JEE Main 2026 Apr 8 Shift 2, Physics Q7: Rolling Motion

JEE Main2026Apr 8, Shift 2Physics

A solid cylinder having radius $R$ and length $L$ is slipping on a rough horizontal plane. At time $t = 0$ the cylinder has a translational velocity $v_{0} = 49$ m/s, perpendicular to its axis and a rotational velocity $v_{0} /4 R$ about the centre. The time taken by the cylinder to start rolling is ________ seconds. (coefficient of kinetic friction $μ_{K} = 0.25$ and $g = 9.8$ m/s $^{2}$ )

A
$15$
B
$5$
C
$10$
D
$7.5$

Solution

Initially $v_{0} ≫ ω_{0} R$ (since $ω_{0} R = v_{0} /4$ ), so the contact point slips forward. Kinetic friction acts backward on the cylinder, decelerating translation and angularly accelerating rotation.

Linear: $a = - μ_{K} g$ , so $v (t) = v_{0} - μ_{K} g t$ .

Angular (torque about centre $= μ_{K} m g R$ , moment of inertia $= \frac{1}{2} M R^{2}$ ):

$α = \frac{μ _{K} m g R}{\frac{1}{2} M R ^{2}} = \frac{2 μ _{K} g}{R}$ , so $ω (t) = \frac{v _{0}}{4 R} + \frac{2 μ _{K} g}{R} t$ .

Rolling condition: $v = ω R$ .

$v_{0} - μ_{K} g t = \frac{v _{0}}{4} + 2 μ_{K} g t ⟹ \frac{3 v _{0}}{4} = 3 μ_{K} g t$

$t = \frac{v _{0}}{4 μ _{K} g} = \frac{49}{4 \cdot 0.25 \cdot 9.8} = \frac{49}{9.8} = 5$ s.

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JEE Main 2026 Apr 8 Shift 2, Physics Q7: Rolling Motion

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