Question

The potential energy of a particle of mass m at a distance r from a fixed point O is given by $V\left(r\right)=\frac{k{r}^2}{2}$, 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)

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

As the particle is moving in a circular orbit there must be some centripetal force acting on in in radial direction.
Centripetal force = $\frac{m{v}^2}{r}$
There must be some force acting on this particle that is providing this centripetal force.
We know that $F=-\frac{dU}{dr}$
$⟹F=-\frac{d\left(k{r}^2\right)}{2}$
$⟹F=-kr$
This $F$ must be equal to centripetal force experienced by the particle
$⟹-kr=\frac{m{v}^2}{r}$
$\therefore v=\sqrt{\frac{k}{m}}R$
We know that $L=mvr$
Using the value of $v$ from above we get
$L=\sqrt{mk}{R}^2$