CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
Question

A uniform solid cylinder of mass m and radius 2R rests on a horizontal table. A string attached to it passes over a pulley (disc) of mass m and radius R that is mounted on a frictionless axle through its centre. A block of mass m is suspended from the free end of the string. The string does not slip over the pulley surface, and the cylinder rolls without slipping on the table. Force of friction acting on the cylinder is
1221032_33cae63b40e746b685f757a414e15fae.png

A
2mg3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
3mg2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
mg3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
mg6
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D mg6
Given :- Mass of cylinder = m
Radius of cylinder = 2R
Mass of pulley = m
Radius of pulley = R
Mass of block = m

To Find :- Frictional force ( f ) acting on cylinder

Solution :- According to diagram ,
On Balancing force for block we get ,
mgτ1=ma
τ1=mgma (1)
On Balancing force for cylinder we get ,
τ2f=ma
τ2=ma+f (2)
On Balancing force for pulley we get ,
(τ1τ2)R=Ipαp (let τ1>τ2)
From (1) and (2) we get ,
(mg2maf)R=mR22αp (Ip=mR22)
αp=2(mg2maf)mR (3)
Now , Acceleration of pulley (a)=Rαp
αp=aR
2(mg2maf)mR=aR ( from 3 )
ma=2(mgf)5 (4)
Now , On equalling the Torque for cylinder we get ,
2Rf=Icαc
αc=fmR (5) (Ic=2mR2)
Now , Acceleration of cylinder (a) = 2Rαc
αc=a2R
fmR=a2R ( from 5 )
ma=2f
2(mgf)5=2f ( from 4 )
f=mg6–––––––
Hence , Option D(mg6) is correct.–––––––––––––––––––––––––––



1711012_1221032_ans_1778faa9ab704caa941315c0184d86fd.jpg

flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon