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Question

A thin circular ring of mass M and radius r is rotating about its axis 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
ωMM+m
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B
ω(M2m)M+2m
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C
ωMM+2m
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D
ω(M+2m)M
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Solution

The correct option is C ωMM+2m
Mass of circular ring =M
Radius of circular ring =r
Angular velocity =ω
Mass of each object =m


Initial angular velocity =ω
Initial angular momentum Li=Iω=Mr2ω ...(i)


After placing masses,final angular velocity of system =ω
(1, 3 represents masses and 2 for the ring)
Final angular momentum
Lf=(I1+I2+I3)ω
Lf=(mr2+Mr2+mr2)ω
Lf=(M+2m)r2ω ...(ii)
As the masses were placed gently on ring, all interactions between ring and masses being internal for the system of (ring+masses) hence the external torque is zero.
τext=0
Applying angular momentum conservation:
Li=Lf
Or, Mr2ω=(M+2m)r2ω
ω=MωM+2m
New angular velocity ω of the ring is MωM+2m

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