}+\braq{\velang{B'}{R}\times \OQvec}{B}=\vector{\ddot x \spsi+\dot x\dot\psi \cpsi}{\ddot x \cpsi-\dot x\dot\psi \spsi +\Ls\ddot\psi}{0}+\vector{0}{0}{\dot\psi}\times\vector{\dot x \spsi}{\dot x \cpsi+\Ls\dot\psi}{0}=\vector{\ddot x \spsi - \Ls \dot\psi^2}{\ddot x \cpsi+\Ls\ddot\psi}{0}
</math>
</center>

[[File:C2-Ex3-1-6-neut.png|thumb|center|500px|link=]]

::The calculation of the radius of curvature in the general configuration is cumbersome. As it is a planar motion, and the velocity and the acceleration only have two components, the third component will not be shown. The vector basis is B (but the same result would be obtained through the vector basis B’).

<center>
<math>
\braq{\vel{Q}{R}}{B} = \vecdosd{\dot x + \Ls\dot\psi \cpsi}{\Ls \dot\psi \spsi}, \braq{\acc{Q}{R}}{B} = \vecdosd{\ddot x + \Ls \ddot\psi cos \psi -\Ls \dot\psi^2\spsi}{\Ls\ddot\psi \spsi + \Ls \dot\psi^2 \cpsi}
</math>
<math>
\abs{\accn{Q}{R}}=\abs{\acc{Q}{R}\times\frac{\vel{Q}{R}}{\abs{\vel{Q}{R}}}}
</math>
<math>
\abs{\accn{Q}{R}}=\abs{\frac{1}{\sqrt{({\dot x + \Ls\dot\psi \cpsi)^2+(\Ls \dot\psi \spsi)^2}}}\vecdosd{\ddot x + \Ls \ddot\psi cos \psi -\Ls \dot\psi^2\spsi}{\Ls\ddot\psi \spsi + \Ls \dot\psi^2 \cpsi}\times\vecdosd{\dot x + \Ls\dot\psi \cpsi}{\Ls\dot\psi \spsi}}
</math>
</center>
<center>
<math>
\abs{\accn{Q}{R}}=\abs{
\frac
{
(\ddot x + \Ls \ddot\psi cos \psi -\Ls \dot\psi^2\spsi)\Ls \dot\psi \spsi-(\Ls\ddot\psi \spsi + \Ls \dot\psi^2 \cpsi)(\dot x + \Ls\dot\psi \cpsi)
}
{
\sqrt{\dot x^2+(\Ls\dot\psi)^2+2\Ls \dot x\dot\psi \cpsi}
}}
</math>
</center>
<center>
<math>
\abs{\accn{Q}{R}}=
\abs{
\frac
{
\Ls\ddot x\dot\psi \spsi-\Ls\dot x\ddot\psi \spsi-L^2\dot\psi^3-\Ls\dot x\dot\psi^2\cpsi
}
{
\sqrt{\dot x^2+(\Ls\dot\psi)^2+2\Ls \dot x\dot\psi \cpsi}
}
}=
\abs{
\frac
{
\Ls(\ddot x\dot\psi-\dot x\ddot\psi)\spsi-\Ls\dot\psi^2(\Ls\dot\psi+\dot x \cpsi)
}
{
\sqrt{\dot x^2+(\Ls\dot\psi)^2+2\Ls \dot x\dot\psi \cpsi}
}
}
</math>
</center>
<center>
<math>
\Re_\Rs(\Qs)=\frac{\textrm{v}^2_\Rs(\Qs)}{\abs{\accn{Q}{R}}}=
\frac
{
\left( \dot x^2+(\Ls\dot\psi)^2+2\Ls\dot x\dot\psi \cpsi\right)^{3/2}
}
{
\abs{
\Ls(\ddot x\dot\psi-\dot x\ddot\psi)\spsi-\Ls\dot\psi^2(\Ls\dot\psi+\dot x \cpsi)
}
}
</math>
</center>
</div>

---------
---------

==C2.4 Angular velocity of a rigid body==
The [[C1. Configuration of a mechanical system#C1.2 Configuration of a rigid body|'''configuration of a rigid body''']] S relative to a reference frame R is totally defined through the position of a point <math>\Qs</math> of the rigid body and the orientation of S relative to R (described, for instance, by means of [[C1. Configuration of a mechanical system#C1.3 Orientation of a rigid body with planar motion|'''Euler angles''']]). Similarly, the evolution of the configuration relative to R can be described through the velocity of a point <math>\Qs</math> of the rigid body <math>\vel{Q}{R}</math>, and the '''angular velocity''' of the rigid body <math>\velang{S}{R}</math> (rate of change of orientation with time). When the orientation relative to R is constant with time, we say that the rigid body has a '''translational motion''' <math>\left(\velang{S}{R}=0\right)</math>.

===Simple rotation===

The orientation of a rigid body with planar motion relative to a reference frame R [[C1. Configuration of a mechanical system#C1.3 Orientation of a rigid body with planar motion|'''is totally defined by an angle <math>\psi</math>''']].If that orientation changes, <math>\dot\psi\neq0</math> .

Giving the value of <math>\dot\psi</math> <math>[rad/s]</math> is not enough to define how the orientation of a rigid body changes when its motion is a planar one.

====✏️ EXAMPLE C2-4.1: wheel with planar motion====
---------

:{|
|-
|
[[File:C2-Ex4-1-1-eng.png|250px|thumb|link=]]
|| The wheel has a planar motion relative to R. Its center <math>\Cs</math> is fix fixed in R, and its orientation changes with a rate <math>\dot\psi</math> <math>[rad/s]</math>. With just that information, we cannot infer the motion it describes. For instance, that information might correspond to any of the following cases:
|}
[[File:C2-Ex4-1-2-neut.png|400px|thumb|center|link=]]

:::* Case (a): angle <math>\psi</math> defined on the horizontal plane; the plane of motion is horizontal.
:::* Case (b): angle <math>\psi</math> defined on a vertical plane; the plane of motion is vertical.

::If nothing is said about the plane where the angle has been defined (and that is equivalent to giving a direction: the direction perpendicular to the plane), the motion is not defined univocally.

------------

Thus, the movement associated with a change in orientation is defined by the rate of change of the angle plus a direction. The mathematical object including those two features is a vector. Hence, the angular velocity <math>\velang{S}{R}</math> is a vector. The convention to associate a direction to that vector is the screw rule (or the [[Vector calculus#V.2 Operations between vectors with geometric representation|'''right hand rule''']], or the corkscrew rule,).

====✏️ EXAMPLE C2-4.2: wheel with planar motion====
---------

::The angular velocity associated with movements (a) and (b) in the previous example is:
[[File:C2-Ex4-2-1-eng.png|450px|thumb|center|link=]]

-------------

===Rotation in space===
The orientation of a rigid body moving in space relative to a reference frame R may be given through three [[C1. Configuration of a mechanical system#C1.3 Orientation of a rigid body with planar motion|'''Euler angles''']] <math>(\psi,\theta,\varphi)</math>. We may associate an angular velocity to the change of each of those angles.

====✏️ EXAMPLE C2-4.3: gyroscope====
---------
<div>

::The [[C1. Configuration of a mechanical system#C1.4 Orientation of a rigid body moving in space|'''orientation of a gyroscope''']] relative to the ground (R) may be given through three Euler angles. The angular velocities associated with <math>(\dot\psi,\dot\theta,\dot\varphi)</math> have the following interpretation:

::<math>\vecdot\psi=\velang{fork}{R}</math>, <math>\vecdot\theta=\velang{arm}{fork}</math>, <math>\vecdot\varphi=\velang{disk}{arm}</math>.

[[File:C2-Ex4-3-1-eng-jpg.jpg|thumb|400px|center|link=]]
::Those angular velocities can be projected on any vector basis suggested by the problem:
:::* Vector basis <math>\Bs_\Rs</math> fixed to the reference frame
:::* Vector basis <math>\Bs</math> fixed to the fork (it can be generated from <math>\Bs_\Rs</math> through the <math>\dot\psi</math> rotation)
:::* Vector basis <math>\Bs'</math> fixed to the arm (it can be generated from <math>\Bs</math> through the <math>\dot\theta</math> rotation)
:::* Vector basis <math>\Bs_\textrm{V}</math> fixed to the disk

[[File:C2-Ex4-3-2-eng-jpg.jpg|thumb|400px|center|link=]]

::Nevertheless, it is advisable to choose a vector basis where the maximum number of rotations have the direction of one of the axes in the basis, in order to minimize the projections. As the axes of the three rotations do not correspond to an orthogonal trihedral, it will always be necessary to project at least one of the angular velocities (<math>\vec{\dot{\psi}}, \vec{\dot{\theta}}, \vec{\dot{\varphi}}</math>). With a proper choice of the vector basis, the angular velocities to be projected will be contained on a plane defined by two axes of the vector basis, and that simplifies the operation. Hence, the best choices are B or B’. The angular velocities that will have two components will be <math>\vec{\dot{\varphi}}</math>, when we choose B, and <math>\vec{\dot{\psi}}</math> when we choose B’:

::{|
|<math>\braq{\velang{fork}{R}}{B}=\vector{0}{0}{\dot\psi}, \braq{\velang{arm}{fork}}{B}=\vector{\dot{\theta}}{0}{0}, \braq{\velang{disk}{arm}}{B}=\vector{0}{\dot{\varphi}\cth}{\dot{\varphi}\sth}</math>

<math>\braq{\velang{fork}{R}}{B'}=\vector{0}{\dot{\psi}\sth}{\dot{\psi}\cth}, \braq{\velang{arm}{fork}}{B'}=\vector{\dot{\theta}}{0}{0}, \braq{\velang{disk}{arm}}{B'}=\vector{0}{\dot{\varphi}}{0}</math>
|[[File:C2-Ex4-3-3-neut.png|thumb|right|200px|link=]]
|}

</div>
---------
---------

==C2.5 Angular acceleration of a rigid body==

The '''angular acceleration''' of a rigid body S relative to a reference frame R (<math>\accang{S}{R}</math>) is the time derivative of its angular velocity relative to R:

<center><math>\accang{S}{R}= \dert{\velang{S}{R}}{R}</math></center>

The description of the angular velocity <math>\velang{S}{R}</math> may be any (rotations about fixed axes, Euler rotations...). When the rigid body has a planar motion relative to R, the direction of its angular velocity <math>\velang{S}{R}</math> is constant (it is always perpendicular to the plane of motion). In this case, the angular acceleration comes exclusively from the change of value of <math>\velang{S}{R}</math>, and it is parallel to <math>\velang{S}{R}</math>. In general motions in space, if <math>\velang{S}{R}</math> is described through Euler rotations, <math>\accang{S}{R}</math> may come from the change of values of (<math>\vecdot\psi</math>, <math>\vecdot\theta</math>,<math>\vecdot\varphi</math>) iand the change of direction of de <math>\vecdot\theta</math> and <math>\vecdot\varphi</math> (<math>\vecdot\psi</math> has always a constant direction in R).

==== ✏️ EXAMPLE C2-5.1: gyroscope====
---------
<div>

::The fork of a [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-4.3: gyroscope|'''gyroscope''']] has a planar motion relative to the ground (R), and its angular velocity is vertical: <math>\velang{fork}{R}=\vecdot\psi</math> Its angular acceleration is also vertical, with value <math>\ddot{\psi}: \accang{S}{R}=\vec{\ddot{\psi}}</math>.

::The angular acceleration of the disk is more complicated. It can be obtained through the geometric time derivative of <math>\velang{disk}{R}=\vecdot\psi+\vecdot\theta+\vecdot\varphi</math>. The rotation <math>\vecdot\varphi</math> can be decomposed in a vertical component with value <math>\dot\varphi\textrm{sin}\theta</math>, and a horizontal one with value <math>\dot\varphi\textrm{cos}\theta</math>. The vertical component can only change its value, whereas the horizontal its value and its direction (because of <math>\vecdot\psi</math>).

[[File:C2-Ex5-1-neut.png|thumb|center|400px|link=]]

<center>
{|
|
Time derivative of the vertical components

[[File:C2-Ex5-3-neut.png|thumb|center|350px|link=]]

|
:::Time derivative of the horizontal components

:::[[File:C2-Ex5-4-neut.png|thumb|center|350px|link=]]
|}</center>

=====Analytical calculation ➕=====
::The same result is obtained if the time derivative is performed analytically through the vector basis rotating with <math>\vecdot\psi</math> relative to R or that rotating with <math>\vecdot\psi+\vecdot\theta</math> (also relative to R):

<center>
<math>\braq{\velang{}{R}}{B}=\vector{\dot\theta}{\dot\varphi \cth}{\dot\psi+\dot\varphi \sth},</math>          <math>\braq{\accang{disk}{R}}{B}=\braq{\dert{\velang{disk}{R}}{R}}{B}=\frac{\textrm{d}}{\textrm{dt}}\braq{\velang{disk}{R}}{B}+\braq{\velang{B}{R}\times\velang{disk}{R}}{B}</math>

<math>\braq{\accang{disk}{R}}{B} = \vector{\ddot\theta}{\ddot\varphi \cth-\dot\varphi\dth \sth}{\ddot\psi+\ddot\varphi \sth+\dot\varphi\dth \cth}+\vector{0}{0}{\dot\psi}\times\vector{\dot\theta}{\dot\varphi \cth}{\dot\psi+\dot\varphi \sth} = \vector{\ddot\theta-\dot\psi\dot\varphi \cth}{\ddot\varphi \cth-\dot\varphi\dth \sth+\dot\psi\dot\theta}{\ddot\psi+\ddot\varphi \sth+\dot\varphi\dth \cth}</math>

<math>\braq{\velang{disk}{R}}{B'}=\vector{\dot\theta}{\dot\varphi+\dot\psi \sth}{\dot\psi \cth},</math>          <math>\braq{\accang{disk}{R}}{B'}=\braq{\dert{\velang{disk}{R}}{R}}{B'}=\frac{\textrm{d}}{\textrm{dt}}\braq{\velang{disk}{R}}{B'}+\braq{\velang{B'}{R}\times\velang{disk}{R}}{B'}</math>

<math>\braq{\accang{disk}{R}}{B'} = \vector{\ddot\theta}{\ddot\varphi+\ddot\psi \sth+\dot\psi\dot\theta \cth}{\ddot\psi \cth-\dot\psi\dot\theta \sth} + \vector{\dth}{\dot\psi\sth}{\dot\psi\cth}\times\vector{\dot\theta}{\dot\varphi+\dot\psi \sth}{\dot\psi \cth} = \vector{\ddth-\dot\psi\dot\varphi\cth}{\ddot\varphi+\ddot\psi \sth+\dot\psi\dth\cth}{\ddot\psi \cth-\dot\psi\dth\sth+\dot\varphi\dth}</math>
</center>

</div>

----------
----------

==C2.6 Particle kinematics VS rigid body kinematics==
Particle (point) and rigid body are two very different models. From a kinematic point of view, the second one is richer because it includes the concept of rotation (not applicable to particles, as they cannot be orientated because they have no dimensions). Because of rotations, points of a same rigid boy may describe different trajectories.

One has to bear that in mind in order not to use erroneously concepts that only apply to one of the models when talking about the other. The following examples illustrate some wrong statements and some correct ones.

