D7. Examples of 3D dynamics

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EN CONSTRUCCIÓ

In this unit, the systematic procedure proposed in 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:

[math]\displaystyle{ \ddot{\qs}_{i}\xrightarrow[]{\int{\ds\ts}}\dot{\qs}_\is\xrightarrow[]{\int{\ds\ts}}{\qs_\is} }[/math]

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]\displaystyle{ (\dot{\qs}_{\text{j,eq}} = 0) }[/math], it remains at rest [math]\displaystyle{ (\ddot{\qs}_{\text{j,eq}} = 0) }[/math]. Therefore, the value of the coordinates in equilibrium is given by:

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

[math]\displaystyle{ \downarrow(\dot{\qs}_{j,eq} = 0, \ddot{\qs}_{j, eq} = 0) }[/math]

[math]\displaystyle{ 0 = {\fs}_{i,eq}({\qs}_{j, eq}, \text{ dynamic parameters, geometric parameters}), i = 1...N }[/math]

The equation that defines the [math]\displaystyle{ \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]\displaystyle{ (\qs_\js = \qs_{\text{j,eq}} + \varepsilon_\js, }[/math] with [math]\displaystyle{ \varepsilon_\js \lt \lt 1 }[/math], hence [math]\displaystyle{ \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]\displaystyle{ \qs = \qs_{\text{eq}} + \varepsilon }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ (\qs_\js=\theta) }[/math] and sine and cosine functions appear:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \As\ddot\varepsilon + \Bs\varepsilon = 0 }[/math]. The general solution is: [math]\displaystyle{ \varepsilon(\ts) = \as\text{sin}(\omega\ts + \varphi) }[/math], with [math]\displaystyle{ \omega = \sqrt{\Bs/\As} }[/math]. The integration constants ([math]\displaystyle{ \as, \varphi }[/math]) depend on the initial conditions [math]\displaystyle{ \varepsilon(\ts = 0)\equiv \varepsilon_0, \dot\varepsilon(\ts = 0)\equiv\dot\varepsilon_0 }[/math].

💭 Proof ➕

The solution [math]\displaystyle{ \varepsilon(\ts) }[/math] of the equation [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ -\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]\displaystyle{ \qs_{\es\qs} }[/math] whose angular frequency [math]\displaystyle{ (\omega) }[/math] depends on system parameters.

The integration constants, on the other hand, depend on the initial conditions ([math]\displaystyle{ \varepsilon(\ts = 0)\equiv \varepsilon_0 = \as\text{sin}(\varphi) }[/math], [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]

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]\displaystyle{ \As\ddot\varepsilon + \Bs\varepsilon = 0\Rightarrow\ddot\varepsilon = -(\Bs/\As)\varepsilon }[/math]

• [math]\displaystyle{ (\Bs/\As) \gt 0 \Rightarrow\ddot\varepsilon\lt 0\Rightarrow\varepsilon(\ts) }[/math] decreases, and the system returns to the equilibrium configuration [math]\displaystyle{ \qs_{\es\qs} }[/math]. It is a STABLE behaviour.

• [math]\displaystyle{ (\Bs/\As) \lt 0 \Rightarrow\ddot\varepsilon \gt 0\Rightarrow\varepsilon(\ts) }[/math] increases, and the system moves away from the equilibrium configuration [math]\displaystyle{ \qs_{\es\qs} }[/math]. It is UNSTABLE behaviour.

D7.2 General examples

✏️ EXAMPLE D7.1: rotating rectangular plate

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The homogeneous rectangular plate, with mass m, is articulated to a massless fork that rotates with constant angular velocity [math]\displaystyle{ \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]\displaystyle{ \Omega_0 }[/math].

Kinematic description:

[math]\displaystyle{ \boxed{\begin{align} \text{AB:ground}\\ \text{REL:fork} \end{align}} }[/math]

[math]\displaystyle{ \velang{plate}{E}=\velang{plate}{AB} = \velang{plate}{REL} + \velang{}{ar} = (\Uparrow\Omega_0) + (\odot\dth) }[/math]

[math]\displaystyle{ \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]\displaystyle{ \vel{G}{E} }[/math] is rigid body kinematics (rigid body: plate):

[math]\displaystyle{ \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

It is a two-DoF system (forced [math]\displaystyle{ \Omega_0 }[/math], free [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \dth }[/math]. Therefore, the systems where [math]\displaystyle{ \ddth }[/math] will appear when applying the vector theorems are: plate, plate + fork.

