physics
posted by physics on .
A coin is placced 12.0 cm from the axis of a roatating turntable of variable speed. When the speed of the turntable is slowly incraeased, the coin remains fixed on the turntable until a rate of 50 mph is reached at which point the coin slides off. What is the coefficent of static friction between the turntable and the coin?
I got this for my equation but it game the wrong answer
Mu s = (rg)^1 v^2
could you please tell me how to do this problem

centrifugal force, Fc
= mrω², or
= mv²/r
where
m=mass, kg
ω = rotational velocity in radians/s.
v=tangential speed m/s
normal force, Fn
=mg
Coefficient of friction, μ
=Fc/Fn
Note: I have a doubt about the value 50 mph for the tangential velocity. It is not a normal unit for a turntable. Please check. 
Net Force Radial = m a radial = F radial = Ffr
ma radial = Us Fn
ma radial = Us m g
cancel out mass
a radial = Us g
a radial = r^1 v^2
r^1 v^2 = Us g
Us = (rg)^1 V^2
don\'t see what I did wrong 
50 mph is in the question in the book

Us = (rg)^1 V^2
Us = (.12 m (9.80 kg^1 N))^1 (22.35 s^1 m)^2
Us = 420
back of book says .34 
If you work back from μ=0.34, r=0.12 and g=9.8, you will get v=0.632 m/s, or when divided by r, ω=5.27 radians/s, or 1.42 mph for v.
I believe there is something wrong with the question.