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Friction on a rotating record

  1. Nov 16, 2008 #1
    1. The problem statement, all variables and given/known data
    A small coin is placed on a record that is rotating at 33.3 rpm. If the coefficient of static friction between the coin and the record is 0.1, how far from the center of the record can the coin be placed without having it slipped off?


    2. Relevant equations
    w = 33.3 rpm = 1.11 rad/sec
    Force due to friction = Centripetal force


    3. The attempt at a solution
    I tried this...
    Force due to friction = .1mg = .98m
    Centripetal force = ma = m((r*w^2)/r) = mrw^2 = 1.2321mr
    soo...
    .98m = 1.2321mr
    => .98 = 1.2321r
    => r = .795 m

    Unfortunately, the answer should be .08 meters.
    I thought at first that I messed up a decimal place somewhere but I can't see where. :-\
    I may be going insane...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 16, 2008 #2

    LowlyPion

    User Avatar
    Homework Helper

    Welcome to PF.

    What acceleration will the coin tolerate without slipping?

    If the coefficient of friction is u and the mass is m then

    u*m*g = m*v2/r

    v = w*r =2*π*f*r

    u*g/w2 = r = u*g/(2*π*f)2
     
  4. Nov 16, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    Fix that conversion.
     
  5. Nov 16, 2008 #4
    ooh thanks.
    It should be 3.487 rad/sec making the answer .0806 meters
     
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