Homework Help: Friction on a rotating record

1. Nov 16, 2008

pucr

1. The problem statement, all variables and given/known data
A small coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.1, how far from the center of the record can the coin be placed without having it slipped off?

2. Relevant equations
w = 33.3 rpm = 1.11 rad/sec
Force due to friction = Centripetal force

3. The attempt at a solution
I tried this...
Force due to friction = .1mg = .98m
Centripetal force = ma = m((r*w^2)/r) = mrw^2 = 1.2321mr
soo...
.98m = 1.2321mr
=> .98 = 1.2321r
=> r = .795 m

Unfortunately, the answer should be .08 meters.
I thought at first that I messed up a decimal place somewhere but I can't see where. :-\
I may be going insane...
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Nov 16, 2008

LowlyPion

Welcome to PF.

What acceleration will the coin tolerate without slipping?

If the coefficient of friction is u and the mass is m then

u*m*g = m*v2/r

v = w*r =2*π*f*r

u*g/w2 = r = u*g/(2*π*f)2

3. Nov 16, 2008

Staff: Mentor

Fix that conversion.

4. Nov 16, 2008

ooh thanks.