Practice Exam 1

The figure displayed below shows two rigid bodies interconnected in the following manner:

A shaft denoted 1 is in rotation about axis \((O, \boldsymbol{\hat{z}}_0)\) relative to body 0 which houses a number of spheres such as that denoted as 2. Let \(\boldsymbol{\omega}_{1/0}= \omega_{1}\boldsymbol{\hat{z}}_0\) the angular velocity of shaft 1 relative to the housing 0 (\(\omega_1\) is assumed known).
Sphere 2 (of center \(C\) and radius \(r\)) is interposed between 0 and 1. It is in contact with housing 0 at points \(I\) and \(J\), and with shaft 1 at point \(K\). The position of center \(C\) relative to 0 is defined by angle \(\theta (t)\) and the corresponding unit vectors \((\boldsymbol{\hat{u}}, \boldsymbol{\hat{v}})\).

Assume that the no-slip condition holds at all contact points \(I\), \(J\), and \(K\).

Denote distance \(|OC|\) as \(R\). Line \(CK\) is perpendicular to line \(IJ\).

The goal of this problem is to analyze the kinematics of sphere 2.

Figure 1

Question 1. From the knowledge of the instantaneous screw axis \(\Delta_{2/0}\) of 2 relative to 0, deduce that angular velocity \(\boldsymbol{\omega}_{2/0}\) can be written in the following form \[ \boldsymbol{\omega}_{2/0} = \omega_2 \frac{\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0}{\sqrt{2}} \]

(hints: identify \(\Delta_{2/0}\). This will give you the expression of the corresponding angular velocity. However do not yet determine \(\omega_2\)).

Question 2. Write the kinematic screw loop equation \(\{{\cal V}_{2 /0 }\} = \{{\cal V}_{2 / 1 }\}+ \{{\cal V}_{1 / 0 }\}\) resolved at point \(K\). From this equation, deduce the expression of \(\omega_2\) in terms of \(\omega_1\) and the parameters \(R\) and \(r\).

(hints: (i) for \(\{ {\cal V}_{2 / 1 } \}\) account for the condition at \(K\), (ii) for \(\{ {\cal V}_{2 / 0} \}\) account for the result of q.1).

Question 3. Find the velocity of point \(C\) (relative to 0). Then deduce the expression of angular velocity \(\dot{\theta}\) in terms of \(\omega_1\) and the parameters \(R\) and \(r\).

Question 4. With a method of your choice, find the location of screw axis \(\Delta_{2/1}\). Carefully sketch the precise location of \(\Delta_{2/1}\).

(hint: identify 2 points through which \(\Delta_{2/1}\) must pass.)

Formula Sheet

1. Time-Derivative Formula: Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), and an arbitrary vector function \(\boldsymbol{U}(t)\), \[ \Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal E}=\Big(\frac{d\boldsymbol{U}}{dt} \Big)_{\cal F}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{U} \] 2. Kinematic Screw: Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), for any two points \(P\) and \(Q\), and at any given time, \[ \boldsymbol{v}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{v}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} \] \[ \boldsymbol{a}_{Q \in {\cal F}/{\cal E}}= \boldsymbol{a}_{P \in {\cal F}/{\cal E}}+ \boldsymbol{\alpha}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ} + \boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times (\boldsymbol{\omega}_{{\cal F}/ {\cal E}}\times \boldsymbol{r}_{PQ}) \] 3. Change of Referential Formulas: Given two rigid bodies (or referentials) \({\cal E}\) and \({\cal F}\), for any particle \(P\), and at any given time, \[ \boldsymbol{v}_{P/{\cal E}} = \boldsymbol{v}_{P/{\cal F}} + \boldsymbol{v}_{P\in {\cal F}/ {\cal E}} \] \[ \boldsymbol{a}_{P/{\cal E}} = \boldsymbol{a}_{P/{\cal F}} + \boldsymbol{a}_{P\in {\cal F}/ {\cal E}} + 2 \boldsymbol{\omega}_{{\cal F}/{\cal E}} \times \boldsymbol{v}_{P/{\cal F}} \] 4. Loop formulas: Given 3 rigid bodies \({\cal A}\), \({\cal B}\), \({\cal C}\), and an arbitrary point \(P\), \[ \boldsymbol{\omega}_{{\cal A}/{\cal B}}+ \boldsymbol{\omega}_{{\cal B}/{\cal C}}+ \boldsymbol{\omega}_{{\cal C}/{\cal A}} = \boldsymbol{0} \] \[ \boldsymbol{v}_{P\in {\cal A}/ {\cal B}}+ \boldsymbol{v}_{P\in {\cal B}/ {\cal C}}+ \boldsymbol{v}_{P\in {\cal C}/ {\cal A}}=\boldsymbol{0} \]

