Q117 / 150 · MHT CET 2024 (09 May Shift 1) · Physics
A thin uniform rod of length ' $L$ ' and mass ' $M$ ', is swinging freely along a horizontal axis passing through its centre. Its maximum angular speed is ' $\omega$ '. Its centre of mass rises to a maximum height of [g= gravitational acceleration]
- A$\frac{\omega^2 \mathrm{~L}^2}{12 \mathrm{~g}^2}$
- B$\frac{\omega^2 L^2 g}{6}$
- C$\frac{\omega^2 g}{12 \mathrm{~L}^2}$
- Correct answerD$\frac{\omega^2 \mathrm{~L}^2}{24 \mathrm{~g}}$
Solution
The moment of inertia of rod about its centre is $\frac{\mathrm{ML}^2}{12}$.
Its kinetic energy K.E. $=\frac{1}{2} \mathrm{I} \omega^2=\frac{1}{2} \times \frac{\mathrm{ML}^2}{12} \times \omega^2$
$=\frac{M L^2 \omega^2}{24}$
If its centre of mass rises by h, then it will gain potential energy. $\therefore \quad$ Potential energy, $\mathrm{Mgh}=\frac{\mathrm{ML}^2 \omega^2}{24}$ $\therefore \quad \mathrm{h}=\frac{\mathrm{L}^2 \omega^2}{24 \mathrm{~g}}$.
If its centre of mass rises by h, then it will gain potential energy. $\therefore \quad$ Potential energy, $\mathrm{Mgh}=\frac{\mathrm{ML}^2 \omega^2}{24}$ $\therefore \quad \mathrm{h}=\frac{\mathrm{L}^2 \omega^2}{24 \mathrm{~g}}$.
Hence, the correct answer is option: 'D'.