A man stands on a rotating platform with his arms stretched holding a `5 kg` weight in each hand. The angular speed of the platform is `1.2 rev s^-1`. The moment of inertia of the man together with the platform may be taken to be constant and equal to `6 kg m^2`. If the man brings his arms close to his chest with the distance `n` each weight from the axis changing from `100 cm` to`20 cm`. The new angular speed of the platform is.

Options

A. `2 rev s^-1`
B. `3 rev s^-1`
C. `5 rev s^-1`
D. `6 rev s^-1`

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. The total angular momentum before the man brings his arms in must equal the total angular momentum after he brings his arms in, as there is no external torque acting on the system.

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Initial angular speed, \( \omega_1 = 1.2 \, \text{rev/s} \)
- Moment of inertia of the man and platform, \( I_m = 6 \, \text{kg m}^2 \)
- Mass of each weight, \( m = 5 \, \text{kg} \)
- Initial distance of weights from the axis, \( r_1 = 100 \, \text{cm} = 1.0 \, \text{m} \)

2. **Calculate Initial Moment of Inertia:**
- The moment of inertia of the weights when arms are stretched is:
\[ I_1 = I_m + 2 \cdot m \cdot r_1^2 = 6 + 2 \cdot 5 \cdot (1.0)^2 = 6 + 10 = 16 \, \text{kg m}^2 \]

3. **Identify Final Conditions:**
- Final distance of weights from the axis, \( r_2 = 20 \, \text{cm} = 0.2 \, \text{m} \)

4. **Calculate Final Moment of Inertia:**
- The moment of inertia of the weights when arms are brought in is:
\[ I_2 = I_m + 2 \cdot m \cdot r_2^2 = 6 + 2 \cdot 5 \cdot (0.2)^2 = 6 + 2 \cdot 5 \cdot 0.04 = 6 + 0.4 = 6.4 \, \text{kg m}^2 \]

5. **Apply Conservation of Angular Momentum:**
- According to the conservation of angular momentum:
\[ L_{initial} = L_{final} \]
\[ I_1 \cdot \omega_1 = I_2 \cdot \omega_2 \]
- Rearranging gives:
\[ \omega_2 = \frac{I_1 \cdot \omega_1}{I_2} \]

6. **Substitute Values:**
- Substitute \( I_1 = 16 \, \text{kg m}^2 \), \( \omega_1 = 1.2 \, \text{rev/s} \), and \( I_2 = 6.4 \, \text{kg m}^2 \):
\[ \omega_2 = \frac{16 \cdot 1.2}{6.4} \]

7. **Calculate Final Angular Speed:**
- Performing the calculation:
\[ \omega_2 = \frac{19.2}{6.4} = 3 \, \text{rev/s} \]

### Final Answer:
The new angular speed of the platform is \( 3 \, \text{rev/s} \).

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