C2. Movement of a mechanical system

[math]\displaystyle{ \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]\displaystyle{ \Qs }[/math] (or a point that belongs to a rigid body) relative to a reference frame R, [math]\displaystyle{ \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]\displaystyle{ \overline{\Or\Qs} }[/math], [math]\displaystyle{ \overline{\Os'_\Rs\Qs} }[/math] ) yield the same velocity because points [math]\displaystyle{ \Os_\Rs }[/math] and [math]\displaystyle{ \Os'_\Rs }[/math] are mutually fixed and fixed to the reference frame, hence [math]\displaystyle{ \overline{\Os_\Rs\Os'_\Rs} }[/math] is constant in R:

[math]\displaystyle{ \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]

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 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]\displaystyle{ \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

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The platform (RP) rotates about an axis perpendicular to the ground (R). The movement of a point [math]\displaystyle{ \Qs }[/math] on the platform periphery depends on whether it is observed from the ground or from the platform.

The center of the platform ([math]\displaystyle{ \Os }[/math]) is fixed to both reference frames. Hence, [math]\displaystyle{ \vec{\Os\Qs} }[/math] is a position vector for point [math]\displaystyle{ \Qs }[/math] both in R and RP. It is evident that [math]\displaystyle{ \vvec_\Rs(\Qs)\neq \vec{0} }[/math] i [math]\displaystyle{ \vvec_{\Rs\Ps}(\Qs)= \vec{0} }[/math], though the vector whose time derivative is being calculated is the same.

As [math]\displaystyle{ \abs{\OQvec} }[/math] is the platform radius r, its value is constant. Hence, the time derivative of [math]\displaystyle{ \abs{\OQvec} }[/math] can only be associated with a change of direction.

To assess the change of orientation of [math]\displaystyle{ \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]\displaystyle{ \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).

[math]\displaystyle{ \psi(t)\neq\psi(t+dt) \implies \OQvec }[/math] changes its direction relative to R [math]\displaystyle{ \implies \textcolor{blau}{\vvec_\Rs(}\Qs\textcolor{blau}{) \neq \vec{0}} }[/math]

As seen in section V.1, [math]\displaystyle{ \textcolor{blau}{\vvec_\Rs(}\Qs\textcolor{blau}{)} }[/math] is perpendicular to [math]\displaystyle{ \OQvec }[/math], and its value is that of [math]\displaystyle{ \OQvec\:(\textrm{r}) }[/math] times the rate of change of orientation of [math]\displaystyle{ \OQvec }[/math] relative to R [math]\displaystyle{ (\dot{\psi}) }[/math]:

Velocity of [math]\displaystyle{ \Qs }[/math] relative to the platform (RP):

[math]\displaystyle{ \psi(t)=\psi(t+dt) \implies \OQvec }[/math] does not change its direction relative to RP [math]\displaystyle{ \textcolor{verd}{\implies \vvec_\Rs(}\Qs\textcolor{verd}{) = \vec{0}} }[/math]

Analytical calculation ➕

The two logical vector bases for the calculation are:

• Basis B (1,2,3) fixed in R (thus moving in RP): [math]\displaystyle{ \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]\displaystyle{ \velang{B'}{RP}=\vec{0},\velang{B'}{R} = -\vec{\dot{\psi}} }[/math]

Projection of the position vector [math]\displaystyle{ \OQvec }[/math] on both bases:

[math]\displaystyle{ \braq{\OQvec}{B}=\vector{r\cpsi}{r\spsi}{0}, \: \: \braq{\OQvec}{B'}=\vector{r}{0}{0} }[/math]

Velocity of [math]\displaystyle{ \Qs }[/math] relative to R:

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

Velocity of [math]\displaystyle{ \Qs }[/math] relative to RP:

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

✏️ EXAMPLE C2-1.2: Euler pendulum

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The endpoint [math]\displaystyle{ \Qs }[/math] of Euler pendulum describes a circular motion relative to the block. The corresponding velocity [math]\displaystyle{ \vel{Q}{BL} = \dert{\vecbf{CQ}}{BL} }[/math] can be obtained in a similar way as that used in the previous example.

The angle [math]\displaystyle{ \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]\displaystyle{ \Qs }[/math] relative to the ground can be obtained as the time derivative of vector [math]\displaystyle{ \vec{\Or\Qs} (=\vec{\Or\Cbf}+\vecbf{CQ}) }[/math] relative to the ground:

[math]\displaystyle{ \vel{Q}{R} = \dert{\vec{\Or\Qs}}{R} = \dert{\vec{\Or\Cbf}}{R}+ \dert{\vec{\Cbf\Qs}}{R} }[/math]

Vector [math]\displaystyle{ \vec{\Or\Cbf} }[/math] has a constant direction in R but a variable value. Hence, its time derivative is parallel to [math]\displaystyle{ \vec{\Or\Cbf} }[/math] with value [math]\displaystyle{ \dot x }[/math]. Vector [math]\displaystyle{ \vec{\Cbf\Qs} }[/math], however, has a constant value (L) but variable direction. Consequently, its time derivative is perpendicular to [math]\displaystyle{ \vec{\Cbf\Qs} }[/math], and its value is that of[math]\displaystyle{ \vec{\Cbf\Qs} }[/math] times the rate of change of orientation of [math]\displaystyle{ \vec{\Cbf\Qs} }[/math] relative to R ([math]\displaystyle{ \dot\psi }[/math]):

The [math]\displaystyle{ \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]\displaystyle{ \dot x }[/math] and [math]\displaystyle{ L\dot\psi }[/math], whose directions do correspond to one of those singular directions.

