# Confusion on dynamics/kinematics of a conical pendulum

I have a question regarding the dynamics and kinematics of a conical pendulum. Let's say I rotate a mass on a string so that I make a system that resembles a Conical pendulum.

Why does the radius of rotation around the vertical and the height of the mass above a certain reference increase if I rotate the mass around the vertical faster, i.e. The rotational velocity increases?

The mass undergoes circular motion about the vertical where the centripetal force is the radial component of the tension, $$T$$. If the string is at angle $$\phi$$ to the vertical, then resolving vertically gives

$$T\sin(\phi)=mg$$ so $$T=mg\operatorname{cosec}(\phi)$$

The radial component of $$T$$ is $$mg\cot(\phi)$$. The equation of circular motion is

$$\frac{mv^2}{r}=mg\cot(\phi)$$

If $$L$$ is the length of the string, $$r=L\sin(\phi)$$. This means $$\sin(\phi)=\frac{r}{L}$$ and therefore $$\cot(\phi)=\sqrt{\frac{L^2}{r^2}-1}$$

$$\frac{v^2}{r}=g\cot(\phi)$$

which gives

$$\frac{v^2}{g}=r\sqrt{\frac{L^2}{r^2}-1}=\sqrt{L^2-r^2}$$

which squared is

$$\frac{v^4}{g^2}=L^2-r^2$$

so

$$r=\sqrt{\frac{v^4}{g^2}-L^2}$$

Edit:

Crikey I resolved it wrong.

$$\frac{mv^2}{r}=mg\tan(\phi)$$ where $$\tan(\phi)=\frac{r}{\sqrt{L^2-r^2}}$$

$$\frac{v^2}{g}=\frac{r^2}{\sqrt{L^2-r^2}}$$

$$\frac{v^4}{g^2}=\frac{r^4}{L^2-r^2}$$

$$\frac{g^2}{v^4}=\frac{L^2-r^2}{r^4}=\frac{L^2}{r^4}-\frac{1}{r^2}$$

Use the quadratic formula to solve for $$\frac{1}{r^2}$$

• It is $T\cos \phi$ which balances $mg$, and not $T \sin \phi$ (where $\phi$ is the angle that the string makes from the vertical). Thus your final answer ($r=\sqrt{\frac{v^4}{g^2}-L^2}$) is incorrect. You can see that if we put $v=0$ (which is the trivial case of the pendulum just hanging and not moving, you answer yields an imaginary radius, but instead the radius should $0$. – user243267 Dec 4 '19 at 9:19
• Thank you for the correction :) – bemjanim Dec 4 '19 at 16:19