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 clock) then the kinetic energy of the particle is

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

Answer 1

The correct option is C $\frac{L^2}{2m{r}^2}$

Kinetic energy $K.E.=\frac{1}{2}I{w}^2$

We know that $w=\frac{L}{I}$

$⟹K.E.=\frac{L^2}{2I}$

Now, moment of inertial of a particle moving in circular path of radius $r$, $I=m{r}^2$

$⟹K.E.=\frac{L^2}{2m{r}^2}$