Step 1

We are given a center of the cylinder which is moving to the left with constant velocity ${{\rm{v}}_o}$.

We are asked to determine the angular velocity $\omega $ and angular acceleration $\alpha $ of the bar.

 
Step 2

The kinematic diagram of the system is given below:

Here, $\omega $ represents the angular velocity of the bar, $\alpha $ is the angular acceleration of the bar, and ${\rm{v}}$ is the velocity of the cylinder.

 
Step 3

At the instant, by using the relation for the rotational motion, the distance between the fixed pin of the rod and the center of the cylinder $O$ is given by,

\[\begin{array}{c} \tan \frac{\theta }{2} = \frac{r}{x}\\ x = \frac{r}{{\tan \frac{\theta }{2}}}\\ x = r\cot \frac{\theta }{2} \end{array}\]
 
Step 4

Differentiating the above expression with respect to the time, we have,

\[\begin{array}{c} \frac{d}{{dt}}\left( x \right) = \frac{d}{{dt}}\left( {r\cot \frac{\theta }{2}} \right)\\ \dot x = r\left( { - \cos e{c^2}\frac{\theta }{2}} \right)\left( {\frac{{\dot \theta }}{2}} \right)\\ \dot x = \frac{r}{2}\left( { - \cos e{c^2}\frac{\theta }{2}} \right)\dot \theta \end{array}\]

Here, $\dot x = - {{\rm{v}}_o}$ represents the constant velocity of the cylinder moving to the left and $\dot \theta = \omega $ represents the angular velocity of the bar.

On substituting the values in the above expression, we get,

\[\begin{array}{c} - {{\rm{v}}_o} = \frac{r}{2}\left( { - \cos e{c^2}\frac{\theta }{2}} \right)\omega \\ \omega = \frac{{2{{\rm{v}}_o}}}{{r\left( {\cos e{c^2}\frac{\theta }{2}} \right)}}\\ \omega = \frac{{2{{\rm{v}}_o}}}{r}\left( {{{\sin }^2}\frac{\theta }{2}} \right) \end{array}\]
 
Step 5

Differentiating the velocity expression again with respect to the time, we get,

\[\begin{array}{c} \frac{d}{{dt}}\left( {\dot x} \right) = \frac{d}{{dt}}\left( {\frac{r}{2}\left( { - \cos e{c^2}\frac{\theta }{2}} \right)\dot \theta } \right)\\ \ddot x = - \frac{r}{2}\left[ {2\cos ec\frac{\theta }{2}\left( { - \cos ec\frac{\theta }{2}\cot \frac{\theta }{2}} \right)\left( {\frac{{\dot \theta }}{2}} \right)\dot \theta + \left( {\cos e{c^2}\frac{\theta }{2}} \right)\ddot \theta } \right]\\ \ddot x = \frac{r}{2}\left[ {\left( {\cos e{c^2}\frac{\theta }{2}\cot \frac{\theta }{2}} \right){{\dot \theta }^2} - \left( {\cos e{c^2}\frac{\theta }{2}} \right)\ddot \theta } \right]\\ \ddot x = \frac{{r\cos e{c^2}\frac{\theta }{2}}}{2}\left[ {\left( {\cot \frac{\theta }{2}} \right){{\dot \theta }^2} - \ddot \theta } \right] \end{array}\]

Here, $\ddot x = 0$ represents the linear acceleration of the cylinder moving to the left, its value is zero for constant linear velocity of the cylinder and $\ddot \theta = \alpha $ represents the angular acceleration of the bar.

On substituting the values in the above expression, we get,

\[\begin{array}{c} 0 = \frac{{r\cos e{c^2}\frac{\theta }{2}}}{2}\left[ {\left( {\cot \frac{\theta }{2}} \right){{\left( {\frac{{2{{\bf{v}}_{\bf{o}}}}}{r}{{\sin }^2}\frac{\theta }{2}} \right)}^2} - \alpha } \right]\\ \alpha = \left( {\cot \frac{\theta }{2}} \right){\left( {\frac{{2{{\bf{v}}_{\bf{o}}}}}{r}{{\sin }^2}\frac{\theta }{2}} \right)^2}\\ \alpha = \left( {\frac{{\cos \frac{\theta }{2}}}{{\sin \frac{\theta }{2}}}} \right)\left( {\frac{{4{{\bf{v}}_{\bf{o}}}^2}}{{{r^2}}}} \right)\left( {{{\sin }^4}\frac{\theta }{2}} \right)\\ \alpha = \frac{{4{{\bf{v}}_{\bf{o}}}^2}}{{{r^2}}}\left( {{{\sin }^3}\frac{\theta }{2}} \right)\left( {\cos \frac{\theta }{2}} \right) \end{array}\]

Rearranging the above expression, we get,

\[\begin{array}{l} \alpha = \frac{{2{{\bf{v}}_{\bf{o}}}^2}}{{{r^2}}}\left( {2\sin \frac{\theta }{2}\cos \frac{\theta }{2}} \right)\left( {{{\sin }^2}\frac{\theta }{2}} \right)\\ \alpha = \frac{{2{{\bf{v}}_{\bf{o}}}^2}}{{{r^2}}}\left( {\sin \theta } \right)\left( {{{\sin }^2}\frac{\theta }{2}} \right) \end{array}\]