JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Rotational Mechanics

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A solid sphere is rolling on a horizontal plane without slipping. If the ratio of angular momentum about axis of rotation of the sphere to the total energy of moving sphere is

π : 22

then, the value of its angular speed will be ______

rad / s

Answer: 4

Solution

For a solid sphere rolling without slipping, the moment of inertia about the center of mass is

I_{com} = \frac{2}{5} M R^{2}

, and

v_{com} = R ω

.

Angular momentum about the axis of rotation:

L = I_{com} ω = \frac{2}{5} M R^{2} ω = \frac{2}{5} M R v_{com}

Total kinetic energy:

K = \frac{1}{2} I_{com} ω^{2} + \frac{1}{2} M v_{com}^{2} = \frac{1}{5} M R^{2} ω^{2} + \frac{1}{2} M v_{com}^{2} = \frac{7}{10} M v_{com}^{2}

The ratio:

\frac{L}{K} = \frac{\frac{2}{5} M R v _{com}}{\frac{7}{10} M v _{com}^{2}} = \frac{4 R}{7 v _{com}} = \frac{4}{7 ω}

Given

\frac{L}{K} = \frac{π}{22}

\frac{4}{7 ω} = \frac{π}{22} ⟹ ω = \frac{4 \times 22}{7 π} = \frac{88}{7 π} \approx 4 rad/s

Therefore,

ω = 4

rad/s.

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer

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JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Rotational Mechanics

Solution

More JEE Main questions on Rotational Mechanics