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

Since disc does not slip, friction is static and static friction can have any value between 0 and `mu N`. Component of mg parallel to the plane is mg sin `theta` which is opposite to the direction of motion of the centre of the disc, and hence speed of the centre of mass decreases. Therefore `omega` must decrease. Only friction can provide a toque about the centre.
Torque due to friction must be opposite to the `vec(omega)`. Therefore, frictional force will act up the plane.
Now, for translational motion
`mg sin theta - f = ma_(cm)" "...(i)`
For rotational motion
fR = I `alpha`, where I = M.I. of the disc about centre.
`= I (a_(cm))/(R),"as a"_(cm) = alpha R`
`rArr" "a_(cm) = (fR^(2))/(I)" "...(ii)`
From (i) and (ii) we get,
`f = (mg sin theta)/(1+(mR^(2))/(I))`
Putting the value of `theta` and I, we get
f = mg/6