Question

A uniform rod of length $l$ and mass $m$ is free to rotate in a vertical plane about $A$. The rod initially in horizontal position is released. The initial angular acceleration of the rod is (Moment of inertia of rod about $A$ is $\frac{m{l}^2}{3}$)

• A. $\frac{3g}{2l}$
• B. $\frac{2l}{3g}$
• C. $\frac{3g}{2l^2}$
• D. $mg\frac{l}{2}$

Answer 1

The correct option is A $\frac{3g}{2l}$

Torque about A is $\tau =mg\left(\frac{l}{2}\right)$
We know that,
$\tau =I\alpha =mg\left(\frac{l}{2}\right)$
$\frac{m{l}^2}{3}\alpha =\frac{mgl}{2}$
$\alpha =\frac{3g}{2l}$