Question

A uniform rod $AB$ 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. $\frac{3g}{2\ell }$
• D. $\frac{2g}{3\ell }$

Answer 1

Answer: (C)

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