AIEEE 2009 Solved Paper

The moment of inertia of the rod about $O$ is $\frac{1}{2} m \ell^2$. The maximum angular speed of the rod is when the rod is instantaneously vertical. The energy of the rod in this conditions is $\frac{1}{2} I \omega^2$ where $I$ is the moment of inertia of the rod about $O$. when the rod is in its extreme portion, its angular velocity is zero momentarily. In this case, the center of mass is raised through $h$, so the increase in potential energy is $m g h$. This is equal to kinetic energy $\frac{1}{2} I \omega^2$. $\therefore m g h = \frac{1}{2} I \omega^2 = \frac{1}{2} \left( \frac{1}{3} m l^2 \right) \omega^2$$\Rightarrow h = \frac{\ell^2 \omega^2}{6 g}$