A uniform horizontal circular platform of mass 200kg is rotating at 5 rpm about vertical axis passing through its centre. A boy of mass 80kg is standing at its edge. if boy moves to centre of platform, find out final angular speed

Answer

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum of a system remains constant if no external torque acts on it.

### Step-by-Step Solution:

1. **Identify the Initial Moment of Inertia (I)**:
- The moment of inertia of the platform (I_platform) is given by the formula for a solid disk:
\[ I_{\text{platform}} = \frac{1}{2} M R^2 \]
where \(M = 200 \, \text{kg}\) is the mass of the platform and \(R\) is its radius.
- The moment of inertia of the boy (I_boy) when he is at the edge of the platform is:
\[ I_{\text{boy}} = m R^2 \]
where \(m = 80 \, \text{kg}\) is the mass of the boy.
- Therefore, the total initial moment of inertia (I_initial) is:
\[ I_{\text{initial}} = I_{\text{platform}} + I_{\text{boy}} = \frac{1}{2} (200) R^2 + (80) R^2 = 100 R^2 + 80 R^2 = 180 R^2 \]

2. **Identify the Final Moment of Inertia (I')**:
- When the boy moves to the center of the platform, his distance from the axis of rotation becomes zero, so his moment of inertia becomes:
\[ I'_{\text{boy}} = 80 \times 0^2 = 0 \]
- The new total moment of inertia (I_final) is:
\[ I_{\text{final}} = I_{\text{platform}} + I'_{\text{boy}} = \frac{1}{2} (200) R^2 + 0 = 100 R^2 \]

3. **Use Conservation of Angular Momentum**:
- According to the conservation of angular momentum:
\[ I_{\text{initial}} \omega_{\text{initial}} = I_{\text{final}} \omega_{\text{final}} \]
- We know the initial angular speed (\(\omega_{\text{initial}}\)) is given as 5 rpm. Converting this to radians per second:
\[ \omega_{\text{initial}} = 5 \times \frac{2\pi}{60} \, \text{rad/s} = \frac{5\pi}{30} \, \text{rad/s} = \frac{\pi}{6} \, \text{rad/s} \]
- Now substituting the values into the conservation equation:
\[ (180 R^2) \left(\frac{\pi}{6}\right) = (100 R^2) \omega_{\text{final}} \]

4. **Solve for Final Angular Speed (\(\omega_{\text{final}}\))**:
- Cancel \(R^2\) from both sides:
\[ 180 \left(\frac{\pi}{6}\right) = 100 \omega_{\text{final}} \]
- Rearranging gives:
\[ \omega_{\text{final}} = \frac{180 \left(\frac{\pi}{6}\right)}{100} = \frac{180\pi}{600} = \frac{3\pi}{10} \, \text{rad/s} \]
- To convert back to rpm:
\[ \omega_{\text{final}} = \frac{3\pi}{10} \times \frac{60}{2\pi} = \frac{3 \times 60}{20} = 9 \, \text{rpm} \]

### Final Answer:
The final angular speed of the platform after the boy moves to the center is **9 rpm**.