====✏️ EXAMPLE C2-6.1: particle inside a circular guide====
------------

::{|
|[[File:C2-Ex6-1-neut_REV01.png|thumb|left|180px|link=]]
|Particle <math>\Ps</math> rotates relative to R: '''WRONG'''

Vector <math>\vec{\textbf{OP}}</math> rotates relative to R: '''CORRECT'''

Particle <math>\Ps</math> describes a circular trajectory relative to R (or has a circular motion relative to): '''CORRECT'''
|}


====✏️ EXAMPLE C2-6.2: particle on an incline====
------------

::{|
|[[File:C2-Ex6-2-neut.png|thumb|left|200px|link=]]
|Particle <math>\Ps</math> has a translational motion relative to R: '''WRONG'''

Particle <math>\Ps</math> describes a rectilinear trajectory relative to R (or has a rectilinear motion relative to R): '''CORRECT'''
|}


====✏️ EXAMPLE C2-6.3: wheel with a nonsliding contact with the ground and with planar motion====
------------

[[File:C2-Ex6-3-neut.png|thumb|center|540px|link=]]
::Points on the wheel rotate relative to R: '''WRONG'''

::The wheel rotates relative to R: '''CORRECT'''

::The center of the wheel has a translational motion relative to R: '''WRONG'''

::The center of the wheel has a rectilinear motion relative to R: '''CORRECT'''

<center>'''Some points on a rotating rigid body may have rectilinear motion.'''</center>


<center><html><iframe width="560" height="315" src="https://www.youtube-nocookie.com/embed/ED3LXV6JWCA?start=11" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" allowfullscreen></iframe></html></center>
<center>'''Video C2.1''' Visualització de les trajectòries de punts</center>

====✏️ EXAMPLE C2-6.4: motion of a ferris wheel====
------------

::{|
|[[File:C2-Ex6-4-1-eng.png|thumb|left|200px|link=]]
|The ring rotates relative to R: '''CORRECT'''

The cabin rotates relative to R: '''WRONG''' if we neglect the pendulum motion, the ground and the ceiling of the cabin are always parallel to te ground, so it does not rotate).

The cabin has a translational motion relative to R '''CORRECT'''
|-
|[[File:C2-Ex6-4-2-neut.png|thumb|left|200px|link=]]
|In this case, all points in the cabin have circular motions with the same radius relative to R, but different center of curvature.

In such a case, we may combine a concept from rigid body kinematics (translational motion) with a concept from particle kinematics (circular motion) to describe the motion of the cabin:

The cabin has a '''translational circular motion''' relative to R.
|}

<center>'''Points in a rigid body with a translational motion may describe curvilinear trajectories.'''</center>


--------
--------

==C2.7 Degrees of freedom==
As we have seen through various examples in this unit, the velocities of the points in a mechanical system depend on a set of scalar variables with dimensions or . The minimum set of scalar variables of this sort needed to describe the system motion is the set of the '''degrees of freedom''' (DOF) of the system.

When the system is just a free rigid body moving in space (without any contact with material objects), [[C4. Rigid body kinematics#C4.1 Velocity distribution|'''the number of DOF is 6''']]: three associated with the motion of one point (for instance, <math>(\dot{\textrm{x}}, \dot{\textrm{y}}, \dot{\textrm{z}})</math>) and three associated with the change of orientation of the rigid body (for instance, <math>(\dot{\psi}, \dot{\theta}, \dot{\varphi})</math>).

In mechanical engineering, the usual mechanical systems are multibody systems: sets of rigid bodies [[C2. Movement of a mechanical system#C2.8 Usual constraints in mechanical systems|'''linked''']] through revolute joints, spherical joints... Because of these links (or constraints), the mechanical state of each rigid body (that is, its configuration in space and its motion) is related to that of the other rigid bodies: in a multibody System with N rigid bodies, the number of DOF is lower than 6N.

------------
------------

==C2.8 Usual constraints in mechanical systems==

'''Falta paragraf versio catala'''

<center>
{| class="wikitable" style="background-color:white; text-align:left"
|-
|<center>[[File:C2-8-eng.png|thumb|center|175px|link=]]</center>||
'''With sliding:''' 
It allows three independent rotations between the rigid bodies (about the normal direction and the two tangential directions), and two independent translational motions (along the two tangential directions) 
'''Without sliding:''' 
It allows three independent rotations between the rigid bodies (about the normal direction and the two tangential directions)

|-
|<center>[[File:C2-8-revolucio-neut.png|thumb|center|220px|link=]]</center>||
'''revolute joint''' 
Allows a rotation between the two rigid bodies about axis 1
|-
|<center>[[File:C2-8-cilindric-neut.png|thumb|center|220px|link=]]</center>||
'''cylindrical joint ''' 
Allows a rotation between the two rigid bodies about axis 1, and a translational motion (displacement without rotation) along axis 1.
|-
|<center>[[File:C2-8-prismatic-neut.png|thumb|center|220px|link=]]</center>||
'''prismatic joint ''' 
Allows a translational motion between the two rigid bodies along axis 1.
|-
|<center>[[File:C2-8-esferic-neut.png|thumb|center|220px|link=]]</center>||
'''spherical joint ''' 
Allows three independent rotations between the two rigid bodies about axes 1, 2, 3.
|-
|<center>[[File:C2-8-helicoidal-neut.png|thumb|center|220px|link=]]</center>||
'''helical joint (screw)''' 
Allows a rotation between the two rigid bodies about axis 3; this rotation provokes a displacement along axis 3. The relationship between the rotation and the displacement is given by the screw pitch e [mm/volta].
|-
|<center>[[File:C2-8-Cardan-rev.png|thumb|center|220px|link=]]</center>||
'''Cardan joint (universal joint)''' 
Allows two independent rotations between the two rigid bodies about axes 1, 3.
|}
</center>

<center><html><iframe width="560" height="315" src="https://www.youtube-nocookie.com/embed/067F1MQVICs" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" allowfullscreen></iframe></html></center>
<center>'''Video C2.2''' Junta Cardan (junta universal o de creueta)</center>

<center><html><iframe width="560" height="315" src="https://www.youtube-nocookie.com/embed/K-xIHJErByk" title="YouTube video player" frameborder="0" allow="accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture; web-share" allowfullscreen></iframe></html></center>
<center>'''Video C2.3''' Graus de Llibertat d'una roda emb moviment pla i contacte amb el terra</center>

====✏️ EXAMPLE C2-8.1: gyroscope====
------------
{|:

::In a [[C2. Movement of a mechanical system#✏️ EXAMPLE C2-4.3: gyroscope|'''gyroscope''']], the support does not move relative to the ground (R). There are revolute joints between the fork and the support, between the arm and the fork, and between the disk and the arm. All that can be represented through a simplified diagram:
[[File:C2-Ex8-1-eng.png|thumb|center|600px|link=]]

::The position of point <math>\Os</math> relative to the ground is constant. Hence, the gyroscope configuration is totally defined by the three angles <math>(\psi,\theta,\varphi)</math>: the gyroscope has 3 IC relative to the ground.

::Regarding its motion, as the variation of any of those angles does not imply that of the other two, their time evolutions are independent: the gyroscope has 3 DOF relative to the ground, and they may be described through <math>(\dot\psi,\dot\theta,\dot\varphi)</math>.
|}


====✏️ EXEMPLE C2-8.2: tricycle====
------------
{|:

::The tricycle is a system with 5 rigid bodies: the chassis, the handlebar and the three wheels. There is no element fixed to the ground. There are revolute joints between the rear wheels and the chassis, between the handlebar and the chassis, and between the front wheel and the handlebar. Moreover, the wheels are in contact with the ground: that too is a restriction (or a constraint). If it moves on horizontal ground without sliding, that contact may be idealized as a single-point contact without sliding (whether a contact is a sliding or a nonsliding one depends on the system dynamics; in kinematics, sliding or nonsliding is a hypothesis).
[[File:C2-Ex8-2-1-eng.png|thumb|650px|center|link=]]
[[File:C2-Ex8-2-2-neut.png|thumb|450px|center|link=]]
::A good way to determine the number of DOF of a system relative to a reference frame is to count up how many motions have to be blocked to reach a complete rest. In a tricycle, if we block the motion of point <math>\Os</math> (that can only be in the longitudinal direction of the wheels do not skid), the chassis would still be able to rotate about a vertical axis through <math>\Os</math>. If we block that rotation <math>(\dot\psi=0)</math>, the rear wheels are blocked, but the handlebar and the front wheel may still rotate about the vertical axis through the wheel center <math>(\dot\psi'\neq 0)</math>. If we blocked that last motion, the tricycle is at rets. We have blocked three motions, hence the tricycle has 3 DOF.
|}


====✏️ EXAMPLE C2-8.3: spherical shell on a platform====
------------
{|:

::The system contains 4 rigid bodies: the platform, the shell, the arm and the fork. There are revolute joints between the platform and the ground, between the shell and the arm, between the arm and the fork, and between the fork and the ceiling (or the ground). Moreover, between shell and platform there is a single-point contact without sliding.

[[File:C2-Ex8-3-eng.png|thumb|500px|center|link=]]
::The DOF of the System relative to the ground (R) can be discovered by blocking different motions until reaching a total rest:
:::* block the rotation of the platform relative to the ground
:::* block the rotation of the fork relative to the ground
::Under those conditions, though the revolute joint between shell and arm allows a rotation, that rotation would provoke a sliding motion between shell and platform, and that is not consistent with the hypothesis of nonsliding contact. Hence, the system is at rest: it has 2 DOF relative to the ground.
|}

-------------
-------------

==C2.E General examples==
====🔎 Example C2-E.1: rotating pendulum====
---------
::{|:

The plate is articulated at point <math>\Os</math> O to a fork, which rotates with constant angular velocity <math>\psio</math> relative to the ground (T). Between fork and ground (ceiling), and between plate and fork there are revolute joints.

[[File:C2-E.Ex1-1-eng.png|thumb|400px|center|link=]]

=====1. How many degrees of freedom (DoF) has the system? Describe them.=====
<div>
:The fork has a simple rotation relative to the ground about a vertical axis.

:Independently from that rotation, the plate may rotate about the horizontal axis of the fork.

:Those two motions are independent because, if we block one of them, the other one may still take place.

:Hence, the system has two degrees of freedom.
</div>

=====2. Find the angular velocity and the angular acceleration of the plate relative to the ground. =====
<div>
:The angular velocity of the plate is the superposition of <math>\vecdot\psi_0</math> (1st Euler rotation, axis fixed to the ground) and <math>\vecdot\theta</math> (2nd Euler rotation, axis rotating with <math>\vecdot\psi_0</math>relative to the ground): <math>\velang{plate}{E}=\vecdot\psi_0+\vecdot\theta</math>

:'''Geometric calculation:'''

:<math>\velang{plate}{E}=\vecdot\psi_0+\vecdot\theta=(\Uparrow \psio)+(\odot \dot{\theta})</math>

[[File:C2-E.Ex1-2-neut.png|thumb|right|200px|link=]]

:<math>\accang{plate}{E}=\dert{\velang{plate}{E}}{E}=\dert{\vecdot\psi_0}{E}+\dert{\vecdot\theta}{E}=\dert{(\Uparrow \psio)}{E}+\dert{(\odot \dot{\theta})}{E}</math>

:As <math>\vecdot\psi_0</math> has constant value and direction, the angular acceleration will be associated only to the change of value and direction of <math>\vecdot\theta</math>.It is a vector with variable value which rotates about a vertical axis because of the 1st Euler rotation <math>(\Omegavec^{\vecdot\theta}_\textrm{T} =\vecdot\psi_0)</math>.

:<math>\accang{plate}{E}=\dert{(\odot\dot{\theta})}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=</math>
:<math>=\left[\ddot{\theta}\frac{\vecdot{\theta}}{|\vecdot{\theta}|}\right]+[\velang{$\vecdot{\theta}$}{$\Ts$}\times\vecdot{\theta}]=[\odot\ddot{\theta}]+[(\Uparrow\psio)\times(\odot\dot{\theta})]=(\odot\ddot{\theta})+(\Rightarrow\psio\dot{\theta})</math>

:'''Analytical calculation:'''
:The time derivative of the angular velocity can also be done analytically. The vector basis where the <math>\velang{plate}{E}</math> projection is straightforward is the vector basis fixed to the fork <math>(\velang{B}{E}=\vecdot{\psi}_0)</math>:

:<math>\braq{\velang{plate}{E}}{B}=\vector{\dot{\theta}}{0}{\psio}</math>

:<math>\braq{\accang{plate}{E}}{B}=\braq{\dert{\velang{plate}{E}}{E}}{B}=\frac{\ds}{\ds\ts}\braq{\velang{plate}{E}}{B}+\braq{\velang{B}{E}}{B}\times\braq{\velang{plate}{E}}{B}=\vector{\ddot{\theta}}{0}{0}+\vector{0}{0}{\psio}\times\vector{\dot{\theta}}{0}{\psio}=\vector{\ddot{\theta}}{\psio\dot{\theta}}{0}</math>
</div>

=====3. Find the velocity and the acceleration of point G of the plate relative to the ground. . =====
<div>
:As point <math>\Os</math> is fixed to the ground, <math>\OGvec</math> vector can be taken as position vector in the ground frame. Its value L is constant, but its direction is not because of <math>\psio</math> and <math>\vecdot{\theta}</math>: <math>\OGvec=(\searrow\Ls)^{*}</math>.

:'''Geometric calculation:'''
:<math>\vel{G}{E}=\dert{\OGvec}{E}=[\text{change of direction}]_\Es=\velang{$\OGvec$}{E}\times\OGvec=(\vec{\dot{\psi}}_0+\vec{\dot{\theta}})\times\OGvec=\left[(\Uparrow\psio)+(\odot\dot{\theta})\right]\times(\searrow\Ls)=(\Uparrow\psio)\times(\rightarrow\Ls\text{sin}\theta)+(\odot\dot{\theta})\times(\searrow\Ls)=(\otimes\Ls\psio\text{sin}\theta)+(\nearrow\Ls\dot{\theta})</math>

[[File:C2-E.Ex1-3-eng.png|thumb|right|300px|link=]]

That velocity has variable value and direction, thus the acceleration has both parallel component and orthogonal component to the velocity.

:<math>\acc{G}{E}=\dert{\vel{G}{E}}{E}=\dert{(\otimes\Ls\psio\text{sin}\theta)}{E}+\dert{(\nearrow\Ls\dot{\theta})}{E}</math>

:The <math>(\otimes\Ls\psio\text{sin}\theta)</math> vector rotates relative to the ground just because of <math>\psio</math>, whereas the <math>(\nearrow\Ls\dot{\theta})</math> vector rotates because of <math>\vec{\psio}</math> and <math>\vec{\dot{\theta}}</math>:

:<math>\dert{(\otimes\Ls\psio\text{sin}\theta)}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=</math>
:<math>\hspace{3.1cm}=[\otimes\Ls\psio\dot{\theta}\text{cos}\theta]+\left[(\Uparrow\psio)\times(\otimes\Ls\psio\text{sin}\theta)\right]=[\otimes\Ls\psio\dot{\theta}\text{cos}\theta]+\left[\leftarrow\Ls\psio^2\text{sin}\theta\right]</math>

:<math>\dert{(\nearrow\Ls\dot{\theta})}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=</math>
<math>\hspace{2.9cm}=\left[\nearrow\Ls\ddot{\theta}\right]+\left[((\Uparrow\psio)+(\odot\dot{\theta}))\times(\nearrow\Ls\dot{\theta})\right]=\left[\nearrow\Ls\ddot{\theta}\right]+\left[(\otimes\Ls\psio\dot{\theta}\text{cos}\theta)+(\nwarrow\Ls\dot{\theta}^2)\right]
</math>
:Finally: <math>\acc{P}{E}=(\otimes 2\Ls\psio\dot{\theta}\text{cos}\theta)+(\leftarrow\Ls\psio^2\text{sin}\theta)+(\nwarrow\Ls\dot{\theta}^2)+(\nearrow\Ls\ddot{\theta})</math>

:'''Analytical calculation:'''
:The whole calculation can be done analytically. The vector basis where the projection of <math>\OGvec</math>is straightforward is fixed to the plate (base B’). That vector basis changes its orientation whenever the values of <math>\psi</math> and <math>\theta</math> change. Hence, the angular velocity of the vector basis is <math>\velang{B'}{E}=\vec{\psio}+\vec{\dot{\theta}}</math>:

:<math>\braq{\OGvec}{B'}=\vector{0}{0}{-L}</math>

:<math>\braq{\vel{G}{E}}{B'}=\braq{\dert{\OGvec}{E}}{B'}=\frac{\ds}{\ds\ts}\braq{\OGvec}{B'}+\braq{\velang{B'}{E}}{B'}\times\braq{\OGvec}{B'}=\vector{0}{0}{0}+\vector{\dot{\theta}}{\psio\text{sin}\theta}{\psio\text{cos}\theta}\times\vector{0}{0}{-\Ls}=\vector{-\Ls\psio\text{sin}\theta}{\Ls\dot{\theta}}{0}</math>

:<math>\braq{\acc{G}{E}}{B'}=\braq{\dert{\vel{G}{E}}{E}}{B'}=\frac{\ds}{\ds\ts}\braq{\vel{G}{E}}{B'}+\braq{\velang{B'}{E}}{B'}\times\braq{\vel{G}{E}}{B'}=\vector{-\Ls\psio\dot{\theta}\text{cos}\theta}{\Ls\ddot{\theta}}{0}+\vector{\dot{\theta}}{\psio\text{sin}\theta}{\psio\text{cos}\theta}\times\vector{-\Ls\psio\text{sin}\theta}{\Ls\dot{\theta}}{0}=\vector{-2\Ls\psio\dot{\theta}\text{cos}\theta}{\Ls\ddot{\theta}-\Ls\psio^2\text{sin}\theta\text{cos}\theta}{\Ls\dot{\theta}^2+\Ls\psio^2\text{sin}^2\theta}</math>
</div>
|}

====🔎 Example C2-E.2: rotating articulated plate====
---------
::{|:

The rectangular plate is joined to a rotation support through two bars with revolute joints at their endpoints. A third bar is joined to the plate through a [[C2. Movement of a mechanical system#C2.8 Usual constraints in mechanical systems|'''spherical joint ''']] at <math>\Ps</math>, and to the support through a [[C2. Movement of a mechanical system#C2.8 Usual constraints in mechanical systems|'''cylindrical joiny ''']]. ). The support rotates with the variable angular velocity <math>\vecdot{\psi}</math> vrelative to the ground (E).

