Solution:
1. We start with the determination of \(\boldsymbol{v}_{A\in 2}\): \[ \boldsymbol{v}_{A\in 2}= \boldsymbol{v}_{B\in 2} + \omega_2 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{BA} = -v_B \boldsymbol{\hat{\jmath}}+ \omega_2 \boldsymbol{\hat{k}}\times l(-\cos\theta\boldsymbol{\hat{\jmath}}-\sin\theta\boldsymbol{\hat{\imath}}) = l\omega_2 \cos\theta\boldsymbol{\hat{\imath}}+ (-v_B - l\omega_2 \sin\theta)\boldsymbol{\hat{\jmath}} \] Then we relate \(\boldsymbol{v}_A =\boldsymbol{v}_{A\in 2}= \boldsymbol{v}_{A\in 1}\) to \(\boldsymbol{v}_{I\in 1}= {\bf 0}\) (taking into account the no-slip condition of the wheel): \[ \boldsymbol{v}_{A\in 1}= \omega_1 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{IA} = \omega_1 \boldsymbol{\hat{k}}\times 2r \boldsymbol{\hat{\jmath}}= - 2r \omega_1 \boldsymbol{\hat{\imath}} \] Imposing \(\boldsymbol{v}_{A\in 2}=\boldsymbol{v}_{A\in 1}\) leads to two equations solving for \((\omega_1, \omega_2)\) \[ \begin{cases} l\omega_2 \cos\theta= - 2r \omega_1 \\ -v_B - l\omega_2 \sin\theta= 0 \end{cases} \Longrightarrow \boxed{\omega_{2} = - \frac{v_B}{l\sin\theta}, \quad \omega_1 = \frac{v_B}{2r} \cot\theta} \]
2. We cannot take the derivative of \(\omega_{1/0}= \frac{v_B}{2r}\cot\theta\) to find \(\alpha_{1/0}\). We also cannot impose \(\boldsymbol{a}_{I\in 1}= {\bf 0}\). Instead we can use the fact that the path of center \(C\) is actually known. We start with \(\boldsymbol{a}_A\) using \(\boldsymbol{a}_B ={\bf 0}\): \[ \boldsymbol{a}_{A\in 2}= \boldsymbol{a}_{B\in 2} + \alpha_2 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{BA} - \omega_2^2 \boldsymbol{r}_{BA} = \alpha_2 \boldsymbol{\hat{k}}\times l(-\cos\theta\boldsymbol{\hat{\jmath}}-\sin\theta\boldsymbol{\hat{\imath}}) -l \omega_2^2 (-\cos\theta\boldsymbol{\hat{\jmath}}-\sin\theta\boldsymbol{\hat{\imath}}) \] We then find \(\boldsymbol{a}_{C\in 1}\): \[ \boldsymbol{a}_{C\in 1} = \boldsymbol{a}_{A\in 1} + \alpha_1 \boldsymbol{\hat{k}}\times (-r \boldsymbol{\hat{\jmath}}) - \omega_1^2 (-r\boldsymbol{\hat{\jmath}}) = l\alpha_2 (\cos\theta\boldsymbol{\hat{\imath}}-\sin\theta\boldsymbol{\hat{\jmath}}) -l \omega_2^2 (-\cos\theta\boldsymbol{\hat{\jmath}}-\sin\theta\boldsymbol{\hat{\imath}}) + r\alpha_1 \boldsymbol{\hat{\imath}}+ r \omega_1^2 \boldsymbol{\hat{\jmath}} \] Now we need to use the fact that \(C\) moves along a circular path of center \(O\) and radius \(R-r\): we use the \(NT\) components of \(\boldsymbol{a}_C\) (with the unit vectors \(\boldsymbol{\hat{e}}_T = -\boldsymbol{\hat{\imath}}\) and \(\boldsymbol{\hat{e}}_N = \boldsymbol{\hat{\jmath}}\) at the time of the configuration): \[ \boldsymbol{a}_C = - \dot{v}_C \boldsymbol{\hat{\imath}}+ \frac{v_C^2}{R-r} \boldsymbol{\hat{\jmath}} \] We also need the speed \(v_C\): we know that \(\boldsymbol{v}_C = \omega_1 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{IC} = -r \omega_1 \boldsymbol{\hat{\imath}}\). This gives \(v_C = r\omega_1\) (always valid) and \(\dot{v}_C = r \alpha_1\). Now using these two expressions of acceleration \(\boldsymbol{a}_C\) we obtain 2 equations solving for \((\alpha_1, \alpha_2)\): \[ \begin{cases} -r \alpha_1 = l\alpha_2 \cos\theta+l \omega_2^2\sin\theta+ r\alpha_1 \\ \frac{r^2}{R-r} \omega_1^2 = -l\alpha_2 \sin\theta+ l \omega_2^2\cos\theta+ r \omega_1^2 \end{cases} \] We now can solve for \(\alpha_1\): \[ \boxed{ -2r \alpha_1 \sin\theta= l \omega_2^2 + (r - \frac{r^2}{R-r}) \omega_1^2 \cos\theta} \]
3. We first identify IC \(I_1\) of body 1 as contact point \(I\): this is a consequence of the no-slip condition of body 1 with fixed support 0.
To build the location of \(I_2\), the IC of body 2, we use two velocity vectors of points of body 2: we have \(\boldsymbol{v}_B = \boldsymbol{v}_{B\in 3}\) (given) and \(\boldsymbol{v}_A = \boldsymbol{v}_{A\in 1}\) (we use the IC \(I_1\) to find the direction of \(\boldsymbol{v}_A\)). Then we build the two normal lines: \(I_2\) is at the intersections of these 2 lines. See the Figure below.
The final step is to show the direction of \(\boldsymbol{v}_A\), \(\boldsymbol{v}_C\), \(\boldsymbol{\omega}_{1}\) and \(\boldsymbol{\omega}_{2}\) in a manner compatible with the given direction of \(\boldsymbol{v}_B\): this gives \(\boldsymbol{\omega}_1\) in the ccw direction and \(\boldsymbol{\omega}_2\) in the cw direction. We also sketch the velocity triangle between \(\boldsymbol{v}_A\) and \(\boldsymbol{v}_C\) (since both normals at \(A\) and \(C\) coincide).
We now can find angular speed \(\omega_1\) and \(\omega_2\). We start with point \(B\) on body 2: we have \(v_B = |I_2 B|\omega_2\) with $ |I_2 B|= l$. This gives \[ \boxed{\omega_2 = \frac{v_B}{l\sin\theta} } \quad (ccw) \] Next we work with point \(A\): we have \(v_{A\in 2}= |I_2 A| \omega_2 = v_{A\in 1} = |I_1 A|\omega_1\). With \(|I_2 A|= l \cos \theta\) and \(|I_1 A|=2 r\) we find \[ \boxed{\omega_1 = \omega_2 \frac{|I_2 A|}{|I_1 A|} = \frac{v_B}{2r}\cot\theta} \quad (cw) \] These expressions match the results of q.1.
