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....

We have a circular platform of mass $$M = 200 \text{ kg}$$ that can be treated as a uniform solid disc. For a solid disc of radius $$R$$, the moment of inertia about the central axis is given by the standard formula

$$I_{\text{disc}} = \frac{1}{2} M R^{2}.$$

The person standing on the rim has mass $$m = 80 \text{ kg}$$. A human can be modelled as a point mass for rotational calculations, so when the person is at a distance $$R$$ from the axis, the moment of inertia contributed by the person is

$$I_{\text{person (initial)}} = m R^{2}.$$

Initially, therefore, the total moment of inertia of the system “platform + person” is the sum of the two individual contributions:

$$$I_{1} = I_{\text{disc}} + I_{\text{person (initial)}} = \frac{1}{2} M R^{2} + m R^{2} = R^{2}\!\left(\frac{1}{2} M + m\right).$$$

The platform is rotating at an initial speed of $$$n_{1} = 5 \text{ revolutions per minute}$$$ (rpm). Converting to angular speed in rad s−1 is unnecessary because the same unit cancels out, but we shall denote the initial angular velocity by $$\omega_{1}$$, so

$$\omega_{1} = 5 \text{ rpm}.$$

Now the person walks from the rim straight to the centre. When the person reaches the centre, his distance from the axis becomes practically zero, so his moment of inertia becomes

$$I_{\text{person (final)}} \approx 0.$$

Therefore the final moment of inertia of the system is only that of the disc:

$$I_{2} = I_{\text{disc}} = \frac{1}{2} M R^{2}.$$

Because no external torque acts on the system, angular momentum is conserved. We state the law first:

$$$\text{Angular momentum } L = I \omega \quad\text{remains constant if the external torque is zero.}$$$

So we set initial angular momentum equal to final angular momentum:

$$I_{1}\,\omega_{1} = I_{2}\,\omega_{2}.$$

Substituting the expressions for $$I_{1}$$ and $$I_{2}$$, we get

$$$\biggl[R^{2}\!\left(\dfrac{1}{2} M + m\right)\biggr] \omega_{1} \;=\; \bigl[\tfrac{1}{2} M R^{2}\bigr] \omega_{2}.$$$

We observe that $$R^{2}$$ cancels from both sides, giving

$$\left(\frac{1}{2} M + m\right) \omega_{1} \;=\; \frac{1}{2} M \,\omega_{2}.$$

Solving for $$\omega_{2}$$, we write

$$\omega_{2} \;=\; \frac{\left(\dfrac{1}{2} M + m\right)}{\dfrac{1}{2} M}\;\omega_{1}.$$

Now we substitute the numerical values $$M = 200 \text{ kg}$$ and $$m = 80 \text{ kg}$$. First compute the denominator:

$$\frac{1}{2} M = \frac{1}{2} \times 200 = 100.$$

Then compute the numerator:

$$\frac{1}{2} M + m = 100 + 80 = 180.$$

Hence the required ratio is

$$\frac{180}{100} = 1.8.$$

Substituting this factor and the initial speed $$\omega_{1} = 5 \text{ rpm}$$, we obtain

$$\omega_{2} = 1.8 \times 5 \text{ rpm} = 9 \text{ rpm}.$$

Therefore, when the person arrives at the centre, the platform-person system speeds up to 9 revolutions per minute.

So, the answer is $$9 \text{ rpm}$$.