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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

  • physics -

    centrifugal force, Fc
    = mrω², or
    = mv²/r
    m=mass, kg
    ω = rotational velocity in radians/s.
    v=tangential speed m/s
    normal force, Fn
    Coefficient of friction, μ

    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.

  • physics -

    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

  • physics -

    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

  • physics -

    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.

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