Question

20)
A coin just remains on a disc rotating at 120
r.p.m.
when kept at the distance of 1.5 cm from the axis
of rotation. Find the coefficient of friction
between the coin and the disc.
.PL​

Answer 1

Answer:

Given : r=1.5cm=

100

1.5

m

Frequency of rotation ν=120 r.p.m=

60

120

rps=2 rps

Angular velocity of rotation of disc ω= 2πν = 2π×2 =4π rad/sec

Equate the Centripetal force and force of friction, we get

mω

2

r=μmg

Or ω

2

r=μg

Putting the values, we get

100

(4π)

2

×1.5

=μ×9.8

⟹μ=0.2414

Hence Option (B) is correct .