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 \(\boldsymbol{r}_{IJ}\).
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 sketch below). 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 \(\boldsymbol{r}_{HK}\)).
© 2025 R. Valéry Roy.