5 constraint unk. + [math]\displaystyle{ \ddth = }[/math] 6 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\:\:\:\: }[/math] 5 constraint unk. + [math]\displaystyle{ \Gamma + \ddth = }[/math] 7 unk.

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:

The characterization of the constraint torsor of the fork on the plate at point [math]\displaystyle{ \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]\displaystyle{ \Os }[/math], the constraint force will not appear, and component 3 (or 3’) will be free of constraint unknowns.

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

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

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

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \I{large} }[/math] and [math]\displaystyle{ \I{low} }[/math] are substituted and by the values given in the tables, the equation becomes:

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \frac{\gs}{2\Ls\Omega_0^2} \leq 1 }[/math], and this is true only if the angular velocity [math]\displaystyle{ \Omega_0 }[/math] is above the critical value [math]\displaystyle{ \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]\displaystyle{ (\varepsilon) }[/math] about an equilibrium configuration [math]\displaystyle{ \theta_\text{eq} }[/math] can be done by approximating the trigonometric functions (section D7.1):

[math]\displaystyle{ \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]\displaystyle{ \theta_\text{eq} = 0 }[/math], , the equation of motion is [math]\displaystyle{ \frac{10}{3}\ddot{\varepsilon} + \left(\frac{\gs}{\Ls} - 2\Omega_0^2\right)\varepsilon = 0 }[/math].

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

• For [math]\displaystyle{ \Omega_0^2 \gt \frac{\gs}{2\Ls},\: \ddot\varepsilon \gt 0\Rightarrow \varepsilon }[/math] increases. It is an UNSTABLE configuration, and no oscillation around this configuration is possible.

• For [math]\displaystyle{ \Omega_0^2 \lt \frac{\gs}{2\Ls},\: \ddot\varepsilon \lt 0\Rightarrow \varepsilon }[/math] decreases. . The movement is an oscillation about a STABLE. configuration. The angular frequency [rad/s] is [math]\displaystyle{ \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]\displaystyle{ \theta(\ts = 0) = 0 }[/math] and [math]\displaystyle{ \dot\theta(\ts = 0) = 0 }[/math], the pendulum motion does not appear, and the system moves only according with the rotation [math]\displaystyle{ \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:

[math]\displaystyle{ (\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]\displaystyle{ \theta_\text{eq} = 0 }[/math]:

[math]\displaystyle{ (\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]\displaystyle{ \Omega_0 }[/math], the coefficient [math]\displaystyle{ \left[2\ms\gs\Ls + (\I{large} - \I{low})\Omega_0^2\right] }[/math] is positive, hence the [math]\displaystyle{ \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:

[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: }[/math]10 constraint unk. + [math]\displaystyle{ \Gamma = }[/math] 11 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\:\:\:\: }[/math] 5 constraint unk. + [math]\displaystyle{ \Gamma = }[/math] 6 unk.

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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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

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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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \theta }[/math] coordinate can be simply [math]\displaystyle{ \ddth = 0 }[/math] and that the [math]\displaystyle{ \Omega_0 }[/math] rotation remains constant)..

General diagram of interactions

It is a 2-DoF system (forced [math]\displaystyle{ \omega_0 }[/math], free [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \dth }[/math]. Therefore, the systems in which [math]\displaystyle{ \ddth }[/math] would appear in the application of the vector theorems are: rigid body, rigid body + fork.

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]\displaystyle{ \Os }[/math], the constraint force will not appear. Hence

[math]\displaystyle{ \boxed{\text{Roadmap: SYSTEM (rigid body), AMT at} \Os} }[/math]

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

[math]\displaystyle{ \Os\in }[/math] rigid body [math]\displaystyle{ \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]\displaystyle{ \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 Steiner corrections to move to [math]\displaystyle{ \Os }[/math]:

[math]\displaystyle{ [\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]

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

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

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \Os }[/math] external to the rigid body is the constraint moment associated with the revolute joint:

