A uniform rod of length l is free to rotate in a vertical plane...

When the rod rotates through an angle $\theta$, the centre of gravity falls through a distance h. From $\triangle B C G^{\prime}$$\begin{aligned} & \cos \theta=\frac{(l / 2)-h}{\mu_2} \\ & or, h=\frac{1}{2}(1-\cos \theta)\end{aligned}$Decrease in PE $=m g \frac{1}{2}(1-\cos \theta)...(i)$ The decrease in P.E. is equal to the kinetic energy of rotation $\left(\frac{1}{2} I \omega^2\right) (\mathrm{KE})_{\text{rotational }}=\frac{1}{2}\left(\frac{m l^2}{3}\right) \omega^2...(ii)$ From Eqs. (i) and (ii), we get\begin{aligned} \frac{1}{2}\left(\frac{m l^2}{3}\right) \omega^2 & =m g \frac{1}{2}(1-\cos \theta) \\ \omega & =\sqrt{\frac{6 g}{l}} \sin \frac{\theta}{2}\end{aligned}$$