Question

A uniform rod of length $L$ and mass $M$ is pivoted freely at one end and placed in vertical position
(a) What is the angular acceleration of the rod when it is at an angle $\theta$ with the vertical?
(b) What is the tangential linear acceleration of the free end when the rod is horizontal?

Answer 1

Figure shows the rod at an angle $\theta$ to the vertical. If we take torque about the pivot we need not be concerned with the force due to the pivot
The torque due to the weight about $O$
${\tau }_{mg}=\frac{mgL}{2}\sin \theta$
Using torque equation about $O$. Here we apply torque equation about fixed axis passing through $O$. We can apply torque equation about fixed axis or about centre of mass only
(a) $\tau =I\alpha \Rightarrow \frac{mgL}{2}\sin \theta =\frac{M{L}^2}{3}\alpha$
Thus, $\alpha =\frac{3g\sin \theta }{2L}$
Wnen the rod is horizontal $\theta =\frac{\pi }{2};\alpha =\frac{3g}{2L}$
(b) The tangential linear acceleration of the free end is $a_t=\alpha L=\frac{3g}{2}$