A fly wheel of moment of inertia I is rotating at n revolutions per sec. The work needed to double the frequency would be -

Options

A. `2 pi^2 In^2`
B. `4pi^2 In^2`
C. `6pi^2 In^2`
D. `8 pi^2 In^2`

Answer

To solve the problem of finding the work needed to double the frequency of a flywheel with a moment of inertia \( I \) rotating at \( n \) revolutions per second, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- The flywheel has a moment of inertia \( I \).
- It is initially rotating at \( n \) revolutions per second.

2. **Convert Revolutions per Second to Radians per Second**:
- The angular velocity \( \omega \) in radians per second can be calculated using the formula:
\[ \omega = 2\pi n \]

3. **Calculate Initial Kinetic Energy**:
- The rotational kinetic energy \( KE \) of the flywheel is given by the formula:
\[ KE = \frac{1}{2} I \omega^2 \]
- Substituting \( \omega = 2\pi n \):
\[ KE_{\text{initial}} = \frac{1}{2} I (2\pi n)^2 = \frac{1}{2} I (4\pi^2 n^2) = 2 I \pi^2 n^2 \]

4. **Determine the New Frequency**:
- If the frequency is doubled, the new frequency becomes \( 2n \).

5. **Calculate New Angular Velocity**:
- The new angular velocity \( \omega' \) is:
\[ \omega' = 2\pi (2n) = 4\pi n \]

6. **Calculate New Kinetic Energy**:
- The new kinetic energy \( KE' \) is:
\[ KE_{\text{new}} = \frac{1}{2} I (4\pi n)^2 = \frac{1}{2} I (16\pi^2 n^2) = 8 I \pi^2 n^2 \]

7. **Calculate the Work Done**:
- The work done \( W \) to change the kinetic energy is the difference between the new and initial kinetic energy:
\[ W = KE_{\text{new}} - KE_{\text{initial}} = (8 I \pi^2 n^2) - (2 I \pi^2 n^2) = 6 I \pi^2 n^2 \]

### Final Answer:
The work needed to double the frequency of the flywheel is:
\[ W = 6 I \pi^2 n^2 \]