JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Rotational Mechanics

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A solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is :

(A) $\frac{2}{5}$
(B) $\frac{5}{2}$ ✓ Correct
(C) $\frac{3}{4}$
(D) $\frac{4}{3}$

Answer: (B)

Solution

For a body of mass

m

and radius

R

rolling without slipping, the rolling condition gives

v_{c m} = ω R

.

Translational kinetic energy:

K E_{trans} = \frac{1}{2} m v_{c m}^{2}

.

Rotational kinetic energy:

K E_{rot} = \frac{1}{2} I ω^{2}

, with

I = \frac{2}{5} m R^{2}

for a solid sphere.

Taking the ratio and substituting

v_{c m} = ω R

\frac{K E _{trans}}{K E _{rot}} = \frac{\frac{1}{2} m v _{c m}^{2}}{\frac{1}{2} ( \frac{2}{5} m R ^{2} ) ω ^{2}} = \frac{m ( ω R ) ^{2}}{\frac{2}{5} m R ^{2} ω ^{2}} = \frac{1}{2/5} = \frac{5}{2} .

JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Single correct option

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JEE Main · 2025 · 24 Jan 2025 · Shift 2 · Rotational Mechanics

Solution

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