# Circular motion at end of string in horizontal plane

1. Jul 19, 2010

### fouquoit

Suppose a mass m suspended from a string of length l is undergoing uniform circular motion in a horizontal plane, with angular velocity ω. Calculate the centripetal acceleration a.

If T is the tension in the string, and θ the angle the string makes with the vertical, then the vertical and horizontal components are:

T cos θ = mg
T sin θ = mrω²

Dividing the latter by the former gives: $$\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sqrt{1-\sin^{2}\theta}}=\frac{r \omega^{2}}{g}$$

If r is the radius of the circle, then $$\sin \theta=\frac{r}{l}$$, and substituting for sin θ gives:

$$\frac{\frac{r}{l}}{\sqrt{1-(\frac{r}{l})^{2}}}=\frac{r \omega^{2}}{g}$$

Rearranging gives $$a=r\omega^{2}=\sqrt{l^{2}\omega^{4}-g^2}}$$

Now, the thing that I find strange about this result is that, to be able to take the square root, we need $$l^{2}\omega^{4}\geq g^2$$, but physically there seems to be no such restriction, since l and ω are arbitrary and can be as small as one likes.

Where have I gone wrong?

2. Jul 19, 2010

### Staff: Mentor

How did you do this rearrangement?

3. Jul 19, 2010

### rcgldr

The parameters may not be so arbritrary. If l is small then w may end up large because of the effect of gravity and the the relatively small radius r.

4. Jul 19, 2010

### fouquoit

OK, thanks. I think I've got it. There is actually nothing wrong with the equation I derived.

As the prevois poster pointed out, ω is not arbitrary. In fact ω just has a lower limit given by $$\omega=\sqrt{\frac{g}{l}}$$, which corresponds to the frequency of a pendulum oscillating at low amplitude.

5. Jul 19, 2010

### Staff: Mentor

$$\omega=\sqrt{\frac{g}{l\cos\theta}}$$

Which has the limit for small angles as you point out. Good stuff!