Question

Centripetal acceleration of a particle of mass $m$ moving in a circle of radius $r$ is $4/r^2$. What is the angular momentum of the particle?

Answer 1

$a=\frac{4}{r^2}=\frac{v^2}{r}$

$v=\sqrt{\frac{4}{r}}$

$Angularmomentu=L=mvr=m\sqrt{4r}=2m\sqrt{r}$