A uniform rod of length  I is free to rotate in a vertical plane about a fixed horizontal axis through B.
The rod begins rotating from rest from its unstable equilibrium position.

When it has turned through an angle $\theta$, its angular velocity $\omega$ is given by ,

When the rod rotates through an angle $\theta$, the centre of gravity falls through a distance h.

$\mathrm{In} \Delta BCG,\cos \theta =\frac{(l/2)-h}{l/2}\text{ or }h=\frac{l}{2}(1-\cos \theta )$

Decrease in potential energy =  $mg\frac{l}{2}(1-\cos \theta )$      …(i)

The decrease in potential energy is equal to the kinetic energy of rotation.

$\therefore (\mathrm{KE}{)}_{\text{rotational }}=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2$          ….(ii)     $\left(\because I=\frac{m{l}^2}{3}\right)$

From Eqs. (i) and (ii), we get

$\frac{1}{2}\left(\frac{m{l}^2}{3}\right){\omega }^2=mg\frac{l}{2}(1-\cos \theta )\Rightarrow \omega =\sqrt{\frac{6g}{l}}\sin \frac{\theta }{2}$