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 Initial Conditions:**
- Mass of the platform (M) = 200 kg
- Mass of the boy (m) = 80 kg
- Initial angular speed (ω_initial) = 5 rpm

2. **Convert Angular Speed to Radians per Second:**
- Since 1 rpm = \( \frac{2\pi}{60} \) rad/s, we convert 5 rpm to rad/s:
\[ \omega_{\text{initial}} = 5 \times \frac{2\pi}{60} = \frac{5\pi}{30} = \frac{\pi}{6} \text{ rad/s} \]

3. **Calculate Initial Moment of Inertia:**
- The moment of inertia of the platform (I_platform) is given by:
\[ I_{\text{platform}} = \frac{1}{2} M R^2 \]
- The boy is initially at the edge, so his moment of inertia (I_boy_initial) is:
\[ I_{\text{boy initial}} = m R^2 \]
- Total initial moment of inertia (I_initial):
\[ I_{\text{initial}} = I_{\text{platform}} + I_{\text{boy initial}} = \frac{1}{2} M R^2 + m R^2 = \left(\frac{1}{2} \times 200 + 80\right) R^2 = (100 + 80) R^2 = 180 R^2 \]

4. **Calculate Initial Angular Momentum:**
- Angular momentum (L_initial) is given by:
\[ L_{\text{initial}} = I_{\text{initial}} \cdot \omega_{\text{initial}} = 180 R^2 \cdot \frac{\pi}{6} = 30\pi R^2 \]

5. **Calculate Final Moment of Inertia:**
- When the boy moves to the center, his moment of inertia becomes zero (I_boy_final = 0). Therefore, the final moment of inertia (I_final) is:
\[ I_{\text{final}} = I_{\text{platform}} + I_{\text{boy final}} = \frac{1}{2} M R^2 + 0 = \frac{1}{2} \times 200 R^2 = 100 R^2 \]

6. **Apply Conservation of Angular Momentum:**
- According to the conservation of angular momentum:
\[ L_{\text{initial}} = L_{\text{final}} \]
- Thus,
\[ 30\pi R^2 = 100 R^2 \cdot \omega_{\text{final}} \]

7. **Solve for Final Angular Speed (ω_final):**
- Dividing both sides by \( R^2 \) (assuming \( R \neq 0 \)):
\[ 30\pi = 100 \omega_{\text{final}} \]
- Therefore,
\[ \omega_{\text{final}} = \frac{30\pi}{100} = \frac{3\pi}{10} \text{ rad/s} \]

### Final Answer:
The final angular speed of the platform after the boy moves to the center is \( \frac{3\pi}{10} \) rad/s.