A uniform rod AB of length l, 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 \(\frac{ml^2}{3}\), the initial angular acceleration of the rod will be : 

\(\frac{3g}{2l}\)

Weight of the rod will produce torque,

\(\tau = mg × \frac{l}{2}\)

Also, \(\tau = I\alpha\)

and I = \(\frac{ml^2}{3}\)

\(\frac{ml^2}{3} \alpha = mg × \frac{l}{2}\)

\(\alpha = \frac{3g}{2l}\)