Question

A coin is placed 14 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 30 rpm is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

Answer 1

Fc = centrifugal force = mV^2/r ---- 1
where,
m = mass of the coin
V = velocity of the the turntable
r = turning radius = 14 cm.

w = rotational speed of turntable = (30 rev/min)*(2pi rad/rev)*(1 min/60 sec)

w = pi = 3.14 rad/sec.

Therefore,

V = pi * 0.14 = 0.44 m/sec.

Substituting this in Equation 1,

Fc = m(0.44)^2/(0.14)

Fc = 1.383(m) N

Frictional force = (mu)(m)(g)

where

mu = coeff of friction
m = mass of coin
g = acceleration due to gravity = 9.8 m/sec^2

Substituting values,

1.383(m) = (mu)(m)(9.8)

Since "m" appears on both sides of the equation, it will cancel out and solving for "mu",

mu = 1.383/9.8

mu = 0.14