A uniform disc of mass `m` and radius `R` is rolling up a rough inclined plane which makes an angle of `30^@` with the horizontal. If the coefficients of static and kinetic friction are each equal to `mu` and the only force acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is and its direction is .(write up or down) the inclined plane.

Explanation

`(mg//6," up")`
Given that the body is rolling up the inclined plane. Therefore, the velocity of the centre of mass in up the inclined plane. The component of weight ( ) is trying to move the point of contact downwards.
Therefore, frictional force will act upwards. From force diagram.

`mg sin theta -f =ma_( c)`........(i)
For rotational motion about the centre of mass of disc O,
`f xx R = Ialpha rArr` But `a_(C) = alphaR`
`rArr a_(0)=(fR^(2))/I = (fR^(2))/(1/2mR^(2)) = (2f)/m`......(ii)
From Eqs. (i) and (ii), `rArr 3f =mg sin theta = mg sin 30^(@) rArr f=(mg)/6`.