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Homework Help: Circular Motion Problem

  1. Jan 22, 2008 #1
    [SOLVED] Circular Motion Problem

    1. The problem statement, all variables and given/known data

    A coin is placed 0.11 meters from the center of a rotating turntable. The speed of the turntable is slowly increased; the coin remains fixed on the turntable until 36 rpm is reached and the coin slides off. What is the coefficient of the static friction between the coin and the turntable?

    2. Relevant equations

    T= (2*pi*r)/v
    T= tau, which I believe is revolutions per second

    (Fnormal)(coefficient of static friction) = (Ffriction)

    Ac = v^2/r
    Ac = centripetal acceleration

    3. The attempt at a solution

    First I divided the rpm by 60 to get the revolutions per second, or 0.6 rps. I put that as T in the tau equation and solved for v: 1.15 m/s. Then, because I was unsure of where to go next because of a lack of a mass value, I did something sort of stupid where I solved for Ffriction using the wrong equation or something (I'm reviewing an old test, not sure what my rationale was):

    F = (coefficient)(Fnormal)
    except for Fnormal part I did mass times centripetal acceleration
    F = (coefficient)(m)(v^2/r)
    and got a coefficient of 0.8123, which was wrong. At least I'm getting close; I believe coefficients of static friction must be less than 1.

    Can anyone please give me a hand?
  2. jcsd
  3. Jan 22, 2008 #2
    Your answer should end up independent of mass, as friction problems usually are.

    T is the period, which is seconds per revolution, not revolutions per second (that's frequency).

    You're on the right track using the centripetal acceleration though. However, you don't simply plug that as the normal force. The normal force is simply mg.

    Try writing out newton's 2nd law for the coin.
    Last edited: Jan 22, 2008
  4. Jan 22, 2008 #3
    OK so this is what I tried:

    Ffriction = Fc
    (coefficient)(Fnormal mg) = (mass)(v^2/r)
    I cancel out the masses and solve for the coefficient and I get 1.226, which I believe is impossible because it's greater than one. I follow your logic, but maybe I did something wrong in the math...?
  5. Jan 22, 2008 #4
    Think about how you determined v.

    [tex]v = \frac{2 \pi r} {T}[/tex]

    You plugged in T = 0.6 revolutions/second, which makes no sense in this equation. Check the units on T.
  6. Jan 22, 2008 #5
    Wait, now I'm really confused. How do I convert 36 revolutions per minute to "X" seconds per revolution?
  7. Jan 22, 2008 #6
    Simple, it's just T = 1/f, where f = 0.6 rev/s. You'll get units of [s/rev] or just .
  8. Jan 22, 2008 #7
    Ah! Thanks! I took that answer, plugged it into all my other equations and solved for the coefficient of static friction, getting 0.15976. It seems right, but I have no way of checking. I appreciate the help!
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