When the rod is released from the horizontal position, the torque is produced by its weight $mg$ acting at its center of mass, which is at a distance of $l/2$ from the pivot point A. Torque $\tau = \text{Force} \times \text{perpendicular distance} = mg \times \frac{l}{2}$ We know that $\tau = I\alpha$, where $I$ is the moment of inertia about point A and $\alpha$ is the angular acceleration. Given $I = \frac{ml^2}{3}$ Equating the two expressions for torque: $mg \frac{l}{2} = \left(\frac{ml^2}{3}\right) \alpha$ $\alpha = \frac{mg \frac{l}{2}}{\frac{ml^2}{3}} = \frac{3g}{2l}$