Two balls of the same mass $m$ are connected to each other with rope of length $l$. One of the balls is also connected to the ceiling with a rope of the same length $l$. The balls are spinning around the axis which intersects the point of the connection of the rope in the ceiling. As a result, they create angles $\alpha$ and $\beta$ with the verticals. Masses of the two ropes can be neglected. What is the angular velocity of the system?

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So I made a free body diagram:

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And the equations are:

For the top ball:

$T_1 \cos \alpha = mg + T_2 \cos \beta\\ T_1 \sin \alpha - T_2 \sin \beta = m (l \sin \alpha) \omega^2$

For the bottom one:

$T_2 \cos \beta = mg\\ T_2 \sin \beta = m (l \sin \alpha + l \sin \beta ) \omega^2$

The process:

$T_1 \cos \alpha = 2mg\\ T_1 \sin \alpha = m (l \sin \alpha) \omega ^2+T_2 \sin \beta = m (l \sin \alpha) \omega ^2 + m(l \sin \alpha + l \sin \beta) \omega ^2=\omega ^2 m l(2\sin \alpha + \sin \beta)$

$2 mg \tan \alpha =\omega ^2 m l(2\sin \alpha + \sin \beta)$

And so:

$$\omega = \sqrt{\frac{2g \tan \alpha}{l(2\sin \alpha + \sin \beta)}}$$

However, according to the book the answer is:

$$\omega = \sqrt{\frac{g \tan \beta}{l(\sin \alpha + \sin \beta)}}$$

I'm quite stuck on that. Where am I wrong?


The book answer seems correct (it can be obtained from the equations for the bottom body). I don't know how you obtained your result, but there is a possibility that it is also correct and actually coincides with the book answer, just because the equations for the top body provide an extra relation between $\alpha$ and $\beta$.

  • $\begingroup$ I've plugged in some values for check, and the value of the expression that I got differs from the book's expression. $\endgroup$ – www Jun 1 '13 at 20:04
  • $\begingroup$ I've added my steps. $\endgroup$ – www Jun 1 '13 at 20:09
  • $\begingroup$ You cannot plug in arbitrary values of $\alpha$ and $\beta$, as they are indeed related under the conditions of the problem. I found how to get your answer, and yes, it seems it's also correct, so it probably coincides with the book answer. $\endgroup$ – akhmeteli Jun 1 '13 at 20:09
  • $\begingroup$ What do you mean by "conditions of the problem"? The only thing that I see this answer is dependent on is whether $\alpha$ smaller than $\beta$. I've assumed this is true. And even so, how can I obtain the book's answer? $\endgroup$ – www Jun 1 '13 at 20:13
  • $\begingroup$ As I said, you can get the book answer from the two equations for the bottom body (just eliminate $T_2$ from them). $\endgroup$ – akhmeteli Jun 1 '13 at 20:15

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