[math]\displaystyle{ \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]\displaystyle{ (\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]\displaystyle{ \dth }[/math] of the frame relative to the fork but keeping a constant [math]\displaystyle{ \Omega_0 }[/math]) is not possible. The reason is the [math]\displaystyle{ \vec{\Hs}_{\text{RTO}}(\Os) }[/math] horizontal component, which is the one that generates [math]\displaystyle{ \dot{\vec{\Hs}}_{\text{RTO}}(\Os)\neq\vec 0 }[/math]. If [math]\displaystyle{ \vec{\Hs}_{\text{RTO}}(\Os) }[/math] were strictly vertical (parallel to[math]\displaystyle{ \vec\Omega_0 }[/math]), then [math]\displaystyle{ \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]\displaystyle{ \Ms_1 = \Ms_2 = 0 }[/math]. In other words: if the direction of the angular velocity were a principal direction of inertia for point [math]\displaystyle{ \Os }[/math], keeping it constant would be possible without the need for an external moment.

Constant vertical rotation [math]\displaystyle{ \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:

The value of the two forces is different, but the direction of the moment they exert about [math]\displaystyle{ \Os }[/math] is the same.

As long as the finger introduces one of these forces, [math]\displaystyle{ \Omega_0 }[/math] remains constant without changing the frame orientation relative to the horizontal plane (without the appearance of [math]\displaystyle{ \dth }[/math]). According to the 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:

ANIMACIONS

ALTERNATIVE

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

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \left.\sum\vec{\Ms}_\text{ext}(\Os)\right]_3 = \ms\gs\sqrt{2}\Ls\sth }[/math].

The equation of motion is:

[math]\displaystyle{ \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]\displaystyle{ (\ddth_{eq} = 0) }[/math] are the solutions of the transcendental equation:

[math]\displaystyle{ \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]\displaystyle{ \Omega_0\neq 0 }[/math], [math]\displaystyle{ \theta_\text{eq} = 0 }[/math] is not one of them.

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

[math]\displaystyle{ \frac{20}{3}\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{3}{20}\Omega_0^2\gt 0 }[/math].

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

✏️ EXAMPLE D7.3: rotating frame with particles

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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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \theta }[/math] coordinate movement can be simply [math]\displaystyle{ \ddth = 0 }[/math] and that the [math]\displaystyle{ \Omega_0 }[/math] rotation remains constant..

General diagram of interactions

It is the same kind of system as in example D7.2: it has 2 DoF (forces [math]\displaystyle{ \Omega_0 }[/math], free [math]\displaystyle{ \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]\displaystyle{ \dth }[/math]. Therefore, the systems in which [math]\displaystyle{ \ddth }[/math] would appear in the application of the vector theorems are: rigid body, rigid body + fork. As in example D7.2, a suitable roadmap is:

[math]\displaystyle{ \boxed{\text{Roadmap: SYSTEM (rigid body), AMT at }\Os} }[/math]

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

[math]\displaystyle{ \Os\in }[/math] rigid body: [math]\displaystyle{ \vec{\Hs}_{\text{RTO}}(\Os) = \Is\Is(\Os)\velang{r.body}{RTO = E} = \Is\Is(\Os)(\Uparrow\Omega_0) }[/math]

The inertia tensor in the B vector basis fixed to the frame is straightforward:

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

[math]\displaystyle{ \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]

The time derivative can also be obtained analytically:

[math]\displaystyle{ \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]\displaystyle{ \Os }[/math], and therefore the only moment about point [math]\displaystyle{ \Os }[/math] external to the rigid body is the constaint moment associated with the revolute joint:

[math]\displaystyle{ \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]\displaystyle{ (\Rightarrow\Ms_1) + (\Uparrow\Ms_2)\neq(\otimes 2\ms\Ls^2\Omega_0^2) }[/math]

As in example D7.2, the intended motion (without rotation of the frame relative to the fork but keeping [math]\displaystyle{ \Omega_0 }[/math] constant) is not possible because the direction of the angular velocity is not a principal direction of inertia.

Constant vertical rotation [math]\displaystyle{ \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:

The values of the two forces are different, but the direction of the moment they generate about [math]\displaystyle{ \Os }[/math] is the same.

While one of these forces is introduced, [math]\displaystyle{ \Omega_0 }[/math] remains constant without changing the orientation of the frame relative to the horizontal plane. By the 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:

ANIMACIONS

ALTERNATIVE

The initial direction of the deviation can be investigated from the equation of motion for the [math]\displaystyle{ \theta }[/math] coordinate , which can be found according to the following roadmap: [math]\displaystyle{ \boxed{\text{Roadmap: SYSTEM (rigid body), AMT at } \left.\Os\right]_3} }[/math]:

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

The equation of motion is:

[math]\displaystyle{ 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]\displaystyle{ (\ddth_{\es\qs} = 0) }[/math] are the solutions of the transcendent equation:

[math]\displaystyle{ \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]\displaystyle{ \theta = 0 }[/math] is not a solution, and that therefore the frame will necessarily acquire a pendulum motion.

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

[math]\displaystyle{ 3\ddth\:(\ts=0) - \Omega_0^2 = 0\Rightarrow\ddth\:(\ts=0) = \frac{1}{3}\Omega_0^2\gt 0 }[/math]

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

ANIMATIONS

✏️ EXAMPLE D7.4: rotating ball

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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]\displaystyle{ (\mu_\text{rad} = 0) }[/math]. The aim is to investigate whether rotation [math]\displaystyle{ \Omega_0 }[/math] can cause the loss of contact between the ball and the ground.

Kinematic description

The ISA of the ball relative to the ground is the straight line [math]\displaystyle{ \Os\Js }[/math], and the angular velocity can be decomposed into two Euler rotations:

General diagram of interactions

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]\displaystyle{ \Rightarrow }[/math] 18 unknowns

3 rigid bodies [math]\displaystyle{ \times\frac{6\text{ eqs.}}{\text{r. body}} }[/math]= 18 equations

The description of the system can be simplified by treating the arm as CAE, since it has no mass and it only undergoes constraint interactions:

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]\displaystyle{ (\Omega_3,\Omega_1) }[/math] relative to the fork. The problem remains determinate:

(11 constraint unk., 1DoF)[math]\displaystyle{ \Rightarrow }[/math] 12 unknowns

2 rigid bodies [math]\displaystyle{ \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.

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]\displaystyle{ \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]\displaystyle{ \boxed{\text{Roadmap: SYSTEM (ball), AMT at } \Os]_{\perp\text{ drawing}}} }[/math]

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

[math]\displaystyle{ \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]\displaystyle{ \vec{\Omega_0} }[/math]:

[math]\displaystyle{ \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]\displaystyle{ }[/math]

The normal force increases with the angular velocity, hence contact with the ground is always maintained.

Alternative:

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

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

[math]\displaystyle{ \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]\displaystyle{ \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

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The ring, with mass m and radius R, rotates with constant angular velocity [math]\displaystyle{ \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]\displaystyle{ \dot\psi }[/math]. Initially, [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \dot\varphi_0 }[/math] and [math]\displaystyle{ \dot\psi }[/math] are independent, and the description of the system velocities and the GDI (EXAMPLE D3.18) are:

[math]\displaystyle{ {\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]

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) (section D3.6):

In both cases, it is a determinate problem:

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

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

OPTION 1

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:

Roadmap for the equation of motion of the [math]\displaystyle{ \psi }[/math] coordinate

Both the ring and the stator(P1) motion depend on [math]\displaystyle{ \psi }[/math]. Hence, there are three systems to which the vector theorems may be applied: ring, P1, ring+P1.

6 constraint unk. + [math]\displaystyle{ \ddot\psi = }[/math] 7 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\: }[/math] 9 constraint unk. + [math]\displaystyle{ \Gamma + \ddot\psi = }[/math] 11 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\: }[/math] 5 constraint unk. + [math]\displaystyle{ \ddot\psi = }[/math] 6 unk.

