CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A hollow cylinder of mass M and radius R is pulled by a horizontal force F acting at its centre of mass on a horizontal surface where the coefficient of friction is μ. The maximum acceleration of the cylinder so that it may not start slipping is :

A
2μg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
μg
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
23μg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
32μg
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B μg
If slipping does not occur, point P is an instant center of rotation for the cylinder.
The equation of motion of the cylinder about point P is
τP=IPα where
τP is the torque acting about point P
IP is MI of the cylinder about point P.
α is the angular acceleration about point P.
FR=(IO+MR2)α .....(1) by parallel axis theorem, IP=IO+MR2, IO is MI of the cylinder about cm O.
α=FRIO+MR2 ......(2)

Frictional forceFfr is responsible for angular acceleration about cm O.
τfr=FfrR=IOα .....(3)
Substituting α from (2) in (3)
FfrR=IOFRIO+MR2
Ffr=IOFIO+MR2 .....(4)
As seen from (4) The applied force F and the frictional forceFfr are proportional to each other. If F increases, Ffr also increases until it reaches its possible max value.
This limiting value is given by (Ffr)maxμN=μMg .....(5)
Combining (4) and (5)
μMg=IOFmaxIO+MR2
Fmax=μMg(1+MR2IO) ......(6)

For a hollow cylinder IO=MR2
Substituting IO=MR2 in (6) we get Fmax=μMg(1+MR2MR2)=2μMg
Using (2) αmax=FmaxRIO+MR2=2μMgRMR2+MR2=μgR .......(7)
amax=αmaxR=μgRR=μg

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