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Turntable Problem

  1. Nov 16, 2004 #1
    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 in to start this.
     
  2. jcsd
  3. Nov 16, 2004 #2
    you need to find the centripetal force [tex] Fc=(4 pi^2 r m)/T^2 [/tex]
    Then use Fc=μmg to solve for μ
     
  4. Nov 16, 2004 #3
    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.
     
  5. Nov 16, 2004 #4

    NateTG

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    Hmm, I got something slightly different. Perhaps you should check your rounding, or use a more precise value for [tex]g[/tex]?
     
  6. Nov 16, 2004 #5
    r=0.1m
    g=9.8m/s^2
    v=0.377m/s

    so it's about .116
     
  7. Nov 16, 2004 #6
    that answer is also wrong, i have no idea what is wrong i'm almost sure i have done it correctly
     
  8. Nov 16, 2004 #7
    ack, sorry, I got 0.145 just as you did.
     
  9. Nov 16, 2004 #8
    Yea, I dont know why its saying its wrong :X
     
  10. Nov 16, 2004 #9
    What is the actual answer?
     
  11. Nov 16, 2004 #10
    I have one more submission left
     
  12. Nov 16, 2004 #11
    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.
     
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