Rotational Motion Question 36

Question: A ring of mass M and radius R is rotating about its axis with angular velocity ω . Two identical bodies each of mass m are now gently attached at the two ends of a diameter of the ring. Because of this, the kinetic energy loss will be:

[JEE ONLINE 25-04-2013]

Options:

A) m(M+2m)Mω2R2

B) Mm(M+m)ω2R2

C) Mm(M+2m)ω2R2

D) (M+m)M(M+2m)ω2R2

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Answer:

Correct Answer: C

Solution:

  • [c] Kinetic energy Kinetic energy Solving we get loss in K.E.

By using angular momentum conservation

MR2×ω=(MR2+MR2+MR2)ω

ω=MωM+2m



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