A person of $80 \mathrm{~kg}$ mass is standing on the rim of a circular platform of mass $200 \mathrm{~kg}$ rotating about its axis at 5 revolutions per minute $(\mathrm{rpm})$. 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___________
$(9.00)$
Here $M_{0}=200 \mathrm{~kg}, m=80 \mathrm{~kg}$
Using conservation of angular momentum, $L_{i}=L_{f}$
$I_{1} \omega_{1}=I_{2} \omega_{2}$
$I_{1}=\left(I_{M}+I_{m}\right)=\left(\frac{M_{0} R^{2}}{2}+m R^{2}\right)$
$I_{2}=\frac{1}{2} M_{0} R^{2}$ and $\omega_{1}=5 \mathrm{rpm}$
$\therefore \omega_{2}=\left(\frac{M_{0} R^{2}}{2}+m R^{2}\right) \times \frac{5}{\frac{M_{0} R^{2}}{2}}$
$=\frac{5 R^{2}}{R^{2}} \times \frac{(80+100)}{100}=9 \mathrm{rpm}$
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