Circular motion at end of string in horizontal plane

  • Thread starter fouquoit
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  • #1
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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?
 

Answers and Replies

  • #2
Doc Al
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Rearranging gives [tex]a=r\omega^{2}=\sqrt{l^{2}\omega^{4}-g^2}}[/tex]
How did you do this rearrangement?
 
  • #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.
 
  • #4
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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.
 
  • #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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