Question 1. To connect the motion of crank-arm \(1\) to the motion of wheel \(3\), we proceed as follows:
Step 1: we find velocity of \(B\) as a point of body \(1\): \[ \boldsymbol{v}_B = \boldsymbol{v}_{B\in 1} = \omega_1 \boldsymbol{\hat{k}}\times \boldsymbol{r}_{AB} = 2r \omega_1 \boldsymbol{\hat{k}}\times (\cos\theta\boldsymbol{\hat{\imath}}+\sin\theta\boldsymbol{\hat{\jmath}}) = 2 r \omega_1 (\cos\theta\boldsymbol{\hat{\jmath}}- \sin\theta\boldsymbol{\hat{\imath}}) \] Step 2: we find velocity of \(C\) as a point of body \(2\): \[\begin{align*} \boldsymbol{v}_C & = \boldsymbol{v}_{C\in 2} =\boldsymbol{v}_{B\in 2} + \boldsymbol{\omega}_2 \times \boldsymbol{r}_{BC}= 2 r \omega_1 (\cos\theta\boldsymbol{\hat{\jmath}}- \sin\theta\boldsymbol{\hat{\imath}}) + 2r \omega_2 \boldsymbol{\hat{k}}\times (\cos\theta\boldsymbol{\hat{\imath}}- \sin\theta\boldsymbol{\hat{\jmath}}) \\ & = 2 r \omega_1 (\cos\theta\boldsymbol{\hat{\jmath}}- \sin\theta\boldsymbol{\hat{\imath}})+2 r \omega_2 (\cos\theta\boldsymbol{\hat{\jmath}}+ \sin\theta\boldsymbol{\hat{\imath}}) \end{align*}\] Step 3: we impose the no-slip condition at \(I\) of wheel \(3\): \[ \boldsymbol{v}_{C\in 3} = \boldsymbol{\omega}_3 \times\boldsymbol{r}_{IC} = - r\omega_3 \boldsymbol{\hat{\imath}} \] Step 4: we impose \(\boldsymbol{v}_{C\in 3}= \boldsymbol{v}_{C\in 2}\) to obtain 2 equations: \[ -2r\omega_1\sin\theta+2r\omega_2\sin\theta= -r\omega_3 \] \[ 2r\omega_1\cos\theta+ 2r\omega_2\cos\theta=0 \] Step 4: we solve for angular velocities \((\boldsymbol{\omega}_2 = \omega_2\boldsymbol{\hat{k}}, \boldsymbol{\omega}_3 =\omega_3\boldsymbol{\hat{k}})\) in terms of \(\omega_1\): \[ \boxed{ \omega_1 = -\omega_2 } \qquad \boxed{ \omega_3 = 4\omega_1 \sin\theta } \qquad\qquad{[1-2]} \] We can then find \(\boldsymbol{v}_B\) and \(\boldsymbol{v}_C\) in terms of \(\omega_1\): \[ \boxed{ \boldsymbol{v}_B = 2 r \omega_1 (\cos\theta\boldsymbol{\hat{\jmath}}- \sin\theta\boldsymbol{\hat{\imath}}) }, \qquad \boxed{ \boldsymbol{v}_C = - 4r\omega_1\sin\theta\,\boldsymbol{\hat{\imath}} } \]
Question 2. We cannot take the time-derivative of (1-2) to obtain angular accelerations \(\boldsymbol{\alpha}_2\) and \(\boldsymbol{\alpha}_{3}\) since these expressions are only valid at the time of the configuration. To proceed we work as in question 1. First, we find \(\boldsymbol{a}_B = \boldsymbol{a}_{B\in 1}\): \[ \boldsymbol{a}_B= -\omega_1^2 \boldsymbol{r}_{AB} = -2r \omega_1^2 (\cos\theta\boldsymbol{\hat{\imath}}+\sin\theta\boldsymbol{\hat{\jmath}}) \] Then we find \(\boldsymbol{a}_C= \boldsymbol{a}_{C\in 2}\): \[\begin{align*} \boldsymbol{a}_C & = \boldsymbol{a}_B + \boldsymbol{\alpha}_2 \times \boldsymbol{r}_{BC} - \omega_2^2 \boldsymbol{r}_{BC} \\ & = -2r \omega_1^2 (\cos\theta\boldsymbol{\hat{\imath}}+\sin\theta\boldsymbol{\hat{\jmath}}) + 2r \alpha_2 \boldsymbol{\hat{k}}\times (\cos\theta\boldsymbol{\hat{\imath}}- \sin\theta\boldsymbol{\hat{\jmath}}) -2r\omega_2^2 (\cos\theta\boldsymbol{\hat{\imath}}- \sin\theta\boldsymbol{\hat{\jmath}}) \\ & =-4r\omega_1^2\cos\theta\boldsymbol{\hat{\imath}}+2r\alpha_2 (\cos\theta\boldsymbol{\hat{\jmath}}+ \sin\theta\boldsymbol{\hat{\imath}}) \end{align*}\] Next we use the fact that \(C\) has a circular trajectory of center \(O\) and radius \((R-r)\): \[ \boldsymbol{a}_C = \dot{v}_C \boldsymbol{\hat{e}}_T + \frac{v_C^2}{R-r} \boldsymbol{\hat{e}}_N \] From question 1, we know that \(v_C= r\omega_3\), \(\dot{v}_C= r\alpha_3\), \(\boldsymbol{\hat{e}}_T=-\boldsymbol{\hat{\imath}}\) and \(\boldsymbol{\hat{e}}_N= \boldsymbol{\hat{\jmath}}\): \[ \boldsymbol{a}_C = -r\alpha_3 \boldsymbol{\hat{\imath}}+ \frac{r^2\omega_3^2}{R-r} \boldsymbol{\hat{\jmath}} \] We have obtained 2 expressions for \(\boldsymbol{a}_C\): this gives us 2 equations to solve for \(\alpha_2\) and \(\alpha_3\): \[ -4r\omega_1^2\cos\theta+2r\alpha_2 \sin\theta= -r \alpha_3 \] \[ 2r\alpha_2 \cos\theta= \frac{r^2\omega_3^2}{R-r} \] The solution gives \[ \boxed{ \alpha_2 = \frac{8 r\omega_1^2 \sin^2\theta}{(R-r)\cos\theta} } \qquad \boxed{ \alpha_3 = 4\omega_1^2 \cos\theta-2 \alpha_2 \sin\theta } \]
Question 3. To construct the position of \(I_2\), we need the velocities of two points of body \(2\):
velocity \(\boldsymbol{v}_{B\in 2}=\boldsymbol{v}_{B\in 1}\) is normal to line \(AB\),
velocity of \(\boldsymbol{v}_{C\in 2}=\boldsymbol{v}_{C\in 3}\) is normal to line \(IC\).
We conclude that \(I_2\) is at the intersection of lines \(AB\) and \(IC\).
Point \(I_3\) is contact point \(I\) due to the no-slip condition at \(I\).
On the sketch displayed below, we indicate the directions of \(\boldsymbol{v}_B\), \(\boldsymbol{v}_C\), \(\boldsymbol{\omega}_2\), and \(\boldsymbol{\omega}_3\) compatible to the given direction of \(\boldsymbol{\omega}_1\):
Then we use the knowledge of \(I_2\) and \(I_3\) to write \[ v_B = 2r \omega_1 = |I_2 B| \omega_2, \qquad v_C = r \omega_3 = |I_2 C| \omega_2 \] In triangle \(I_2 BC\) we find \[ \frac{|I_2 B|}{\sin(\frac{\pi}{2}-\theta) } =\frac{|I_2 C|}{\sin(2\theta) } = \frac{2r}{\sin(\frac{\pi}{2}-\theta) } \] giving \[ |I_2 B|= 2r, \qquad |I_2 C|= 4r \sin\theta \] Finally we find the angular speeds \(\omega_2\) and \(\omega_3\): \[ \boxed{\omega_2 = \omega_1} , \qquad \boxed{\omega_3 =4 \sin\theta\omega_1} \] These expressions are identical to those obtained in question 1.