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I have a mechanics question (Source: Kleppner,Kolenkow; 2nd ed., p.104, problem 2.11), which states the following (along with an answer): [![One of the classic problems in which a mass attached to a vertically rotating poll with symmetrically strings that are both at 45 degree angle to the rod, and are of the length l. // In this set up, the poll is left of the mass. Both mass and poll are rotating.][1]][1]

In this image, using the hint provided that $$\ell\omega^2 = \sqrt{2}g\tag{1.1}$$ and that $T_{up} = \sqrt{2}mg$. However, when I started out to solve with the Newtonian Laws and the FBD, I got (note: $T_{up} = T_1, T_{low} = T_2$) $$\sum F_y = T_1 cos(\theta)-T_2cos(\theta) - mg = 0\tag{2.1}$$ $$\sum F_x = -T_1 sin(\theta)-T_2sin(\theta) = -m\ell sin(\theta)\omega^2\tag{2.2}$$ or $$T_1 - T_2 = \frac{mg}{cos(\theta)}\tag{3.1}$$ $$T_1 + T_2 = m\ell\omega^2\tag{3.2}$$ which when solved for the tension forces, gives $$T_1 = \frac{1}{\sqrt{2}}mg + \frac{1}{2}m\ell\omega^2\tag{4.1}$$ $$T_2 = \frac{1}{2}m\ell\omega^2 - \frac{1}{\sqrt{2}}mg\tag{4.2}$$

So, would that mean that $T_2 = 0$? And if so, when is it true (for instance, should $\omega$ be within a certain range)?

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  • $\begingroup$ The image can be accessed via the link: i.stack.imgur.com/AlZWf.png $\endgroup$ – Sam Lee Jun 17 at 3:32
  • $\begingroup$ The hint is not saying that $\ell \omega^2 = \sqrt{2} g$ in general, it's saying that in the case that $\ell \omega^2 = \sqrt{2} g$, then $T_1 = \sqrt{2} m g$. It's just a way to check your answer without giving the actual solution. $\endgroup$ – d_b Jun 17 at 3:39
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$\omega$ can take on any value as long as it is large enough to allow the strings to not have any slack in them.

Yes, in the case of $l\omega^2=\sqrt 2g$ we would have $T_2=0$. For $\omega$ smaller than this we would get slack in the bottom string as the mass would come closer to the pole. For $\omega$ larger then this we would just get larger tensions. There is no issue with any of this here.

The given value of $l\omega^2=\sqrt 2g$ is just something to check you work. It's not saying this must always be true.

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