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 horizontal position. Given that the moment of inertia of the rod about $A$ is $\frac{m{l}^2}{3}$. 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

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

Let the initial angular acceleration of the rod be $\alpha$

Torque $\tau =F\times r$
where, $F=mg$ and $r=l/2$
$\tau =l/2\times mg\tau =\frac{mgl}{2}.....\left(1\right)$

$\tau =I\times \alpha .....\left(2\right)$
Equating (1) and (2),
$\frac{mgl}{2}=I\times \alpha \frac{mgl}{2}=\frac{m{l}^2}{3}\times \alpha \alpha =\frac{3g}{2l}$