A uniform rod of length l and mass m is free to rotate in a vertical plane about A. The rod initially in horizontal position is released. The initial angular acceleration of the rod is : (Moment of inertia of rod about A is (ml2/3))

3g/2l 2l/3g 3g/2l2 mg(l/2)

3g/2l

\(\tau = I \alpha\) mg x \(\frac{l}{2}\) = \(\frac{ml^2}{3}\)(α) α = \(\frac{3g}{2l}\)