JEE Main 2025 Jan 24 Shift 2, Physics Q4: Rolling Motion

JEE Main2025Jan 24, Shift 2Physics

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
$\frac{5}{2}$
C
$\frac{3}{4}$
D
$\frac{4}{3}$

Solution

For a solid sphere about its centre, $I = \frac{2}{5} M R^{2}$ . Rolling without slipping requires $v_{c m} = ω R$ .

Translational KE: $K_{T} = \frac{1}{2} M v_{c m}^{2}$

Rotational KE: $K_{R} = \frac{1}{2} I ω^{2} = \frac{1}{2} (\frac{2}{5} M R^{2}) (\frac{v _{c m}}{R})^{2} = \frac{1}{5} M v_{c m}^{2}$

Taking the ratio:

$\frac{K _{T}}{K _{R}} = \frac{\frac{1}{2} M v _{c m}^{2}}{\frac{1}{5} M v _{c m}^{2}} = \frac{5}{2}$

Correct option: (2)

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JEE Main 2025 Jan 24 Shift 2, Physics Q4: Rolling Motion

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