Q.

A ball rolls without slipping. 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 rotational energywill be

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a

K2K2+R2

b

R2K2+R2

c

K2+R2R2

d

K2R2

answer is A.

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Detailed Solution

Moment of inertia,  I=mK2∵  Ball is rolling without slipping, v=RωTranslational kinetic energy,  KR=12Iω2=mK2v22R2∴   Total kinetic energy  =12mv2+mK2v22R2Required fraction  =mK2v22R212mv2+mK2v22R2=K2R2+K2
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