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A ball rolls without slippping. The radius of gyration of the ball about an axis passing through its centre of mass is k. If radius of the ball be R, then the fraction of total energy associated with its rotation will be

a
k2+R2R2
b
k2R2
c
k2k2+R2
d
R2k2+R2

detailed solution

Correct option is C

TOTAL K.E    = K.E. of translation + K.E. of rotation        = 12Mv2+12Iω2        = 12Mv2+12Mk2v2R2(I = Mk2 and v = ωR)       = 12Mv2(1+k2R2)K.E. of rotationTotal K.E. = 12Mv2K2R212Mv2(1+k2R2)                          = k2R21+k2R2 = k2k2+R2

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