The first term of the expression [math]\displaystyle{ \vel{Q}{R} = \dert{\vec{\Or\Cbf}}{R}+\dert{\vecbf{CQ}}{R} }[/math] corresponds to the velocity of [math]\displaystyle{ \Cs }[/math] relative to the ground [math]\displaystyle{ \left(\vel{Q}{R} = \dert{\vec{\Or\Cbf}}{R} \right) }[/math], whereas the second one has no physical interpretation: point [math]\displaystyle{ \Cs }[/math] is not fixed in R, thus it is not a position vector in that reference frame.

Analytical calculation ➕

The two logical vector bases for the calculation are:

• Basis B (1,2,3) fixed relative to R and BL [math]\displaystyle{ \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]\displaystyle{ \velang{P}{B'}=\vec{0} }[/math], [math]\displaystyle{ \velang{RL}{B'} = -\vec{\dot{\psi}} }[/math]

Projection of the position vector [math]\displaystyle{ \OQvec }[/math] in both bases:

[math]\displaystyle{ \braq{\OQvec}{B} = \vector{x+\Ls \spsi}{-\Ls \cpsi}{0}\:\:\:\:\:\:\:\:\:\: \braq{\OQvec}{B'} = \vector{\Ls+ x \spsi}{x\cpsi}{0} }[/math]

Velocity of [math]\displaystyle{ \Qs }[/math] relative to R:

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

If we want to calculate the velocity of [math]\displaystyle{ \Qs }[/math] relative to BL, the position vector to be differentiated is [math]\displaystyle{ \vecbf{CQ} }[/math]:

[math]\displaystyle{ \braq{\vecbf{CQ}}{B} = \vector{\Ls sin \psi}{-\Ls cos \psi}{0}, \:\:\:\:\:\:\:\:\:\: \braq{\vecbf{CQ}}{B'}=\vector{\Ls}{0}{0} }[/math]

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C2.2 Acceleration of a particle

The acceleration of a particle [math]\displaystyle{ \Qs }[/math] (or of a point belonging to a rigid body) relative to a reference frame R, [math]\displaystyle{ \acc{Q}{R} }[/math], is the rate of change of its velocity with time:

[math]\displaystyle{ \acc{Q}{R} = \dert{\vel{Q}{R}}{R} }[/math]

✏️ EXAMPLE C2-2.1: rotating platform

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In the circular motion of point [math]\displaystyle{ \Qs }[/math] of the platform relative to the ground , the acceleration [math]\displaystyle{ \acc{Q}{R} }[/math] comes both from the change of value and the change of orientation of [math]\displaystyle{ \vel{Q}{R} }[/math]. As [math]\displaystyle{ \vel{Q}{R} }[/math] is always perpendicular to [math]\displaystyle{ \OQvec }[/math], its rate of change of orientation is [math]\displaystyle{ \dot\psi }[/math], the same as that of [math]\displaystyle{ \OQvec }[/math] :

The [math]\displaystyle{ \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]\displaystyle{ \rs\ddot\psi }[/math] and [math]\displaystyle{ r\dot\psi^2 }[/math] , whose directions do correspond to one of those singular directions.

Analytical calculation ➕

The vector bases B and B’ are the same as in example C2-1.1.

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

✏️ EXAMPLE C2-2.2: Euler pendulum

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The calculation of the acceleration of [math]\displaystyle{ \Qs }[/math] relative to the ground (R) is laborious because the velocity [math]\displaystyle{ \vel{Q}{R} }[/math] comes from the addition of two terms:

[math]\displaystyle{ \vel{Q}{R} = \dert{\vec{\Os_\Rs\Cbf}}{R} + \dert{\vecbf{CQ}}{R} }[/math].

• [math]\displaystyle{ \dert{\vec{\Os_\Rs\Cbf}}{R} }[/math]: constant direction (horizontal), variable value [math]\displaystyle{ (\dot x) }[/math]. Thus, its time derivative [math]\displaystyle{ \ddert{\vec{\Os_\Rs\Cbf}}{R} }[/math] is horizontal with value [math]\displaystyle{ \ddot x }[/math].
• [math]\displaystyle{ \dert{\vecbf{CQ}}{R} }[/math]: direction perpendicular to the bar, thus variable; variable value [math]\displaystyle{ \Ls\dot\psi }[/math]. Thus, its time derivative [math]\displaystyle{ \ddert{\vecbf{CQ}}{R} }[/math] has a component perpendicular to [math]\displaystyle{ \dert{\vecbf{CQ}}{R} }[/math] (parallel to the bar) with value [math]\displaystyle{ \Ls\dot\psi\cdot\dot\psi }[/math] , and another one parallel to [math]\displaystyle{ \dert{\vecbf{CQ}}{R} }[/math] (perpendicular to the bar) with value [math]\displaystyle{ \Ls\ddot\psi }[/math].

