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Tetherball physics problem

  1. Jun 24, 2004 #1
    i'm driving myself insane here...ok, you have a ball wrapping around a pole (was initialy spinning with a horizontal velocity). Ok, Can someone tell me what is wrong with the formula dWork = -(mv^2/lsinX)*dxsinX + mgdlsinX where X is the angle the rope makes with the pole, v is current velocity and l is length of string? I am making the assumption that the pole is small so the velocity is assumed to be horizontal.

    I don't want a solution to the problem, just a reason why that formula is wrong.

  2. jcsd
  3. Jun 24, 2004 #2


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    I think there's an "l" missing in the first term : should be -(mv^2/lsinX)*lsinX*dX
  4. Jun 24, 2004 #3


    should read:
    dW = -(mv^2/lsinX)*sinXdl + mgdlcosX

  5. Jun 25, 2004 #4
    I'm confused as to the question. Can you state the problem?
  6. Jun 25, 2004 #5
    Elucidation of the problem

    Ok, you have a tetherball apparatus. Initially (before the rope wraps around the pole), the ball is moving with a horizontal velocity of Vo m/s (causing an angle Xi with the pole). Now, as the ball begins to wrap around, the height of the ball will increase (since it is accelerating upwards). I need to find what the final height of the ball is versus the initial height when the ball finishes wrapping around the pole. So far I have y = Integral(cosxdl) and tanx^2=v^2/(lgsinX) by F=ma. I tried dW = -(mv^2/lsinX)*sinXdl - mgcosXdl
    (ie. dW = -Tx*dr - Ty*dy). But this formula isn't producing the right results. I then set dW=d(1/2*mv^2) = mvdv and so I have vdv=-(v^2/l)*dl - gcosXdl.
    This however doesn't seem to work.
  7. Jun 25, 2004 #6
    how will the height of the ball increase, wouldent it decrease??
  8. Jun 26, 2004 #7


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    When you say horizontal velocity, I assume you are referring to the ball's tangential velocity relative to the pole. And maybe i'm off base here (or your description of the problem is), but I'm pretty sure that there is no vertical acceleration of the ball since all its acceleration takes place via the tension force in the centripital direction, and the only vertical force acting on the ball countering the force of gravity is the vertical component of of the tension force, making the total force acting vertically zero, thereby making the acceleration vertically zero.
    Last edited: Jun 26, 2004
  9. Jun 26, 2004 #8

    clearly the tension in the string would be >= to the force of gravity. Hence if the lenght of the string shortens, it will move up! I don't want to argue about that pointer further, because i'm sure about that at least.
  10. Jun 26, 2004 #9


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    Read my appended post above. And I was just curious of how you obtained
    dWork = -(mv^2/lsinX)*dxsinX + mgdlsinX ? I don't need a detailed derivation, just your assumptions regaurding the problem.
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