The speed of rolling of a ring of mass m changes from V to 3V . What is the change in its kinetic energy

Options

A. `3MV^(2)`
B. `4MV^(2)`
C. `6MV^(2)`
D. `8MV^(2)`

Answer

To find the change in kinetic energy of a ring of mass \( m \) when its speed changes from \( V \) to \( 3V \), we need to consider both the translational and rotational kinetic energy of the ring.

### Step-by-step Solution:

1. **Identify the Kinetic Energy Components**:
The total kinetic energy \( K \) of the ring is the sum of its translational kinetic energy \( K_{trans} \) and its rotational kinetic energy \( K_{rot} \):
\[ K = K_{trans} + K_{rot} \]

2. **Calculate the Initial Kinetic Energy**:
- The translational kinetic energy when the speed is \( V \):
\[ K_{trans, initial} = \frac{1}{2} m V^2 \]
- The moment of inertia \( I \) of a ring about its center is \( I = mR^2 \).
- The angular velocity \( \omega \) can be expressed in terms of \( V \):
\[ \omega = \frac{V}{R} \]
- The rotational kinetic energy is given by:
\[ K_{rot, initial} = \frac{1}{2} I \omega^2 = \frac{1}{2} (mR^2) \left(\frac{V}{R}\right)^2 = \frac{1}{2} m V^2 \]
- Therefore, the total initial kinetic energy \( K_{initial} \) is:
\[ K_{initial} = K_{trans, initial} + K_{rot, initial} = \frac{1}{2} m V^2 + \frac{1}{2} m V^2 = m V^2 \]

3. **Calculate the Final Kinetic Energy**:
- The translational kinetic energy when the speed is \( 3V \):
\[ K_{trans, final} = \frac{1}{2} m (3V)^2 = \frac{1}{2} m (9V^2) = \frac{9}{2} m V^2 \]
- The angular velocity at this speed is:
\[ \omega = \frac{3V}{R} \]
- The rotational kinetic energy is:
\[ K_{rot, final} = \frac{1}{2} I \omega^2 = \frac{1}{2} (mR^2) \left(\frac{3V}{R}\right)^2 = \frac{1}{2} (mR^2) \left(\frac{9V^2}{R^2}\right) = \frac{9}{2} m V^2 \]
- Therefore, the total final kinetic energy \( K_{final} \) is:
\[ K_{final} = K_{trans, final} + K_{rot, final} = \frac{9}{2} m V^2 + \frac{9}{2} m V^2 = 9 m V^2 \]

4. **Calculate the Change in Kinetic Energy**:
The change in kinetic energy \( \Delta K \) is given by:
\[ \Delta K = K_{final} - K_{initial} = 9 m V^2 - m V^2 = 8 m V^2 \]

### Final Answer:
The change in kinetic energy is \( \Delta K = 8 m V^2 \).

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