Question

A coin is sitting on a record as sketched in the figure below. It is found that the coin slips off the record when the rotation rate is 0.42 rev/s. What is the coefficient of static friction between the coin and the record?

the cd is 15 cm or .15 m

Answer 1

The coin slips when the necessary centripetal force to keep the object moving in a circle can no longer be applied. In this case, the centripetal force is caused solely by the force of friction of the record on the coin.

In other words, the coin will slip when centripetal force is greater than frictional force.

We know the following
centripetal force is given by Fc = m*w^2*r, where m is the mass, w is the angular velocity, and r is raidus
Friction force is given by Ff = mu*n where n is the normal force. Because the record is flat, n = m*g

We then set these equal to each other
m*w^2*r = m*mu*g we can cancel out the mass
w^2*r = mu*g divide by g on both sides
mu = (w^2 * r)/g

It is given that r = 15cm, which is the same as r = .15m
We also need to convert rotation rate to radians/sec. w = 2*pi*rotation_rate

w = 2.63 rad/s
r = .15m
g = 9.8 m/s^2

Plugging all these in to mu = (w^2 * r)/g
we get mu = ((2.63rad/s)^2 * .15m) / (9.81m/s^2)
mu = .106