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]\displaystyle{ \Os }[/math], the constraint force associated with the indirect constraint between P1 and the ceiling is avoided, and only two constraint unknowns [math]\displaystyle{ (\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]\displaystyle{ \boxed{\text{Roadmap: SYSTEM (ring+P1), AMT at }\Os} }[/math]

The angular momentum can be calculated fomr the inertia tensor at [math]\displaystyle{ \Os }[/math] since [math]\displaystyle{ \Os }[/math] is fixed to the ring:

[math]\displaystyle{ \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]\displaystyle{ 2\Is = \ms\Rs^2 }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \Os }[/math] come from the weight, the forces at [math]\displaystyle{ \Js }[/math] and the [math]\displaystyle{ \Ms_1 }[/math] moment associated with the indirectt constraint between P1 and gthe floor:

[math]\displaystyle{ \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]\displaystyle{ \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:

[math]\displaystyle{ \boxed{\ddot\psi = \frac{2\mu}{\Rs^2+2\Ls^2}(\gs\Ls+\Rs^2\dot\psi\dot\varphi_0)} }[/math]

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

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

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \psi }[/math] coordinate and the normal force have already been determined, the number of unknowns in these two cases is:

5 constraint unk. + [math]\displaystyle{ \Gamma = }[/math] 6 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\: }[/math] 9 constraint unk. + [math]\displaystyle{ \Gamma = }[/math] 10 unk.

The best option, then, is to apply the vector theorems to the ring. The external interactions on that rigid body are:

The first component of the AMT at [math]\displaystyle{ \Os }[/math] is free of constraint unknowns (it contains the moment of the friction force, but N is no longer an unknown). Therefore:

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

The kinetic moment at [math]\displaystyle{ \Os }[/math] is that of the ring, and the first component of its time derivative is zero:

[math]\displaystyle{ \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]

OPTION 2

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

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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]\displaystyle{ \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

It is a system with 3 DoF: the pendulum motion [math]\displaystyle{ \dot\theta }[/math], the motion of the support raltive to the guide (that will be denoted by [math]\displaystyle{ \dot\xs }[/math]) and the forced vertical rotation [math]\displaystyle{ \dot\psi_0 }[/math].

The motion of the ring center [math]\displaystyle{ \Gs }[/math] relative to the ground can be found through a double composition:

[math]\displaystyle{ \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]\displaystyle{ \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

It is a determinate problem:

(15 constraint unk., 3DoF) [math]\displaystyle{ \Rightarrow }[/math] 18 unknowns

3 rigid bodies [math]\displaystyle{ \times\frac{\text{6 eqs.}}{\text{r. body}} }[/math] = 18 equations

Roadmap for the equation of motion for the [math]\displaystyle{ \theta }[/math] coordinate

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

5 constraint unk. + [math]\displaystyle{ \ddth + \ddot x = }[/math] 7 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\: }[/math] 5 constraint unk. + [math]\displaystyle{ \ddth + \ddot x = }[/math] 7 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\: }[/math] 5 constraint unk. + [math]\displaystyle{ \ddth + \ddot x + \Gamma = }[/math] 8 unk.

The third option is the least suitable. As for the other two, the external interactions to be taken into account are:

If the AMT at [math]\displaystyle{ \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]\displaystyle{ \dth }[/math] (and therefore the change in its value [math]\displaystyle{ \ddth }[/math]) appears. In the (ring+support) option, that component includes a constraint moment. Therefore:

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

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

On the other hand, [math]\displaystyle{ \Os }[/math] is a point fixed to the ring: [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ 2\Is = \ms\Rs^2 }[/math]

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

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \ddot\xs }[/math]. This means that the degrees of freedom [math]\displaystyle{ \dth }[/math] and [math]\displaystyle{ \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]\displaystyle{ \Os }[/math] for the ring system is the only one where [math]\displaystyle{ \dth }[/math] and [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \theta }[/math] has been determined, [math]\displaystyle{ \ddth }[/math] is no longer an unknown.

5 constraint unk. + [math]\displaystyle{ \ddot x = }[/math] 6 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\:\:\:\:\ }[/math] 5 constraint unk. + [math]\displaystyle{ \ddot x = }[/math] 6 unk.

10 constraint unk. + [math]\displaystyle{ \ddot x = }[/math] 11 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\: }[/math] 10 constraint unk. + [math]\displaystyle{ \ddot x + \Gamma= }[/math] 12 unk.[math]\displaystyle{ \:\:\:\:\:\:\:\:\:\: }[/math] 5 constraint unk. + [math]\displaystyle{ \ddot x + \Gamma = }[/math] 7 unk.

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]\displaystyle{ \boxed{\text{Roadmap: SYSTEM (ring + support), }\left.\As\Ms\Ts\right]_{3'}} }[/math].