[[File:C4-E-Ex2-1-eng.png|thumb|center|450px|link=]]

[[File:C2-E.Ex2-2-eng.png|thumb|center|450px|link=]]

=====1. How many degrees of freedom (DoF) has the system? Describe them.=====
<div>
:The support may rotate freely about a vertical axis fixed to the ground (simple rotation). If we block that motion, the system may still move.
:The <math>\OCvec</math> bars may have a simple rotation, relative to the support, about the horizontal axis through <math>\Os</math> orthogonal to the bars. If we block the motion of one of those bars relative to the support, the plate, the <math>\OCvec</math> bars and the bended bar cannot move. Alternatively, if the vended bar is blocked (if ts vertical translational motion relative to the support is blocked), neither the plate nor the <math>\OCvec</math> bars may move relative to the support.
:Hence, the system has two degrees of freedom.

</div>

=====2. Find the angular velocity and the angular acceleration of the plate relative to the ground.=====
<div>
:The angular velocity of the plate is the superposition of <math>\vec{\dot{\psi}}</math> (1st Euler rotation, axis fixed to the ground) and <math>\vec{\dot{\theta}}</math> (2nd Euler rotation, axis rotation because of <math>\vec{\dot{\psi}}</math>):

:'''Geometric calculation:'''

:<math>\velang{plate}{E}=\vec{\dot{\psi}}+\vec{\dot{\theta}}=(\Uparrow\dot{\psi})+(\otimes\dot{\theta})</math>

:The angular acceleration is associated to the change of value of <math>\vec{\dot{\psi}}</math>, and the change of direction of <math>\vec{\dot{\theta}}</math>:

:<math>\accang{plate}{E}=\dert{\velang{plate}{E}}{E}=\dert{(\Uparrow\dot{\psi})}{E}=\dert{(\otimes\dot{\theta})}{E}</math>

:<math>\dert{(\Uparrow\dot{\psi})}{E}=[\text{change of value}]=\ddot{\psi}\frac{\vec{\dot{\psi}}}{|\vec{\dot{\psi}}|}=(\Uparrow\ddot{\psi})</math>

:<math>\dert{(\otimes\dot{\theta})}{E}=[\text{change of value}]+[\text{change od direction}]_\Es=\left[\ddot{\theta}\frac{\vec{\dot{\theta}}}{|\vec{\dot{\theta}}|}\right]+\left[\velang{$\vec{\dot{\theta}}$}{E}\times\vec{\dot{\theta}}\right]=[\otimes\ddot{\theta}]+\left[(\Uparrow\dot{\psi})\times(\otimes\dot{\theta})\right]</math>

:Hence, <math>\accang{plate}{E}=(\Uparrow\ddot{\psi})+(\otimes\ddot{\theta})+(\Leftarrow\dot{\psi}\dot{\theta})</math>

:'''Analytical calculation:'''

[[File:C4-E-Ex2-3-neut.png|thumb|right|150px|link=]]

:The time derivative of the angular velocity can also be done analytically. The vector basis B where the <math>\velang{plate}{E}</math> projection is straightforward is the one fixed to the support <math>(\velang{B}{E}=\vec{\dot{\psi}})</math>:

:<math>\braq{\velang{plate}{E}}{B}=\vector{0}{\dot{\theta}}{\dot{\psi}}</math>

:<math>\braq{\accang{plate}{E}}{B}=\braq{\dert{\velang{plate}{E}}{E}}{B}=\frac{\ds}{\ds\ts}\braq{\velang{plate}{E}}{B}+\braq{\velang{B}{E}}{B}\times\braq{\velang{plate}{E}}{B}=</math>
:<math>=\vector{0}{\ddot{\theta}}{\ddot{\psi}}+\vector{0}{0}{\dot{\psi}}\times\vector{0}{\dot{\theta}}{\dot{\psi}}=\vector{-\dot{\psi}\dot{\theta}}{\ddot{\theta}}{\ddot{\psi}}</math>

</div>

=====3. Find the velocity and the acceleration of point Q of the plate relative to the ground. =====
<div>
:As point <math>\Os</math> is fixed to the ground, <math>\OQvec</math> is a position vector in the ground frame. Its value is <math>2\Ls\text{cos}\theta</math>, , and its direction is always horizontal. The velocity of <math>\Qs</math>comes both from the change of that value (as <math>\theta</math> is variable) and the change of its direction relative to the ground (because of the support rotation <math>\vec{\dot{\psi}}</math>).

:'''Geometric calculation:'''

[[File:C2-E.Ex2-4-neut.png|thumb|right|230px|link=]]

:<math>\OQvec=(\rightarrow 2\Ls\text{cos}\theta)</math>

:<math>\vel{Q}{E}=\dert{\OQvec}{E}=\dert{(\rightarrow 2\Ls\text{cos}\theta)}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=</math>
:<math>=\left[\rightarrow -2\Ls\dot{\theta}\text{sin}\theta\right]+\left[(\Uparrow\dot{\psi})\times(\rightarrow 2\Ls\text{cos}\theta)\right]=\left[\leftarrow 2\Ls\dot{\theta}\text{sin}\theta\right]+\left[\otimes 2\Ls\dot{\psi}\text{cos}\theta\right]</math>

:The acceleration of <math>\Qs</math> comes from the change of value and direction (associated with <math>\vec{\dot{\psi}}</math>) of both terms in the velocity:
:<math>\acc{Q}{E}=\dert{\vel{Q}{E}}{E}=\dert{(\leftarrow 2\Ls\dot{\theta}\text{sin}\theta)}{E}+\dert{\otimes 2\Ls\dot{\psi}\text{cos}\theta}{E}</math>

:<math>\dert{(\leftarrow 2\Ls\dot{\theta}\text{sin}\theta)}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=\left[\leftarrow 2\Ls(\ddot{\theta}\text{sin}\theta+\dot{\theta}^2\text{cos}\theta)\right]+\left[(\Uparrow\dot{\psi})\times(\leftarrow 2\Ls\dot{\theta}\text{sin}\theta)\right]=\left[\leftarrow 2\Ls(\ddot{\theta}\text{sin}\theta+\dot{\theta}^2\text{cos}\theta)\right]+\left[\odot 2\Ls\dot{\psi}\dot{\theta}\text{sin}\theta\right]</math>

:<math>\dert{(\otimes 2\Ls\dot{\psi}\text{cos}\theta)}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=\left[\otimes 2\Ls(\ddot{\psi}\text{cos}\theta-\dot{\psi}\dot{\theta}\text{sin}\theta)\right]+\left[(\Uparrow\dot{\psi})\times(\otimes 2\Ls\dot{\psi}\text{cos}\theta)\right]=\left[\otimes 2\Ls(\ddot{\psi}\text{cos}\theta-\dot{\psi}\dot{\theta}\text{sin}\theta)\right]+\left[\leftarrow 2\Ls\dot{\psi}^2\text{cos}\theta\right]</math>

:Finally, <math>\acc{Q}{E}=(\leftarrow 2\Ls(\ddot{\theta}\text{sin}\theta+(\dot{\psi}^2+\dot{\theta}^2)\text{cos}\theta))+(\odot 4\Ls\dot{\psi}\dot{\theta}\text{sin}\theta)+(\otimes 2\Ls\ddot{\psi}\text{cos}\theta)</math>

:'''Analytical calculation:'''
[[File:C4-E-Ex2-3-neut.png|thumb|right|150px|link=]]
:The tome derivative can also be done analytically. The vector basis B where the <math>\OPvec</math> projection is straightforward is the one fixed to the support <math>(\velang{B}{E}=\vec{\dot{\psi}})</math>:

:<math>\braq{\OQvec}{B}=\vector{2\Ls\text{cos}\theta}{0}{0}</math>

:<math>\braq{\vel{Q}{E}}{B}=\braq{\dert{\OQvec}{E}}{B}=\frac{\ds}{\ds\ts}\braq{\OQvec}{B}+\braq{\velang{B}{E}}{B}\times\braq{\OQvec}{B}=</math>
:<math>=\vector{-2\Ls\dot{\theta}\text{sin}\theta}{0}{0}+\vector{0}{0}{\dot{\psi}}\times\vector{2\Ls\text{cos}\theta}{0}{0}=\vector{-2\Ls\dot{\theta}\text{sin}\theta}{2\Ls\dot{\psi}\text{cos}\theta}{0}</math>

:<math>\braq{\acc{Q}{E}}{B}=\braq{\dert{\vel{Q}{E}}{E}}{B}=\frac{\ds}{\ds\ts}\braq{\vel{Q}{E}}{B}+\braq{\velang{B}{E}}{B}\times\braq{\vel{Q}{E}}{B}=2\Ls\vector{-\ddot{\theta}\text{sin}\theta-\dot{\theta}^2\text{cos}\theta}{\ddot{\psi}\text{cos}\theta-\dot{\psi}\dot{\theta}\text{sin}\theta}{0}+\vector{0}{0}{\dot{\psi}}\times 2\Ls\vector{-\dot{\theta}\text{sin}\theta}{\dot{\psi}\text{cos}\theta}{0}=2\Ls\vector{-\ddot{\theta}\text{sin}\theta-(\dot{\psi}^2+\dot{\theta}^2)\text{cos}\theta}{\ddot{\psi}\text{cos}\theta-2\dot{\psi}\dot{\theta}\text{sin}\theta}{0}</math>

</div>
|}

====🔎 Example C2-E.3: rotating pendulum with oscillating articulation point====
---------
::{|:

The ring-shaped pendulum is articulated to the support, which is linked to the guide through a prismatic joint. The guide is articulated to the ceiling, and its angular velocity relative to the ceiling <math>(\vec{\psio})</math> is constant. The spring between support and guide guarantees that the former does not fall to the ground when the system is at rest.

[[File:C2-E.Ex3-1-eng.png|thumb|center|600px|link=]]

=====1. How many degrees of freedom (DoF) has the system? Describe them.=====
<div>
:The guide may rotate about the vertical axis through <math>\Os '</math> (simple rotation).

:Independently form that rotation, the support may have a translational motion along the guide (rectilinear translational motion).

:Finally, if those two motions are blocked, the ring may still have a simple rotation about the horizontal axis through <math>\Os '</math>, which is perpendicular to the ring plane and is fixed to the support.

:Hence, the system has 3 degrees of freedom.

</div>

=====2. Find the angular velocity and the angular acceleration of the ring relative to the ground. =====
<div>
:'''Geometric calculation:'''

:The angular velocity of the ring is the superposition of <math>(\vec{\psio})</math> (1st Euler rotation, axis fixed to the ground) and <math>\vec{\dot{\theta}}</math> (2nd Euler rotation, axis rotating with <math>\vec{\dot{\psi}}</math>):

[[File:C2-E.Ex3-2-neut.png|thumb|right|300px|link=]]

:<math>\velang{ring}{E}=\vec{\psio}+\vec{\dot{\theta}}=(\Uparrow\psio)+(\odot\dot{\theta})</math>

:<math>\accang{ring}{E}=\dert{\velang{ring}{E}}{E}=\dert{(\vec{\psio}+\vec{\dot{\theta}})}{E}=\dert{\vec{\psio}}{E}+\dert{\vec{\dot{\theta}}}{E}</math>
:<math>=\dert{(\Uparrow\psio)}{E}+\dert{(\odot\dot{\theta})}{E}</math>

:The angular acceleration comes exclusively from the change of value and direction of <math>\vec{\dot{\theta}}</math>, as <math>\vec{\psio}</math> has both constant value and constant direction.

:<math>\accang{ring}{E} = \dert{\velang{ring}{E}}{E} = \dert{(\odot\dot{\theta})}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=</math>
:<math>=\left[\ddot{\theta}\frac{\vec{\dot{\theta}}}{|\vec{\dot{\theta}}|}\right]+\left[\velang{$\vec{\dot{\theta}}$}{T}\times\vec{\dot{\theta}}\right]=[\odot\ddot{\theta}]+\left[(\Uparrow\dot{\psi})\times(\odot\dot{\theta})\right]=(\odot\ddot{\theta})+(\Rightarrow\psio\dot{\theta})</math>

:'''Analytical calculation:'''

:The time derivative of the angular velocity of the ring may be done analytically. The vector basis where the projection of <math>\velang{ring}{E}</math> és immediata és la fixa al suport <math>(\velang{B}{E}=\vec{\dot{\psi}})</math>:

:<math>\braq{\velang{ring}{E}}{B}=\vector{0}{\psio}{\dot{\theta}}</math>

:<math>\braq{\accang{ring}{E}}{B}=\braq{\dert{\velang{ring}{E}}{E}}{B}=\frac{\ds}{\ds\ts}\braq{\velang{ring}{E}}{B}+\braq{\velang{B}{E}
}{B}\times\braq{\velang{ring}{E}}{B}=\vector{0}{0}{\ddot{\theta}}+\vector{0}{\psio}{0}\times\vector{0}{\psio}{\dot{\theta}}=\vector{\psio\dot{\theta}}{0}{\ddot{\theta}}</math>

</div>

=====3. Find the velocity and the acceleration of point G of the ring relative to the ground.=====
<div>
:'''Geometric calculation:'''

:<math>\vec{\Os'\Gs}</math> is a position vector for <math>\Gs</math> in the ground frame, as <math>\Os'</math> is a point fixed to the ground.