Analytical calculation ➕

The vector bases B and B’ are the same as in example C2-1.2.

Acceleration of [math]\displaystyle{ \Qs }[/math] relative to BL:

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

Acceleration of [math]\displaystyle{ \Qs }[/math] relative to R:

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

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C2.3 Intrinsic directions. Intrinsic components of the acceleration

A simple drawing shows that the velocity of a point [math]\displaystyle{ \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.

Figure C2.1 The velocity vector is always tangent to the trajectory

In a general case, the velocity [math]\displaystyle{ \vel{Q}{R} }[/math] changes both its value and its direction. Hence, the acceleration [math]\displaystyle{ \acc{Q}{R} }[/math] has two components, one associated with the change of value (parallel to [math]\displaystyle{ \vel{Q}{R} }[/math]) and another one associated with the change of direction (orthogonal to [math]\displaystyle{ \vel{Q}{R} }[/math]). Those components are the intrinsic components of the acceleration, and they are called tangential component [math]\displaystyle{ \accs{Q}{R} }[/math] and normal component [math]\displaystyle{ \accn{Q}{R} }[/math], respectively:

[math]\displaystyle{ \acc{Q}{R}=\accs{Q}{R}+\accn{Q}{R} }[/math]

In a 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):

Figure C2.2 Intrinsic components of the acceleration in a circular motion

That result may be used locally for any other movement. Indeed, as the calculation of the velocity of a point [math]\displaystyle{ \Qs }[/math] with respect to a reference frame R ([math]\displaystyle{ \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]\displaystyle{ \acc{Q}{R} }[/math] calls for three:

[math]\displaystyle{ \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]

The calculation of vector [math]\displaystyle{ \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]\displaystyle{ \vvec_\Rs(\textbf{Q},\textrm t) }[/math] and the first point to calculate [math]\displaystyle{ \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]\displaystyle{ \textrm{CC}_\textrm{R}(\textbf{Q}) }[/math] and [math]\displaystyle{ \Re_\textrm{R}(\textbf Q) }[/math] respectively). The results obtained for the circular motion may be used locally to calculate [math]\displaystyle{ \Re_\textrm{R}(\textbf Q) }[/math] (Figure C2.3).

Figure C2.3 local geometry of the trajectory of a particle Q relative to a reference frame R

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]\displaystyle{ \vecbf{s} }[/math] ([math]\displaystyle{ \vecbf{s}=\velo{R}/|\velo{R}|=\accso{R}/|\accso{R}| }[/math]) and the normal unit vector [math]\displaystyle{ \vecbf{n} }[/math] ([math]\displaystyle{ \vecbf{n}=\accno{R}/|\accno{R}| }[/math]) may be completed with a third unit vector [math]\displaystyle{ \vecbf{b} }[/math] orthogonal to the other two (binormal unit vector, [math]\displaystyle{ \vecbf{b}\equiv\vecbf{s}\times\vecbf{n} }[/math]), and constitute the intrinsic basis or Frenet basis for the motion of [math]\displaystyle{ \Qs }[/math] in the reference frame R.

✏️ EXAMPLE C2-3.1: Euler pendulum

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In the circular motion of the endpoint [math]\displaystyle{ \Qs }[/math] of the bar relative to the block, the two intrinsic components of the acceleration [math]\displaystyle{ \acc{Q}{BL} }[/math] are nonzero. Their values and directions are those of the circular motion:

• tangential acceleration [math]\displaystyle{ \accs{Q}{BL} }[/math]: parallel to [math]\displaystyle{ \vel{Q}{BL} }[/math] with value L[math]\displaystyle{ \ddot\psi }[/math].
• normal acceleration [math]\displaystyle{ \accn{Q}{BL} }[/math] : perpendicular to [math]\displaystyle{ \vel{Q}{BL} }[/math] with value L[math]\displaystyle{ \dot\psi^2 }[/math].

Though it is evident that the radius of curvature of the trajectory of [math]\displaystyle{ \Qs }[/math] relative to BL is L (it is a circular motion), it can also be obtained as [math]\displaystyle{ \frac{\vecbf{v}_{\textrm{BL}}^2(\Qs)}{|\accn{Q}{BL}|}=\frac{(\Ls\dot\psi)^2}{\Ls\dot\psi^2}=\Ls }[/math].

