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

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ma = mg sinθ – f . . . (i)

Rotation , Torque f.r = Iα

for rolling , α = a/r

$\displaystyle f = \frac{I a}{r^2}$ . . . (ii)

for disc, $I = \frac{m r^2}{2}$

Solving (i) & (ii)

$\displaystyle a = \frac{g sin\theta}{1+\frac{I}{m r^2}}$

f = mg sinθ – m a

$\displaystyle f = m g sin\theta – \frac{m g sin\theta}{1+\frac{I}{m r^2}}$

$\displaystyle f = \frac{m g sin\theta }{3} $

$\displaystyle f = \frac{m g sin30 }{3} = \frac{m g}{6} $