Question

# A solid cylinder of mass 10kg is rolling perfectly on a plane of inclination 300. The force of friction between the cylinder and surface of the inclined plane is :

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

## The correct option is C 493 NUsing Newton's 2nd Law of motion,Mgsinθ−f=Ma ......(1) along the inclined plane where f is the frictional force up the incline.Mgcosθ=N .........(2) perpendicular to the inclined plane.Substituting (2) in (1) we get,Mgsinθ−μN=Ma∴Mgsinθ−μMgcosθ=MaOnly frictional force exerts torque since weight Mg and normal reaction N pass through the centre of the cylinder and do not exert any torque.Torque due to frictional force is τ=fR where R is the radius of the cylinder.Using equation for rotational motion,τ=fR=Iα .....(3) where I is the MI of the cylinder about its center and α is the angular acceleration.Since there is no slipping, linear acceleration a=RαSubstituting a=Rα in (3) we get,f=IαR=12MR2×aR2=Ma2 .....(4)Substituting for Ma from eqn(1) in eqn(4)f=Mgsinθ−f2⇒3f=Mgsinθ⇒f=Mgsinθ3=10×9.8×sin3003=493N

Suggest Corrections
0
Join BYJU'S Learning Program
Select...
Related Videos
Rolling
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program
Select...