The acceleration [math]\displaystyle{ \acc{Q}{R} }[/math] has been described in example C2-2.2 as the addition of three terms (the two horizontal ones corresponding to [math]\displaystyle{ \acc{Q}{BL} }[/math] plus a permanently horizontal one with value [math]\displaystyle{ \ddot x }[/math]). Identifying in that case which is the tangential component (parallel to [math]\displaystyle{ \vel{Q}{R} }[/math]) and which is the normal one (orthogonal to [math]\displaystyle{ \vel{Q}{R} }[/math]) is not straightforward, as the [math]\displaystyle{ \vel{Q}{R} }[/math] direction is not that of a singular direction of the problem example C2-1.2.

That identification is straightforward in two particular configurations where the [math]\displaystyle{ \vel{Q}{R} }[/math] direction (which is the tangential direction) is horizontal:

The radius of curvature of the pendulum endpoint relative to the ground for the [math]\displaystyle{ \psi=0 }[/math] configuration is:

The center of curvature is always above [math]\displaystyle{ \Qs }[/math] because the normal acceleration points in that direction.

Particular cases:

The dotted circular lines correspond to the approximation of the trajectory in the neighbourhood of the [math]\displaystyle{ \psi=0 }[/math] configuration for those two particular cases.

Though it is a laborious, it is possible to calculate [math]\displaystyle{ \re{Q}{R} }[/math] for a general configuration if we remember that only the parallel components participate in the scalar product [math]\displaystyle{ \vel{Q}{R}\cdot\acc{Q}{R} }[/math] (and so [math]\displaystyle{ \accs{Q}{R} }[/math]), and that only the orthogonal components participate in the cross product [math]\displaystyle{ \vel{Q}{R}\times\acc{Q}{R} }[/math], (and so [math]\displaystyle{ \accn{Q}{R} }[/math]) (example C2-3.1 analytical). The result is:

[math]\displaystyle{ \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]

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]\displaystyle{ \dot x=0 }[/math] permanently (that is, [math]\displaystyle{ \ddot x=0 }[/math]), the trajectory of [math]\displaystyle{ \Qs }[/math] relative to R is circular with radius L:

[math]\displaystyle{ \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]\displaystyle{ \dot\psi=0 }[/math] permanently ([math]\displaystyle{ \ddot\psi=0 }[/math]), the trajectory of [math]\displaystyle{ \Qs }[/math] relative to R is rectilinear, and the radius of curvature has to be infinite:

[math]\displaystyle{ \re{Q}{R}\big]_{\dot\psi=0, \ddot\psi=0}=\frac{(\dot x^2)^{3/2}}{0}\rightarrow\infty }[/math]

Analytical calculation ➕

The vector bases B and B’ are the same as in example C2-2.1.

Acceleration of [math]\displaystyle{ \Qs }[/math] relative to BL:

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

Acceleration of [math]\displaystyle{ \Qs }[/math] relative to R:

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

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’).

[math]\displaystyle{ \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]\displaystyle{ \abs{\accn{Q}{R}}=\abs{\acc{Q}{R}\times\frac{\vel{Q}{R}}{\abs{\vel{Q}{R}}}} }[/math] [math]\displaystyle{ \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]

[math]\displaystyle{ \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]

[math]\displaystyle{ \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]

[math]\displaystyle{ \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]

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C2.4 Angular velocity of a rigid body

The configuration of a rigid body S relative to a reference frame R is totally defined through the position of a point [math]\displaystyle{ \Qs }[/math] of the rigid body and the orientation of S relative to R (described, for instance, by means of Euler angles). Similarly, the evolution of the configuration relative to R can be described through the velocity of a point [math]\displaystyle{ \Qs }[/math] of the rigid body [math]\displaystyle{ \vel{Q}{R} }[/math], and the angular velocity of the rigid body [math]\displaystyle{ \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]\displaystyle{ \left(\velang{S}{R}=0\right) }[/math].

Simple rotation

The orientation of a rigid body with planar motion relative to a reference frame R is totally defined by an angle [math]\displaystyle{ \psi }[/math].If that orientation changes, [math]\displaystyle{ \dot\psi\neq0 }[/math] .

Giving the value of [math]\displaystyle{ \dot\psi }[/math] [math]\displaystyle{ [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

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The wheel has a planar motion relative to R. Its center [math]\displaystyle{ \Cs }[/math] is fix fixed in R, and its orientation changes with a rate [math]\displaystyle{ \dot\psi }[/math] [math]\displaystyle{ [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:

• Case (a): angle [math]\displaystyle{ \psi }[/math] defined on the horizontal plane; the plane of motion is horizontal.
• Case (b): angle [math]\displaystyle{ \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.

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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]\displaystyle{ \velang{S}{R} }[/math] is a vector. The convention to associate a direction to that vector is the screw rule (or the right hand rule, or the corkscrew rule,).

✏️ EXAMPLE C2-4.2: wheel with planar motion

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The angular velocity associated with movements (a) and (b) in the previous example is:

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Rotation in space

The orientation of a rigid body moving in space relative to a reference frame R may be given through three Euler angles [math]\displaystyle{ (\psi,\theta,\varphi) }[/math]. We may associate an angular velocity to the change of each of those angles.

