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

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

Author: Rajesh Jha

QuantumStudy.com for CBSE|IIT-JEE|NEET

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