# Turntable Problem

1. Nov 16, 2004

### mattx118

A coin is placed 10.0 cm from the axis of a rotating turntable of variable speed. When the speed of the turntable is slowly increased, the coin remains fixed on the turntable until a rate of 36 rpm is reached, at which point the coin slides off. What is the coefficient of static friction between the coin and the turntable?

I know that MueK, is going to be Fs / Fn, but I'm having trouble getting those, if anyone can give me a clue in to start this.

2. Nov 16, 2004

### UrbanXrisis

you need to find the centripetal force $$Fc=(4 pi^2 r m)/T^2$$
Then use Fc=μmg to solve for μ

3. Nov 16, 2004

### mattx118

i came up with the formula, Mue = v^2 / (g*r), i am getting .15 as the answer, however it is saying it is wrong.

4. Nov 16, 2004

### NateTG

Hmm, I got something slightly different. Perhaps you should check your rounding, or use a more precise value for $$g$$?

5. Nov 16, 2004

### UrbanXrisis

r=0.1m
g=9.8m/s^2
v=0.377m/s

6. Nov 16, 2004

### mattx118

that answer is also wrong, i have no idea what is wrong i'm almost sure i have done it correctly

7. Nov 16, 2004

### UrbanXrisis

ack, sorry, I got 0.145 just as you did.

8. Nov 16, 2004

### mattx118

Yea, I dont know why its saying its wrong :X

9. Nov 16, 2004

### UrbanXrisis

10. Nov 16, 2004

### mattx118

I have one more submission left

11. Nov 16, 2004

### mattx118

Well then haha, it was .145 I got it right, maybe it was looking for a special amount of sig figs. I was rounding it off.