Solution:
- 1. We assume angles \(\theta\) and \(\phi\) arbitrary and use a vectorial method: we define basis \((\boldsymbol{\hat{\imath}},\boldsymbol{\hat{\jmath}},\boldsymbol{\hat{k}})\) as shown in the sketch of Figure 1 (left).
Step 1: We first express the no-slip velocity at \(I\) to find the angular velocity of the wheel 1: \(\boldsymbol{\omega}_1 = \omega_1 \boldsymbol{\hat{k}}\) \[ \boldsymbol{v}_C = v_C \boldsymbol{\hat{\imath}}= \omega_1 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{IC} = - r \omega_1 \boldsymbol{\hat{\imath}} \] This gives \(v_C = -r \omega_1\).
Step 2: Next we find \(\boldsymbol{v}_{B\in 1}\) from \(\boldsymbol{v}_{C\in 1}\): \[ \boldsymbol{v}_{B\in 1} = \boldsymbol{v}_{C\in 1} + \omega_1 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{CB} = v_C \boldsymbol{\hat{\imath}}- v_C \boldsymbol{\hat{k}}\times (\cos\theta\boldsymbol{\hat{\jmath}}- \sin\theta\boldsymbol{\hat{\imath}}) = v_C (1+\cos\theta)\boldsymbol{\hat{\imath}}+ v_C \sin\theta\boldsymbol{\hat{\jmath}} \]
Step 3: Next we find \(\boldsymbol{v}_{A\in 2}\) from \(\boldsymbol{v}_{B\in 2}=\boldsymbol{v}_{B\in 1}\):
\[ \boldsymbol{v}_{A\in 2} = v_A \boldsymbol{\hat{\imath}}= \boldsymbol{v}_{B\in 2} + \omega_2 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{BA} = \boldsymbol{v}_{B\in 1} - L\omega_2\boldsymbol{\hat{k}}\times (\cos\phi \boldsymbol{\hat{\imath}}+ \sin\phi \boldsymbol{\hat{\jmath}}) \] which gives \[ v_A \boldsymbol{\hat{\imath}}= v_C (1+\cos\theta)\boldsymbol{\hat{\imath}}+ v_C \sin\theta\boldsymbol{\hat{\jmath}}- L\omega_2 (\cos\phi \boldsymbol{\hat{\jmath}}- \sin\phi \boldsymbol{\hat{\imath}}) \] This gives two equations for the unknowns \((v_A, \omega_2)\): \[ v_A = v_C (1+\cos\theta) + L\omega_2\sin\phi , \quad 0= v_C \sin\theta- L\omega_2 \cos\phi \]
Step 4: We solve for angular velocity \(\boldsymbol{\omega}_2 =\omega_2\boldsymbol{\hat{k}}\) \[ \boxed{ \boldsymbol{\omega}_2 = \frac{v_C}{L} \frac{\sin\theta}{\cos\phi} \boldsymbol{\hat{k}} } \qquad{(1)} \] and velocity \(\boldsymbol{v}_A=\boldsymbol{v}_{A\in 2} =v_A \boldsymbol{\hat{\imath}}\) \[ \boxed{ v_A = v_C (1+\cos\theta) + v_C \sin\theta\tan\phi } \qquad{(2)} \] Note that angle \(\phi\) is a function of \(\theta\): \(L \sin\phi = r + r\cos\theta\).
Step 5: If we set \(\theta=0\) in (1-2), we find \[ \boxed{\boldsymbol{\omega}_2 = \boldsymbol{0}, \qquad \boldsymbol{v}_A = 2 \boldsymbol{v}_C} \] We can explain this result: when \(\theta\) reaches the value \(\theta =0\), \(\boldsymbol{v}_B\) becomes parallel to \(\boldsymbol{v}_A\) (both are directed along \(\boldsymbol{\hat{\imath}}\)). Since both \(A\) and \(B\) are 2 points of body 2, this implies that \(\boldsymbol{\omega}_2 = \boldsymbol{0}\) and \(\boldsymbol{v}_A =\boldsymbol{v}_B\). The IC \(I_2\) of body is at infinity. At this instant, the rod is in instantaneous translation. We also know that \(B\) is in instantaneous rotation about contact point \(I\): this imposes \(\boldsymbol{v}_B = 2\boldsymbol{v}_C\). See Figure 1 (right).
