A solid cylinder is rolling down a rough inclined plane of inclination `theta`. Then

Options

A. the friction force is dissipative
B. the friction force is necessarily changing
C. the friction force will aid rotation but hinder translation
D. the friction force is reduced if `theta` is reduced

Answer

C, D

Explanation

As shown in the figure, the component of weight `Mgsintheta` tends to slide the poitn of contact (of the clinder with inclined plane) along its direction. The slicding friction acts in the opposite direction to this relative motion. Because of frictional force the cylinder rolls. Thus frictional force aids rotation but hinders translational motion.

Appling `F_("net")=ma` along the directon of inclined plane, we get
`Mgsintheta-f=Ma_(c)`
Where `a_(c)=` acceleration of centre of mass of the cylinder
`:. f=Mgsintheta-Ma_(c)`........i
But `a_(c)=(gsintheta)/(1+(I_(c))/(MR^(2)))=(gsintheta)/(1+(M(R^(2))/2)/(MR^(2)))`
`=2/3gsintheta`.........ii
From eqn i and ii
`f=(Mgsintheta)/3`
If `theta` is reduced, then friction force is reduced.