JEE Main · 2026 · 24 Jan 2026 · Shift 2 · Rotational Mechanics

The pivot lies $L/3$ above the lower end, so its distance from the centre of mass (at $L/2$) is $d=\frac{L}{2}-\frac{L}{3}=\frac{L}{6}$. The moment of inertia about the pivot is $I=\frac{M{L}^2}{12}+M{\left(\frac{L}{6}\right)}^2=\frac{M{L}^2}{12}+\frac{M{L}^2}{36}=\frac{M{L}^2}{9}$. Standing vertically the centre of mass is $L/6$ above the pivot, and when the rod turns to horizontal the centre of mass drops to pivot level, a fall of $L/6$. Equating the loss in potential energy to the rotational kinetic energy, $Mg\frac{L}{6}=\frac{1}{2}\cdot \frac{M{L}^2}{9}{\omega }^2$, which gives ${\omega }^2=\frac{3g}{L}$. The angular velocity as the rod strikes the table is $\sqrt{\frac{3g}{L}}$.