Question

The potential energy of a particle of mass $m$ at a distance $r$ from a fixed point $O$ is given by $V\left(r\right)=k{r}^2/2$, where $k$ is a positive constant of appropriate dimensions. This particle is moving in a circular orbit of radius $R$ about the point $O$. If $v$ is the speed of the particle and $L$ is the magnitude of its angular momentum about $O$, which of the following statements is (are) true ?

• A. $v=\sqrt{\frac{k}{2m}}R$
• B. $v=\sqrt{\frac{k}{m}}R$
• C. $L=\sqrt{mk}{R}^2$
• D. $L=\sqrt{\frac{mk}{2}}{R}^2$

Answer 1

The correct option is B $v=\sqrt{\frac{k}{m}}R$

Given potential energy $V=\frac{k{r}^2}{2}$
$F=-kr$ (towards centre) $[F=-\frac{dV}{dr}]$
At $r=R$,
$kR=\frac{m{v}^2}{R}$ [Centripetal force]
where $v$ is the speed of the particle

$\Rightarrow$ velocity $v=\sqrt{\frac{k{R}^2}{m}}=\sqrt{\frac{k}{m}}R$
$L=mvR=m\sqrt{\frac{k}{m}}{R}^2$