Question

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

A
m(M+2m)Mω2R2
B
Mm(M+m)ω2R2
C
Mm(M+2m)ω2R2
D
(M+m)(M+2m)ω2R2

Solution

## The correct option is C $$\dfrac{Mm}{(M + 2m)}\omega^2 R^2$$By using angular momentum conservation$$MR^2 \times \omega = (MR^2 + MR^2 + MR^2) \omega '$$$$\omega ' = \dfrac{M\omega}{M + 2m}$$Loss in kinetic energy$$= \dfrac{1}{2}mR^2 \times \omega^2 - \dfrac{1}{2}(M + 2m)R^2 \times \dfrac{M^2 \omega^2}{(M + 2m)^2} = \dfrac{Mn\omega^2 R^2}{(M + 2m)}$$Physics

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