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.

Options

A. `(mg)/2`
B. `(mg)/4`
C. `(Mg)/6`
D. `Mg`

Answer

C

Explanation

under the given condition only posibility is that friction is upward and it accelerated downwards as shown below:
the equation of motion are :
`a=(mg sin theta-1)/m=(mg sin 30^(@)-f)/m=g/2-1/m....(1)`
`alpha=(tau)/I=(fR)/I=(fR)/((mR^(2))/2)=(2f)/(mR)......(2)`
For rolling (no spelling) ltbrltgt `a=R alpha` or `g//2-f//m=2f//m`
`:. (3f)/m=g//2` or `f=mg//6`