(adsbygoogle = window.adsbygoogle || []).push({}); 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

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# Homework Help: Friction on a rotating record

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