A uniform disc of mass m and radius R is rolling down a rough inclined plane which makes an angle 30 degree with…..

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

rotation

Solving (i) (ii) and (iii)

a =  g sinθ/(1+ I/mR²)

f = mgsinθ/(1+mR²/I)

= mgsin30°/(1+2)

= mg/6 Upward

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