If the angular momentum of a body increases by 50%, its kinetic energy of rotation increases by

Options

A. `100%`
B. `125%`
C. `225%`
D. `25%`

Answer

To solve the problem, we need to determine how much the kinetic energy of rotation increases when the angular momentum of a body increases by 50%.

### Step-by-Step Solution:

1. **Understanding Angular Momentum and Kinetic Energy**:
- The angular momentum \( L \) of a body is given by the formula:
\[ L = I \omega \]
where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
- The kinetic energy \( K \) of rotation is given by:
\[ K = \frac{1}{2} I \omega^2 \]

2. **Assuming Constant Moment of Inertia**:
- For this problem, we will assume that the moment of inertia \( I \) remains constant.

3. **Calculating the Change in Angular Momentum**:
- If the angular momentum increases by 50%, we can express the new angular momentum \( L' \) as:
\[ L' = L + 0.5L = 1.5L \]

4. **Relating Angular Momentum to Angular Velocity**:
- Since \( L = I \omega \), we can express the new angular velocity \( \omega' \) in terms of the new angular momentum:
\[ L' = I \omega' \implies 1.5L = I \omega' \implies \omega' = \frac{1.5L}{I} \]

5. **Finding the New Kinetic Energy**:
- The new kinetic energy \( K' \) can be expressed as:
\[ K' = \frac{1}{2} I (\omega')^2 = \frac{1}{2} I \left(\frac{1.5L}{I}\right)^2 \]
- Simplifying this, we have:
\[ K' = \frac{1}{2} I \cdot \frac{(1.5L)^2}{I^2} = \frac{1}{2} \cdot \frac{2.25L^2}{I} = \frac{2.25}{2} \cdot \frac{L^2}{I} = \frac{1.125 L^2}{I} \]

6. **Calculating the Original Kinetic Energy**:
- The original kinetic energy \( K \) is:
\[ K = \frac{1}{2} I \omega^2 = \frac{1}{2} I \cdot \left(\frac{L}{I}\right)^2 = \frac{L^2}{2I} \]

7. **Finding the Ratio of New Kinetic Energy to Original Kinetic Energy**:
- Now, we can find the ratio of the new kinetic energy to the original kinetic energy:
\[ \frac{K'}{K} = \frac{\frac{1.125 L^2}{I}}{\frac{L^2}{2I}} = \frac{1.125}{\frac{1}{2}} = 2.25 \]

8. **Calculating the Percentage Increase**:
- The percentage increase in kinetic energy is given by:
\[ \frac{K' - K}{K} \times 100 = \left(2.25 - 1\right) \times 100 = 1.25 \times 100 = 125\% \]

### Final Answer:
The kinetic energy of rotation increases by **125%**.