Q112 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane is I. It is rotating with angular velocity $\omega$. Another identical ring is gently placed on it so that their centres coincide. If both rings are rotating about the same axis then loss in kinetic energy is
- A$I\omega^2$
- B$\dfrac{I\omega^2}{2}$
- Correct answerC$\dfrac{I\omega^2}{4}$
- D$\dfrac{I\omega^2}{8}$
Solution
Initial angular momentum of the system is $L_i = I\omega$
Initial kinetic energy is $K_i = \dfrac{1}{2}I\omega^2$
When an identical ring is placed on it, the new moment of inertia becomes $I_f = I + I = 2I$
By conservation of angular momentum, $I_f\omega_f = I_i\omega_i$
$2I\omega_f = I\omega \Rightarrow \omega_f = \dfrac{\omega}{2}$
Final kinetic energy is $K_f = \dfrac{1}{2}(2I)\left(\dfrac{\omega}{2}\right)^2 = \dfrac{1}{4}I\omega^2$
Loss in kinetic energy is $\Delta K = K_i - K_f = \dfrac{1}{2}I\omega^2 - \dfrac{1}{4}I\omega^2 = \dfrac{I\omega^2}{4}$
Answer: $\dfrac{I\omega^2}{4}$
Initial kinetic energy is $K_i = \dfrac{1}{2}I\omega^2$
When an identical ring is placed on it, the new moment of inertia becomes $I_f = I + I = 2I$
By conservation of angular momentum, $I_f\omega_f = I_i\omega_i$
$2I\omega_f = I\omega \Rightarrow \omega_f = \dfrac{\omega}{2}$
Final kinetic energy is $K_f = \dfrac{1}{2}(2I)\left(\dfrac{\omega}{2}\right)^2 = \dfrac{1}{4}I\omega^2$
Loss in kinetic energy is $\Delta K = K_i - K_f = \dfrac{1}{2}I\omega^2 - \dfrac{1}{4}I\omega^2 = \dfrac{I\omega^2}{4}$
Answer: $\dfrac{I\omega^2}{4}$
Hence, the correct answer is option: 'C'.