A particle of mass $m$ moves in two dimensions in a potential $V(r)=\frac{1}{2}\left(\frac{\alpha}{r^2}+\beta r^2 \right)$ where $r$ is the radial distance from the origin in polar coordinates, and $\alpha$ and $\beta$ are positive constants.

I have thus calculated that the force corresponding to this potential is $\mathbf{F}=\left(\frac{\alpha}{r^3}-\beta r \right)\mathbf{e_r}$.

If the particle moves in a circle of radius $r=R$, find $R$ in terms of $m$, $\alpha$, $\beta$ and the angular momentum $L$.

This is what I have so far:

By Newton's Second Law,

$\mathbf{F}=\left(\frac{\alpha}{r^3}-\beta r \right)\mathbf{e_r}=m((\ddot{r}-r\dot{\theta}^2)\mathbf{e_r} + ({2\dot{r}\dot{\theta}}+r\ddot{\theta})\mathbf{e_\theta}) \\\frac{\alpha}{r^3}-\beta r=m(\ddot{r}-r\dot{\theta}^2)=m(0-r\frac{L^2}{m^2r^4})=-\frac{L^2}{mr^3}$

Therefore $r = R = \left(\frac{1}{\beta}\left(\alpha+\frac{L^2}{m}\right)\right)^\frac{1}{4}$

Is my reasoning correct?


1 Answer 1


I think it is correct.

A faster approach:

Use $$-F_r=\frac{mv^2}{r}$$ $$v=\frac{L}{mr}$$

to solve for $r$.

p.s. it’s better to post it on Physics SE to receive response more quickly.


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