# Kleppner/Kolenkow: Conical Pendulum & Angle with Vertical

1. Feb 14, 2016

### Cosmophile

1. The problem statement, all variables and given/known data
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.

2. Relevant equations
There are only two forces acting on the mass:

1) $T$, the tension of the string.
2) $W$, the weight of the mass.

3. The attempt at a solution

For starters, I find $$\Sigma F_y = T\cos\alpha - W = 0 \qquad. (1)$$ 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
$$T\sin\alpha = mr \omega ^2. \qquad (2)$$ From here, I substitute $r = l \sin\theta$ and get $$T\sin\alpha = Ml \omega^2 \sin\alpha \qquad (3)$$
$$T = Ml \omega^2 \qquad \qquad (4)$$

Substituting $(4)$ into $(1)$ we get $Ml\omega^2 \cos\alpha = W$. Since $W = Mg$, we have $$\cos \alpha = \frac {g}{\omega^2l}$$.

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: $$\cos \alpha = 1$$ $$\cos\alpha = \sqrt{\frac{g}{\omega^2 l}}$$

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?

Last edited: Feb 14, 2016
2. Feb 14, 2016

### haruspex

Typo in your first equation, you meant cos, not sin.
It took me a while to deduce the rod is vertical. But that means the rotation of the rod is irrelevant, it might as well be hanging from a fixed point.
But to answer your question, yes it can be written as a single solution, by not cancelling the sin: $\sin(\alpha)\cos(\alpha)=\sin(\alpha)\sqrt{\frac g{\omega^2l}}$, but that does not alter the conclusion that there are two solutions for $\alpha$ in one range of $\omega$ and only one solution in another.

3. Feb 14, 2016

### Cosmophile

Fixed the typo, thanks! How did you deduce the rod is vertical?

4. Feb 14, 2016

### haruspex

From this:
Ok, it could be that the rod is horizontal and the string is attached at its axis of rotation, but either way there is no contribution to the radius from the rod length. Thus, simply hanging vertically is always a solution.