Question

A uniform rod $A B$ of length $\ell$ 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 $\frac{m \ell^{2}}{3}$, the initial angular acceleration of the rod will be:-

• A. $mg \frac{\ell}{2}$
• B. $\frac{3}{2} g \ell$
• C. $\frac{3 g}{2 \ell}$
• D. $\frac{2 g}{3 \ell}$

Answer 1

Answer: (C)

Given that moment of inertia about $A$ is
$I = ml ^{3} / 3$
Now, torque about $A$ is given as
$\tau= F \times r $
$\tau= mg \frac{1}{2}$
$ I \alpha= mg \frac{1}{2} $
$\frac{ ml ^{2}}{3} \alpha= mg \frac{1}{2} $
$\alpha=\frac{3 g }{2 l }$