Question

A particle of mass $m$ is describing a circular path of radius $r$ with uniform speed. If $L$ is the angular momentum of the particle about the axis of the circle the kinetic energy of the particle is given by

• A. $L^2/m{r}^2$
• B. $L^2/2m{r}^2$
• C. $2L^2/m{r}^2$
• D. $m{r}^{2}L$

Answer 1

The correct option is B $L^2/2m{r}^2$

$K.E.=1/2m{v}^2$

$L=mvr$

Squareing,

$L^2=m^2{v}^2{r}^2$

Divide by 2mr^2 on both sides

$L^2/2m{r}^2=m{v}^2/2=K.E.$

So, Option B is correct.