- #1

- 2

- 0

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: [tex]\frac{\sin \theta}{\cos \theta}=\frac{\sin \theta}{\sqrt{1-\sin^{2}\theta}}=\frac{r \omega^{2}}{g}[/tex]

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

[tex]\frac{\frac{r}{l}}{\sqrt{1-(\frac{r}{l})^{2}}}=\frac{r \omega^{2}}{g}[/tex]

Rearranging gives [tex]a=r\omega^{2}=\sqrt{l^{2}\omega^{4}-g^2}}[/tex]

Now, the thing that I find strange about this result is that, to be able to take the square root, we need [tex]l^{2}\omega^{4}\geq g^2[/tex], 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?