:<math>\vec{\Os'\Gs}=\vec{\Os'\Os}+\vec{\Os\Gs}=(\downarrow \textrm{x})+(\searrow \Ls)^*</math>

:<math>\vel{G}{E}=\dert{\vec{\Os'\Gs}}{E}=\dert{\vec{\Os'\Os}}{E}+\dert{\vec{\Os\Gs}}{E}=\dert{(\downarrow \textrm{x})}{E}+\dert{(\searrow \Ls)}{E}</math>

[[File:C2-E.Ex3-3-eng.png|thumb|right|250px|link=]]

:The term <math>(\downarrow \text{x})</math> has a variable value but a constant orientation, whereas the term <math>(\searrow \Ls)</math>, with constant value, changes its orientation relative to the ground because of <math>\vec{\psio}</math> and <math>\vec{\dot{\theta}}</math>:

:<math>\dert{\vec{\Os'\Os}}{E}=\dert{(\downarrow \textrm{x})}{E}=[\text{change of value}]=(\downarrow\dot{\text{x}})</math>

:<math>\dert{\vec{\Os\Gs}}{E}=\dert{(\searrow \Ls)}{E}=[\text{change of direction}]_\Es=\velang{$\OGvec$}{E}\times\OGvec=((\Uparrow\psio)+(\odot\dot{\theta}))\times(\searrow \Ls)=</math>
:<math>=(\Uparrow\psio)\times(\rightarrow\Ls\text{sin}\theta)+(\odot\dot{\theta})\times(\searrow \Ls)=(\otimes\Ls\psio\text{sin}\theta)+(\nearrow\Ls\dot{\theta})</math>

:Thus, <math>\vel{G}{E}=(\downarrow\dot{\text{x}})+(\otimes\Ls\psio\text{sin}\theta)+(\nearrow\Ls\dot{\theta})</math>

:<math>\acc{Q}{E}=\dert{\vel{G}{E}}{E}=\dert{(\downarrow\dot{\text{x}})}{E}+\dert{(\otimes\Ls\psio\text{sin}\theta)}{E}+\dert{(\nearrow\Ls\dot{\theta})}{E}</math>

:The three terms of the velocity have variable value, but just the last two rotate (change their orientation) relative to the ground. The second one, which is perpendicular to the ring plane, rotates jst because of <math>\vec{\psio}</math>, whereas the third one rotates because of <math>\vec{\psio}</math> and <math>\vec{\dot{\theta}}</math>.The time derivatives of those terms are:

:<math>\dert{(\downarrow\dot{\text{x}})}{E}=[\text{change of value}]=(\downarrow\ddot{x})</math>

:<math>\dert{(\otimes\Ls\psio\text{sin}\theta)}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=[\otimes\Ls\psio\dot{\theta}\text{cos}\theta]+[(\Uparrow\psio)\times(\otimes\Ls\psio\text{sin}\theta)]=[\otimes\Ls\psio\dot{\theta}\text{cos}\theta]+[\leftarrow\Ls\psio^2\text{sin}\theta]</math>

:<math>\dert{(\nearrow\Ls\dot{\theta})}{E}=[\text{change of value}]+[\text{change of direction}]_\Es=[\nearrow\Ls\ddot{\theta}]+[((\Uparrow\psio)+(\odot\dot{\theta}))\times(\nearrow\Ls\dot{\theta})]=[\nearrow\Ls\ddot{\theta}]+[(\nwarrow\Ls\dot{\theta}^2)+(\otimes\Ls\psio\dot{\theta}\text{cos}\theta)]</math>

:Hence, <math>\acc{G}{E}=(\downarrow\ddot{x})+(\leftarrow\Ls\psio^2\text{sin}\theta)+(\nearrow\Ls\ddot{\theta})+(\nwarrow\Ls\dot{\theta}^2)+(\otimes 2\Ls\psio\dot{\theta}\text{cos}\theta)</math>

[[File:C4-E-Ex3-3-neut.png|thumb|right|300px|link=]]

:'''Analytical calculation: '''

:The whole calculation can be done analytically. The first term in <math>\OGvec=\vec{\Os\Os'}+\vec{\Os'\Gs}</math> is vertical, hence its projection on the vector basis B fixed to the support <math>(\velang{B}{E}=\vec{\psio})</math> is straightforward; however, the second term can be easily projected on the B’ vector basis fixed to the ring <math>(\velang{B'}{E}=\vec{\psio}+\vec{\dot{\theta}})</math>. Any of these two vector bases is suitable.

:Calculation with the B vector basis

:<math>\braq{\OGvec}{B}=\vector{\Ls\text{sin}\theta}{-\text{x}-\Ls\text{cos}\theta}{0}</math>

:<math>\braq{\vel{G}{E}}{B}=\braq{\dert{\OGvec}{E}}{B}=\frac{\ds}{\ds\ts}\braq{\OGvec}{B}+\braq{\velang{B}{E}}{B}\times\braq{\OGvec}{B}=</math>
:<math>=\vector{\Ls\dot{\theta}\text{cos}\theta}{-\dot{x}+\Ls\dot{\theta}\text{sin}\theta}{0}+\vector{0}{\psio}{0}\times\vector{\Ls\text{sin}\theta}{-x-\Ls\text{cos}\theta}{0}=\vector{\Ls\dot{\theta}\text{cos}\theta}{-\dot{x}+\Ls\dot{\theta}\text{sin}\theta}{-\Ls\psio\text{sin}\theta}</math>

:<math>\braq{\acc{G}{E}}{B}=\braq{\dert{\vel{G}{E}}{E}}{B}=\frac{\ds}{\ds\ts}\braq{\vel{G}{E}}{B}+\braq{\velang{B}{E}}{B}\times\braq{\vel{G}{E}}{B}=\vector{\Ls(\ddot{\theta}\text{cos}\theta-\dot{\theta}^2\text{sin}\theta)}{-\ddot{x}+\Ls(\ddot{\theta}\text{sin}\theta-\dot{\theta}^2\text{cos}\theta)}{-\Ls\psio\dot{\theta}\text{cos}\theta}+\vector{0}{\psio}{0}\times\vector{\Ls\dot{\theta}\text{cos}\theta}{-\dot{x}+\Ls\dot{\theta}\text{sin}\theta}{0}=\vector{\Ls(\ddot{\theta}\text{cos}\theta-\dot{\theta}^2\text{sin}\theta)}{-\ddot{x}+\Ls(\ddot{\theta}\text{sin}\theta+\dot{\theta}^2\text{cos}\theta)}{-2\Ls\psio\dot{\theta}\text{cos}\theta}</math>

:Calculation in the B’ vector basis

:<math>\braq{\OGvec}{B}=\vector{ x\text{sin}\theta}{- x\text{cos}\theta-\Ls}{0}</math>

:<math>\braq{\vel{G}{E}}{B'}=\braq{\dert{\OGvec}{E}}{B'}=\frac{\ds}{\ds\ts}\braq{\OGvec}{B'}+\braq{\velang{B'}{E}}{B'}\times\braq{\OGvec}{B'}=\vector{-\dot{x}\text{sin}\theta- x\dot{\theta}\text{cos}\theta}{-\dot{x}\text{cos}\theta+ x\dot{\theta}\text{sin}\theta}{0}+\vector{\psio\text{sin}\theta}{\psio\text{cos}\theta}{\dot{\theta}}\times\vector{- x\text{sin}\theta}{- x\text{cos}\theta-\Ls}{0}=\vector{-\dot{x}\text{sin}\theta+\Ls\dot{\theta}}{-\dot{x}\text{cos}\theta}{-\Ls\psio\text{sin}\theta}</math>

:<math>\braq{\acc{G}{E}}{B'}=\braq{\dert{\vel{G}{E}}{E}}{B'}=\frac{\ds}{\ds\ts}\braq{\vel{G}{E}}{B'}+\braq{\velang{B'}{E}}{B'}\times\braq{\vel{G}{E}}{B'}=\vector{-\ddot{x}\text{sin}\theta-\dot{x}\dot{\theta}\text{cos}\theta+\Ls\ddot{\theta}}{-\ddot{x}\text{cos}\theta+\dot{x}\dot{\theta}\text{sin}\theta}{-\Ls\psio\dot{\theta}\text{cos}\theta}+\vector{\psio\text{sin}\theta}{\psio\text{cos}\theta}{\dot{\theta}}\times\vector{-\dot{x}\text{sin}\theta+\Ls\dot{\theta}}{-\dot{x}\text{cos}\theta}{-\Ls\psio\text{sin}\theta}=\vector{-\ddot{x}\text{sin}\theta+\Ls(\ddot{\theta}-\psio^2\text{sin}\theta\text{cos}\theta)}{-\ddot{x}\text{cos}\theta+\Ls(\dot{\theta}^2+\psio^2\text{sin}^2\theta)}{-2\Ls\psio\dot{\theta}\text{cos}\theta}</math>

</div>

|}

-------------

'''*NOTE:''' In this web (for lack of more precise symbols), though the arrows s <math>\nearrow</math>, <math>\swarrow</math>, <math>\nwarrow</math> and <math>\searrow</math> seem to indicate that the vectors form a 45° angle with the vertical direction, this does not have to be the case. The arrows must be interpreted qualitatively, observing the figure that is always included when using this type of notation. For instance, in section 3 of exercise C2-E.1, the <math>\OPvec</math> vector forms a generic <math>\theta</math> angle with the vertical direction. If the value of <math>\theta</math> is less than 90° (as in the following figure), the <math>\OPvec</math> vector has a downward and rightward component.

© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |All rights reserved]]
-------------
-------------

<center>
[[C1. Configuration of a mechanical system|<<< C1. Configuration of a mechanical system]]

[[C3. Composition of movements|C3. Composition of movements >>>]]
</center>

D7. Examples of 3D dynamics

2026-02-26T16:25:38Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\mat}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{anella} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|690px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-02-26T16:24:55Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\mat}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{anella} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|500px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

File:D7-Ex5-5-eng.png

2026-02-26T16:24:04Z

Eantem: Eantem uploaded a new version of File:D7-Ex5-5-eng.png

D7-Ex5-5-eng

D7. Examples of 3D dynamics

2026-02-26T16:22:58Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
\newcommand{\Qs}{\textbf{Q}}
\newcommand{\Cs}{\textbf{C}}
\newcommand{\Ps}{\textbf{P}}
\newcommand{\Ss}{\textbf{S}}
\newcommand{\P}{\textrm{P}}
\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\mat}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
\newcommand{\vecdosd}[2]{
\begin{Bmatrix}
{#1}\\
{#2}
\end{Bmatrix}}
\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
\newcommand{\acc}[2]{\vecbf{a}_{\textrm{#2}} (\textbf{#1})}
\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{anella} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the P1 and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|450px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-02-26T16:21:05Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
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\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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\newcommand{\vec}[1]{\overline{#1}}
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\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
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\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
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\begin{Bmatrix}
{#1}\\
{#2}
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\newcommand{\vel}[2]{\vvec_{\textrm{#2}} (\textbf{#1})}
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\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
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\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
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\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{anella} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the arm and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|450px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

File:D7-Ex5-4-eng.png

2026-02-26T16:20:17Z

Eantem: Eantem uploaded a new version of File:D7-Ex5-4-eng.png

D7-Ex5-4-eng

D7. Examples of 3D dynamics

2026-02-25T17:48:36Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
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</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3._Interactions_between_rigid_bodies#D3.6_Interactions_through_linear_and_rotatory_actuators|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{anella} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the arm and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|350px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

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D7. Examples of 3D dynamics

2026-02-25T17:45:18Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
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\newcommand{\accs}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{s}} (\textbf{#1})}
\newcommand{\accn}[2]{\vecbf{a}_{\textrm{#2}}^{\textrm{n}} (\textbf{#1})}
\newcommand{\velo}[1]{\vvec_{\textrm{#1}}}
\newcommand{\accso}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{s}}}
\newcommand{\accno}[1]{\vecbf{a}_{\textrm{#1}}^{\textrm{n}}}
\newcommand{\re}[2]{\Re_{\textrm{#2}}(\textbf{#1})}
\newcommand{\psio}{\dot{\psi}_0}
\newcommand{\Pll}{\textbf{P}_\textrm{lliure}}
\newcommand{\agal}[1]{\vecbf{a}_{\textrm{Gal}} (#1)}
\newcommand{\angal}[1]{\vecbf{a}_{\textrm{NGal}} (#1)}
\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
\newcommand{\cth}{\text{cos}\theta}
\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3. Interaccions entre sòlids rígids#✏️ Exemple D3.18: actuador rotacional entre dos sòlids|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{anella} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the arm and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|350px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
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D7. Examples of 3D dynamics

2026-02-25T17:45:09Z

Eantem: /* ✏️ EXAMPLE D7.5: rotating ring */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
\newcommand{\Rs}{\textrm{R}}
\newcommand{\Ts}{\textrm{T}}
\newcommand{\Ls}{\textrm{L}}
\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
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\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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\newcommand{\vec}[1]{\overline{#1}}
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</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3. Interaccions entre sòlids rígids#✏️ Exemple D3.18: actuador rotacional entre dos sòlids|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

<center><math>{\vvec}_{\Ts}(J) = {\vvec}_{\Ts}(\textrm{G})+\vec{\Omega}_{\Ts}^\text{anella} \times \vec{\textrm{GJ}} = (\odot L \dot{\psi})+[(\Uparrow\dot{\psi})+(⇒\dot{\varphi_0})] \times (↓R) = (\odot L \dot{\psi}) + (\otimes R \dot{\varphi_0})</math></center>

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the arm and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|350px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

File:D7-Ex5-2-eng.png

2026-02-25T17:45:00Z

Eantem: Eantem uploaded a new version of File:D7-Ex5-2-eng.png

D7-Ex5-2-eng

D7. Examples of 3D dynamics

2026-02-25T12:18:49Z

Eantem: /* D7.1 Analysis of the equations of motion */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
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\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
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\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
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\newcommand{\vec}[1]{\overline{#1}}
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</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3. Interaccions entre sòlids rígids#✏️ Exemple D3.18: actuador rotacional entre dos sòlids|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the arm and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|350px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
-------------

<center>
[[D6. Exemples de dinàmica 2D|<<< D6. Exemples de dinàmica 2D]]

''EN CONSTRUCCIÓ'' >>>
</center>

D7. Examples of 3D dynamics

2026-02-25T12:17:27Z

Eantem: /* D7.1 Analysis of the equations of motion */

<div class="noautonum">__TOC__</div>
<math>\newcommand{\uvec}{\overline{\textbf{u}}}
\newcommand{\vvec}{\overline{\textbf{v}}}
\newcommand{\evec}{\overline{\textbf{e}}}
\newcommand{\Omegavec}{\overline{\mathbf{\Omega}}}
\newcommand{\velang}[2]{\Omegavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\Alfavec}{\overline{\mathbf{\alpha}}}
\newcommand{\accang}[2]{\Alfavec^{\textrm{#1}}_{\textrm{#2}}}
\newcommand{\As}{\textrm{A}}
\newcommand{\as}{\textrm{a}}
\newcommand{\ds}{\textrm{d}}
\newcommand{\ps}{\textrm{p}}
\newcommand{\Hs}{\textrm{H}}
\newcommand{\hs}{\textrm{h}}
\newcommand{\Ns}{\textrm{N}}
\newcommand{\Fs}{\textrm{F}}
\newcommand{\ms}{\textrm{m}}
\newcommand{\Ms}{\textrm{M}}
\newcommand{\ts}{\textrm{t}}
\newcommand{\cs}{\textrm{c}}
\newcommand{\us}{\textrm{u}}
\newcommand{\vs}{\textrm{v}}
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\newcommand{\Ts}{\textrm{T}}
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\newcommand{\Bs}{\textrm{B}}
\newcommand{\es}{\textrm{e}}
\newcommand{\fs}{\textrm{f}}
\newcommand{\is}{\textrm{i}}
\newcommand{\Is}{\textrm{I}}
\newcommand{\ks}{\textrm{k}}
\newcommand{\js}{\textrm{j}}
\newcommand{\qs}{\textrm{q}}
\newcommand{\rs}{\textrm{r}}
\newcommand{\ss}{\textrm{s}}
\newcommand{\Os}{\textbf{O}}
\newcommand{\Js}{\textbf{J}}
\newcommand{\Gs}{\textbf{G}}
\newcommand{\gs}{\textrm{g}}
\newcommand{\Cbf}{\textbf{C}}
\newcommand{\Or}{\Os_\Rs}
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\newcommand{\Ps}{\textbf{P}}
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\newcommand{\Q}{\textrm{Q}}
\newcommand{\deg}{^\textsf{o}}
\newcommand{\xs}{\textsf{x}}
\newcommand{\ys}{\textsf{y}}
\newcommand{\zs}{\textsf{z}}
\newcommand{\dert}[2]{\left.\frac{\ds{#1}}{\ds\ts}\right]_{\textrm{#2}}}
\newcommand{\ddert}[2]{\left.\frac{\ds^2{#1}}{\ds\ts^2}\right]_{\textrm{#2}}}
\newcommand{\vec}[1]{\overline{#1}}
\newcommand{\vecbf}[1]{\overline{\textbf{#1}}}
\newcommand{\vecdot}[1]{\overline{\dot{#1}}}
\newcommand{\CJvec}{\vec{\Cs\Js}}
\newcommand{\OQvec}{\vec{\Os\Qs}}
\newcommand{\QPvec}{\vec{\Qs\Ps}}
\newcommand{\OPvec}{\vec{\Os\textbf{P}}}
\newcommand{\OCvec}{\vec{\Os\Cs}}
\newcommand{\OGvec}{\vec{\Os\Gs}}
\newcommand{\abs}[1]{\left|{#1}\right|}
\newcommand{\braq}[2]{\left\{{#1}\right\}_{\textrm{#2}}}
\newcommand{\mat}[9]{
\begin{bmatrix}
{#1} & {#2} & {#3}\\
{#4} & {#5} & {#6}\\
{#7} & {#8} & {#9}
\end{bmatrix}}
\newcommand{\vector}[3]{
\begin{Bmatrix}
{#1}\\
{#2}\\
{#3}
\end{Bmatrix}}
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\begin{Bmatrix}
{#1}\\
{#2}
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\newcommand{\psio}{\dot{\psi}_0}
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\newcommand{\vgal}[1]{\vecbf{v}_{\textrm{Gal}} (#1)}
\newcommand{\fvec}[2]{\overline{\mathbf{F}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\mvec}[2]{\overline{\mathbf{M}}_{\textrm{#1}\rightarrow\textrm{#2}}}
\newcommand{\dth}{\dot{\theta}}
\newcommand{\ddth}{\ddot{\theta}}
\newcommand{\sth}{\text{sin}\theta}
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\newcommand{\sqth}{\text{sin}^2\theta}
\newcommand{\cqth}{\text{cos}^2\theta}
\newcommand{\I}[1]{\Is_\text{#1}}
</math>
<center>''EN CONSTRUCCIÓ''</center>

In this unit, the systematic procedure proposed in [[D6. Exemples de dinàmica 2D#D6.4 Diagrama general d’interaccions (DGI)|'''section D6.4''']] is applied to 3D dynamics examples to obtain equations of motion and motor torques. A systematic analysis of the equations of motion is also presented.

