# Coin on Turntable

1. Feb 27, 2009

### julz3216

1. The problem statement, all variables and given/known data
A coin is placed on a turntable that is rotating at 45.0 rpm. If the coefficient of static friction between the coin and the turntable is 0.186, how far from the center of the turntable can the coin be placed without having it slip off?

2. Relevant equations

mv^2/r <= usmg
us= coefficient of static friction

3. The attempt at a solution
I tried using the equation and I converted 45 rpms to rad/sec and I got r=12.17 but this is wrong so I don't know what to do. Maybe im using the wrong equation?

2. Feb 27, 2009

### hage567

3. Feb 27, 2009

### julz3216

mv^2/r =usmg
v^2/r = usg
(4.71^2)/r <= .186g
22.184 = 1.8228r
r= 12.17

This seems logical to me but it says it is wrong.

4. Feb 27, 2009

### hage567

Note: the 4.71 rad/s, is angular velocity. You need to convert that to m/s, which is linear velocity. Do you know the relationship between these two? It should be in your textbook.

5. Feb 27, 2009

### julz3216

I think v = omega*r

Where v is the velocity, omega is angular speed (rad/s), and r is the radius.
so v= 4.71r
then (4.71r)^2/r= 22.18r = 1.8228
so r= .082

which is right! thank you!!

6. Feb 27, 2009

### hage567

You're welcome.