Question

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 the rings are rotating about the same axis, then loss in kinetic energy is

• A. \( I\omega^2 \)
• B. \( \frac{I\omega^2}{2} \)
• C. \( \frac{I\omega^2}{4} \)
• D. \( \frac{I\omega^2}{3} \)

Hint:

When two objects with the same moment of inertia rotate about the same axis, the total kinetic energy is the sum of their individual kinetic energies. The loss in kinetic energy can be calculated by subtracting the initial energy from the final energy.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
When a second ring is placed on a rotating ring, the moment of inertia of the system changes, which affects the angular velocity due to conservation of angular momentum. This redistribution of mass results in energy loss.
Step 2: Key Formula or Approach:
1. Conservation of Angular Momentum: \( I_1 \omega_1 = I_2 \omega_2 \).
2. Initial Rotational KE: \( K_i = \frac{1}{2} I \omega^2 \).
3. Final Rotational KE: \( K_f = \frac{1}{2} I_{total} \omega_f^2 \).
Step 3: Detailed Explanation:
Initial moment of inertia \( I_1 = I \), initial angular velocity \( \omega_1 = \omega \).
Final moment of inertia \( I_2 = I + I = 2I \) (since rings are identical).
Applying conservation of angular momentum:
\[ I \omega = (2I) \omega_f \Rightarrow \omega_f = \frac{\omega}{2} \]
Initial Kinetic Energy: \( K_i = \frac{1}{2} I \omega^2 \).
Final Kinetic Energy:
\[ K_f = \frac{1}{2} (2I) \left( \frac{\omega}{2} \right)^2 = \frac{1}{2} \cdot 2I \cdot \frac{\omega^2}{4} = \frac{1}{4} I \omega^2 \]
Loss in Kinetic Energy \( \Delta K \):
\[ \Delta K = K_i - K_f = \frac{1}{2} I \omega^2 - \frac{1}{4} I \omega^2 = \frac{1}{4} I \omega^2 \]
Step 4: Final Answer:
The loss in kinetic energy is \( \frac{I\omega^2}{4} \).