Coin on Turntable

[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?

[hage567]

Can you show more details of your calculations? We can't see what you've done wrong if you don't show your work.

[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.

[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.

[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!!

[hage567]

You're welcome. :smile: