Merlion Audio

All The Best Sound

Speed Turntable

A coin is placed 11.5 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 38 rpm is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable? HELP.

Centrifugal force will make the coin slide off the turntable. And for the coin to slide off, the centrifugal force must be able overcome the force of friction trying to hold off the coin from slipping.

Fc = centrifugal force = mV^2/r --- call this Equation 1

where

m = mass of the coin
V = velocity of the the turntable
r = turning radius = 11.5 cm.

w = rotational speed of turntable = (38 rev/min)*(2pi rad/rev)*(1 min/60 sec)

w = 3.98 rad/sec.

Therefore,

V = 3.98 * 0.115 = 0.458 m/sec.

Substituting this in Equation 1,

Fc = m(0.458)^2/(0.115)

Fc = 1.824(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,

1.824(m) = (mu)(m)(9.8)

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

mu = 1.824/9.8

mu = 0.186

Hope this helps.

« Previous12345...1516Next »

Technosonic TCD-98WE USB Turntable 3 Speed Record Player