A body of mass `m`, radius `R` and moment of inertia `I` (about an axis passing through the centre of mass and perpendicular to plane of motion) is released from rest over a sufficiently rough ground (to provide accelerated pure rolling) find linear acceleration of the body.

Explanation

Let linear acceleration is `a` and angular acceleration `alpha` for accelerated pure rolling
`alpha=(a)/(R)`
After time t, displacement of centre of mass along the plane `s=(1)/(2)at^(2)`
`therefore` height fallen by centre of mass along the plane `s=(1)/(2)at^(2)`
`therefore` height fallen by centre of mass
`h=(s)(sintheta)=(1)/(2)at^(2)sintheta`
Linear velocity `v=at`
angular velocity `omega=alphat=(at)/(R)`
From energy conservation principle, decrease in potential energy `=` increase in translational and rotaional kinetic energy
or `mgh=(1)/(2)mv^(2)+(1)/(2)Iomega^(2)`
Substituting the value we have
`mg((1)/(2)at^(2)sintheta)=(1)/(2)m(at)^(2)+(1)/(2)I((at)/(R))^(2)`
Solving this equation we get `a=(gsintheta)/(1+(I)/(mR^(2)))`