If a rigid bady of radius 'R' starts from rest and rolls down an inclined plane of inclination `theta` then linear acceleration of body rolling down the plane is ______

Explanation

Consider a circularly symmetric rigid body, such as a sphere or a whell or a disc, rolling with friction down a plane inclined at an angle `theta` to the horizontal. If the frictional force on the body is large enough , the body rolls withoput slipping.

Let M and R be the mass and radius of the body. Let I be its moment of inertia for rotation about an axis through its centre. Let the body start from rest at the top of the incline at an height h. Let v be teh translational speed of the centre of mass at the bottom of the incline and `omega` the corresponding angular speed. Then, its kinetic energy at the energy at the bottom of the incline is
`E=(1)/(2)Mv^(2)[1+(I)/(MR^(2))]=(1)/(2)Mv^(2)(1+c)" "`...........(1)
where `c=(I)/(MR^(2))`.
By conservation of energy (igoring the work done against the frictional force), `(KE+PE)_("initial")=(KE+PE)_("final")" "`.......(2)
`:. 0+ltgh=(1)/(2)Mv^(2)(1+c)+0`
`:. Mgh=(1)/(2)Mv^(2)(1+c)" "`.......(3)
`:. v^(2)=(2gh)/(1+c)`
`:. v=sqrt((2gh)/(1+c))=sqrt((2gh)/(1+(I//MR^(2))))" "`......(4)
Since h=L sin `theta, v=sqrt((2gL sin theta)/(1+(I//MR^(2))))" "`......(5)
Let a be the accelaration of the centre of mass of the body along the inclined plane. Since the body starts from rest,
`v^(2)=2aL :. a=(v^(2))/(2L)" "`......(6)
`:. a=(2gL sin theta)/(1+c).(1)/(2L)=(g sin theta)/(1+c)=(g sin theta)/(1+(L//MR^(2)))" "`.........(7)