Discuss the effect of rolling on inclined plane and derive the expression for the acceleration.

Explanation

Let us assume a round object of mass m ad radius R is rolling down an inclined plane without slipping as shown in the figure (Rolling on inclined plane). There are two forces acting on the object along the inclined plane.
(i) They are the component of gravitational force (mg `sin theta)`
(ii) The static frictionakl force `(f)`. The other component of gravitational force `(mg cos theta)` is cancelled by the normal force (N) exerted by the plane. As the motion is happened along the incline, let us write the equation for motion from the free body diagram (FBP) of the object.
For translational motion, `mg sin theta` is the suporting force and f is teh opposing force,
`m g sin theta -f=ma`
For rotational motion, let us take the torque with respect to the centre of the object. THen `mg sin theta` cannot cause torque as it passes through it but the frictional force f can set torque of Rf.
`Rf=I alpha`
By using the relation `a=r alpha`, and moment of inertia `I=mK^(2)` we get
`Rf=mK^(2)a/R, f=ma((K^(2))/(R^(2)))`
Now equation KE `=1/2MV_(CM)^(2)(1+(K^(2))/(R^(2)))`
becomes
`mg sin theta -ma ((K^(2))/(R^(2)))m=ma`
`mg sin theta=ma+ma((K^(2))/(R^(2)))`
`a(1+(K^(2))/(R^(2)))=g sin theta`
After rewriting it for acceleration we get
`a=(g sin theta)/((1+(K^(2))/(R^(2))))`