The condition for pure rolling of a cylinder on an inclined plane is that the point of contact does not slip with respect to the incline. So friction will be static friction
So acceleration of point of contact is zero.
So Mgsinθ - f = Ma
and fR = I alpha (I = 0.5MR^2)
and acceleration of point of contact = a -R*alpha = zero
Solving these we get the value of friction = f = Mgsinθ/3
and Normal by inclined plane must be Mgcosθ
So the condition f <= mu*N must also be satisfied . ( Here mu is coefficient of static friction)
If this fails then friction will be kinetic friction
and therefore friction will be (coefficient of kinetic friction)*N
so the cylinder will accelerate downwards with constant acceleration =
gsinθ- (coefficient of kinetic friction)*gcosθ.
Therefore the cylinder will rotate due to torque of friction with angular acceleration = 2*(mu)gcosθ/R. (mu = coefficient of kinetic friction).
But this will not be in pure rolling as the point of contact is not at rest with respect to the inclined plane