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

• A

  $m g \frac{l}{2}$
• Correct answer
  B

  $\frac{3 g}{2 l}$
• C

  $\frac{2 l}{3 g}$
• D

  $\frac{3 g}{2 l^2}$

Solution

Here torque
$$ \tau=m g\left(\frac{1}{2}\right) $$ M. I. of $\operatorname{rod}$ about $A$ is: $$ I=\frac{m l^2}{3} $$ $\therefore$ Angular acceleration of the rod is $$ \alpha=\frac{\tau}{I} $$ $$ \alpha=\frac{m g\left(\frac{1}{2}\right)}{\frac{m l^2}{3}}=\frac{3 g}{2 l} . $$

Hence, the correct answer is option: 'B'.