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 need to apply the principle of conservation of angular momentum. The angular momentum before the man brings his arms in must equal the angular momentum after he brings his arms in, since there is no external torque acting on the system.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of each weight, \( m = 5 \, \text{kg} \)
- Initial distance from the axis of rotation, \( r_i = 100 \, \text{cm} = 1 \, \text{m} \)
- Final distance from the axis of rotation, \( r_f = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Initial angular speed, \( \omega_i = 1.2 \, \text{rev/s} \)
- Moment of inertia of the man and platform, \( I_m = 6 \, \text{kg m}^2 \)

2. **Calculate Initial Moment of Inertia:**
The total moment of inertia before the man brings his arms in is the sum of the moment of inertia of the man/platform and the moment of inertia of the weights.
\[ I_i = I_m + 2 \cdot m \cdot r_i^2 \]
\[ I_i = 6 \, \text{kg m}^2 + 2 \cdot 5 \, \text{kg} \cdot (1 \, \text{m})^2 = 6 + 10 = 16 \, \text{kg m}^2 \]

3. **Calculate Final Moment of Inertia:**
The total moment of inertia after the man brings his arms in is:
\[ I_f = I_m + 2 \cdot m \cdot r_f^2 \]
\[ I_f = 6 \, \text{kg m}^2 + 2 \cdot 5 \, \text{kg} \cdot (0.2 \, \text{m})^2 = 6 + 0.08 = 6.4 \, \text{kg m}^2 \]

4. **Apply Conservation of Angular Momentum:**
Since angular momentum is conserved, we have:
\[ I_i \cdot \omega_i = I_f \cdot \omega_f \]
Rearranging gives us:
\[ \omega_f = \frac{I_i \cdot \omega_i}{I_f} \]

5. **Substituting Values:**
\[ \omega_f = \frac{16 \, \text{kg m}^2 \cdot 1.2 \, \text{rev/s}}{6.4 \, \text{kg m}^2} \]
\[ \omega_f = \frac{19.2}{6.4} = 3 \, \text{rev/s} \]

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