Physics Q32 – JEE-MAIN 2026

Home Physics Rotational MotionQuestion #437

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}

)

Clue 1 of 3: Understand the forces and torques

Identify the kinetic friction force acting on the cylinder and the torque it produces about the center of mass.

🧠 Full Solution Path

Step 1: Determine linear and angular accelerations✦ Active

The kinetic friction force acting on the cylinder is $f_{k} = μ_{K} N = μ_{K} M g$ . This force opposes the translational motion. The linear acceleration is $a = - f_{k} / M = - μ_{K} g$ . The torque due to friction about the center of mass is $τ = f_{k} R = μ_{K} M g R$ . For a solid cylinder, the moment of inertia is $I = \frac{1}{2} M R^{2}$ . The angular acceleration is $α = τ / I = \frac{μ_{K} M g R}{\frac{1}{2} M R^{2}} = \frac{2 μ_{K} g}{R}$ . Assuming the initial translational velocity $v_{0}$ is positive and the initial angular velocity $ω_{0} = v_{0} / (4 R)$ is in the direction that causes slipping (i.e., $v_{0} > R ω_{0}$ ), friction acts to decrease $v$ and increase $ω$ .

a = - μ_{K} g

α = \frac{2 μ_{K} g}{R}

Step 2: Formulate velocity equations○ Expand

The translational velocity at time $t$ is given by $v (t) = v_{0} + a t$ . The angular velocity at time $t$ is given by $ω (t) = ω_{0} + α t$ . Substituting the accelerations and initial angular velocity:

v (t) = v_{0} - μ_{K} g t

ω (t) = \frac{v_{0}}{4 R} + \frac{2 μ_{K} g}{R} t

💡 Teacher's Secret Hint

Ensure the signs of acceleration are consistent with the direction of friction and initial motion.

Step 3: Apply pure rolling condition and solve for time○ Expand

Pure rolling begins when the translational velocity equals $R$ times the angular velocity, i.e., $v (t) = R ω (t)$ . Substituting the expressions for $v (t)$ and $ω (t)$ :

v_{0} - μ_{K} g t = R (\frac{v_{0}}{4 R} + \frac{2 μ_{K} g}{R} t)

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

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

\Rightarrow \frac{3 v_{0}}{4} = 3 μ_{K} g t

\Rightarrow t = \frac{v_{0}}{4 μ_{K} g}

Substitute the given values: $v_{0} = 49$ m/s, $μ_{K} = 0.25$ , and $g = 9.8$ m/s $^{2}$ .

t = \frac{49}{4 \times 0.25 \times 9.8} = \frac{49}{1 \times 9.8} = \frac{49}{9.8} = 5 seconds

💡 Teacher's Secret Hint

Double-check the arithmetic and unit consistency in the final calculation.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Physics Question 32 – JEE-MAIN 2026

Clue 1 of 3: Understand the forces and torques

🧠 Full Solution Path