Two wheels A and B are mounted on the same axle. Moment of inertia of A is 6 kg `m^(2)` and is rotated at 600 rpm, when B is at rest. What will be moment of inertia of B, if their combined speed is 400 rpm?

Options

A. 8 kg `m^2`
B. 4 kg `m^2`
C. 3 kg `m^2`
D. 5 kg `m^2`

Answer

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum before and after the interaction must be equal.

### Step-by-Step Solution:

1. **Identify the given values:**
- Moment of inertia of wheel A, \( I_A = 6 \, \text{kg m}^2 \)
- Initial angular velocity of wheel A, \( \omega_A = 600 \, \text{rpm} \)
- Final combined angular velocity, \( \omega_f = 400 \, \text{rpm} \)
- Moment of inertia of wheel B, \( I_B \) (unknown)

2. **Convert angular velocities from rpm to rad/s:**
- To convert from rpm to rad/s, use the conversion factor \( \frac{2\pi \, \text{rad}}{1 \, \text{rev}} \) and \( \frac{1 \, \text{min}}{60 \, \text{s}} \).
- \( \omega_A = 600 \, \text{rpm} = 600 \times \frac{2\pi}{60} = 62.83 \, \text{rad/s} \)
- \( \omega_f = 400 \, \text{rpm} = 400 \times \frac{2\pi}{60} = 41.89 \, \text{rad/s} \)

3. **Apply the conservation of angular momentum:**
- Before interaction: \( L_{initial} = I_A \cdot \omega_A + I_B \cdot 0 \)
- After interaction: \( L_{final} = (I_A + I_B) \cdot \omega_f \)
- Setting the two equal gives us:
\[ I_A \cdot \omega_A = (I_A + I_B) \cdot \omega_f \]

4. **Substituting the known values:**
\[ 6 \cdot 62.83 = (6 + I_B) \cdot 41.89 \]

5. **Calculate the left side:**
\[ 6 \cdot 62.83 = 376.98 \]

6. **Set up the equation:**
\[ 376.98 = (6 + I_B) \cdot 41.89 \]

7. **Divide both sides by \( 41.89 \):**
\[ \frac{376.98}{41.89} = 6 + I_B \]
\[ 9 = 6 + I_B \]

8. **Solve for \( I_B \):**
\[ I_B = 9 - 6 = 3 \, \text{kg m}^2 \]

### Final Answer:
The moment of inertia of wheel B is \( I_B = 3 \, \text{kg m}^2 \).

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