Question

A uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A$. The rod is released from rest in the horizontal position. Given that the moment of inertia of the rod about $A$ is $\frac{ml^2}{3}$ the initial angular acceleration of the rod will be

• A. $\frac{2g}{3l}$
• B. $\frac{mgl}{2}$
• C. $\frac{3gl}{2}$
• D. $\frac{3g}{2l}$

Answer 1

Answer: (D)

Torque about $A$
$\tau=mg \frac{l}{2}$
also, $\tau=I \alpha$
Angular acceleration $\alpha=\frac{\tau}{I}=\frac{mgl /2}{ml^{2}/ 3}=\frac{3g}{2l}$
${\text{(Given}} \, I=\frac{ml^{2}}{3})$