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

Answer: (B), (C)

$L=\sqrt{mk}{R}^2$


Given,
Potential energy of the particle
$V=\frac{k{r}^2}{2}$
then total force acting on partical
$[F=-\frac{dV}{dr}]$

$\Rightarrow F=\frac{d\left(\frac{k{r}^2}{2}\right)}{dr}F=-kr$
this force is acting towards the centre.

At $r=R$ the acting force will equals to centripetal force.
$kR=\frac{m{v}^2}{R}$
where $v$ is the speed of the particle.
then the speed is given by
$\Rightarrow$ $v=\sqrt{\frac{k{R}^2}{m}}=\sqrt{\frac{k}{m}}R$
Hence option B is correct.

Where, in case of angular momentum $L$ it equals to $mvr.$
so,
$L=mvRL=m\left(R\sqrt{\frac{k}{m}}\right)R\Rightarrow m\sqrt{\frac{k}{m}}{R}^2=R^2\sqrt{mk}$
Hence, option C is also correct.