Q: A uniform disc of mass m and radius R is rolling down a rough inclined plane which makes an angle 30º with the horizontal. If the coefficients of static and kinetic friction are each equal to μ and the only forces acting are gravitational and frictional, then the magnitude of the frictional force acting on the disc is

(A) (mg/3) upward

(B) (mg/3) downward

(C) (mg/6) upward

(D) (mg/6) downward

Solution :Laws of motion , Translational :

ma = mg sinθ – f . . . (i)

Rotation ,Torque f.r = Iα . . . (ii)

for rolling α = a/r . . . (iii) for disc, I = mR²/2

Solving (i) (ii) and (iii)

a = g sinθ/(1+ I/mR²)

f = mgsinθ/(1+mR²/I)

= mgsin30°/(1+2)

= mg/6 Upward