JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Rotational Mechanics

For a rolling spherical shell, the ratio of rotational kinetic energy and total kinetic energy is

\frac{x}{5}

. The value of

x

is ____ .

Answer: 2

Solution

For a rolling spherical shell, the moment of inertia is

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

.

Rotational kinetic energy:

K_{rot} = \frac{1}{2} I ω^{2} = \frac{1}{2} \cdot \frac{2}{3} m R^{2} \cdot \frac{v ^{2}}{R ^{2}} = \frac{1}{3} m v^{2}

.

Translational kinetic energy:

K_{trans} = \frac{1}{2} m v^{2}

.

Total kinetic energy:

K_{total} = K_{trans} + K_{rot} = \frac{1}{2} m v^{2} + \frac{1}{3} m v^{2} = \frac{5}{6} m v^{2}

.

Ratio:

\frac{K _{rot}}{K _{total}} = \frac{\frac{1}{3} m v ^{2}}{\frac{5}{6} m v ^{2}} = \frac{2}{5}

.

Therefore,

x = 2

JEE Main · 2023 · 12 Apr 2023 · Shift 1 · Integer answer

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