A wheel is rotating with frequency of 500 rpm on a shaft, second identical wheel initially at rest is suddenly coupled on same shaft. The frequency of the resultant combination is

Options

A. 250 rps
B. 500 rps
C. 250 rpm
D. 500rpm

Explanation

To solve the problem, we need to determine the frequency of the resultant combination when a second identical wheel, initially at rest, is coupled to a rotating wheel.

### Step-by-Step Solution:

1. **Identify the initial conditions**:
- The first wheel is rotating at a frequency of \( f_1 = 500 \) rpm.
- The second identical wheel is initially at rest, so its frequency \( f_2 = 0 \) rpm.

2. **Understand the concept of coupling**:
- When two identical wheels are coupled on the same shaft, they will share the same angular velocity after coupling.
- The total angular momentum of the system must be conserved.

3. **Calculate the initial angular momentum**:
- The angular momentum \( L \) of a rotating wheel can be expressed as \( L = I \omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular velocity.
- For the first wheel, the angular velocity \( \omega_1 \) can be calculated from its frequency:
\[ \omega_1 = 2\pi \times \frac{f_1}{60} = 2\pi \times \frac{500}{60} \text{ rad/s} \]

4. **Calculate the moment of inertia**:
- Since both wheels are identical, let the moment of inertia of one wheel be \( I \).
- The total initial angular momentum of the system is:
\[ L_{initial} = I \omega_1 + I \omega_2 = I \omega_1 + 0 = I \omega_1 \]

5. **Determine the final angular momentum**:
- After coupling, both wheels will rotate together at a new frequency \( f_f \). The angular velocity \( \omega_f \) will be:
\[ \omega_f = 2\pi \times \frac{f_f}{60} \]
- The total moment of inertia after coupling is \( 2I \) (since there are two identical wheels).
- The final angular momentum is:
\[ L_{final} = 2I \omega_f \]

6. **Apply conservation of angular momentum**:
- According to the conservation of angular momentum:
\[ L_{initial} = L_{final} \]
- Substituting the expressions we have:
\[ I \omega_1 = 2I \omega_f \]
- We can cancel \( I \) from both sides (assuming \( I \neq 0 \)):
\[ \omega_1 = 2 \omega_f \]

7. **Solve for the final angular velocity**:
- Rearranging gives:
\[ \omega_f = \frac{\omega_1}{2} \]

8. **Convert back to frequency**:
- Since \( \omega_f = 2\pi \times \frac{f_f}{60} \), we can substitute:
\[ \frac{500 \times 2\pi}{60} = 2 \left(2\pi \times \frac{f_f}{60}\right) \]
- Simplifying gives:
\[ f_f = \frac{500}{2} = 250 \text{ rpm} \]

### Final Answer:
The frequency of the resultant combination after coupling the two wheels is **250 rpm**.