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