As per the given question,
When a cylinder is rolling on the inclined plane, then the frictional force is acting on the block in the upward direction.
"mg\\sin\\theta" is working downwards, which will try to rotate this cylinder due to which it's speed will start to increase that is called the rotational acceleration.
"I\\alpha= mg \\sin \\theta R"
"\\Rightarrow \\alpha =\\frac{R mg \\sin \\theta }{I}"
if the block which is on the inclined plane is not a cylindrical, then the frictional force is working which is greater than the frictional force on the cylinder. and the downward acceleration of the block will be "mg \\sin \\theta-f_s" that will lead decrease the acceleration of the block with respect to rotational acceleration.
Comments
But sir the question is about the constraints on the mechanical system. You haven't explained it.
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