[math]\displaystyle{ \sum\vec\Fs_{\es\xs\ts} = \ms\acc{G}{Gal} }[/math]

Calculation of the [math]\displaystyle{ \Gs }[/math] acceleration:

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

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

Option 2: through rigid body kinematics. .

[math]\displaystyle{ \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]\displaystyle{ \velang{ring}{E} = (\Uparrow\dot\psi_0) + (\odot\dth) }[/math]

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

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

Formulation of the spring force

The vertical translational motion of the support [math]\displaystyle{ (\dot\xs) }[/math] is associated with the variation of an [math]\displaystyle{ \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]\displaystyle{ \xs=0 }[/math] for the equilibrium configuration in the absence of rotation [math]\displaystyle{ \dot\psi_0 }[/math], it is clear that the spring will have to exert an attraction force [math]\displaystyle{ \Fs_0 }[/math] on the pendulum to counteract the weight: [math]\displaystyle{ \Fs_0 = \ms\gs }[/math]. The general formulation of the spring attraction force will then be [math]\displaystyle{ \Fs_\ss^{\as\ts} = \ms\gs + \ks\Delta\rho }[/math].

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

[math]\displaystyle{ \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]\displaystyle{ \xs(\ts=0) }[/math], [math]\displaystyle{ \dot\xs(\ts=0) }[/math], [math]\displaystyle{ \theta(\ts=0) }[/math], [math]\displaystyle{ \dth(\ts=0) }[/math] under which the movement is started determine the DoF that will appear.

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

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

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

[math]\displaystyle{ \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]\displaystyle{ \dot\psi_0 = 0 }[/math]) are obtained from the equations of motion by imposing [math]\displaystyle{ \ddot\xs_\text{eq} = 0 }[/math], [math]\displaystyle{ \dot\xs_\text{eq} = 0 }[/math], [math]\displaystyle{ \ddth_\text{eq} = 0 }[/math], [math]\displaystyle{ \dth_\text{eq} = 0 }[/math]:

[math]\displaystyle{ \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]\displaystyle{ (\xs = \xs_\text{eq} + \varepsilon_\xs, \theta = \theta_\text{eq} + \varepsilon_\theta) }[/math], the equations can be linearized. For the configuration [math]\displaystyle{ (\xs_\text{eq}, \theta_\text{eq}) = (0,0) }[/math]:

[math]\displaystyle{ \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 \lt 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 \lt 0 \end{cases} }[/math]

Since [math]\displaystyle{ \ddot\varepsilon_\xs \lt 0 }[/math], [math]\displaystyle{ \ddot\varepsilon_\theta \lt 0 }[/math], it is a STABLE configuration.

For the [math]\displaystyle{ (\xs_\text{eq}, \theta_\text{eq}) = (0,180\deg) }[/math] configuration:

[math]\displaystyle{ \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 \gt 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 \gt 0 \end{cases} }[/math]

Since [math]\displaystyle{ \ddot\varepsilon_\xs \gt 0 }[/math], [math]\displaystyle{ \ddot\varepsilon_\theta \gt 0 }[/math], it is an UNSTABLE configuration.

Study of the equilibrium configurations under rotation

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

[math]\displaystyle{ (\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]\displaystyle{ \dot\psi_0^2 }[/math] since the [math]\displaystyle{ \text{cos}\theta_\text{eq} }[/math] function is bounded between -1 and +1:

[math]\displaystyle{ |\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]\displaystyle{ (\xs_\text{eq}, \theta_\text{eq}) = (0,0) }[/math] configuration, the linearization of the equation of motion for [math]\displaystyle{ \theta }[/math] leads to:

[math]\displaystyle{ \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]\displaystyle{ \dot\psi_0 \lt \dot\psi_{\cs\rs} }[/math] and [math]\displaystyle{ \ddot\varepsilon_\theta \lt 0 }[/math], the configuration is STABLE. If [math]\displaystyle{ \dot\psi_0 \gt \dot\psi_{\cs\rs} }[/math] and [math]\displaystyle{ \ddot\varepsilon_\theta \gt 0 }[/math], the configuration is UNSTABLE.

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

[math]\displaystyle{ \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 \gt 0 }[/math]

The configuration is UNSTABLE for all [math]\displaystyle{ \dot\psi_0^2 }[/math] values.

The study of configurations [math]\displaystyle{ \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. Tots els drets reservats

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