Question 48: Centre of Mass and Rotational Motion

Question

A thin circular ring of mass ,$$M$$ and radius $$R$$ rotates about an axis through its centre and perpendicular to its plane, with a constant angular velocity $$\omega$$. Four small spheres each of mass $$m$$ (negligible radius) are kept gently to the opposite ends of two mutually perpendicular diameters of the ring. The new angular velocity of the ring will be

Accepted Answer

According to conservation of angular momentum,
$$I \omega=	ext { constant }$$
i.e. we can write,
$$\begin{aligned}
I_1 \omega_1 & =I_2 \omega_2 \
	ext { or } \quad & M R^2 \omega =(M+4 m) R^2 \omega_2 \quad (\because \omega_1=\omega)\
	ext { or } \quad & \omega_2=\left(\frac{M}{M+4 m}ight) \omega\end{aligned}$$

Answer

$$\left(\frac{M+4 m}{M}\right) \omega$$

Answer

$$\frac{M}{4 m} \omega$$

Answer

$$\left(\frac{M}{M+4 m}\right) \omega$$

Answer

$$\left(\frac{M}{M-4 m}\right) \omega$$

Centre of Mass and Rotational Motion

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