1. Use conservation of angular momentum

Since no external torque acts on the person-platform system,

$L_i=L_f$

That is,

$I_i\omega_i=I_f\omega_f$

2. Initial moment of inertia

The platform is a uniform circular disc, so its moment of inertia about its axis is

$I_{\text{disc}}=\frac{1}{2}MR^2$

Given:

• Mass of platform, $M=200\,\text{kg}$
• Mass of person, $m=80\,\text{kg}$

Initially, the person stands at the rim, so the person's moment of inertia is

$I_{\text{person}}=mR^2$

Hence,

$I_i=\frac{1}{2}MR^2+mR^2$

$I_i=\frac{1}{2}(200)R^2+80R^2$

$I_i=100R^2+80R^2=180R^2$

3. Final moment of inertia

When the person reaches the centre, his distance from the axis becomes zero, so his moment of inertia becomes zero.

Thus,

$I_f=\frac{1}{2}MR^2=100R^2$

4. Apply conservation of angular momentum

Initial angular speed is $5\,\text{rpm}$.

$I_i\omega_i=I_f\omega_f$

$180R^2\cdot 5=100R^2\cdot \omega_f$

Cancel $R^2$:

$180\cdot 5=100\omega_f$

$\omega_f=\frac{180\cdot 5}{100}=9\,\text{rpm}$

5. Final answer

The rotational speed of the platform when the person reaches the centre is

$\boxed{9\,\text{rpm}}$