The moment of inertia of the uniform rod about an axis through one end and perpendicular to its length is 

$=\frac{m{l}^2}{3}$

where, $m$ is the mass and $l$ its length.

Torque $\left(\tau =I\alpha \right)$ acting on centre of gravity of rod is given by 

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

also, $\frac{m{l}^2}{3}·\alpha =mg\frac{l}{2}$

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