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Circular motion at end of string in horizontal plane

  1. Jul 19, 2010 #1
    Suppose a mass m suspended from a string of length l is undergoing uniform circular motion in a horizontal plane, with angular velocity ω. Calculate the centripetal acceleration a.

    If T is the tension in the string, and θ the angle the string makes with the vertical, then the vertical and horizontal components are:

    T cos θ = mg
    T sin θ = mrω²

    Dividing the latter by the former gives: [tex]\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sqrt{1-\sin^{2}\theta}}=\frac{r \omega^{2}}{g}[/tex]

    If r is the radius of the circle, then [tex]\sin \theta=\frac{r}{l}[/tex], and substituting for sin θ gives:

    [tex]\frac{\frac{r}{l}}{\sqrt{1-(\frac{r}{l})^{2}}}=\frac{r \omega^{2}}{g}[/tex]

    Rearranging gives [tex]a=r\omega^{2}=\sqrt{l^{2}\omega^{4}-g^2}}[/tex]

    Now, the thing that I find strange about this result is that, to be able to take the square root, we need [tex]l^{2}\omega^{4}\geq g^2[/tex], but physically there seems to be no such restriction, since l and ω are arbitrary and can be as small as one likes.

    Where have I gone wrong?
     
  2. jcsd
  3. Jul 19, 2010 #2

    Doc Al

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    Staff: Mentor

    How did you do this rearrangement?
     
  4. Jul 19, 2010 #3

    rcgldr

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    The parameters may not be so arbritrary. If l is small then w may end up large because of the effect of gravity and the the relatively small radius r.
     
  5. Jul 19, 2010 #4
    OK, thanks. I think I've got it. There is actually nothing wrong with the equation I derived.

    As the prevois poster pointed out, ω is not arbitrary. In fact ω just has a lower limit given by [tex]\omega=\sqrt{\frac{g}{l}}[/tex], which corresponds to the frequency of a pendulum oscillating at low amplitude.
     
  6. Jul 19, 2010 #5

    Doc Al

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    [tex]\omega=\sqrt{\frac{g}{l\cos\theta}}[/tex]

    Which has the limit for small angles as you point out. Good stuff!
     
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