A uniform rod of length L and mass M is pivoted freely at one end and hangs vertically. It is displaced slightly and released. The angular velocity o…

By conservation of mechanical energy, loss in potential energy = gain in rotational kinetic energy. The center of mass falls by a height $h = L/2$. Thus, $Mg(L/2) = \frac{1}{2} I \omega^2$. For a rod pivoted at one end, $I = \frac{1}{3}ML^2$. Substituting this: $Mg \frac{L}{2} = \frac{1}{2} \left(\frac{1}{3}ML^2\right) \omega^2 \implies gL = \frac{1}{3}L^2 \omega^2 \implies \omega = \sqrt{\frac{3g}{L}}$.