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Mass on a turntable - Force

  1. May 8, 2010 #1
    1. The problem statement, all variables and given/known data

    A 50 gram coin is on a horz. turntable. it rides without slipping making 1 revolution every 1.8 seconds. it is 10 cm from the center. the coefficient of static are 1 and the coefficient of kinetic is 0.5.

    the size of the force of friction on the coin is

    2. Relevant equations

    F = ma

    mu_s = F/N

    3. The attempt at a solution

    I dont think i have to use either static or kinetic coefficients for this problem but i may be wrong.

    anyways i did

    F = ma

    F = m(v2/R)

    v = 1rev/1.8s (2pi rad / 1rev)
    v = 3.49 rad/s

    F = .05kg(3.49^2)/(.1m)

    F = 6.09 N

    The answer is 60.9 mN so it just seems like i am off a few decimal places but i cant seem to find out where i went wrong or if this is even the right way to approach the problem.

    Thanks for any help :)
     
  2. jcsd
  3. May 8, 2010 #2

    Doc Al

    User Avatar

    Staff: Mentor

    That 'v' should be the tangential velocity--you found the angular velocity, ω.

    How would you find the tangential velocity of the coin?
     
  4. May 8, 2010 #3
    Thank you :)

    tangential velocity is v = 2piR / T

    v = .349

    F = .05kg (.349^2 / .1)

    F = .0609 N or 60.9 mN
     
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