Question

A uniform disc of mass M and radius R rolls without slipping down a plane inclined at an angle θ with the horizontal. 

If the disc is replaced by a ring of the same mass M and the same radius R, the ratio of the frictional force on the ring to that on the disc will be 

(A) 3/2 

(B) 2 

(C) √2 

(D) 1

Answer 1

Correct option: (A) 3/2

Explanation:

Τ = Iα

f ∙ R = Iα

f =  [(Iα) / R] = [(Ia) / R²]

for ring, I = MR² hence

f = [(MR^2­a) / R²] = Ma         (4)

from (1) Mg sinθ = f + Ma

= ma + ma

Mg sinθ = 2ma

a = [(g sinθ) / 2]

from (4) f = Ma = [(Mg sinθ) / 2]

from (3) & (4)

[(f_ring) / (f_disc)] = [{(Mg sinθ) / 2} / {(Mg sinθ) / 3}] = (3/2)