A uniform rod of length l and mass m is free to rotate in a...

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

$= \dfrac{ml^{2}}{3}$

where, $m is the mass and$l$its length.</p><p>Torque \left(τ = I\alpha\right)$~acting on centre of gravity of rod is given by&

$τ = mg\dfrac{l}{2}$

also,~$\dfrac{ml^{2}}{3} \cdot \alpha = mg\dfrac{l}{2}$

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