✏️ EXAMPLE C2-4.3: gyroscope

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The orientation of a gyroscope relative to the ground (R) may be given through three Euler angles. The angular velocities associated with [math]\displaystyle{ (\dot\psi,\dot\theta,\dot\varphi) }[/math] have the following interpretation:

[math]\displaystyle{ \vecdot\psi=\velang{fork}{R} }[/math], [math]\displaystyle{ \vecdot\theta=\velang{arm}{fork} }[/math], [math]\displaystyle{ \vecdot\varphi=\velang{disk}{arm} }[/math].

File:C2-Ex4-3-1-eng-jpg.jpg

Those angular velocities can be projected on any vector basis suggested by the problem:

• Vector basis [math]\displaystyle{ \Bs_\Rs }[/math] fixed to the reference frame
• Vector basis [math]\displaystyle{ \Bs }[/math] fixed to the fork (it can be generated from [math]\displaystyle{ \Bs_\Rs }[/math] through the [math]\displaystyle{ \dot\psi }[/math] rotation)
• Vector basis [math]\displaystyle{ \Bs' }[/math] fixed to the arm (it can be generated from [math]\displaystyle{ \Bs }[/math] through the [math]\displaystyle{ \dot\theta }[/math] rotation)
• Vector basis [math]\displaystyle{ \Bs_\textrm{V} }[/math] fixed to the disk

File:C2-Ex4-3-2-eng-jpg.jpg

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]\displaystyle{ \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]\displaystyle{ \vec{\dot{\varphi}} }[/math], when we choose B, and [math]\displaystyle{ \vec{\dot{\psi}} }[/math] when we choose B’:

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

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C2.5 Angular acceleration of a rigid body

The angular acceleration of a rigid body S relative to a reference frame R ([math]\displaystyle{ \accang{S}{R} }[/math]) is the time derivative of its angular velocity relative to R:

[math]\displaystyle{ \accang{S}{R}= \dert{\velang{S}{R}}{R} }[/math]

The description of the angular velocity [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \velang{S}{R} }[/math], and it is parallel to [math]\displaystyle{ \velang{S}{R} }[/math]. In general motions in space, if [math]\displaystyle{ \velang{S}{R} }[/math] is described through Euler rotations, [math]\displaystyle{ \accang{S}{R} }[/math] may come from the change of values of ([math]\displaystyle{ \vecdot\psi }[/math], [math]\displaystyle{ \vecdot\theta }[/math],[math]\displaystyle{ \vecdot\varphi }[/math]) iand the change of direction of de [math]\displaystyle{ \vecdot\theta }[/math] and [math]\displaystyle{ \vecdot\varphi }[/math] ([math]\displaystyle{ \vecdot\psi }[/math] has always a constant direction in R).

✏️ EXAMPLE C2-5.1: gyroscope

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The fork of a gyroscope has a planar motion relative to the ground (R), and its angular velocity is vertical: [math]\displaystyle{ \velang{fork}{R}=\vecdot\psi }[/math] Its angular acceleration is also vertical, with value [math]\displaystyle{ \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]\displaystyle{ \velang{disk}{R}=\vecdot\psi+\vecdot\theta+\vecdot\varphi }[/math]. The rotation [math]\displaystyle{ \vecdot\varphi }[/math] can be decomposed in a vertical component with value [math]\displaystyle{ \dot\varphi\textrm{sin}\theta }[/math], and a horizontal one with value [math]\displaystyle{ \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]\displaystyle{ \vecdot\psi }[/math]).

Time derivative of the vertical components

Time derivative of the horizontal components

Analytical calculation ➕

The same result is obtained if the time derivative is performed analytically through the vector basis rotating with [math]\displaystyle{ \vecdot\psi }[/math] relative to R or that rotating with [math]\displaystyle{ \vecdot\psi+\vecdot\theta }[/math] (also relative to R):

[math]\displaystyle{ \braq{\velang{}{R}}{B}=\vector{\dot\theta}{\dot\varphi \cth}{\dot\psi+\dot\varphi \sth}, }[/math]          [math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \braq{\velang{disk}{R}}{B'}=\vector{\dot\theta}{\dot\varphi+\dot\psi \sth}{\dot\psi \cth}, }[/math]          [math]\displaystyle{ \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]\displaystyle{ \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]

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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

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Particle [math]\displaystyle{ \Ps }[/math] rotates relative to R: WRONG Vector [math]\displaystyle{ \vec{\textbf{OP}} }[/math] rotates relative to R: CORRECT Particle [math]\displaystyle{ \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

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Particle [math]\displaystyle{ \Ps }[/math] has a translational motion relative to R: WRONG Particle [math]\displaystyle{ \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

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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

Some points on a rotating rigid body may have rectilinear motion.

Video C2.1 Visualització de les trajectòries de punts

✏️ EXAMPLE C2-6.4: motion of a ferris wheel

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	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
	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.

Points in a rigid body with a translational motion may describe curvilinear trajectories.

