JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Rotational Mechanics

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A hollow sphere is rolling on a plane surface about its axis of symmetry. The ratio of rotational kinetic energy to its total kinetic energy is

\frac{X}{5}

. The value of

x

is ____.

Answer: 2

Solution

For a hollow sphere,

I = \frac{2}{3} m R^{2}

; rolling without slipping gives

v = R ω

K E_{r o t} = \frac{1}{2} I ω^{2} = \frac{1}{3} m R^{2} ω^{2}, K E_{t r an s} = \frac{1}{2} m v^{2} = \frac{1}{2} m R^{2} ω^{2}

K E_{t o t a l} = \frac{1}{3} m R^{2} ω^{2} + \frac{1}{2} m R^{2} ω^{2} = \frac{5}{6} m R^{2} ω^{2}

\frac{K E _{r o t}}{K E _{t o t a l}} = \frac{1/3}{5/6} = \frac{2}{5}

\frac{x}{5} = \frac{2}{5} ⟹ x = 2

JEE Main · 2024 · 05 Apr 2024 · Shift 2 · Integer answer

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