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Centripetal Motion: Tetherball

  1. Jan 10, 2010 #1
    1. The problem statement, all variables and given/known data

    Tetherball is seen on playgrounds across the globe. It consists of a vertical pole with a ball attached to the top with a rope. The ball has a mass of 1kg and is sent around the pole in a horizontal circular path. If the rope has a length of 2.25 meters and makes an angle of 20.9 degrees with the pole, (not theta).

    Radius of path = .802m i got that part.
    The velocity of the tetherball =
    The period of the tetherball's motion=
    The tension in the rope = I got that also.


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jan 10, 2010 #2

    rl.bhat

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    What are the forces acting on the ball?
    Which force keeps the ball in the circular motion?
     
  4. Jan 10, 2010 #3
    I think i figured it out. I used the square root of radius * 9.8 to find the minimum velocity it can stay in a circular path.
     
  5. Jan 10, 2010 #4

    rl.bhat

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    During the circular motion of the ball, three forces act on the ball.
    If T is the tension in the rope, T*cosθ balances the weight of the ball and T*sinθ provides the centripetal force which keeps the ball in the circular motion.
     
  6. Jan 10, 2010 #5
    thats correct! So if the velocity is not given, nor the time, using the square root of .802 *9.8 to give me 2.8m/s correct?
     
  7. Jan 10, 2010 #6

    rl.bhat

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    No.
    T*sinθ/Τ*cosθ = (m*v^2/R)/mg.
    Simplify and substitute the values to find v.
     
  8. Jan 10, 2010 #7
    ok doing it that way i got 4.53 m/s is that correct?
    Thanks a lot btw
     
  9. Jan 10, 2010 #8

    rl.bhat

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    Check the calculation again.
    After simplification you get
    tanθ = v^2/Rg.
    Now solve for v.
     
  10. Jan 10, 2010 #9
    tan (69.1) = v^2 / .802*9.8
    2.618 = v^2 / 7.8596
    2.057 = v^2
    4.5361 = v

    Are you sure?
     
  11. Jan 10, 2010 #10

    rl.bhat

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    Here θ is the angle made by the rope to the pole. According to your problem that is equal to 20.9 degrees.
     
  12. Jan 10, 2010 #11
    Oh ok, in the past I have been taught to use the theta to the x axis.
     
  13. Jan 10, 2010 #12
    okay
     
    Last edited: Jan 10, 2010
  14. Jan 10, 2010 #13
    Would the tension force be: 18.92 N

    FT = mv^2/r +mgsin69.1


    and are you positive tanθ = v^2/Rg. is correct?

    That would give me 1.732m/s
     
    Last edited: Jan 10, 2010
  15. Jan 10, 2010 #14

    rl.bhat

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    Your answer is correct.
     
  16. Jan 10, 2010 #15
    Both the FN and V?
     
  17. Jan 10, 2010 #16

    rl.bhat

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    Your calculation of v is correct. In the problem you have to find the period.
    t = 2*π*R/v.
     
  18. Jan 11, 2010 #17
    Yes I have figured that one out, because it is in a circular orbit circumference can be used.

    t = 2.91s
     
  19. Jan 11, 2010 #18
    You said the calculation of v is correct, is the calculation of FN correct as well?!?
     
  20. Jan 11, 2010 #19

    rl.bhat

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    What is FN?
     
  21. Jan 11, 2010 #20
    Never mind I figured out everything is right! Thanks for all your help
     
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