Question 12: Centre of Mass and Rotational Motion

Question

A man with a mass of 80 kg is standing on the rim to a circular platform with a mass of 200 kg . The circular platform is rotating at 12 revolutions per minute (rpm) about its axis. The man moves from the rim towards the centre of the platform. The new angular velocity of the system will be (Assuming that the man's moment of inertia at the centre of the plateform is negligible)

Accepted Answer

According to the law of conservation of angular momentum.
Total initial angular momentum = Total final angular momentum
$$ \begin{aligned} \left(I+m R^2ight) \omega & =I \omega^{\prime} \ 	herefore \quad \omega^{\prime} & =\frac{I+m R^2}{I} \omega \Rightarrow \omega^{\prime}=\frac{\frac{M R^2}{2}+m R^2}{\frac{M R^2}{2}} \omega \ 	ext { or } \quad \omega^{\prime} & =\frac{M+2 m}{M}=\frac{200+2 	imes 80}{200} 	imes 12 \ \omega^{\prime} & =\frac{360}{200} 	imes 12=21.6 \mathrm{rpm} \end{aligned} $$

Answer

10 rpm

Answer

12 rpm

Answer

21.6 rpm

Answer

zero

Centre of Mass and Rotational Motion

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