Question

The potential energy of a mass $m$ at a distance $r$ from a fixed point $o$ is given by $v\left(r\right)=\frac{k{r}^2}{2}$, where $k$ is positive constant of appropriate dimension this particle is moving in a circular orbit of radius $R$ about the point $o$. If $v$ is the speed of the particle & $L$ is the magnitude of its angular momentum about $o$? Find L.

Answer 1

$V=\frac{k{r}^2}{2}$

$F=-\frac{dV}{dr}=-kr$

$\therefore \frac{m{v}^2}{r}=kr$

$v=\sqrt{\frac{k}{m}}R$

Angular momentum $l=mvr=m\sqrt{\frac{k}{m}}{R}^2=\sqrt{mk}{R}^2$