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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 ω. 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 ωM(M+2m)
Moments of inertia (MOI) of the ring before attaching the masses I= MR2,
MOI of the ring after attaching the masses I=(M+2m)R2
Let angular momentum after the attaching the masses ω
Since there is no external torque, so we use conservation of angular momentum.
I×ω=I×ω
MR2×ω=MR2×ωM+2m,
ω=ωMM+2m

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