2. Since equations (1-2) are true at all time, they can be time-differentiated. We can find angular acceleration \(\boldsymbol{\alpha}_2 = \alpha_2\boldsymbol{\hat{k}}\) by time-differentiating \(\omega_2\boldsymbol{\hat{k}}\): \[ \alpha_2 = \frac{v_C}{L} \frac{d}{dt}\left(\frac{\sin\theta}{\cos\phi}\right) = \frac{v_C}{L} \frac{\dot{\theta}\cos\theta\cos\phi+ \dot{\phi}\sin\theta\sin\phi}{\cos^2\phi} \] with \(\dot{\theta}= \omega_2 = -v_C /r\) and \(\dot{\phi}= \omega_2\): \[ \boxed{ \alpha_2 = -\frac{v_C^2}{rL} \frac{\cos\theta}{\cos\phi} + \frac{v_C^2}{L^2} \frac{\sin^2\theta\sin\phi}{\cos^3\phi} } \qquad{(3)} \]
3. Now we assume that \(\theta = \pi /2\). We construct the instantaneous center \(I_2\) of rod 2 by first sketching velocity \(\boldsymbol{v}_B\) using the IC \(I_1 = I\) of wheel 1: \(\boldsymbol{v}_B\) is normal to line \(IB\). Then we find the IC \(I_2\) of rod 2 at the intersection of line \((A,\boldsymbol{\hat{\jmath}})\) and line \(IB\): see the sketch in Figure 2.
The angular speed \(\omega_{AB}\) and speed \(v_A\) are then found from the relationships \[ v_A = v_{A\in 2} = |I_2 A| \omega_2, \qquad v_B= v_{B\in 2} = |I_2B| \omega_2 = v_{B\in 1} = |I_1B| \omega_1 , \qquad v_C = v_{C\in 1} = |I_1 C| \omega_1 \qquad{(4)} \] We find the distance \(|I_2A|\) and \(|I_2B|\) by applying the law of sines in triangle \(AI_2 B\): \[ \frac{|I_2A|}{\sin(\frac{\pi}{4}+\phi) } = \frac{|I_2B|}{\sin(\frac{\pi}{2}-\phi) } = \frac{L}{\sin\frac{\pi}{4}} \] This gives \[ \frac{|I_2A|}{|I_2 B|}= \frac{1+ \tan\phi}{\sqrt{2}}, \qquad |I_2B|=\sqrt{2} L \cos\phi \] Then using (4) we find \[ \boxed{ \omega_2 = \frac{v_C}{L \cos\phi} } \qquad{(5)} \] and \[ \boxed{ v_A = v_C (1+\tan\phi) } \qquad{(6)} \] with angle \(\phi\) given by \(\sin\phi= r/L\). We find the same results as given by (1-2) after setting \(\theta=\pi/2\).
4. Assume \(\theta = \pi /2\). To find the radius of curvature \(\rho\) of the trajectory described by \(B\) we use the fact that \(\rho = v_B^2 /(\boldsymbol{a}_B \cdot \boldsymbol{\hat{e}}_N)\) where \(\boldsymbol{\hat{e}}_N\) is the unit vector normal to the path. We found \(\boldsymbol{v}_B = v_C (\boldsymbol{\hat{\imath}}+\boldsymbol{\hat{\jmath}})\) which gives \(\boldsymbol{\hat{e}}_T = (\boldsymbol{\hat{\imath}}+\boldsymbol{\hat{\jmath}})/\sqrt{2}\) and \(\boldsymbol{\hat{e}}_N = (\boldsymbol{\hat{\imath}}-\boldsymbol{\hat{\jmath}})/\sqrt{2}\) (\(\boldsymbol{\hat{e}}_N\) points towards the center of curvature).
We still need to find \(\boldsymbol{a}_B = \boldsymbol{a}_C + \boldsymbol{\alpha}_1 \times \boldsymbol{r}_{CB} - \omega^2_1\boldsymbol{r}_{CB} = r \omega_1^2 \boldsymbol{\hat{\imath}}= (v_C^2/r)\boldsymbol{\hat{\imath}}\) (using \(\boldsymbol{a}_C = 0\) and \(\boldsymbol{\alpha}_1 =0\) as a consequence of \(v_C= Cst\)). Now we can find \(\boldsymbol{a}_B \cdot \boldsymbol{\hat{e}}_N\) \[ \boldsymbol{a}_B \cdot \boldsymbol{\hat{e}}_N =\frac{ v_C^2} {\sqrt{2} r} = \frac{v_B^2}{\rho} \] This gives \(\boxed{\rho = 2 \sqrt{2} r}\).