==D7.1 Analysis of the equations of motion==

Obtaining the equations of motion of the free DoF of multibody systems is not, in general, the objective of dynamics problems, but rather a previous step to their integration in order to know how the coordinates that describe the configuration of the system evolve over time:
<center>
<math>\ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is}</math>
</center>
However, these equations are often nonlinear, and their integration is necessarily numerical. Despite this, there are some aspects of the system's behavior that can be investigated analytically.

Equilibrium configurations

Equilibrium configurations are those configurations for which, if the system is left at rest <math>(\dot{\qs}_{\text{j,eq}} = 0)</math>, it remains at rest <math>(\ddot{\qs}_{\text{j,eq}} = 0)</math>. Therefore, the value of the coordinates in equilibrium is given by:

<center>
<math>\ddot{\qs}_i = f(\qs_j, \dot{\qs}_j, \ddot{\qs}_j, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>

<math>\downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0)</math>

<math>0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N</math>
</center>

The equation that defines the <math>\qs_{\text{j,eq}}</math> may be transcendental and not have an analytical solution. In this case, a numerical or graphical solution may be used.

Analysis of small oscillations about an equilibrium configuration

If the value of the coordinates is considered to be very close to that of an equilibrium configuration <math>(\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js,</math> with <math>\varepsilon_\js << 1</math>, hence <math>\varepsilon_\js^2 \approx 0</math>), the nonlinear functions appearing in the equations of motion can be approximated by the linear terms of their Taylor series expansion. For example:

:* if polynomials of degree greater than 1 appear:

::<math>\qs = \qs_{\text{eq}} + \varepsilon</math>

::<math>\qs^2 = (\qs_{\text{eq}} + \varepsilon)^2 = \varepsilon^2 + 2\qs_{\text{eq}}\varepsilon + \qs_{\text{eq}}^2\approx \varepsilon^2 + 2\qs_{\text{eq}} </math>

::<math>\qs^3 = (\qs_{\text{eq}} + \varepsilon)^3 = \varepsilon^3 + 3\qs_{\text{eq}}\varepsilon^2 + 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3\approx 3\qs_{\text{eq}}^2\varepsilon + \qs_{\text{eq}}^3</math>

:* if it is an angular coordinate <math>(\qs_\js=\theta)</math> and sine and cosine functions appear:

::<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\

\sth = \text{sin}(\theta_\text{eq} + \varepsilon) = \text{sin}\theta_\text{eq}\:\text{cos}\varepsilon + \text{cos}\theta_\text{eq}\:\text{sin}\varepsilon\\

\cth = \text{cos}(\theta_\text{eq} + \varepsilon) = \text{cos}\theta_\text{eq}\:\text{cos}\varepsilon -\text{sin}\theta_\text{eq}\:\text{sin}\varepsilon\\

\text{sin}\varepsilon = \varepsilon - (1/3!)\varepsilon^3 + ... \simeq \varepsilon\\

\text{cos}\varepsilon = 1 - (1/2)\varepsilon^2 + ... \simeq 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\text{sin}(\theta_\text{eq} + \varepsilon) \simeq\text{sin}\theta_\text{eq} + \varepsilon\text{cos}\theta_\text{eq}\\

\text{cos}(\theta_\text{eq} + \varepsilon) \simeq\text{cos}\theta_\text{eq} - \varepsilon\text{sin}\theta_\text{eq}
\end{cases}
</math>

Once linearized, the equation is of the form: <math>\As\ddot\varepsilon + \Bs\dot\varepsilon + \text{C}\varepsilon = 0</math>, where A, B, and C are scalars. In this course, however, the equation is often simpler and contains no first time derivative: <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math>. The general solution is: <math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi)</math>, with <math>\omega = \sqrt{\Bs/\As}</math>. The integration constants (<math>\as, \varphi</math>) depend on the initial conditions <math>\varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0</math>.

<div>
=====💭 Proof ➕=====

:The solution <math>\varepsilon(\ts)</math> of the equation <math>\As\ddot\varepsilon + \Bs\varepsilon = 0</math> must be a function whose second time derivative is proportional to the function before differentiation. The sine, cosine and exponential functions satisfy this condition. If the first one (with two integration constants) is tested:

:<math>\varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi), \:\:\ \dot\varepsilon(\ts) = \as\omega\text{cos}(\omega\ts + \varphi), \:\: \ddot\varepsilon(\ts) = -\as\omega^2\text{sin}(\omega\ts + \varphi)</math>

:If these expressions are substituted into the equation of motion:

:<math>-\As\as\omega^2\text{sin}(\omega\ts + \varphi) +\Bs\as\text{sin}(\omega\ts + \varphi) = 0\Rightarrow\omega = \sqrt{\frac{B}{A}}</math>

:The motion is an oscillation around the equilibrium configuration <math>\qs_{\es\qs}</math> whose angular frequency <math>(\omega)</math> depends on system parameters.

:The integration constants, on the other hand, depend on the initial conditions (<math>\varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi)</math>, <math>\dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 = \as\omega\text{cos}(\varphi)</math>), and therefore, are not intrinsic to the system.

:* position initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0) = 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 90\deg\\
\as = \varepsilon_0
\end{aligned}\right.</math>

:* velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0) = 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\varphi = 0\deg\\
\as = \dot\varepsilon_0/\omega
\end{aligned}\right.</math>

:* position and velocity initial conditions:
:<math>\left.\begin{aligned}
\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0 \\
\dot\varepsilon(\ts = 0)\equiv \varepsilon_0\neq 0
\end{aligned}\right\} \Rightarrow\left\{\begin{aligned}
\text{tan}\varphi = \omega(\varepsilon_0/\dot\varepsilon_0)\\
\as = \sqrt{\varepsilon_0^2 + (\varepsilon_0/\omega)^2}
\end{aligned}\right.</math>

</div>

Analysis of the stability of small oscillations about an equilibrium configuration

Oscillations about an equilibrium configuration are only possible when that configuration is stable.
Stability can be analyzed very easily from the linearized equation of motion:

<math>\As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon</math>
:* <math>(\Bs/\As) > 0 \Rightarrow\ddot\varepsilon<0\Rightarrow\varepsilon(\ts)</math> decreases, and the system returns to the equilibrium configuration <math>\qs_{\es\qs}</math>. It is a '''STABLE ''' behaviour.

:* <math>(\Bs/\As) < 0 \Rightarrow\ddot\varepsilon > 0\Rightarrow\varepsilon(\ts)</math> increases, and the system moves away from the equilibrium configuration <math>\qs_{\es\qs}</math>. It is '''UNSTABLE''' behaviour.

==D7.2 General examples==
====✏️ EXAMPLE D7.1: rotating rectangular plate====
------

:[[File:D7-Ex1-1-eng.png|thumb|left|150px|link=]]

:The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. We want to find the equation of motion associated with the movement between the plate and the fork, and the value of the torque that guarantees a constant <math>\Omega_0</math>.
[[File:D7-Ex1-2-neut.png|thumb|right|200px|link=]]
:Kinematic description:
:<math>\boxed{\begin{align}
\text{AB:ground}\\
\text{REL:fork}
\end{align}}</math>

:<math>\velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth)</math>

:<math>\vel{G}{E} = \vel{G}{AB} = \vel{G}{REL} + \vel{G}{ar} = (\nearrow 2\Ls\dth) + (\otimes 2\Ls\Omega_0\sth)</math>

:Another option to calculate <math>\vel{G}{E}</math> is rigid body kinematics (rigid body: plate):

:<math>\vel{G}{E} = \vel{O}{E} + \OGvec\times\velang{plate}{E} = (\searrow 2\Ls)\times[(\Uparrow\Omega_0) + (\odot\dth)] = (\otimes 2\Ls\Omega_0\sth) + (\nearrow 2\Ls\dth)</math>

:General diagram of interactions
:[[File:D7-Ex1-3-eng.png|thumb|right|220px|link=]]

:It is a two-DoF system (forced <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per solid. The problema is determinate:

:equations: 2 rigid bodies <math>\times\frac{6\text{ eqs}}{\text{r.body}} = 12\text{eqs}</math>

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends on <math>\dth</math>. Therefore, the systems where <math>\ddth</math> will appear when applying the vector theorems are: plate, plate + fork.

:[[File:D7-Ex1-4-eng.png|thumb|center|420px|link=]]
<center> 5 constraint unk. + <math>\ddth = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma + \ddth = </math> 7 unk.</center>
:The six equations generated when applying the vector theorems to the plate allow the calculation of the 6 unknowns, while in the other option, the number of unknowns exceeds the number of equations that can be generated. The external interactions on the plate are:
:[[File:D7-Ex1-5-neut.png|thumb|left|250px|link=]]

:The characterization of the constraint torsor of the fork on the plate at point <math>\Os</math> is straightforward whether the base B or B’ is used.

:If the AMT is applied, the three components include constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

:A good proposal is: <math>\boxed{\text{Roadmap: SYSTEM(plate), AMT at }\Os]_{3=3'}}</math>

:AMT at <math>\Os: \:\: \sum\vec{\Ms}_{\text{ext}}(\Os) = \dot{\vec{\Hs}}_\text{RTO}(\Os)</math>

:<math>\Os\in</math> plate: <math>\vec{\Hs}_\text{RTO}(\Os) = \Is\Is(\Os)\velang{plate}{RTO=E} = \Is\Is(\Os)[(\Uparrow\Omega_0) + \odot\dth]</math>

:In order to have an inertia tensor with constant terms, it is convenient to use the B vector basis fixed to the plate:

:<math>[\Is\Is(\Os)]_\Bs = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}</math>, with <math>\left\{\begin{aligned}
\I{low} = (4/3)\ms\Ls^2 \\
\I{large} = (16/3)\ms\Ls^2
\end{aligned}\right.</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \mat{\I{large}}{0}{0}{0}{\I{low}}{0}{0}{0}{\I{large + low}}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = \vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = (\I{large} + \I{low})\ddth + (\I{large} - \I{low})\Omega_0^2\sth\cth\\
\sum\vec{\Ms}_{\text{ext}}(\Os)]_3 = -\ms\gs 2\Ls\sth
\end{aligned}\right\} \Rightarrow \boxed{(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{low} - \I{large})\Omega_0^2\cth\right]\sth = 0}</math>

:If <math>\I{large}</math> and <math>\I{low}</math> are substituted and by the values given in the tables, the equation becomes:

:<math>\boxed{\frac{10}{3}\ddth + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth\right)\sth = 0}</math>

:Analysis of the equation of motion: equilibrium configurations

:<math>\left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>. This equation has two families of solutions:

:<math>\left\{\begin{aligned}
\sth_\text{eq} = 0\Rightarrow \theta_\text{eq} = 0,180\deg\\
\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq} = 0\Rightarrow \cth_\text{eq} = \frac{\gs}{2\Ls\Omega_0^2}
\end{aligned}\right.</math>

:Since the cosine function is bounded between -1 and +1, the second family only exists if <math>\frac{\gs}{2\Ls\Omega_0^2} \leq 1</math>, and this is true only if the angular velocity <math>\Omega_0</math> is above the critical value <math>\Omega_\text{cr} = \sqrt{\frac{\gs}{2\Ls}}</math>.

:Analysis of the equation of motion: motion of the plate relative to the fork

:Since this is a nonlinear equation (due to the sine function), the general motion is obtained by numerical integration.

:Analytical analysis for small amplitudes <math>(\varepsilon)</math> about an equilibrium configuration <math>\theta_\text{eq}</math> can be done by approximating the trigonometric functions [[D7. Examples of 3D dynamics#D7.1 Analysis of the equations of motion|('''section D7.1''')]]:

:<math>\left.\begin{aligned}
\theta = \theta_\text{eq} + \varepsilon\\
\varepsilon^2\approx 0
\end{aligned}\right\}\Rightarrow \frac{10}{3}\ddot{\varepsilon} + \left[\frac{\gs}{\Ls} + 2\Omega_0^2(\text{sin}^2\theta_\text{eq} - \text{cos}^2\theta_\text{eq})\right]\varepsilon + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\cth_\text{eq}\right)\sth_\text{eq} = 0</math>

:For small amplitudes around <math>\theta_\text{eq} = 0</math>, , the equation of motion is <math>\frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0</math>.

:If the initial conditions are <math>(\dot\varepsilon = 0, \varepsilon\neq 0)</math>, , the time evolution of <math>\varepsilon</math> is given by: <math>\ddot{\varepsilon} = \frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)\varepsilon</math>.

:* For <math>\Omega_0^2 > \frac{\gs}{2\Ls},\: \ddot\varepsilon > 0\Rightarrow \varepsilon</math> '''increases'''. It is an '''UNSTABLE''' configuration, and no oscillation around this configuration is possible.

:* For <math>\Omega_0^2 < \frac{\gs}{2\Ls},\: \ddot\varepsilon < 0\Rightarrow \varepsilon</math> '''decreases'''. . The movement is an oscillation about a '''STABLE'''. configuration. The angular frequency [rad/s] is <math>\omega = 2\pi\fs = \sqrt{\frac{3}{10}\left(2\Omega_0^2 - \frac{\gs}{\Ls}\right)}</math>.

:NOTE: If the initial conditions are <math>\theta(\ts = 0) = 0</math> and <math>\dot\theta(\ts = 0) = 0</math>, the pendulum motion does not appear, and the system moves only according with the rotation <math>\Omega_0</math>.

:ANIMACIO

:Additional comment

:If the plate had been suspended from the fork so that axis 2 was the one with a large moment of inertia and axis 1 was the one with a low moment of inertia, the equation of motion would have been:

[[File:D7-Ex1-6-neut.png|thumb|right|170px|link=]]

:<math>(\I{large} + \I{low})\ddth + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\cth\right]\sth = 0</math>

:If linearized around the configuration <math>\theta_\text{eq} = 0</math>:

:<math>(\I{large} + \I{low})\ddot\varepsilon + \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]\varepsilon = 0</math>

:For all values of <math>\Omega_0</math>, the coefficient <math>\left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right]</math> is positive, hence the <math>\theta_\text{eq} = 0</math> configuration is always '''STABLE'''.

:ANIMACIO

:Roadmap for the motor torque

:There are two options for calculating the motor torque: fork, fork + plate:
[[File:D7-Ex1-7-eng.png|thumb|center|450px|link=]]
<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:</math>10 constraint unk. + <math>\Gamma = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\Gamma = </math> 6 unk.
[[File:D7-Ex1-8-neut.png|thumb|right|120px|link=]]
:The option (fork + plate) is the most suitable. The description of external interactions on this system is shown in the figure.

:Since the unknown is a torque, the AMT is the right theorem to arrive at the solution:

:<math>\boxed{\text{Roadmap: SYSTEM (fork + plate), AMT at }\Os]_\text{vert = 2'}}</math>

:The angular momentum is the same as that calculated before (since the fork has no mass).

