- #1
Cosmophile
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Homework Statement
Mass ##M## hangs from a string of length ##l## which is attached to a rod rotating at constant angular frequency ##\omega##. The mass moves with a steady speed in a circular path of constant radius. Find ##\alpha##, the angle the string makes with the vertical.
Homework Equations
There are only two forces acting on the mass:
1) ##T##, the tension of the string.
2) ##W##, the weight of the mass.
The Attempt at a Solution
[/B]
For starters, I find [tex]\Sigma F_y = T\cos\alpha - W = 0 \qquad. (1) [/tex] Since the radius is constant, ##a_r = \ddot {r} - r \dot {\theta}^2## simplifies to ##-r\dot {\theta}^2 = -r\omega ^2##.
The component of ##T## in the direction of the radius is
[tex] T\sin\alpha = mr \omega ^2. \qquad (2) [/tex] From here, I substitute ##r = l \sin\theta## and get [tex] T\sin\alpha = Ml \omega^2 \sin\alpha \qquad (3) [/tex]
[tex] T = Ml \omega^2 \qquad \qquad (4) [/tex]
Substituting ##(4)## into ##(1)## we get ##Ml\omega^2 \cos\alpha = W##. Since ##W = Mg##, we have [tex] \cos \alpha = \frac {g}{\omega^2l} [/tex].
This makes physical sense if ##\omega > \sqrt{\frac {g}{l}}##. As ##\omega \rightarrow \infty##, ##\cos \alpha \rightarrow 0## and ##\alpha \rightarrow \pi /2##. That being said, when ##\omega## is small, the result breaks down, as it implies ##\cos \alpha \rightarrow \infty##.
I realize that the issue is when I go from ##(3)## to ##(4)##, as I divide by ##\sin\alpha##, but that isn't allowed when ##\omega = \sqrt{\frac {g}{l}}##, as this gives ##\cos \alpha = 1 \implies \sin \alpha = 0##.
K&K explain that in doing the problem, we overlooked a second solution, namely, ##\sin \alpha = 0, T = W##. They then say:
"Physically, for ##\omega \leq \sqrt {g/l}##, the only acceptable solution is ##\alpha = 0, \cos\alpha = 1##. For ##\omega > \sqrt{g/l}##, there are two solutions: [tex] \cos \alpha = 1 [/tex] [tex] \cos\alpha = \sqrt{\frac{g}{\omega^2 l}} [/tex]
I'm having a difficult time understanding why this is the case. Is there no way to come up with one solution which suffices on its own?
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