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JEE Main 2022 · 26 Jun · Shift 2, Physics question 34

A solid spherical ball is rolling on a frictionless horizontal plane surface about its axis of symmetry. The ratio of rotational kinetic energy of the ball to its total kinetic energy is -

Choose one option
Answer and worked solution

Official answer: B

Rolling solid sphere. Krot=12⋅25mR2ω2=15mv2K_{\rm rot}=\frac12\cdot\frac25mR^2\omega^2=\frac15mv^2 and Ktot=12mv2+15mv2=710mv2K_{\rm tot}=\frac12mv^2+\frac15mv^2=\frac7{10}mv^2.

Ratio 27\dfrac27.

(B)

Topics:Rolling motionRotational kinetic energy
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