JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Rotational Mechanics

A solid cylinder of radius

R

rolls without slipping with a center of mass speed

v_{0} = \frac{gR}{3}

on a horizontal surface with a vertical edge, as shown in the figure. Here,

g

is the acceleration due to the gravity. At the moment when the cylinder loses contact with the surface due to rotation around the corner, the speed of its center of mass is:

(A)0
(B) $\frac{5 gR}{7}$ ✓ Correct
(C) $\frac{gR}{15}$
(D) $\frac{3 gR}{7}$

Answer: (B)

Solution

Initially,

ω_{0} = \frac{v _{0}}{R} = \frac{g}{3 R}

.

When the cylinder rotates about the corner, it loses contact when the normal force becomes zero. At this point, the centripetal force equals the component of weight:

m ω^{2} R = m g cos θ

.

Using work-energy theorem from initial position to the point of losing contact:

m g R (1 - cos θ) = \frac{1}{2} \cdot \frac{3}{2} m R^{2} (ω^{2} - ω_{0}^{2})

Substituting

ω^{2} R = g cos θ

and

ω_{0}^{2} = \frac{g}{3 R}

g - g cos θ = \frac{3}{4} R (ω^{2} - \frac{g}{3 R})

Solving:

ω = \frac{5 g}{7 R}

, so

v = R ω = \frac{5 g R}{7}

JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Single correct option

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JEE Advanced · 2026 · 17 May 2026 · Paper 1 · Rotational Mechanics

Solution

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