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. $\frac{l}{2}$

Answer 1

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

Here $\tau$ = $I\alpha$
$\Rightarrow \left(mg\right)\left(\frac{l}{2}\right)$ = $\left(\frac{m{l}^2}{3}\right)\left(\alpha \right)\Rightarrow \alpha$ = $\frac{3g}{2l}$