A person of 80 kg mass is standing on the rim of a circular platform of mass 200 kg rotating about its axis at 5 revolutions per minute (rpm). The person now starts moving towards the centre of the platform. What will be the rotational speed (in rpm) of the platform when the person reaches its centre___________.

Explanation

To solve the problem, we will use the principle of conservation of angular momentum. The angular momentum of the system must remain constant if no external torques act on it.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Mass of the person, \( m = 80 \, \text{kg} \)
- Mass of the platform, \( m_0 = 200 \, \text{kg} \)
- Initial rotational speed of the platform, \( n_1 = 5 \, \text{rpm} \)

2. **Convert RPM to Radians per Second:**
- The angular speed in radians per second is given by:
\[ \omega_1 = \frac{2\pi n_1}{60} = \frac{2\pi \times 5}{60} = \frac{\pi}{6} \, \text{rad/s} \]

3. **Calculate Initial Moment of Inertia:**
- The moment of inertia of the circular platform (treated as a disc) is:
\[ I_0 = \frac{1}{2} m_0 R^2 = \frac{1}{2} \times 200 R^2 = 100 R^2 \]
- The moment of inertia of the person standing at the rim is:
\[ I_p = m R^2 = 80 R^2 \]
- Therefore, the total initial moment of inertia \( I_1 \) is:
\[ I_1 = I_0 + I_p = 100 R^2 + 80 R^2 = 180 R^2 \]

4. **Calculate Final Moment of Inertia:**
- When the person moves to the center, their distance from the axis of rotation becomes zero, so their moment of inertia becomes zero:
\[ I_2 = I_0 = 100 R^2 \]

5. **Apply Conservation of Angular Momentum:**
- According to the conservation of angular momentum:
\[ I_1 \omega_1 = I_2 \omega_2 \]
- Substituting the known values:
\[ 180 R^2 \cdot \frac{\pi}{6} = 100 R^2 \cdot \omega_2 \]

6. **Solve for \( \omega_2 \):**
- Cancel \( R^2 \) from both sides:
\[ 180 \cdot \frac{\pi}{6} = 100 \cdot \omega_2 \]
- Simplifying gives:
\[ 30\pi = 100 \cdot \omega_2 \]
- Thus,
\[ \omega_2 = \frac{30\pi}{100} = \frac{3\pi}{10} \, \text{rad/s} \]

7. **Convert \( \omega_2 \) back to RPM:**
- To convert back to RPM:
\[ n_2 = \frac{\omega_2 \cdot 60}{2\pi} = \frac{\frac{3\pi}{10} \cdot 60}{2\pi} = \frac{3 \cdot 60}{20} = 9 \, \text{rpm} \]

### Final Answer:
The rotational speed of the platform when the person reaches its center is **9 rpm**.