Total energy =21Iω2+21mv2 =21mv2(1+R2K2) Rotational energy =21Iω2=1+R2K2K2+R2 Required fraction =1+K2/R2K2/R2=R2+K2K2
NEET UG 2003 — Physics Mechanics
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 energy will be:
Held on 30 Apr 2003 · Verified 9 Jul 2026.
R2K2+R2
R2K2
K2+R2K2
K2+R2R2
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A bullet of mass 10 g moving with 400 m/s penetrates a wall and comes to rest. The loss in kinetic energy is:
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