Energy conservation, Kinetic Energy,  Work and Power of Rotational Motion Questions in English

(B) By the law of conservation of energy, the potential energy lost by the rod is equal to the rotational kinetic energy gained by it.
Initially, the center of mass of the rod is at a height $h = l/2$ from the ground.
Potential Energy $(PE)$ lost = $mgh = mg(l/2)$.
Rotational Kinetic Energy $(KE)$ gained = $(1/2)I\omega^2$, where $I$ is the moment of inertia of the rod about the hinge $A$.
The moment of inertia of a rod about one end is $I = ml^2/3$.
Equating $PE$ and $KE$: $mgl/2 = (1/2)(ml^2/3)\omega^2$.
$mgl/2 = (ml^2/6)\omega^2$.
$\omega^2 = (mgl/2) \times (6/ml^2) = 3g/l$.
Therefore, $\omega = \sqrt{3g/l}$.