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\}_\Bs = \vector{\I{large}\Omega_0\dth\cth}{-\I{low}\Omega_0\dth\sth}{(\I{large} + \I{low})\ddth} + \vector{\Omega_0\sth}{\Omega_0\cth}{\dth}\times\vector{\I{large}\Omega_0\sth}{\I{low}\Omega_0\cth}{(\I{large} + \I{low})\dth} = \vector{2\I{large}\Omega_0\dth\cth}{-2\I{low}\Omega_0\dth\sth}{...} = \frac{8}{3}\ms\Ls^2\vector{4\Omega_0\dth\cth}{-\Omega_0\dth\sth}{...}</math>

:<math>\left.\begin{aligned}
\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_\text{vert} = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 \sth + \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_2 \cth\\

\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_\text{vert} = \Gamma
\end{aligned}\right\}

\Rightarrow \boxed{\Gamma =\frac{8}{3}\ms\Ls^2\Omega_0\dth(4\text{cos}^2\theta - \text{sin}^2\theta)}</math>


====✏️ EXAMPLE D7.2: rotating bars====
-----

[[File:D7-Ex2-1-eng.png|thumb|left|170px|link=]]
:The system consists of a massless frame and two identical homogeneous bars attached to the frame. The part is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The revolute joint between the frame and the fork allows a free DoF (rotation <math>\sth</math> with axis orthogonal to the frame), but we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of motion for the <math>\theta</math> coordinate can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant)..
[[File:D7-Ex2-2-eng.png|thumb|right|170px|link=]]
:General diagram of interactions
:It is a 2-DoF system (forced <math>\omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. On the other hand, it contains 2 rigid bodies, and the two vector theorems generate 6 equations per rigid body. It is a determinate problem:

:equations: 2 rigid bodies <math>\times\frac{6\text{eqs.}}{\text{r. body}} = 12</math> eqs.

:unknowns: 2 associated with the DoF + 10 constraint unk. = 12 unk.

:Roadmap for the equation of motion

:The plate is the only element whose movement depends <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork.

[[File:D7-Ex2-3-eng.png|thumb|center|350px|link=]]

:The six equations generated by applying the vector theorems to the rigid body allow the calculation of the 6 unknowns, while in the other option the number of unknowns exceeds the number of equations that can be generated. The external interactions on the piece are shown in the figure.
:If the AMT is applied, all three components contain constraint unknowns. If the AMT is applied at <math>\Os</math>, the constraint force will not appear. Hence

<center><math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os}</math></center>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO=E}= \Is\Is(\Os)(\Uparrow\Omega_0)</math>

:In order to have an inertia tensor of constant terms, it is convenient to use the B vector basis fixed to the frame. Since we assumed that the movement corresponds only to the forced GL, that vector basis rotates relative to the ground with angular velocity <math>\vec\Omega_0</math>.

:The qualitative analysis of the inertia tensor can be done by first considering the tensor of each bar at its center of inertia and then adding the [[D5. Geometria de masses#D5.4 Algunes propietats rellevants del tensor d’inèrcia|'''Steiner''']] corrections to move to <math>\Os</math>:

[[File:D7-Ex2-4-neut.png|thumb|center|350px|link=]]
:<math>[\Is\Is(\Os)] = \mat{\Is}{+|\Is_{12}|}{0}{+|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \mat{\Is}{-|\Is_{12}|}{0}{-|\Is_{12}|}{\Is}{0}{0}{0}{2\Is} + \ms\Ls^2\mat{9/2}{-3/2}{0}{-3/2}{1/2}{0}{0}{0}{5} + \ms\Ls^2\mat{1/2}{1/2}{0}{1/2}{1/2}{0}{0}{0}{1} = \mat{2\Is + 5\ms\Ls^2}{-\ms\Ls^2}{0}{-\ms\Ls^2}{2\Is + \ms\Ls^2}{0}{0}{0}{4\Is + 6\ms\Ls^2}</math>

[[File:D7-Ex2-5-neut.png|thumb|right|150px|link=]]

: Taking into account that <math>2\Is = \frac{1}{3}\ms\Ls^2 :</math> <math>[\Is\Is(\Os)] = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{0}{\Omega_0}{0} = \frac{1}{3}\ms\Ls^2\vector{-3\Omega_0}{4\Omega_0}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has a constant value and is contained in the frame plane. Therefore, it rotates with <math>\vec\Omega_0</math> relative to the ground and sweeps a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative of comes from this change in direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Leftarrow\ms\Ls^2\Omega_0) + (\Uparrow\I{large}\Omega_0)] = (\odot\ms\Ls^2\Omega_0^2)</math>

:The time derivative can also be calculated analyically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \{\velang{B}{RTO}\}\times\{\vec{\Hs}_{\text{RTO}}(\Os)\} = \vector{0}{\Omega_0}{0}\times\vector{-\ms\Ls^2\Omega_0}{\I{large}\Omega_0}{0} = \vector{0}{0}{\ms\Ls^2\Omega_0}</math>

:The only moment about point <math>\Os</math> external to the rigid body is the constraint moment associated with the revolute joint:

:<math>\sum\vec{\Ms}_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>.

:None of those two componentes is consistent with the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\odot\ms\Ls^2\Omega_0^2)❗️❗️</math>

:Conclusion: the motion that we were looking for (without the rotation <math>\dth</math> of the frame relative to the fork but keeping a constant <math>\Omega_0</math>) is not possible. The reason is the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> horizontal component, which is the one that generates <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0</math>. If <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> were strictly vertical (parallel to<math>\vec\Omega_0</math>), then <math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os)=\vec 0</math>, and the application of the AMT would lead to zero value of the two moment components <math>\Ms_1 = \Ms_2 = 0</math>. In other words: if the direction of the angular velocity were a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|''' principal direction of inertia''']] for point <math>\Os</math>, keeping it constant would be possible without the need for an external moment.

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force on the frame that generates the required moment. For example, the following forces could be applied with a single finger:
[[File:D7-Ex2-7-neut.png|thumb|center|350px|link=]]

:The value of the two forces is different, but the direction of the moment they exert about <math>\Os</math> is the same.

:As long as the finger introduces one of these forces, <math>\Omega_0</math> remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of <math>\dth</math>). According to the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts on the finger the same force but in opposite direction (that is, the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a clockwise direction:
[[File:D7-Ex2-8-neut.png|thumb|center|400px|link=]]

ANIMACIONS
<div>
=====ALTERNATIVE=====

:[[File:D7-Ex2-9-neut.png|thumb|left|150px|link=]]

:The initial direction of the deflection can be investigated from the equation of motion for the <math>\theta</math> coordinate, which can be found with the <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at}\Os]_3}</math>:

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\mat{16}{-3}{0}{-3}{4}{0}{0}{0}{20}\vector{-\Omega_0\sth}{\Omega_0\cth}{-\dth}</math>

:<math>\left\{\vec{\Hs}_\text{RTO}(\Os)\right\} = \frac{1}{3}\ms\Ls^2\vector{-\Omega_0(16\sth + 3\cth)}{\Omega_0(3\sth + 4\cth)}{-20\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = -\frac{20}{3}\ms\Ls^2\ddth + \ms\Ls^2\Omega_0^2(4\sth\cth + \text{cos}^2\theta - \text{sin}^2\theta)</math>
:<math>\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth</math>.

:The equation of motion is:

:<math>\frac{20}{3}\ddth - \Omega_0^2\:[2\text{sin}(2\theta) + \text{cos}(2\theta)] + \frac{\gs}{\Ls}\sqrt{2}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{eq} = 0)</math> are the solutions of the transcendental equation:

:<math>\Omega_0^2\:[2\text{sin}(2\theta_\text{eq}) + \text{cos}(2\theta_\text{eq})] = \frac{\gs}{\Ls}\sqrt{2}\text{sin}\theta_\text{eq}</math>, and it is evident that, if <math>\Omega_0\neq 0</math>, <math>\theta_\text{eq} = 0</math> is not one of them.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting them into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>\frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2>0</math>.

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the deviation <math>\theta</math> that has been represented in the previous figure. Therefore, a clockwise rotation appears.

</div>


====✏️ EXAMPLE D7.3: rotating frame with particles====
-----
:
:[[File:D7-Ex3-1-eng.png|thumb|left|150px|link=]]
:The system consists of a massless frame and two identical particles attached to it. The frame is articulated to a massless fork which rotates with constant angular velocity <math>\Omega_0</math> relative to the ground under the action of a motor. The articulation between the frame and the fork allows a free DoF (<math>\dth</math> rotation with axis orthogonal to the frame), and we want to investigate whether it is possible for this movement not to occur (therefore, investigating whether the equation of the <math>\theta</math> coordinate movement can be simply <math>\ddth = 0</math> and that the <math>\Omega_0</math> rotation remains constant..

:General diagram of interactions

:It is the same kind of system as in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']]: it has 2 DoF (forces <math>\Omega_0</math>, free <math>\dth</math>) with 10 constraint unknowns. The number of equations that can be generated if the vector theorems are applied to the two rigid bodies is 12: it is a determinate problem.

:Roadmap for the equation of motion
:The rigid body (frame + particles) is the only element whose movement would depend on <math>\dth</math>. Therefore, the systems in which <math>\ddth</math> would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in [[D7. Examples of 3D dynamics#✏️ EXAMPLE D7.2: rotating bars|'''example D7.2''']], a suitable roadmap is:

[[File:D7-Ex3-2-neut.png|thumb|right|150px|link=]]

:<math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os}</math>

:AMT at <math>\Os:\:\sum\vec{\Ms}_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os) </math>
:<math>\Os\in</math> rigid body: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0)</math>

[[File:D7-Ex3-3-neut.png|thumb|right|150px|link=]]
:The inertia tensor in the B vector basis fixed to the frame is straightforward:

:<math>[\Is\Is(\Os)] = [\Is\Is^\text{up. part.}(\Os)] + [\Is\Is^\text{low. part.}(\Os)] = \ms\Ls^2\left(\mat{0}{0}{0}{0}{1}{0}{0}{0}{1} + (\mat{4}{2}{0}{2}{1}{0}{0}{0}{5}\right)</math>
:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2 \mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{0}{\Omega_0}{0} = 2\ms\Ls^2\Omega_0\vector{1}{1}{0}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> has constant value, is contained in the frame plane, and rotates relative to the ground with <math>\vec\Omega_0</math> while sweeping a conical surface. The <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> time derivative comes from this change of direction:

:<math>\dot{\vec{\Hs}}_{\text{RTO}}(\Os) = \velang{H($\Os$)}{RTO}\times\vec{\Hs}_{\text{RTO}}(\Os) = (\Uparrow\Omega_0)\times[(\Rightarrow2\ms\Ls^2\Omega_0) + (\Uparrow\ms\Ls^2\Omega_0)] = (\otimes 2\ms\Ls^2\Omega_0^2)</math>
[[File:D7-Ex3-4-neut.png|thumb|right|150px|link=]]
:The time derivative can also be obtained analytically:

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \left\{ \velang{B}{RTO}\right\}\times\left\{ \vec{\Hs}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times 2\ms\Ls^2\Omega_0\vector{1}{1}{0} = \vector{0}{0}{-2\ms\Ls^2\Omega_0^2}</math>

:The inertia center of the rigid body is located on the vertical line through <math>\Os</math>, and therefore the only moment about point <math>\Os</math> external to the rigid body is the constaint moment associated with the revolute joint:

:<math>\sum\vec\Ms_\text{ext}(\Os) = (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)</math>

:Neither of these two components can provide the time derivative of the angular momentum:

:<math>(\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2)</math>

:As in [[D7. Examples of 3D dynamics#EXAMPLE D7.2: rotating bars|'''example D7.2''']], the intended motion (without rotation of the frame relative to the fork but keeping <math>\Omega_0</math> constant) is not possible because the direction of the angular velocity is not a [[D5. Geometria de masses#D5.3 Eixos principals d'inèrcia|'''principal direction of inertia''']].

:Constant vertical rotation <math>\Omega_0</math> can be achieved by applying a force to the frame that generates the required moment. For example, the following forces could be applied with just a finger:

[[File:D7-Ex3-5-neut.png|thumb|center|350px|link=]]

:The values of the two forces are different, but the direction of the moment they generate about <math>\Os</math> is the same.

:While one of these forces is introduced, <math>\Omega_0</math> remains constant without changing the orientation of the frame relative to the horizontal plane. By the [[D1. Lleis fundacionals de la mecànica newtoniana#D1.6 Tercera llei de Newton (principi d’acció i reacció)|'''principle of action and reaction''']], the frame exerts the same force on the finger but in the opposite direction (i.e., the frame “rests” on the finger). If at any time the finger is removed, the frame is left without support and deviates from its initial orientation in a counterclockwise direction:

[[File:D7-Ex3-6-neut.png|thumb|center|400px|link=]]

ANIMACIONS

<div>
=====ALTERNATIVE=====
:[[File:D7-Ex3-7-neut.png|thumb|left|150px|link=]]
:The initial direction of the deviation can be investigated from the equation of motion for the <math>\theta</math> coordinate , which can be found according to the following roadmap: <math>\boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3}</math>:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = 2\ms\Ls^2\mat{2}{1}{0}{1}{1}{0}{0}{0}{3}\vector{\Omega_0\sth}{\Omega_0\cth}{\dth} = 2\ms\Ls^2\vector{\Omega_0(2\sth + \cth)}{\Omega_0(\sth + \cth)}{3\dth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_3 = 3\ms\Ls^2\ddth + 2\ms\Ls^2\Omega_0(\sqth - \cqth - \sth\cth)</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_3 = -2\ms\gs\Ls\sth</math>

:The equation of motion is:
:<math>3\ddth - \Omega_0^2\left[\text{cos}(2\theta) + \frac{1}{2}\text{sin}(2\theta)\right] + \frac{\gs}{\Ls}\sth = 0</math>

:The equilibrium configurations <math>(\ddth_{\es\qs} = 0)</math> are the solutions of the transcendent equation:

:<math>\Omega_0^2\left[\text{cos}(2\theta_{\es\qs}) + \frac{1}{2}\text{sin}(2\theta_{\es\qs})\right] = \frac{\gs}{\Ls}\text{sin}\theta_{\es\qs}</math>, and it is evident that <math>\theta = 0</math> is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

:The initial conditions are: <math>\theta(\ts=0) = 0</math>, <math>\dth\:(\ts=0) = 0</math>. Substituting those values into the equation of motion, the initial acceleration <math>\ddth\:(\ts=0)</math> can be determined:

:<math>3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2>0</math>

:The fact that <math>\ddth\:(\ts=0)>0</math> indicates that it has the same direction as the <math>\theta</math> deviation that has been shown in the previous figure. Hence, a counterclockwise rotation appears.

ANIMATIONS
</div>



====✏️ EXAMPLE D7.4: rotating ball====
-----

:
:[[File:D7-Ex4-1-eng.png|thumb|left|200px|link=]]

:The ball, with mass m and radius r, keeps a single-point contact with the ground without sliding, and is articulated to a horizontal arm. The arm is articulated to a fork that rotates with constant angular velocity under the action of a motor. The arm and fork have negligible mass. The coefficient of friction in the radial direction between the ball and the ground is zero <math>(\mu_\text{rad} = 0)</math>. The aim is to investigate whether rotation <math>\Omega_0</math> can cause the loss of contact between the ball and the ground.

:Kinematic description 
: The [[C4. Rigid body kinematics#C4.3 Geometry of the velocity distribution: Instantaneous Screw Axis (ISA)|'''ISA''']] of the ball relative to the ground is the straight line <math>\Os\Js</math>, and the angular velocity can be decomposed into two Euler rotations:
[[File:D7-Ex4-2-eng.png|thumb|center|400px|link=]]

:General diagram of interactions 
:[[File:D7-Ex4-3-eng.png|thumb|left|200px|link=]]
:t is a system with one DoF and 17 constraint unknowns. As it consists of three rigid bodies, it is a determinate problem:

:(17 constraint unk., 1DoF)<math>\Rightarrow</math> 18 unknowns 
:3 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math>= 18 equations

:The description of the system can be simplified by treating the arm as [[D3. Interaccions entre sòlids rígids#D3.5 Interaccions d’enllaç indirectes: Sòlids Auxiliars d’Enllaç|'''CAE''']], since it has no mass and it only undergoes constraint interactions:

[[File:D7-Ex4-4-eng.png|thumb|center|400px|link=]]

:The number of constraint unknowns associated with the indirect constraint between ball and fork through the arm is 4, since the ball has two independent rotations <math>(\Omega_3,\Omega_1)</math> relative to the fork. The problem remains determinate:

:(11 constraint unk., 1DoF)<math>\Rightarrow</math> 12 unknowns 
:2 rigid bodies <math>\times\frac{6\text{ eqs.}}{\text{r. body}}</math> = 12 equations

:Roadmap to study contact loss
:Contact loss implies the suppression of the constaint between the ball and the ground, therefore the cancellation of the normal force N that the ground exerts on the ball. The two systems to which the vector theorems can be applied to calculate N are: ball, ball + (arm) + fork.

