Say somebody is spinning a mass on a string with a period $T$ around. If $T$ is very big, the mass will describe a small circle around the person's waist. If $T$ is very small, the plane of movement should be around shoulder high, with the actual radius being the full length of the arm and string.


The question asks for the angle of equilibrium for a given $T$ and $R$. The angle $\theta$ is measured between the person and his arm. $\theta=0$ means that the string is straight down, and $\theta = \pi/2$ means that the string is parallel to the ground. $R$ is the length of arm and string.

So I thought that the two forces on the mass are the gravity, $F_g=mg$, and the centrifugal force, $F_c=m \omega^2 r=m \omega^2 \sin \theta R$. The two forces have a resulting force, which also acts in an angle $\tan \phi = F_c/F_g$ onto the mass.


In equilibrium, both angles have to be the same. So I say:

$$\tan \theta = \tan \phi = \frac{m 4 \pi^2 \sin \theta R}{m g T^2}$$ $$\frac{\tan \theta}{\sin \theta} = \frac{m 4 \pi^2 R}{m g T^2} $$ $$\cos \theta = \frac{g T^2}{4 \pi^2 R}$$


For the given values, $T=0.45s$ and $R=15cm$ I get an $\theta=1.23$, which seems reasonable to me.

Then the problem asks for $T=0.85s$. The question is worded as asking to trouble, so I do not feel to bad about my function giving me $\theta = 0.618i$.

But how is this supposed to make any sense at all? If I spin the mass slower, the angle has to be lower. If I spin with almost zero speed (infinite period $T$), I should get an angle around $\theta=0$, nothing else. But the angle is exactly zero for a given $R$ and $T$ and then goes imaginary beyond that.

Can anyone enlighten me, or tell me what I did wrong in the derivation?

More thoughts based on Mark Eichenlaub's answer

The forces perpendicular to the rope look like this:


(Sorry, the restoring force has a Sine in it, no Tangent.)

The net force on the mass is therefore the difference:


(If you plot this with Sine instead of Tangent, it comes out at 1.22 like I already had before.)

These are the values for the first part of the problem, and one can see the two points of equilibrium (including zero) clearly.

If the frequency is lower like in the second part, there is no equilibrium other than zero. So no matter how big the angle is, the restoring force is bigger than the centrifugal force.


(This picure is slightly different with the Sine instead of Tangent as well, but it does not change the roots.)

In this light, the imaginary equilibrium point I got previously seems pretty right.


1 Answer 1


Your analysis is correct. At small speeds, the mass hangs straight down.

You might start with your picture; it is quite misleading. (Edit: queueoverflow has now fixed up the picture!) The text of your problem assumes that your arm is held exactly at the same angle as the string, making effectively one long string. Your picture shows a different story.

If your arm is held straight out, then the centrifugal force is $\omega^2(R_{arm} + R_{string}\sin\theta)$, which gives the behavior you expect; $\theta$ goes to zero as $\omega$ goes to zero.

However, when $R_{arm} = 0$, the string is hanging from the axis of rotation. A better picture would have the string attached to the center of a rotating disk on the ceiling.

Such a string/disk combination has two equilibria at high speeds. One is $\theta = 0$; the string can hang straight down. This is a trivial solution to your equation. The other has the string shooting out at an angle that increases with $\omega$. Of these two solutions, you can work out that the second one is stable and the first is unstable.

However, as $\omega$ decreases, the stable angle comes in, and eventually reaches 0 as $\omega^2 r \to g$. At slower speeds, the stable angle is zero.

Why? Think of it like a pendulum. The pendulum has a restoring force that is proportional to its angle for small angles. That's the hallmark of simple harmonic motion. That restoring force is $mg\sin\theta$ or just $mg\theta$ for small angles.

Meanwhile, the centrifugal force has a strength $\omega^2R\sin\theta$ or just $\omega^2R\theta$. If that's smaller than the restoring force from gravity, the string, when displaced slightly from $\theta = 0$, will feel a stronger restoring force than centrifugal force and will return to the origin.

Alternatively, scrap the rotating frame. You just have a pendulum with two degrees of freedom. A slow, circular rotation should be a solution to the equation of motion, but there is a fixed frequency for a given size of the circle. For small angles, that rotation frequency should just be the same as the famous pendulum frequency, $\omega = \sqrt{g/R}$. Set $\cos\theta = 1$ and that's exactly the frequency that comes out of your analysis.

  • $\begingroup$ I amended my question with some graphs. I fully understand it now, thank you very much! $\endgroup$ Nov 27, 2011 at 19:19
  • $\begingroup$ I also updated my picture. $\endgroup$ Nov 27, 2011 at 19:26
  • $\begingroup$ Glad to help. It was a good question. $\endgroup$ Nov 27, 2011 at 22:05

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