Solve Turntable Problem: Find Coefficient of Friction

[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 into start this.

 

[UrbanXrisis]

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

 

[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.

 

[NateTG]

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.

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

 

[UrbanXrisis]

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

so it's about .116

 

[mattx118]

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

 

[UrbanXrisis]

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

 

[mattx118]

Yea, I don't know why its saying its wrong :X

 

[UrbanXrisis]

What is the actual answer?

 

[mattx118]

I have one more submission left

 

[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.