[[File:D7-Ex4-5-eng.png|thumb|center|450px|link=]]

:[[File:D7-Ex4-6-neut.png|thumb|left|200px|link=]]

:The best option is to apply the vector theorems to the ball, since the number of unknowns is equal to the number of equations that can be generated. The external interactions on the ball are:

:If the AMT is applied at <math>\Os</math>, the constraint forces associated with the revolute joint between arm and fork will not appear, and the only moments in the direction perpendicular to the drawing will be associated with the weight and N. Therefore
:<math>\boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}}</math>

:The angular momentum <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> can be calculated from the inertia tensor at <math>\Os</math> (since <math>\Os</math> is fixed to the ball) or through barycentric decomposition. In the latter case, since the ball is a spherical rotor at <math>\Gs</math>, <math>\vec{\Hs}_{\text{RTO}}(\Gs)</math> is parallel to the angular velocity of the ball:

:[[File:D7-Ex4-7-eng.png|thumb|left|200px|link=]]
:<math>\begin{aligned}
\vec{\Hs}_{\text{RTO}}(\Os) &= \vec{\Hs}_{\text{RTG}}(\Gs) + \OGvec\times\ms\vel{G}{RTO} =\\
&= \left(\Uparrow\frac{2}{5}\ms\rs^2\Omega_0\right) + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) + (\rightarrow\Ls)\times(\otimes\ms\Ls\Omega_0)=\\
&=\left[\Uparrow\ms\left(\frac{2}{5}\rs^2 + \Ls^2\right)\Omega_0\right] + \left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right)
\end{aligned} </math>

:The two components are constant in value, but the horizontal component changes direction due to the vertical rotation<math>\vec{\Omega_0}</math>:

:<math>\left.\begin{aligned}
\dot{\vec{\Hs}_{\text{RTO}}(\Os)} = (\Uparrow\Omega_0)\times\left(\Leftarrow\frac{2}{5}\ms\rs\Ls\Omega_0\right) = \left(\odot\frac{2}{5}\ms\rs\Ls\Omega_0^2\right)\\
\left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_{\perp\text{drawing}} = (\odot\Ns\Ls)+(\otimes\ms\gs\Ls) = [\odot(\Ns-\ms\gs)\Ls]
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\frac{2}{5}\rs\Omega_0^2\right)} </math>
<math></math>
:The normal force increases with the angular velocity, hence contact with the ground is always maintained.

:Alternative:

:If the <math>\vec{\Hs}_{\text{RTO}}(\Os)</math> calculation is done from <math>\Is\Is(\Os)</math> and the time derivative is done analytically:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = ([\Is\Is(\Gs)]+[\Is\Is^\oplus(\Os)])\velang{ball}{RTO=E}</math>

:<math>\begin{aligned}\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} &= \left(\frac{2}{5}\ms\rs^2\mat{1}{0}{0}{0}{1}{0}{0}{0}{1} + \ms\Ls^2\mat{0}{0}{0}{0}{1}{0}{0}{0}{1}\right)\vector{(\Ls/\rs)\Omega_0}{\Omega_0}{0}=\\
&=\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2+\Ls^2}{0}
\end{aligned}</math>

<math></math>

:<math>\left.\begin{aligned}
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{0}{\Omega_0}{0}\times\ms\Omega_0\vector{(2/5)\rs\Ls}{(2/5)\rs^2 +\Ls^2}{0} = \vector{0}{0}{-\ms\Ls\Omega_0^2}\\
\left\{\sum\vec{\Ms}_\text{ext}(\Os)\right\} = \vector{\Ms_1}{\Ms_2}{0} + \vector{0}{0}{\ms\gs\Ls} + \vector{-\Ts\rs}{\Ts\Ls}{-\Ns\Ls}
\end{aligned}\right\}\Rightarrow \boxed{\Ns = \ms\left(g+\rs\Omega_0^2\right)} </math>



====✏️ EXAMPLE D7.5: rotating ring ====
-----

:

:[[File:D7-Ex5-1-eng.png|thumb|left|200px|link=]]

:The ring, with mass m and radius R, rotates with constant angular velocity <math>\dot\varphi_0</math> relative to the massless fork driven by a motor that has the stator (P1) articulated to the fork, and the rotor (P2) fixed to the ring. The fork can rotate relative to the ground with angular velocity <math>\dot\psi</math>. Initially, <math>\dot\psi(t=0) = 0</math>. We want to investigate the evolution of this initial condition, whether the sliding between the ring and the ground will disappear at some point, and the value of the motor torque that guarantees a constant <math>\dot\varphi_0</math> while there is sliding.

:Kinematic description and general diagram of interactions 

:For the more general motion in the sliding phase, <math>\dot\varphi_0</math> and <math>\dot\psi</math> are independent, and the description of the system velocities and the GDI ([[D3. Interaccions entre sòlids rígids#✏️ Exemple D3.18: actuador rotacional entre dos sòlids|'''EXAMPLE D3.18''']]) are:

[[File:D7-Ex5-2-eng.png|thumb|center|250px|link=]]

:As for the dynamic description, there are two options depending on whether the direct constraints between the fork and the motor stator (P1), and the stator (P1) and the rotor (P2, fixed to the ring) are considered (option 1), or whether the indirect constraint between the fork and the ring is considered (option 2) ([[D3. Interaccions entre sòlids rígids#D3.6 Interaccions per mitjà d’actuadors lineals i rotacionals|'''section D3.6''']]):

[[File:D7-Ex5-3-eng.png|thumb|center|450px|link=]]

:In both cases, it is a determinate problem:

:Option 1: (16 constraint unk., 2DoF)<math>\Rightarrow</math> 18 unknowns, 3 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 18 equations

:Opció 2: (10 constraint unk., 2DoF)<math>\Rightarrow</math> 12 unknowns, 2 rigid bodies<math>\times\frac{\text{6 eqs.}}{\text{r. body}} =</math> 12 equations

:<big>'''OPTION 1'''</big>
:In this option, the DGI can be simplified by treating the fork as CAE (since it has zero mass and it only undergoes constraint interactions). The indirect constraint between the arm and the ceiling (floor) introduces 4 unknowns, since it allows two independent rotations between them:

[[File:D7-Ex5-4-eng.png|thumb|center|350px|link=]]

:Roadmap for the equation of motion of the <math>\psi</math> coordinate
:Both the ring and the stator(P1) motion depend on <math>\psi</math>. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

[[File:D7-Ex5-5-eng.png|thumb|center|600px|link=]]
<center> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma + \ddot\psi = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot\psi = </math> 6 unk.</center>
:The best option is (ring+P1). The characterization of the indirect constraint between P1 and the ceiling is straightforward: since it allows two independent rotations in directions 2 and 3 of P1 relative to the ground, the torsor will contain three force components and one moment component (in direction 1). The external interactions on the system (ring+P1) are:

:If the AMT is applied at <math>\Os</math>, the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns <math>(\Ns, \Ms_1)</math> will appear. Since there is no component free of link unknowns, it will be necessary to work in principle with all three components to find the equation of motion.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os}</math>

:The angular momentum can be calculated fomr the inertia tensor at <math>\Os</math> since <math>\Os</math> is fixed to the ring:

:<math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=([\Is\Is(\Gs)+\Is\Is^\oplus(\Os)])\velang{ring}{RTO=E},\:\:\:\:\:\: [\Is\Is(\Gs)] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is}</math>, with <math>2\Is = \ms\rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left(\mat{\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2}{0}{0}{0}{(1/2)\ms\rs^2} + \mat{0}{0}{0}{0}{\ms(2\rs)^2}{0}{0}{0}{\ms(2\rs)^2}\right)\vector{-\dot\varphi_0}{0}{\dot\psi}=\frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}
</math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{0}{9\ddot\psi} + \vector{0}{0}{\dot\psi}\times\frac{1}{2}\ms\rs^2\vector{-2\dot\varphi_0}{0}{9\dot\psi} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}
</math>

:The external moments about <math>\Os</math> come from the weight, the forces at <math>\Js</math> and the <math>\Ms_1</math> moment associated with the indirectt constraint between P1 and gthe floor:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs}</math>

:The second component yields <math>\Ns = \ms\gs+\frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>, Substitution of that value into the third component yields the equation of motion:

:<math>\boxed{\ddot\psi = \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)}</math>

:Initially, <math>\dot\psi(\ts=0)=0\Rightarrow\ddot\psi(\ts=0)=\frac{4}{9}\mu\frac{\gs}{\rs}>0</math>, and therefore the <math>\dot\psi</math> angular velocity increases, which increases the normal force. This indicates that there is no risk of losing contact at <math>\Js</math>.

:When <math>\dot\psi</math> reaches the value <math>(\dot\varphi_0/2)</math>, sliding at <math>\Js</math> stops:

:<math>\vel{J}{E} = \vel{C}{E} + \velang{ring}{E}\times\CJvec = (\rightarrow 2\rs\dot\psi) + [(\Uparrow\dot\psi) + (\otimes\dot\varphi_0)] \times(\downarrow\rs) = (\rightarrow 2\rs\dot\psi) + (\leftarrow \rs\dot\varphi_0) = 0</math>

:From this point on, the system has only 1 DoF, but the number of unknowns increases, since two more components of constraint forcé appear at <math>\Js</math> (in addition to N): the problem becomes indeterminate.

:Roadmap for the motor torque 

:Since the motor torque acts between P1 and the ring, the two system options for applying the vector theorems are: ring, P1. Since the equation of motion for the <math>\psi</math> coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

[[File:D7-Ex5-7-eng.png|thumb|center|400px|link=]]
<center> 5 constraint unk. + <math>\Gamma = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 9 constraint unk. + <math>\Gamma = </math> 10 unk.</center>

:The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:
:[[File:D7-Ex5-8-neut.png|thumb|left|200px|link=]]
:The first component of the AMT at <math>\Os</math> is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os]_1}</math>

:The kinetic moment at <math>\Os</math> is that of the ring, and the first component of its time derivative is zero:

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = \left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = 0 \Rightarrow\boxed{\Gamma=\mu\Ns\rs=\mu\ms\rs^2\left(\frac{\gs}{\rs} + \frac{1}{2}\dot\psi\dot\varphi_0\right)}</math>
 
 
 

:<big>'''OPTION 2'''</big>
:Roadmap for the equation of motion of the <math>\psi</math> coordinate

:The ring is the only element whose motion depends on <math>\psi</math>. Therefore, there are two systems to which the vector theorems can be applied: ring, ring+fork.

[[File:D7-Ex5-9-eng.png|thumb|center|470px|link=]]
<center> 5 constraint unk. + <math>\Gamma + \ddot\psi = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 6 constraint unk. + <math>\ddot\psi = </math> 7 unk.</center>
:The two options seem equivalent. Let’s analyze the external interactions that act on each of them in order to choose one option:

[[File:D7-Ex5-10-eng.png|thumb|center|400px|link=]]

:In both cases, if the AMT is applied at <math>\Os</math> three components of the constraint force are avoided, but in direction 1 (which is where <math>\ddot\varphi</math> may appear) there is always a moment (the motor torque in one case, or the <math>\Ms_1</math> moment between the ceiling and the fork in the other). Whether one system or the other is chosen, it will be necessary to work in principle with more than one component of the AMT to find the equation of motion. The two options are considered below.

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\Os}</math>

:The angular momentum at <math>\Os</math> and its time derivative are the same as those calculated in option 1:

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\mat{2}{0}{0}{0}{9}{0}{0}{0}{9}\vector{-\dot\varphi_0}{0}{\dot\psi}, \:\:\:\:\:\:
\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi}</math>

:The external moments about <math>\Os</math> come form the weight, the forces at <math>\Js</math>, the motor torque and the <math>\Ms_3</math> moment:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Gamma}{2(\ms\gs-\Ns)\rs}{2\mu\Ns\rs + \Ms_3}</math>

:The second component yields <math>\Ns</math>: <math>\Ns = \ms\gs + \frac{1}{2}\ms\rs\dot\psi\dot\varphi_0</math>. As that value is always positive, the ground contact at <math>\Js</math> is guaranteed.

:The first component yields the motor torque: <math>\Gamma = \mu\Ns\rs = \mu\ms\rs\left(\gs + \frac{1}{2}\rs\dot\psi\dot\varphi_0\right)</math>.

:But in the third, the <math>\ddot\varphi</math> acceleration is a function of <math>\Ms_3</math>. Therefore, this option does not seem appropriate.

:<math>\boxed{\text{Roadmap: SYSTEM (ring+ fork), AMT at }\Os}</math>

:The external moments are different from the previous case:

:<math>\sum\vec\Ms_\text{ext}(\Os) = \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\Rightarrow\frac{1}{2}\ms\rs^2\vector{0}{-2\dot\psi\dot\varphi_0}{9\ddot\psi} = \vector{\mu\Ns\rs+\Ms_1}{2(\ms\gs-\Ns)\rs + \Ms_2}{2\mu\Ns\rs}</math>

:In this option, the acceleration <math>\ddot\varphi</math> is a function of N, but the other components do not yield the value of this force.

:But if the results for the two options are combined, the equation of motion is straightforward:

:<math>\left.\begin{aligned}
\text{SYSTEM ring, AMT at }\left.\Os\right]_2 \:\:\:\:\:\:\:\: \Ns = \ms\gs +(1/2)\ms\rs\dot\psi\dot\varphi_0\\
\text{SYSTEM (ring + fork), AMT at }\left.\Os\right]_3\:\:\:\:\:\:\:\: (9/2)\ms\rs^2\ddot\psi = 2\mu\Ns\rs
\end{aligned}\right\} \Rightarrow\ddot\psi= \frac{2}{9}\mu\left(2\frac{\gs}{\rs}+\dot\psi\dot\varphi_0\right)</math>

:The end of the sliding phase can be studied exactly in the same way as in option 1.



====✏️ EXAMPLE D7.6: rotating ring pendulum ====
-----

:

:[[File:D7-Ex6-1-eng.png|thumb|left|150px|link=]]
:The system consists of a ring, with mass m and radius R, attached to an arm that is articulated to a support. The support can slide along a smooth guide. A linear spring of constant k connects the support and the guide. The whole system rotates around the vertical axis with constant angular velocity <math>\dot\psi_0</math> under the action of a motor. The mass of all the elements, except the ring, is negligible. We want to find the equations of motion and study the possible equilibrium configurations.
 
:Kinematic description
:[[File:D7-Ex6-2-neut.png|thumb|right|150px|link=]]
:It is a system with 3 DoF: the pendulum motion <math>\dot\theta</math>, the motion of the support raltive to the guide (that will be denoted by <math>\dot\xs</math>) and the forced vertical rotation <math>\dot\psi_0</math>.

:The motion of the ring center <math>\Gs</math> relative to the ground can be found through a double composition:

:<math>\left.\begin{aligned}
\text{AB: guide} \\
\text{REL: support}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs)</math>

:<math>\left.\begin{aligned}
\text{AB: ground} \\
\text{REL: guide}
\end{aligned}\right\} \vel{G}{AB} = \vel{G}{REL} + \vel{G}{tr} = (\nearrow\Ls\dth) + (\downarrow\dot\xs) + (\otimes\Ls\dot\psi_0\cth)</math>
 
:General diagram of interactions
:[[File:D7-Ex6-3-eng.png|thumb|left|200px|link=]]
 
 
: It is a determinate problem:

:(15 constraint unk., 3DoF) <math>\Rightarrow</math> 18 unknowns
:3 rigid bodies <math>\times\frac{\text{6 eqs.}}{\text{r. body}}</math> = 18 equations
 
 
 

:Roadmap for the equation of motion for the <math>\theta</math> coordinate

:The ring motion is the only one that depends on <math>\ddth</math>. Therefore, the appropriate systems for the application of the vector theorems are: ring, ring+support, ring+support+guide.

[[File:D7-Ex6-4-eng.png|thumb|center|700px|link=]]
<center> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x = </math> 7 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddth + \ddot x + \Gamma = </math> 8 unk.</center>
:The third option is the least suitable. As for the other two, the external interactions to be taken into account are:
:[[File:D7-Ex6-5-eng.png|thumb|right|300px|link=]]

:If the AMT at <math>\Os</math> is applied to the ring, component 1 is free of constraint unknowns, and it is precisely in that direction that the angular velocity <math>\dth</math> (and therefore the change in its value <math>\ddth</math>) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

:<math>\boxed{\text{Roadmap: SYSTEM (ring), AMT at }\left.\Os\right]_1}</math>

:As point <math>\Os</math> moves relative to the ground: <math>\sum\vec\Ms_\text{ext}(\Os) + \OGvec\times\ms\acc{O}{Gal} = \dot{\vec{\Hs}}_{\text{RTO}}</math>

:On the other hand, <math>\Os</math> is a point fixed to the ring: <math>\vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{ring}{RTO=E}=\left[\Is\Is(\Gs)+\Is\Is^\oplus(\Os)\right]\left[(\Uparrow\dot\psi_0) + (\odot\dth)\right]
</math>

:<math>\left[\Is\Is(\Os)\right] = \mat{2\Is}{0}{0}{0}{\Is}{0}{0}{0}{\Is} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}</math>, with <math>2\Is = \ms\Rs^2</math>

:<math>\left\{\vec{\Hs}_{\text{RTO}}(\Os)\right\} = \left( \mat{\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2}{0}{0}{0}{(1/2)\ms\Rs^2} + \mat{\ms\Ls^2}{0}{0}{0}{\ms\Ls^2}{0}{0}{0}{0}\right)\vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth} = \vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth} </math>

:<math>\left\{\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right\} = \vector{\ms(\Rs^2 + \Ls^2)\ddth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\dth\sth}{-(1/2)\ms\Rs^2\dot\psi_0\dth\cth} + \vector{\dth}{\dot\psi_0\sth}{\dot\psi_0\cth}\times\vector{\ms(\Rs^2 + \Ls^2)\dth}{\ms\left[(1/2)\Rs^2 + \Ls^2\right]\dot\psi_0\sth}{(1/2)\ms\Rs^2\dot\psi_0\cth}</math>

:<math>\left.\dot{\vec{\Hs}}_{\text{RTO}}(\Os)\right]_1 = \ms\left(\Rs^2 + \Ls^2\right)\ddth-\ms\Ls^2\dot\psi_0^2\sth\cth</math>

:<math>\left.\sum\vec\Ms_\text{ext}(\Os)\right]_1 = -\ms\gs\Ls\sth</math>

:<math>\OGvec\times\ms\acc{O}{Gal} = (\searrow\Ls)\times\ms(\downarrow\ddot\xs) = [(\rightarrow\Ls\sth) + (\downarrow\Ls\cth)]\times\ms(\uparrow\ddot\xs) = (\otimes\ms\Ls\ddot\xs\sth)</math>

:Finally: <math>\boxed{(\Rs^2 + \Ls^2)\ddth + (\gs - \ddot\xs - \Ls\dot\psi_0\cth)\Ls\sth = 0}</math>.

:This equation of motion also includes the variable <math>\ddot\xs</math>. This means that the degrees of freedom <math>\dth</math> and <math>\dot\xs</math> are '''coupled''': although the motion begins with initial conditions that are only nonzero for one of them, the other may appear.

:The component 1 of the AMT at <math>\Os</math> for the ring system is the only one where <math>\dth</math> and <math>\ddot\xs</math> appear. Therefore, the other equation of motion cannot be determined with either of the other two components.

:Roadmap for the equation of motion of the x coordinate 

:The motion of the ring and the support depend on <math>\ddot\xs</math>. Hence, the appropriate systems for the application of the vector theorems are: ring, ring+support, support, support+guide, ring+support+guide. Since the equation of motion for <math>\theta</math> has been determined, <math>\ddth</math> is no longer an unknown.

[[File:D7-Ex6-6-eng.png|thumb|center|450px|link=]]
<center> 5 constraint unk. + <math> \ddot x = </math> 6 unk.<math>\:\:\:\:\:\:\:\:\:\:\:\:\:\</math> 5 constraint unk. + <math>\ddot x = </math> 6 unk.</center>

[[File:D7-Ex6-7-eng.png|thumb|center|650px|link=]]
<center> 10 constraint unk. + <math>\ddot x = </math> 11 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 10 constraint unk. + <math>\ddot x + \Gamma= </math> 12 unk.<math>\:\:\:\:\:\:\:\:\:\:</math> 5 constraint unk. + <math>\ddot x + \Gamma = </math> 7 unk.</center>
 

:As before, the best options are the first two. The first has already been used, so we will apply the vector theorems to the second. From the description of the external interactions acting on the system (ring+support), it can be seen that the vertical component (direction 3’) of the AMT will be free of constraint unknowns. Therefore: <math>\boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}}</math>.

:<math>\sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal}</math>

:Calculation of the <math>\Gs</math> acceleration: 

:'''Option 1''': as the time derivative of the velocity described above, since it corresponds to a general configuration.

:<math>\left\{\vel{G}{E}\right\}_{\Bs'} = \vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}, \:\:\:\:\:\:\:\: \left\{\acc{G}{E}\right\}_{\Bs'} = \vector{\Ls\dot\psi_0\dth\sth}{-\Ls\dth^2\sth}{-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth} + \vector{0}{0}{\dot\psi_0}\times\vector{-\Ls\dot\psi_0\cth}{\Ls\dth\cth}{-\dot\xs + \Ls\dth\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>

:'''Option 2''': through rigid body kinematics. .

:<math>\left.\begin{aligned}
\Gs\in\text{ring} \\
\Os\in\text{ring}
\end{aligned}\right\} \acc{G}{E} = \acc{O}{E} + \velang{ring}{E}\times\velang{ring}{E}\times\OGvec + \accang{ring}{E}\times\OGvec</math>

:<math>\velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth)</math>

:The angular acceleration <math>\accang{ring}{E}</math> is asociated with the change of value and direction of <math>\vec{\dth}</math>: <math>\:\accang{ring}{E} = (\odot\ddth) + (\Rightarrow\dot\psi_0\dth)</math>

:<math>\left\{\acc{G}{E}\right\}_{\Bs'} = \vector{0}{0}{-\ddot\xs} + \vector{\dth}{0}{\dot\psi_0}\times\left(\vector{\dth}{0}{\dot\psi_0}\times\vector{0}{\Ls\cth}{\Ls\sth}\right) + \vector{\ddth}{\dot\psi_0\dth}{0}\times\vector{0}{\Ls\cth}{\Ls\sth}</math>

:<math>\left.\acc{G}{E}\right]_{3'} = -\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth</math>
 
:Formulation of the spring force 
[[File:D7-Ex6-8-neut.png|thumb|right|250px|link=]]

:The vertical translational motion of the support <math>(\dot\xs)</math> is associated with the variation of an <math>\xs</math> coordinate whose origin has not yet been defined. It is usual to choose the origin of the coordinates so that they coincide with equilibrium configurations.

:If we take <math>\xs=0</math> for the equilibrium configuration in the absence of rotation <math>\dot\psi_0</math>, it is clear that the spring will have to exert an attraction force <math>\Fs_0</math> on the pendulum to counteract the weight: <math>\Fs_0 = \ms\gs</math>. The general formulation of the spring attraction force will then be <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho</math>.

:Since the <math>\dot\xs</math> motion has been defined as positive downwards, an increase in <math>\xs</math> implies an increase in the spring length. Therefore <math>\Fs_\ss^{\as\ts} = \ms\gs + \ks\xs</math>.

:<math>\left.\begin{aligned}
\left.\sum\Fs_{\es\xs\ts}\right]_{3'}=\Fs_\ss^\text{at} -\ms\gs = \ks\xs\\
\left.\ms\acc{G}{T}\right]_{3'} = \ms(-\ddot\xs + \Ls\ddth\sth + \Ls\dth^2\cth)
\end{aligned}\right\} \Rightarrow \boxed{\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0}</math>

:Comments on the DoF coupling 

:The initial conditions <math>\xs(\ts=0)</math>, <math>\dot\xs(\ts=0)</math>, <math>\theta(\ts=0)</math>, <math>\dth(\ts=0)</math> under which the movement is started determine the DoF that will appear.

:The initial conditions <math>\xs(\ts=0) = \xs_0</math>, <math>\dot\xs(\ts=0) = \dot\xs_0</math>, <math>\theta(\ts=0)=0</math>, <math>\dth(\ts=0) =0</math> will never succeed in provoking the pendulum motion, since the equations for the initial instant are:

:<math>(\Rs^2 + \Ls^2)\ddth(\ts=0) = 0, \:\:\:\:\:\:\ddot\xs(\ts=0) + \frac{\ks}{\ms}\xs_0 = 0</math>.

:However, the initial conditions <math>\xs(\ts=0) = 0</math>, <math>\dot\xs(\ts=0) = 0</math>, <math>\theta(\ts=0)=\theta_0</math>, <math>\dth(\ts=0) =\dth_0</math> will provoke the vertical motion <math>\dot\xs</math>, since the equation describing the time evolution of <math>\dot\xs</math> it for the initial instant is:

:<math>\ddot\xs(\ts=0) -\Ls\left[\ddth(\ts=0)\text{sin}\theta_0 + \dth_0^2\text{cos}\theta_0\right] = 0</math>.

:'''ANIMACIO'''

:Study of static equilibrium configurations

:The static equilibrium configurations (those that correspond to the system at rest, with <math>\dot\psi_0 = 0</math>) are obtained from the equations of motion by imposing <math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddth_\text{eq} = 0</math>, <math>\dth_\text{eq} = 0</math>:

:<math>\left.\begin{aligned}
\ddot\xs + \frac{\ks}{\ms}\xs -\Ls(\ddth\sth + \dth^2\cth) = 0 \Rightarrow \frac{\ks}{\ms}\xs_{\es\qs} = 0\\
\left(\Rs^2 + \Ls^2\right)\ddth + (\gs - \ddot \xs)\Ls\sth = 0 \Rightarrow \gs\Ls\text{sin}\theta_{\es\qs} = 0
\end{aligned}\right\}\:\: \Rightarrow

\begin{cases}
(\xs_\text{eq}, \theta_\text{eq}) = (0,0)\\
(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)
\end{cases}
</math>

:If we consider a small perturbation of these equilibrium configurations <math>(\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta)</math>, the equations can be [[D7. Exemples de dinàmica 3D#D7.1 Anàlisi d’equacions del moviment|'''linearized''']]. For the configuration <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math>:

:<math>\left.\begin{aligned}
\sth\sim\varepsilon_\theta \\
\cth\sim 1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs < 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta + (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta = 0\Rightarrow \ddot\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = -\frac{\Ls}{\Rs^2 + \Ls^2}(\gs + |\ddot\varepsilon_\xs|)\varepsilon_\theta < 0
\end{cases}
</math>

:Since <math>\ddot\varepsilon_\xs < 0</math>, <math>\ddot\varepsilon_\theta < 0</math>, it is a '''STABLE''' configuration.
 
:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math> configuration:

:<math>\left.\begin{aligned}
\sth\sim -\varepsilon_\theta \\
\cth\sim -1
\end{aligned}\right\} \Rightarrow

\begin{cases}
\ddot\varepsilon_\xs + \frac{\ks}{\ms}\varepsilon_\xs = 0\Rightarrow \ddot\varepsilon_\xs = -\frac{\ks}{\ms}\varepsilon_\xs > 0\\
(\Rs^2 + \Ls^2)\ddot\varepsilon_\theta - (\gs - \ddot\varepsilon_\xs)\Ls\varepsilon_\theta\Rightarrow \ddot\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - \ddot\varepsilon_\xs)\varepsilon_\theta = \frac{\Ls}{\Rs^2 + \Ls^2}(\gs - |\ddot\varepsilon_\xs|)\varepsilon_\theta > 0
\end{cases}</math>

:Since <math>\ddot\varepsilon_\xs > 0</math>, <math>\ddot\varepsilon_\theta > 0</math>, it is an '''UNSTABLE''' configuration.
 
:Study of the equilibrium configurations under rotation

:If <math>\dot\psi_0>0</math>, for (<math>\ddot\xs_\text{eq} = 0</math>, <math>\dot\xs_\text{eq} = 0</math>, <math>\ddot\theta_\text{eq} = 0</math>, <math>\dot\theta_\text{eq} = 0</math>) the equation of motion for x does not change (therefore, <math>\xs_\text{eq}=0</math> is stable: <math>\ddot\varepsilon_\xs < 0</math>), but that of <math>\theta</math> has an extra term, and two families of equilibrium configurations appear:

:<math>
(\gs-\Ls\dot\psi_0^2\text{cos}\theta_\text{eq})\Ls\text{sin}\theta_\text{eq} = 0\Rightarrow
\begin{cases}
\text{sin}\theta_\text{eq} = 0\Rightarrow \theta_\text{eq} = (0,180\deg)\\
\text{cos}\theta_\text{eq} = \frac{\gs}{\Ls\dot\psi_0^2}\Rightarrow\theta_\text{eq}=\text{arcos}\left(\frac{\gs}{\Ls\dot\psi_0^2}\right)
\end{cases}
</math>

:However, the second family only exists above a critical value of <math>\dot\psi_0^2</math> since the <math>\text{cos}\theta_\text{eq}</math> function is bounded between -1 and +1:

:<math>|\text{cos}\theta_\text{eq}| \leq 1\Rightarrow \dot\psi_0\geq\dot\psi_\text{cr}\equiv\sqrt\frac{\gs}{\Ls}</math>

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,0)</math> configuration, the linearization of the equation of motion for <math>\theta</math> leads to:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta + \left(\gs - \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 - \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta</math>

:If <math>\dot\psi_0 < \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta < 0</math>, the configuration is '''STABLE'''. If <math>\dot\psi_0 > \dot\psi_{\cs\rs}</math> and <math>\ddot\varepsilon_\theta > 0</math>, the configuration is '''UNSTABLE'''.

:For the <math>(\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg)</math>, configuration, the linearized equation of motion for <math>\theta</math> is:

:<math>\left(\Rs^2 + \Ls^2\right)\ddot\varepsilon_\theta - \left(\gs + \Ls\dot\psi_0^2\right)\Ls\varepsilon_\theta = 0 \Rightarrow \ddot\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \frac{\gs}{\Ls}\right)\varepsilon_\theta = \frac{\Ls^2}{\Rs^2 + \Ls^2}\left(\dot\psi_0^2 + \dot\psi_{\cs\rs}^2\right)\varepsilon_\theta > 0</math>

:The configuration is '''UNSTABLE''' for all <math>\dot\psi_0^2</math> values.

:The study of configurations <math>\theta_\text{eq} = \text{arcos}\left(\gs/\Ls\dot\psi_0^2\right) = \text{arcos}\left(\dot\psi_{\cs\rs}^2/\dot\psi_0^2\right)</math> can be done in the same way, but it takes much longer. For this reason, it is not done.

'''ANIMACIO'''


© Universitat Politècnica de Catalunya. [[Mecànica:Drets d'autor |Tots els drets reservats]]
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D5. Mass distribution

2026-02-24T18:23:25Z

Eantem: /* D5.4 Steiner’s Theorem */

D5. Mass distribution

2026-02-24T18:23:00Z

Eantem: /* D5.4 Steiner’s Theorem */

D5. Mass distribution

2026-02-24T18:21:49Z

Eantem: /* D5.4 Steiner’s Theorem */

D5. Mass distribution

2026-02-24T18:20:50Z

Eantem: /* D5.4 Steiner’s Theorem */

D5. Mass distribution

2026-02-24T18:02:33Z

Eantem: /* ✏️ Example D5.7: symmetrical rotor */