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Friction constant

  1. Aug 10, 2008 #1
    Hey, i'm having trouble finding the friction constant in the following question:

    Lisa is sitting on a carousel 0.55m from the centre. The carousel rotates at 15laps/min and Lisa weighs 33kg.

    a) what is Lisa's velocity?
    b) Lisa moves away from her spot so that she is rotating 1.2m from the centre. What is the lowest friction constant that will keep her in place?

    Now I calculated the velocity by setting v=s/t=2πr/T, where T = 60s/15laps=4
    so v=0.864 m/s (I'd appreciate if someone could tell me if this is right..)
    I'm trying to pick between 2 methods for finding the friction constant:

    1) calculating the friction constant could be done by v=(μrg)1/2
    which gives μ=0.05.

    2)But then Fμ=μmg, and since Fμ is also = ma, I could calculate the centripetal acceleration by having a=v2/r=0.62 m/s2
    And then putting it into F=ma, giving Fμ=20.5N.
    Putting the force back into Fμ=μmg, giving μ=0.063.

    So which answer is right? I would really appreciate if someone could help me out!
     
  2. jcsd
  3. Aug 11, 2008 #2

    Doc Al

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    Staff: Mentor

    Looks OK. (For part a, not for part b.)
    Those two methods seem identical to me: They both derive from setting static friction equal to centripetal force.

    But neither of your answers are correct. (How did you calculation the speed?)
     
  4. Aug 12, 2008 #3
    Thanks for helping out!
    Is the assumption wrong? So then I should calculate the normal&gravitational force and then split the centripetal force between them? How would you do that?
     
  5. Aug 12, 2008 #4

    Doc Al

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    What assumption? That static friction must equal the centripetal force? No, that's perfectly correct. It's just an application of Newton's 2nd law.

    If you follow through on your method 2 approach symbolically (not by plugging in numbers), you'll get the same equation as in your method 1.
     
  6. Aug 12, 2008 #5
    Oh, I used the wrong velocity! So now I did it with Lisa sitting 1.2m from the centre, v=s/t=2πr/T giving v = 1.88m/s. Plugging into v^2/rg= μ=0.3..?
     
  7. Aug 12, 2008 #6

    Doc Al

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    Staff: Mentor

    Good!
     
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