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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), the number of DOF is 6: three associated with the motion of one point (for instance, [math]\displaystyle{ (\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]\displaystyle{ (\dot{\psi}, \dot{\theta}, \dot{\varphi}) }[/math]).

In mechanical engineering, the usual mechanical systems are multibody systems: sets of rigid bodies 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.

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C2.8 Usual constraints in mechanical systems

Falta paragraf versio catala

	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)
	revolute joint Allows a rotation between the two rigid bodies about axis 1
	cylindrical joint Allows a rotation between the two rigid bodies about axis 1, and a translational motion (displacement without rotation) along axis 1.
	prismatic joint Allows a translational motion between the two rigid bodies along axis 1.
	spherical joint Allows three independent rotations between the two rigid bodies about axes 1, 2, 3.
	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].
	Cardan joint (universal joint) Allows two independent rotations between the two rigid bodies about axes 1, 3.

Video C2.2 Junta Cardan (junta universal o de creueta)

Video C2.3 Graus de Llibertat d'una roda emb moviment pla i contacte amb el terra

✏️ EXAMPLE C2-8.1: gyroscope

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In a 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:

The position of point [math]\displaystyle{ \Os }[/math] relative to the ground is constant. Hence, the gyroscope configuration is totally defined by the three angles [math]\displaystyle{ (\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]\displaystyle{ (\dot\psi,\dot\theta,\dot\varphi) }[/math].

✏️ EXEMPLE C2-8.2: tricycle

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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).

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]\displaystyle{ \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]\displaystyle{ \Os }[/math]. If we block that rotation [math]\displaystyle{ (\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]\displaystyle{ (\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

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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.

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.

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C2.E General examples

🔎 Example C2-E.1: rotating pendulum

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The plate is articulated at point [math]\displaystyle{ \Os }[/math] O to a fork, which rotates with constant angular velocity [math]\displaystyle{ \psio }[/math] relative to the ground (T). Between fork and ground (ceiling), and between plate and fork there are revolute joints.

1. How many degrees of freedom (DoF) has the system? Describe them.

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.

2. Find the angular velocity and the angular acceleration of the plate relative to the ground.

The angular velocity of the plate is the superposition of [math]\displaystyle{ \vecdot\psi_0 }[/math] (1st Euler rotation, axis fixed to the ground) and [math]\displaystyle{ \vecdot\theta }[/math] (2nd Euler rotation, axis rotating with [math]\displaystyle{ \vecdot\psi_0 }[/math]relative to the ground): [math]\displaystyle{ \velang{plate}{E}=\vecdot\psi_0+\vecdot\theta }[/math]

Geometric calculation:

[math]\displaystyle{ \velang{plate}{E}=\vecdot\psi_0+\vecdot\theta=(\Uparrow \psio)+(\odot \dot{\theta}) }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \vecdot\theta }[/math].It is a vector with variable value which rotates about a vertical axis because of the 1st Euler rotation [math]\displaystyle{ (\Omegavec^{\vecdot\theta}_\textrm{T} =\vecdot\psi_0) }[/math].

[math]\displaystyle{ \accang{plate}{E}=\dert{(\odot\dot{\theta})}{E}=[\text{change of value}]+[\text{change of direction}]_\Es= }[/math]

[math]\displaystyle{ =\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]\displaystyle{ \velang{plate}{E} }[/math] projection is straightforward is the vector basis fixed to the fork [math]\displaystyle{ (\velang{B}{E}=\vecdot{\psi}_0) }[/math]:

[math]\displaystyle{ \braq{\velang{plate}{E}}{B}=\vector{\dot{\theta}}{0}{\psio} }[/math]

[math]\displaystyle{ \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]

3. Find the velocity and the acceleration of point G of the plate relative to the ground. .

As point [math]\displaystyle{ \Os }[/math] is fixed to the ground, [math]\displaystyle{ \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]\displaystyle{ \psio }[/math] and [math]\displaystyle{ \vecdot{\theta} }[/math]: [math]\displaystyle{ \OGvec=(\searrow\Ls)^{*} }[/math].

Geometric calculation:

[math]\displaystyle{ \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]

That velocity has variable value and direction, thus the acceleration has both parallel component and orthogonal component to the velocity.

[math]\displaystyle{ \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]\displaystyle{ (\otimes\Ls\psio\text{sin}\theta) }[/math] vector rotates relative to the ground just because of [math]\displaystyle{ \psio }[/math], whereas the [math]\displaystyle{ (\nearrow\Ls\dot{\theta}) }[/math] vector rotates because of [math]\displaystyle{ \vec{\psio} }[/math] and [math]\displaystyle{ \vec{\dot{\theta}} }[/math]:

[math]\displaystyle{ \dert{(\otimes\Ls\psio\text{sin}\theta)}{E}=[\text{change of value}]+[\text{change of direction}]_\Es= }[/math]

[math]\displaystyle{ \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]\displaystyle{ \dert{(\nearrow\Ls\dot{\theta})}{E}=[\text{change of value}]+[\text{change of direction}]_\Es= }[/math]

