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
B
ω(M2m)M+2m
C
ωMM+2m
D
ω(M+2m)M

Solution

## The correct option is D ωMM+2mMass 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+2mPhysics

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