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Frictional coin sliding on turntable

  1. Oct 20, 2007 #1
    A 5.0 g coin is placed 22 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of µs = 0.80 and µk = 0.50. What is the maximum rpm that the turntable could speed up to without the coin sliding off?



    m = .005 kg
    r = .22 m
    µs = 0.8

    Equations found..
    v = angular velocity * r
    Force(net) = m(v)^2 / r

    Inertia > µs N when coin slips (?)

    I believe this gets set equal to mg (Normal Force) but I haven't been able to generate the correct answer multiplying µs as a coefficient of either side.

    I'm not sure what isn't being accounted for, what do I do?
     
  2. jcsd
  3. Oct 20, 2007 #2

    learningphysics

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    Looks like you have the right idea. slipping occurs when force > us*N.

    can you show your calculations for angular velocity?
     
  4. Oct 20, 2007 #3
    m(ω * r)^2 / r

    simplifies to => m(r)(ω)^2

    so,
    mg * µs = m(r)(ω)^2 (?)
     
  5. Oct 20, 2007 #4

    learningphysics

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    yes, solve for w. what do you get? you have to convert to rpm afterwards also.
     
  6. Oct 20, 2007 #5
    (.005)(9.8)(.8) = (.005)(.22)(ω)^2
    => (9.8)(.8) = (.22)(ω)^2
    => 7.84 / .22 = ω^2
    => ω = (7.84 / .22)^(1/2) = 5.969

    then i'm not sure how to convert this to rpms..
    but, 5.969 / r = 27.134, where r = .22, which isn't right
     
    Last edited: Oct 20, 2007
  7. Oct 20, 2007 #6

    learningphysics

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    first convert to revolutions/second. 1 revolution = 2*pi radians.

    then convert to revolutions/minute by multiplying by 60.
     
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