Question

A thin uniform circular ring is rolling down an inclined plane of inclination 30º without slipping, its linear acceleration along the inclined plane will be

Answer 1

There will be three force acting on the disc. These are gravitational force, normal force from the inclined plane and friction. This is shown in the diagram below.


We can resolve weight of the body into two components. The forces normal to the plane are balanced. Then there will be the downward component of weight, mgsinθ and friction f.

In case of no friction, the disc would slide downward. Therefore, we will take direction of friction up the plane. If the disc rolls downward, its center of mass will also moves downward. For this, mgsinθ should be greater than friction force f.

Force equation will be,

mgsinθ − f = ma     ....11

During the downward motion of the disc, the disc would rotates about its center. As a result, torque would act on it. Here, Torque mgsinθ would be zero about the center of the disc.

The torque due to friction about the center of the disc is fR. This torque will responsible for angular acceleration α.

Writing the torque equation, we have
fR = Iα
For the solid disc the moment of inertia about the axis of rotation is mR22mR22 and angular acceleration α=aRα=aR.
So, 
fR=mR22aR⇒f=ma2     .....(2)fR=mR22aR⇒f=ma2     .....2
Adding equation 11 and 22, we get

mgsinθ=ma+ma2⇒mgsinθ=3ma2⇒a=22gsinθPutting this value in equation (2) we havef=ma2=13mgsinθmgsinθ=ma+ma2⇒mgsinθ=3ma2⇒a=22gsinθPutting this value in equation (2) we havef=ma2=13mgsinθ
So, friction on the disc is13mgsinθ

Answer 2

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