JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Rotational Mechanics

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}

)

Answer: (B)

Solution

While the cylinder slips, kinetic friction

f = μ_{K} m g

acts backward at the contact, decelerating the centre at

a = μ_{K} g

and, through its torque about the centre, increasing the spin at

α = \frac{f R}{\frac{1}{2} m R ^{2}} = \frac{2 μ _{K} g}{R}

. Thus

v = v_{0} - μ_{K} g t

and

ω = \frac{v _{0}}{4 R} + \frac{2 μ _{K} g}{R} t

. Rolling begins when

v = ω R

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

, which gives

3 μ_{K} g t = \frac{3 v _{0}}{4}

and

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

. The cylinder starts rolling after

5 s

JEE Main · 2026 · 08 Apr 2026 · Shift 2 · Single correct option