Q64 / 75 · JEE Main 2026 (24 January Shift 2) · Physics
A thin uniform rod $(X)$ of mass $M$ and length $L$ is pivoted at a height $\left(\frac{L}{3}\right)$ as shown in the figure. The rod is allowed to fall from a vertical position and lie horizontally on the table. The angular velocity of this rod when it hits the table top, is $\_\_\_\_$. ($\mathrm{g}=$ gravitational acceleration)
- A$\frac{3}{\sqrt{2}} \sqrt{\frac{g}{L}}$
- B$\frac{1}{\sqrt{2}} \sqrt{\frac{g}{L}}$
- C$\sqrt{\frac{3}{2} \frac{g}{L}}$
- Correct answerD$\sqrt{\frac{3 g}{L}}$
Solution
The rod is pivoted at a distance $\frac{L}{3}$ from one end. The centre of mass is at a distance $\frac{L}{6}$ from the pivot.
Using energy conservation:
$mg\frac{\ell}{6} = \frac{1}{2}I\omega^2$
Moment of inertia about the pivot (using parallel axis theorem):
$I = \frac{m\ell^2}{12} + m\left(\frac{\ell}{6}\right)^2 = \frac{m\ell^2}{12} + \frac{m\ell^2}{36} = \frac{m\ell^2}{9}$
$mg\frac{\ell}{6} = \frac{1}{2} \cdot \frac{m\ell^2}{9} \cdot \omega^2$
$\omega^2 = \frac{9g}{6 \cdot \frac{\ell}{2}} = \frac{3g}{\ell}$
$\omega = \sqrt{\frac{3g}{\ell}}$
Using energy conservation:
$mg\frac{\ell}{6} = \frac{1}{2}I\omega^2$
Moment of inertia about the pivot (using parallel axis theorem):
$I = \frac{m\ell^2}{12} + m\left(\frac{\ell}{6}\right)^2 = \frac{m\ell^2}{12} + \frac{m\ell^2}{36} = \frac{m\ell^2}{9}$
$mg\frac{\ell}{6} = \frac{1}{2} \cdot \frac{m\ell^2}{9} \cdot \omega^2$
$\omega^2 = \frac{9g}{6 \cdot \frac{\ell}{2}} = \frac{3g}{\ell}$
$\omega = \sqrt{\frac{3g}{\ell}}$
Hence, the correct answer is option: 'D'.