Question

A uniform rod of mass $m$ and length $l$ can rotate in vertical plane about a smooth horizontal axis hinged at point $H$.
(i) Find angular acceleration $\alpha$ of the rod just after it is released from initial horizontal position from rest?
(ii) Calculate the acceleration (tangential and radial) of point $A$ at this moment.

Answer 1

$\left(i\right)\tau =I\alpha$

So,

$\frac{mgl}{2}=\frac{1}{3}m{l}^2\left(\alpha \right)\Rightarrow \alpha =\frac{3g}{2l}$

$\left(ii\right)$ Tangential acceleration of poit $A=g$

But as angular velocity of point is zero so, angular acceleration will be zero.