# Rotational Motion of a penny

## Homework Statement

A penny is placed on an LP record that is slowly accelerating up to 78 revolutions per minute. It is found that if the penny is placed at 0.080m or greater from the center, then the penny slides off the edge of the record. Find the coefficient of static friction if the mass of the penny is 0.0032kg.

Ff = mu x n

a = v2/r

## The Attempt at a Solution

I don't know how to start. Can someone guide me please??

Ok. Let us begin with this question. You are given the frequency of the LP (and thus the period as well), and you are given the mass of the penny. As you have written above,
Frictional Force = co-eff * Normal and centripetal acceleration = v^2/r = 4Pi^2r*frequency^2. For the penny to stay on the LP, friction has to provide enough centripetal force. When the radius is too great, the velocity of the penny is too large and it flys off the LP. Thus, we are looking for:
Force Friction = Force centripetal = mass of penny* centripetal acceleration.
The rest is plain math and some unit conversions.

Ok. Let us begin with this question. You are given the frequency of the LP (and thus the period as well), and you are given the mass of the penny. As you have written above,
Frictional Force = co-eff * Normal and centripetal acceleration = v^2/r = 4Pi^2r*frequency^2. For the penny to stay on the LP, friction has to provide enough centripetal force. When the radius is too great, the velocity of the penny is too large and it flys off the LP. Thus, we are looking for:
Force Friction = Force centripetal = mass of penny* centripetal acceleration.
The rest is plain math and some unit conversions.

thanks.

Normal is just equal to mg, right?

yes. So youll notice that the mass does not actually matter in the question since it cancels out.