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 problem.
We are given that the moment of inertia of a ring about an axis passing through its center and perpendicular to its plane is \( I \). The ring is rotating with angular velocity \( \omega \). Another identical ring is placed on the first one so that their centers coincide. Both rings are rotating about the same axis, and we are asked to find the loss in kinetic energy.

Step 2: Finding the total kinetic energy of the system.
The kinetic energy \( K \) of a rotating object is given by:
\[ K = \frac{1}{2} I \omega^2, \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
For the first ring, the kinetic energy is:
\[ K_1 = \frac{1}{2} I \omega^2. \]
For the second identical ring, the kinetic energy is the same:
\[ K_2 = \frac{1}{2} I \omega^2. \]
The total kinetic energy of the system is:
\[ K_{\text{total}} = K_1 + K_2 = \frac{1}{2} I \omega^2 + \frac{1}{2} I \omega^2 = I \omega^2. \]

Step 3: Initial kinetic energy of the system.
Initially, before the second ring is placed, the total kinetic energy is just the kinetic energy of the first ring:
\[ K_{\text{initial}} = \frac{1}{2} I \omega^2. \]

Step 4: Loss in kinetic energy.
The loss in kinetic energy is the difference between the total kinetic energy and the initial kinetic energy:
\[ \Delta K = K_{\text{total}} - K_{\text{initial}} = I \omega^2 - \frac{1}{2} I \omega^2 = \frac{1}{2} I \omega^2. \]
Final Answer:
Thus, the loss in kinetic energy is:
\[ \boxed{\frac{I \omega^2}{2}}. \]