A horizontal force F is applied at the centre of mass of a cylindrical object of mass m and radius R , perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is μ . The center of mass of the object has an acceleration a . The acceleration due to gravity is g . Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

For a solid cylinder, the maximum possible value of a is 2 μ g

For a thin-walled hollow cylinder, a=frac F 2 m

For the same F, the value of a does not depend on whether the cylinder is solid or hollow

The magnitude of the frictional force on the object due to the ground is always μ mg

Physics · Rotational Dynamics

JEE Advanced 2021 — Paper 1 — Question 4

A horizontal force F is applied at the centre of mass of a cylindrical object of mass m and radius R , perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is $\mu$ . The center of mass of the object has an acceleration $a$ . The acceleration due to gravity is g . Given that the object rolls without slipping, which of the following statement(s) is(are) correct?

Option A:
For the same F, the value of $a$ does not depend on whether the cylinder is solid or hollow
Option B:
For a solid cylinder, the maximum possible value of $a$ is $2 \mu \mathrm{~g}$
Correct
Option C:
The magnitude of the frictional force on the object due to the ground is always $\mu \mathrm{mg}$
Option D:
For a thin-walled hollow cylinder, $a=\frac{\mathrm{F}}{2 \mathrm{~m}}$
Correct

Answer: B, D

Step-by-step solution

$\mathrm{F}-\mathrm{f}=\mathrm{ma}$ $\mathrm{fR}=\mathrm{I} \alpha$ (about center of mass) $\mathrm{a}=\mathrm{R} \alpha$

For hollow cylinder $\mathrm{a}=\frac{\mathrm{F}}{2 \mathrm{~m}}, \mathrm{f}=\frac{\mathrm{F}}{2}$ For solid cylinder,

$\mathrm{a}=\frac{2 \mathrm{~F}}{3 \mathrm{~m}}, \mathrm{f}=\frac{\mathrm{F}}{3}$ Also for solid cylinder

$\frac{\mathrm{F}}{2} \leq \mu \mathrm{mg}$ Therefore a $\leq 2 \mu \mathrm{~g}$

Answer key and solution verified before publishing.

Practise Rotational Dynamics

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Advanced 2021
Paper: Paper 1
Subject: Physics
Chapter: Rotational Dynamics
Topic: Rolling Motion

Answer: B, D

Step-by-step solution

Practise Rotational Dynamics

More Rotational Dynamics questions from this paper