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Centripetal acceleration and static friction

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data


    A 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 0.2, how far from the center of the record can the coin be placed without having it slip off?

    2. Relevant equations

    mv^2/r = mg(mu)

    3. The attempt at a solution

    I cant figure out what i'm doing wrong. I converted the 33.3 rpm to radians /sec and pluged that in for v then solved for R.

    I keep getting 6.204254889
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 12, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi shaanyr! Welcome to PF! :smile:

    (have a mu: µ and an omega: ω :wink:)
    Nooo … that's ω, not v isn't it? :cry:

    Use ω2r. :smile:
     
  4. Feb 12, 2009 #3
    im sorry i dont understand

    use

    [tex]\omega^2/r[/tex] = g x [tex]\mu[/tex]
     
  5. Feb 12, 2009 #4

    Delphi51

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    High school physics to the rescue!
    Make a proportion to find the period:
    33 turns/60 s = T turns/1 second
    Once you have the period you can use v = circumference/period.

    This is an interesting problem because r is unknown, yet needed to find v. In this kind of problem, you have to work with the letters and put in numbers only at the last stage when you have an r = formula.

    Show your equations if you would like more help or a check!
     
  6. Feb 13, 2009 #5

    tiny-tim

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    Just got up :zzz: …
    No …

    look at the dimensions (or the units) …

    v = ωr,

    so centripetal acceleration = v2/r = ω2r. :smile:
     
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