JEE Main 2009 — Rotational Motion Question with Solution

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 $$mgh$$. This is equal to kinetic energy $$\frac{1}{2}I{\omega }^2$$.

$$\therefore$$ $$mgh=\frac{1}{2}I{\omega }^2=\frac{1}{2}\left(\frac{1}{3}m{l}^2\right){\omega }^2$$.

$$\Rightarrow h=\frac{{\ell }^2{\omega }^2}{6g}$$