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Question

A uniform thin circular ring of mass 'M' and radius 'R' is rotating about its fixed axis, passing through its centre and perpendicular to its plane of rotation, with a constant angular velocity $$\omega$$. Two objects each of mass m, are attached gently to the opposite ends of a diameter of the ring. The ring now rotates with an angular velocity of :


A
ωM(M+m)
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B
ωM(M+2m)
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C
ωMM2m
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D
ω(M+3m)M
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Solution

The correct option is B $$\displaystyle \frac{\omega M}{(M\, +\, 2m)}$$
Moments of inertia (MOI) of the ring before attaching the masses $$I$$= $$ MR^2$$,
MOI of the ring after attaching the masses $$I^{'}= (M+2m) R^2$$
Let angular momentum after the attaching the masses $$\omega ^{'}$$
Since there is no external torque,  so we use conservation of angular momentum.
$$I \times \omega= I ^ {'} \times \omega ^ {'} $$
$$\Rightarrow MR^2 \times \omega= \dfrac { MR^2 \times \omega ^{'}}{ M + 2m}$$,
$$\Rightarrow \omega ^{'} = \dfrac {\omega M}{M+2m}$$

Physics

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