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System of Particles and Rotational Motion

Rolling motion

A homogeneous solid cylindrical roller of radius R and mass m

Question

Hard

Solving time: 4 mins

A homogeneous solid cylindrical roller of radius $R$ and mass $m$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is
(a) $\frac{F}{2 m R}$
(b) $\frac{2 F}{3 m R}$
(c) $\frac{3 F}{2 m R}$
(d) $\frac{F}{3 m R}$

Views: 5,352 students

Updated on: Jun 7, 2023

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

Found 8 tutors discussing this question

Jack Discussed

A homogeneous solid cylindrical roller of radius

R

and mass

m

is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is
(a)

\frac{F}{2 m R}

(b)

\frac{2 F}{3 m R}

(c)

\frac{3 F}{2 m R}

(d)

\frac{F}{3 m R}

15 mins ago

Discuss this question LIVE

15 mins ago

Text solutionVerified

Exp. (b)
When force

F

is applied at the centre of roller of mass

m

as shown in the figure below
Solution Image

Its acceleration is given by

\frac{( F - f )}{m} = α

.....(i)
where,

f =

force of friction and

m =

mass of roller.
Torque on roller is provided by friction

f

and it is

τ = f R = l α

.....(ii)
where,

I =

moment of inertia of solid cylindrical roller.

= m R^{2} /2

and

α =

angular acceleration of cylinder

= a / R

.
Hence,

τ = \frac{m R ^{2}}{2} \cdot \frac{a}{R} = \frac{ma R}{2}

From eq. (ii), (

τ = f R)

f = \frac{ma}{2}

.....(iii)
From Eqs. (i) and (iii), we get

F = \frac{3}{2} ma

\Rightarrow a = \frac{2 F}{3 m}

So,

α = \frac{2 F}{3 m R}

[∵ α = \frac{a}{R}]

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Question Text	A homogeneous solid cylindrical roller of radius $R$ and mass $m$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is (a) $\frac{F}{2 m R}$ (b) $\frac{2 F}{3 m R}$ (c) $\frac{3 F}{2 m R}$ (d) $\frac{F}{3 m R}$
Updated On	Jun 7, 2023
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 9
Upvotes	877
Avg. Video Duration	11 min

A homogeneous solid cylindrical roller of radius $R$ and mass $m$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is
(a) $\frac{F}{2 m R}$
(b) $\frac{2 F}{3 m R}$
(c) $\frac{3 F}{2 m R}$
(d) $\frac{F}{3 m R}$

Text solutionVerified

Filo tutor solutions (9)

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Text solutionVerified

Filo tutor solutions (9)

Questions from JEE Mains 2019 - PYQs

Practice questions from Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

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Previous year question papers

A homogeneous solid cylindrical roller of radius $R$ and mass $m$ is pulled on a cricket pitch by a horizontal force. Assuming rolling without slipping, angular acceleration of the cylinder is
(a) $\frac{F}{2 m R}$
(b) $\frac{2 F}{3 m R}$
(c) $\frac{3 F}{2 m R}$
(d) $\frac{F}{3 m R}$