A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? [2017]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A rope is wound around a hollow cylinder of mass 3 kg and radius 40 cm. What is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N [2017]

1 $0.25 rad$ $s^{- 2}$

2 $25 rad$ $s^{- 2}$

3 $5 m$ $s^{- 2}$

4 $25 m$ $s^{- 2}$

Ans.
(2)

Here, $m = 3 kg, r = 40 cm = 40 \times 10^{- 2} m, F = 30 N$

Moment of inertia of hollow cylinder about its axis

$= m r^{2} = 3 kg \times {(0.4)}^{2} m^{2} = 0.48 kg$ $m^{2}$

The torque is given by, $τ = I α$

where $I$ = moment of inertia, $α$ = angular acceleration

In the given case, $τ = r F$ , as the force is acting perpendicularly to the radial vector.

$∴$ $α = \frac{τ}{I} = \frac{F r}{m r^{2}} = \frac{F}{m r} = \frac{30}{3 \times 40 \times 10^{- 2}} = \frac{30 \times 100}{3 \times 40}$

$α = 25 rad$ $s^{- 2}$

Want Us to Email you About Special Offers & Updates?