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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

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+2mM)
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Solution

## The correct option is C ωMM+2mAs we can see, there is no external torque on the system. Hence angular momentum of the system will be conserved about the rotational axis. Given, mass of circular ring =M Radius of circular ring =R Angular velocity about rotational axis =ω MOI of circular ring I=MR2 Hence, initial angular momentum Li=Iω Li=(MR2)ω MOI of system (ring + objects) I′=mR2+mR2+MR2 I′=(M+2m)R2 Hence, final angular momentum Lf=(M+2m)R2ω′ As we know, Li=Lf (MR2)ω=(M+2m)R2ω′ i.e ω′=(MωM+2m)

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