[math]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \OGvec }[/math]is straightforward is fixed to the plate (base B’). That vector basis changes its orientation whenever the values of [math]\displaystyle{ \psi }[/math] and [math]\displaystyle{ \theta }[/math] change. Hence, the angular velocity of the vector basis is [math]\displaystyle{ \velang{B'}{E}=\vec{\psio}+\vec{\dot{\theta}} }[/math]:

[math]\displaystyle{ \braq{\OGvec}{B'}=\vector{0}{0}{-L} }[/math]

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

🔎 Example C2-E.2: rotating articulated plate

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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 spherical joint at [math]\displaystyle{ \Ps }[/math], and to the support through a cylindrical joiny . ). The support rotates with the variable angular velocity [math]\displaystyle{ \vecdot{\psi} }[/math] vrelative to the ground (E).

1. How many degrees of freedom (DoF) has the system? Describe them.

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]\displaystyle{ \OCvec }[/math] bars may have a simple rotation, relative to the support, about the horizontal axis through [math]\displaystyle{ \Os }[/math] orthogonal to the bars. If we block the motion of one of those bars relative to the support, the plate, the [math]\displaystyle{ \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]\displaystyle{ \OCvec }[/math] bars may move relative to the support.

Hence, the system has two degrees of freedom.

2. Find the angular velocity and the angular acceleration of the plate relative to the ground.

The angular velocity of the plate is the superposition of [math]\displaystyle{ \vec{\dot{\psi}} }[/math] (1st Euler rotation, axis fixed to the ground) and [math]\displaystyle{ \vec{\dot{\theta}} }[/math] (2nd Euler rotation, axis rotation because of [math]\displaystyle{ \vec{\dot{\psi}} }[/math]):

Geometric calculation:

[math]\displaystyle{ \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]\displaystyle{ \vec{\dot{\psi}} }[/math], and the change of direction of [math]\displaystyle{ \vec{\dot{\theta}} }[/math]:

[math]\displaystyle{ \accang{plate}{E}=\dert{\velang{plate}{E}}{E}=\dert{(\Uparrow\dot{\psi})}{E}=\dert{(\otimes\dot{\theta})}{E} }[/math]

[math]\displaystyle{ \dert{(\Uparrow\dot{\psi})}{E}=[\text{change of value}]=\ddot{\psi}\frac{\vec{\dot{\psi}}}{|\vec{\dot{\psi}}|}=(\Uparrow\ddot{\psi}) }[/math]

[math]\displaystyle{ \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]\displaystyle{ \accang{plate}{E}=(\Uparrow\ddot{\psi})+(\otimes\ddot{\theta})+(\Leftarrow\dot{\psi}\dot{\theta}) }[/math]

Analytical calculation:

The time derivative of the angular velocity can also be done analytically. The vector basis B where the [math]\displaystyle{ \velang{plate}{E} }[/math] projection is straightforward is the one fixed to the support [math]\displaystyle{ (\velang{B}{E}=\vec{\dot{\psi}}) }[/math]:

[math]\displaystyle{ \braq{\velang{plate}{E}}{B}=\vector{0}{\dot{\theta}}{\dot{\psi}} }[/math]

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

3. Find the velocity and the acceleration of point Q of the plate relative to the ground.

As point [math]\displaystyle{ \Os }[/math] is fixed to the ground, [math]\displaystyle{ \OQvec }[/math] is a position vector in the ground frame. Its value is [math]\displaystyle{ 2\Ls\text{cos}\theta }[/math], , and its direction is always horizontal. The velocity of [math]\displaystyle{ \Qs }[/math]comes both from the change of that value (as [math]\displaystyle{ \theta }[/math] is variable) and the change of its direction relative to the ground (because of the support rotation [math]\displaystyle{ \vec{\dot{\psi}} }[/math]).

Geometric calculation:

[math]\displaystyle{ \OQvec=(\rightarrow 2\Ls\text{cos}\theta) }[/math]

[math]\displaystyle{ \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]\displaystyle{ =\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]\displaystyle{ \Qs }[/math] comes from the change of value and direction (associated with [math]\displaystyle{ \vec{\dot{\psi}} }[/math]) of both terms in the velocity:

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

The tome derivative can also be done analytically. The vector basis B where the [math]\displaystyle{ \OPvec }[/math] projection is straightforward is the one fixed to the support [math]\displaystyle{ (\velang{B}{E}=\vec{\dot{\psi}}) }[/math]:

[math]\displaystyle{ \braq{\OQvec}{B}=\vector{2\Ls\text{cos}\theta}{0}{0} }[/math]

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

🔎 Example C2-E.3: rotating pendulum with oscillating articulation point

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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]\displaystyle{ (\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.

1. How many degrees of freedom (DoF) has the system? Describe them.

The guide may rotate about the vertical axis through [math]\displaystyle{ \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]\displaystyle{ \Os ' }[/math], which is perpendicular to the ring plane and is fixed to the support.

Hence, the system has 3 degrees of freedom.

2. Find the angular velocity and the angular acceleration of the ring relative to the ground.

Geometric calculation:

The angular velocity of the ring is the superposition of [math]\displaystyle{ (\vec{\psio}) }[/math] (1st Euler rotation, axis fixed to the ground) and [math]\displaystyle{ \vec{\dot{\theta}} }[/math] (2nd Euler rotation, axis rotating with [math]\displaystyle{ \vec{\dot{\psi}} }[/math]):

[math]\displaystyle{ \velang{ring}{E}=\vec{\psio}+\vec{\dot{\theta}}=(\Uparrow\psio)+(\odot\dot{\theta}) }[/math]

[math]\displaystyle{ \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]\displaystyle{ =\dert{(\Uparrow\psio)}{E}+\dert{(\odot\dot{\theta})}{E} }[/math]

The angular acceleration comes exclusively from the change of value and direction of [math]\displaystyle{ \vec{\dot{\theta}} }[/math], as [math]\displaystyle{ \vec{\psio} }[/math] has both constant value and constant direction.

[math]\displaystyle{ \accang{ring}{E} = \dert{\velang{ring}{E}}{E} = \dert{(\odot\dot{\theta})}{E}=[\text{change of value}]+[\text{change of direction}]_\Es= }[/math]

[math]\displaystyle{ =\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]\displaystyle{ \velang{ring}{E} }[/math] és immediata és la fixa al suport [math]\displaystyle{ (\velang{B}{E}=\vec{\dot{\psi}}) }[/math]:

[math]\displaystyle{ \braq{\velang{ring}{E}}{B}=\vector{0}{\psio}{\dot{\theta}} }[/math]

[math]\displaystyle{ \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]

3. Find the velocity and the acceleration of point G of the ring relative to the ground.

Geometric calculation:

[math]\displaystyle{ \vec{\Os'\Gs} }[/math] is a position vector for [math]\displaystyle{ \Gs }[/math] in the ground frame, as [math]\displaystyle{ \Os' }[/math] is a point fixed to the ground.

[math]\displaystyle{ \vec{\Os'\Gs}=\vec{\Os'\Os}+\vec{\Os\Gs}=(\downarrow \textrm{x})+(\searrow \Ls)^* }[/math]

[math]\displaystyle{ \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]

The term [math]\displaystyle{ (\downarrow \text{x}) }[/math] has a variable value but a constant orientation, whereas the term [math]\displaystyle{ (\searrow \Ls) }[/math], with constant value, changes its orientation relative to the ground because of [math]\displaystyle{ \vec{\psio} }[/math] and [math]\displaystyle{ \vec{\dot{\theta}} }[/math]:

[math]\displaystyle{ \dert{\vec{\Os'\Os}}{E}=\dert{(\downarrow \textrm{x})}{E}=[\text{change of value}]=(\downarrow\dot{\text{x}}) }[/math]

[math]\displaystyle{ \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]\displaystyle{ =(\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]\displaystyle{ \vel{G}{E}=(\downarrow\dot{\text{x}})+(\otimes\Ls\psio\text{sin}\theta)+(\nearrow\Ls\dot{\theta}) }[/math]

[math]\displaystyle{ \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]\displaystyle{ \vec{\psio} }[/math], whereas the third one rotates because of [math]\displaystyle{ \vec{\psio} }[/math] and [math]\displaystyle{ \vec{\dot{\theta}} }[/math].The time derivatives of those terms are:

[math]\displaystyle{ \dert{(\downarrow\dot{\text{x}})}{E}=[\text{change of value}]=(\downarrow\ddot{x}) }[/math]

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

Analytical calculation:

The whole calculation can be done analytically. The first term in [math]\displaystyle{ \OGvec=\vec{\Os\Os'}+\vec{\Os'\Gs} }[/math] is vertical, hence its projection on the vector basis B fixed to the support [math]\displaystyle{ (\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]\displaystyle{ (\velang{B'}{E}=\vec{\psio}+\vec{\dot{\theta}}) }[/math]. Any of these two vector bases is suitable.

Calculation with the B vector basis

[math]\displaystyle{ \braq{\OGvec}{B}=\vector{\Ls\text{sin}\theta}{-\text{x}-\Ls\text{cos}\theta}{0} }[/math]

[math]\displaystyle{ \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]\displaystyle{ =\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]\displaystyle{ \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]\displaystyle{ \braq{\OGvec}{B}=\vector{ x\text{sin}\theta}{- x\text{cos}\theta-\Ls}{0} }[/math]

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

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*NOTE: In this web (for lack of more precise symbols), though the arrows s [math]\displaystyle{ \nearrow }[/math], [math]\displaystyle{ \swarrow }[/math], [math]\displaystyle{ \nwarrow }[/math] and [math]\displaystyle{ \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]\displaystyle{ \OPvec }[/math] vector forms a generic [math]\displaystyle{ \theta }[/math] angle with the vertical direction. If the value of [math]\displaystyle{ \theta }[/math] is less than 90° (as in the following figure), the [math]\displaystyle{ \OPvec }[/math] vector has a downward and rightward component.

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