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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?

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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 $\cot(\phi)=\sqrt{\frac{L^2}{r^2}-1}$

This leads to

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

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 $\frac{1}{r^2}$

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    $\begingroup$ 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 $0$. $\endgroup$ – user243267 Dec 4 '19 at 9:19
  • $\begingroup$ Thank you for the correction :) $\endgroup$ – bemjanim Dec 4 '19 at 16:19

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