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___________.

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answer is B.

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Detailed Solution

Here $M_{0} = 200 k g, m = 80 k g$

Using conservation of angular momentum $L_{i} = L_{f}$

$I_{1} ω_{1} = I_{2} ω_{2}, I_{1} = (I_{M} + I_{m}) = (\frac{M_{0} R^{2}}{2} + m R^{2}), I_{2} = \frac{1}{2} M_{0} R^{2} a n d ω_{1} = 5 r p m$

$∴ w_{2} = (\frac{M_{0} R^{2}}{2} + m R^{2}) \times \frac{5}{\frac{M^{0} R^{2}}{2}} = \frac{5 R^{2}}{R^{2}} \times \frac{(80 + 100)}{100} = 9 r p m .$

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