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RPMs and Static friction

  1. Jun 22, 2008 #1
    My problem is: A coin is placed on a turntable that is rotating at 33.3 rpm. If the coefficient of static friction is .1, how far from the center of the turntable can the coin be placed without having to slip off?



    Am I correct with V= the 33.3 rpms? and that w (angular speed) =1.11 pi rad/s?
    not sure where to go from there
     
  2. jcsd
  3. Jun 22, 2008 #2

    Doc Al

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    33.3 rpm is not a linear velocity, but is just another measure of angular speed. But, yes, you should convert to standard units of rad/s.

    Apply Newton's 2nd law to the coin.
     
  4. Jun 22, 2008 #3
    N = ma
    N = m * 1.11 pi rad/s ?
     
  5. Jun 22, 2008 #4

    Doc Al

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    (1) What's N?
    (2) What forces act on the coin?
    (3) What's the acceleration of the coin?
     
  6. Jun 22, 2008 #5
    N is the pull inward?
    but there is friction holding it on the turntable.
    acceleration is the 1.11 pi rad/s?

    (I'm really struggling with this and appreciate your help)
     
  7. Jun 22, 2008 #6

    tiny-tim

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    Welcome to PF!

    Hi Djbari! Welcome to PF! :smile:

    (have an omega: ω and a pi: π :smile:)

    Yes, you are correct: ω = 1.11π rad/s.

    Now, do you know the formula for centripetal acceleration (acceleration in a circle)? :smile:
     
  8. Jun 22, 2008 #7

    Doc Al

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    Hint: There is only one horizontal force acting on the coin.
    That's the angular speed, not the acceleration. Hint: The coin moves in a circle.
     
  9. Jun 22, 2008 #8
    why is ω - 1.11 π rad/s? thought it is accel?

    Do you mean the formula a=v^2/r ?
     
  10. Jun 22, 2008 #9
    N would be the weight of the coin or m * a?

    accel is v^2/r?
     
  11. Jun 22, 2008 #10

    Doc Al

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    Linear acceleration has units of m/s^2.
    Sure! (You can also use another version of that formula expressed in terms of angular speed.)
     
  12. Jun 22, 2008 #11
    ω = v^2/r ?
     
  13. Jun 22, 2008 #12

    Redbelly98

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    No. If a=v^2/r, then omega must be something else (since omega is not a)

    Do you remember the formula for v in terms of r and omega?
     
  14. Jun 22, 2008 #13
    v = r(w)
     
  15. Jun 22, 2008 #14

    Redbelly98

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    Yes, if by r(w) you mean r times w.

    If you substitute that expression for v in
    a = v^2 / r,
    that will help in solving the problem.
     
  16. Jun 22, 2008 #15
    do I know a or v?
     
  17. Jun 22, 2008 #16
    not v. do I know a?
     
  18. Jun 22, 2008 #17

    Redbelly98

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    No, but that's okay.

    You have one expression for a in terms of r and w.

    Can you get another expression for a, using what you know about friction?
     
  19. Jun 22, 2008 #18

    tiny-tim

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    Hi Djbari! :smile:

    I'm confused. :confused:

    In your first post, you got it right … you said ω is angular speed:
    But later you said acceleration: :confused:
    Anyway … you got two formulas right:
    v = ωr

    and a = v²/r​

    If you combine them, you get a = (ωr)²/r = … ? :wink:

    Then use F = ma. :smile:
     
  20. Jun 22, 2008 #19
    I'm not seeing it. I only see a in terms that have t (time) and I don't have t. or in terms with F net or N and I don't have those do I?
     
  21. Jun 22, 2008 #20
    Hi Tiny -Tim,

    I didn't see the previous note matching the two equations. I have actually gotten to there but thought it was wrong - when maybe it's just I thought it stopped there. Am I right with
    (w*r)^2/r = 3.45......is it rads? How do I put that into F=ma?
     
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