5. Triple product formula: Given 3 vectors \(\boldsymbol{A}\), \(\boldsymbol{B}\) and \(\boldsymbol{C}\)
\[ \boldsymbol{A}\times (\boldsymbol{B}\times \boldsymbol{C})= (\boldsymbol{A}\cdot \boldsymbol{C}) \boldsymbol{B}- (\boldsymbol{A}\cdot \boldsymbol{B}) \boldsymbol{C} \]

Solution

Question 1.

To find the screw axis \(\Delta_{2/0}\), we use the no-slip conditions at \(I\) and \(J\): \[ \boldsymbol{v}_{I\in 2/0} = \boldsymbol{v}_{J\in 2/0} = \boldsymbol{0}. \] This implies that \(\Delta_{2/0}\) is necessarily an instantaneous axis of rotation which passes through points \(I\) and \(J\). Then angular velocity \(\boldsymbol{\omega}_{2/0}\) is necessarily colinear to a unit vector along line \(IJ\): \[ \boldsymbol{\omega}_{2/0} = \omega_2 \frac{\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0}{\sqrt{2}} \]

Remark: This last equation can also be found by writing: \(\boldsymbol{v}_{J\in 2/0} = \boldsymbol{v}_{I\in 2/0} + \boldsymbol{\omega}_{2/0}\times \boldsymbol{r}_{IJ}\) leading to \(\boldsymbol{\omega}_{2/0}\times \boldsymbol{r}_{IJ} = \boldsymbol{0}\). We find that vector \(\boldsymbol{\omega}_{2/0}\) is colinear to \(\boldsymbol{r}_{IJ}\).

Figure 2

Question 2.

We want write the loop equation \[ \{{\cal V} _{1 / 0 }\}+ \{{\cal V} _{2 / 1 }\} = \{{\cal V} _{2 / 0 }\} \] resolved at point \(K\).

We start with \[ \{ {\cal V} _{1 / 0 } \} = \left\{ \begin{array}{c} \omega_1 \boldsymbol{\hat{z}}_0 \\ \boldsymbol{0} \end{array} \right\}_{O} \] which imposes that the motion of body 1 relative to 0 is a rotation about axis \((O,\boldsymbol{\hat{z}}_0)\) at angular velocity \(\omega_1\boldsymbol{\hat{z}}_0\).

Next we find \[ \{ {\cal V} _{2 / 1 } \} = \left\{ \begin{array}{c} \boldsymbol{\omega}_{2/1} \\ \boldsymbol{0} \end{array} \right\}_{K} \] where we have imposed the no-slip condition at \(K\). At this stage, angular velocity \(\boldsymbol{\omega}_{2/1}\) is unknown.

Finally we write \[ \{ {\cal V} _{2 / 0 } \} = \left\{ \begin{array}{c} \frac{\omega_{2}}{\sqrt{2}} (\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0) \\ \boldsymbol{0} \end{array} \right\}_{I} \] according to the result of question 1.

