Q.
A uniform rod $AB$ of length $l$ and mass $m$ is free to rotate about point $A$. The rod is released from rest in 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) $mg\frac{l}{2}$
C) $\frac{3}{2}gl$
D) $\frac{3g}{2l}$
Explanation
(d) Weight of the rod will produce the torque
$\tau = I\alpha \Rightarrow mg \times \frac{l}{2} = \frac{ml^{2}}{3} \times \alpha$
[image:7_image89.png]
Angular acceleration
$\alpha = \frac{3g}{2l}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.