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Ball twirling on string attached to stick

  1. Sep 26, 2009 #1
    So we did this little experiment; we attached a tennis ball to a string which ran through a tube and to a plastic bag with 3 tennis balls. Here's a picture

    http://img7.imageshack.us/img7/6160/twirls.jpg [Broken]

    So in a nutshell we can represent Mg like this:
    [tex]Mg = (m*4pi^2*R)/(T^2)[/tex]
    lol i tried my best with the latex stuff...

    anyways in the image the ball is perfectly perpendicular to the tube. But in our experiment it was moved downwards somewhat due to gravity on the ball, with an angle. However I am told that this has no bearing on the results of the formula. That is the formula works even though it's spinning around on an angle.

    My problem is: How do I prove this?

    PS: this problem won't matter at all but M = 3m
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Sep 26, 2009 #2


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    Welcome to Physics Forums.

    The rotating ball is an example of what is commonly called a conical pendulum (the string traces out a conical surface). Now, let us suppose that the string makes and angle [itex]\theta[/itex] with the horizontal. In this case, the tension will no longer be acting parallel to the radius of the circle transcribed by the ball. Instead, it will be acting towards the pivot, which in this case is the top of the tube. So, what we need to do is look at the components of the tension. If T is the magnitude of the tension in the string, then the components of the tension will be given by

    [tex]T_x = T\cos\theta\;\;\;,\;\;\;T_y = T\sin\theta[/tex]

    Now, what can you say about Tx and Ty, bearing in mind that the tennis ball transcribes a circle in the (x,z) plane and remains at constant height, y, throughout it's motion?
  4. Sep 26, 2009 #3
    So the vector sum of Ty and Tx is equal to T?
  5. Sep 26, 2009 #4


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    Like Hootenanny said, Tx=Tcosθ and Ty=sinθ.

    Now considering the horizontal component Tx, since it is spinning in a circle, Tx provides the centripetal force so Tcosθ = ?

    Now the vertical tension,T, in the string balances out the weight of the mass M, to T equals what now ?
  6. Sep 26, 2009 #5
    3m? I'm not good at this...=P
  7. Sep 26, 2009 #6


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    no no

    you know what is the centripetal force right? If you do then, what is the formula for centripetal force?

    in the Tension-mass system (the part with the string in the tube and the mass M), I don't believe the mass M is moving up or down, so the tension T should balance out the mass M right? So what is T equal to here?
  8. Sep 26, 2009 #7
    I don't know dude...I figure if they balance then the tension force should be equal to Mg...
  9. Sep 26, 2009 #8


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    Right! So T=Mg.

    Since it looks like you don't know centripetal force =mω2R

    so if Tcosθ= mω2R.

    What is Mg equal to ?
  10. Sep 26, 2009 #9
    Mg = mω2R?
    what's that w symbol?
  11. Sep 26, 2009 #10


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    ω is called the angular velocity. Which for a circle is ω=2π/T where T=time period.

    so we have

    [tex]Mg= \frac{m \omega ^2 R}{cos \theta}[/tex]

    Now θ should be small. And what is cosθ approximately equal to for very small angles of θ?
  12. Sep 26, 2009 #11
    So cos([tex]theta[/tex]) = Mg?
  13. Dec 30, 2010 #12
    No, for small angles cos(theta) is approximately 1. Therefore your formula becomes Mg=m(w^2)R. That's your proof since this formula is the same as for the original case with no "dipping" of the tennis ball.
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