A person is standing on the edge of a circular platform, which is moving with constant angular speed about an axis passing through its centre and perpendicular to the plane of platform. If person is moving along any radius towards axis of rotation then the angular velocity will:

Options

A. decrease
B. remain unchanged
C. increase
D. data is insufficient

Answer

To solve the problem, we need to analyze the situation using the principles of rotational motion and the conservation of angular momentum.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a circular platform rotating about its center with a constant angular speed \( \omega \).
- A person is initially standing at the edge of the platform, at a distance \( R \) from the axis of rotation.

2. **Movement of the Person**:
- The person moves along the radius towards the center of the platform, reducing their distance from the axis of rotation to \( r_1 \) (where \( r_1 < R \)).

3. **Conservation of Angular Momentum**:
- Since there are no external torques acting on the system (the platform is rotating at a constant angular speed), the angular momentum of the system must be conserved.
- The angular momentum \( L \) is given by the product of the moment of inertia \( I \) and the angular velocity \( \omega \):
\[ L = I \cdot \omega \]

4. **Calculating Initial and Final Angular Momentum**:
- **Initial Angular Momentum** (\( L_1 \)):
- The moment of inertia of the person at the edge is \( I_1 = mR^2 \) (where \( m \) is the mass of the person).
- Therefore, the initial angular momentum is:
\[ L_1 = I_1 \cdot \omega = mR^2 \cdot \omega \]

- **Final Angular Momentum** (\( L_2 \)):
- When the person moves to a distance \( r_1 \), their moment of inertia becomes \( I_2 = mr_1^2 \).
- The final angular momentum is:
\[ L_2 = I_2 \cdot \omega_2 = mr_1^2 \cdot \omega_2 \]

5. **Setting Initial and Final Angular Momentum Equal**:
- By conservation of angular momentum:
\[ L_1 = L_2 \implies mR^2 \cdot \omega = mr_1^2 \cdot \omega_2 \]

6. **Solving for Final Angular Velocity**:
- Canceling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ R^2 \cdot \omega = r_1^2 \cdot \omega_2 \]
- Rearranging gives:
\[ \omega_2 = \frac{R^2}{r_1^2} \cdot \omega \]

7. **Analyzing the Result**:
- Since \( r_1 < R \), it follows that \( \frac{R}{r_1} > 1 \).
- Therefore, \( \left(\frac{R}{r_1}\right)^2 > 1 \), which implies:
\[ \omega_2 > \omega \]

### Conclusion:
The angular velocity \( \omega_2 \) increases as the person moves towards the center of the platform.