JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Rotational Mechanics

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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 :-

(A) $\frac{2}{5}$
(B) $\frac{2}{7}$ ✓ Correct
(C) $\frac{1}{5}$
(D) $\frac{7}{10}$

Answer: (B)

Solution

For a solid sphere rolling without slipping, the total kinetic energy is

K_{total} = \frac{1}{2} I_{cm} ω^{2} + \frac{1}{2} m v_{cm}^{2}

. The moment of inertia is

I_{cm} = \frac{2}{5} m R^{2}

, and the rolling condition gives

v_{cm} = R ω

. Thus,

K_{rot} = \frac{1}{2} \cdot \frac{2}{5} m R^{2} \cdot \frac{v _{cm}^{2}}{R ^{2}} = \frac{1}{5} m v_{cm}^{2}

, and

K_{total} = \frac{1}{5} m v_{cm}^{2} + \frac{1}{2} m v_{cm}^{2} = \frac{7}{10} m v_{cm}^{2}

. Therefore,

\frac{K _{rot}}{K _{total}} = \frac{1/5}{7/10} = \frac{2}{7}

JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 26 Jun 2022 · Shift 2 · Rotational Mechanics

Solution

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