Question

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

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

Answer 1

Answer: (A)

$\frac{3g}{2l}$

Given that moment of inertia about A is

$I={ml}^3/3$

Now, torque about A is given as

$\tau =F\times r$

$\tau =mg\frac{l}{2}$

$I\alpha =mg\frac{l}{2}$

$\frac{{ml}^2}{3}\alpha =mg\frac{l}{2}$

$\alpha =\frac{3g}{2l}$