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?

Question

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

Answer 1

centrifugal force, Fc
= mrω&sup2, 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.

Answer 2

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

Answer 3

50 mph is in the question in the book

Answer 4

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

Answer 5

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.