This gives the screw equality \[ \left\{ \begin{array}{c} \frac{\omega_{2}}{\sqrt{2}} (\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0) \\ \boldsymbol{0} \end{array} \right\}_{I} = \left\{ \begin{array}{c} \boldsymbol{\omega}_{2/1} \\ \boldsymbol{0} \end{array} \right\}_{K} + \left\{ \begin{array}{c} \omega_1 \boldsymbol{\hat{z}}_0 \\ \boldsymbol{0} \end{array} \right\}_{O} \] or after resolving each screw at point \(K\) \[ \left\{ \begin{array}{c} \frac{\omega_{2}}{\sqrt{2}} (\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0) \\ \frac{\omega_{2}}{\sqrt{2}} (\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0) \times \boldsymbol{r}_{IK} \end{array} \right\}_{K} = \left\{ \begin{array}{c} \boldsymbol{\omega}_{2/1} \\ \boldsymbol{0} \end{array} \right\}_{K} + \left\{ \begin{array}{c} \omega_1 \boldsymbol{\hat{z}}_0 \\ \omega_1 \boldsymbol{\hat{z}}_0 \times \boldsymbol{r}_{OK} \end{array} \right\}_{K} \] with \[ \boldsymbol{r}_{IK} = \boldsymbol{r}_{IC} + \boldsymbol{r}_{CK}= r\boldsymbol{\hat{z}}_0 + \frac{r}{\sqrt{2}} (\boldsymbol{\hat{z}}_0-\boldsymbol{\hat{u}})= \frac{r}{\sqrt{2}}[ (1+\sqrt{2}) \boldsymbol{\hat{z}}_0 - \boldsymbol{\hat{u}}] \] and \[ \boldsymbol{r}_{OK} = \boldsymbol{r}_{OC}+\boldsymbol{r}_{CK}= (R -\frac{r}{\sqrt{2}} )\boldsymbol{\hat{u}}+ \frac{r}{\sqrt{2}} \boldsymbol{\hat{z}}_0. \] Note: to find these expressions, we used the assumption that line \(CK\) is perpendicular to line \(IJ\).

The equality \(\boldsymbol{\omega}_{2/0}\times \boldsymbol{r}_{IK}= \boldsymbol{\omega}_{1/0} \times \boldsymbol{r}_{OK}\) can now be written as \[ \frac{\omega_{2}}{\sqrt{2}} (\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0) \times\frac{r}{\sqrt{2}}[ (1+\sqrt{2}) \boldsymbol{\hat{z}}_0 - \boldsymbol{\hat{u}}] = \omega_1 \boldsymbol{\hat{z}}_0 \times [(R -\frac{r}{\sqrt{2}})\boldsymbol{\hat{u}}+ \frac{r}{\sqrt{2}} \boldsymbol{\hat{z}}_0] \] All terms are directed along unit vector \(\boldsymbol{\hat{v}}\): \[ -r \omega_{2} (1+\frac{1}{\sqrt{2}}) = \omega_1 (R -\frac{r}{\sqrt{2}}) \] \[ \boxed{ \omega_2 = - \omega_1 \frac{\sqrt{2} R -r}{(1 + \sqrt{2})r} } \]

Question 3. To find the velocity of point \(C\), we can either take the derivative of \(\boldsymbol{r}_{OC} = R \boldsymbol{\hat{u}}\) or use the expression of kinematic screw \(\{ {\cal V} _{2 / 0 } \}\): \[ \boldsymbol{v}_{C \in 2/0} = R \dot{\theta}\boldsymbol{\hat{v}} \] or (using point \(I\)) \[ \boldsymbol{v}_{C \in 2/0} = \frac{\omega_2}{\sqrt{2}} (\boldsymbol{\hat{u}}+ \boldsymbol{\hat{z}}_0) \times r\boldsymbol{\hat{z}}_0 = -\frac{r \omega_2}{\sqrt{2}} \boldsymbol{\hat{v}} \] Compare these two results to get \(R\dot{\theta}= - r \omega_2 / \sqrt{2}\) or \[ \boxed{ \dot{\theta}= -\frac{r}{R\sqrt{2}} \omega_2 = \omega_1 {\sqrt{2} R -r \over R (2 + \sqrt{2})} } \]

Question 4.

Axis \(\Delta_{2/1}\) must necessarily pass through point \(K\) since \(\boldsymbol{v}_{K\in 2/1}=\boldsymbol{0}\). It is an instantaneous axis of rotation. Also we note that axes \(\Delta_{1/0}\) and \(\Delta_{2/0}\) intersect at point \(H\) on the axis \((O,\boldsymbol{\hat{z}}_0)\) (see Figure 2). Then we can write at point \(H\): \[ \boldsymbol{v}_{H \in 2/1} = \boldsymbol{v}_{H \in 2/0}+\boldsymbol{v}_{H \in 0/1} = \boldsymbol{0} \] which shows that axis \(\Delta_{2/1}\) must pass though \(H\). Therefore, axis \(\Delta_{2/1}\) is line \(KH\) (this implies that angular velocity \(\boldsymbol{\omega}_{2/1}\) is colinear to \(\boldsymbol{r}_{HK}\)).