Question

A particle of mass $m$ is moving in circular path under action of central force whose potential is $V=-k{r}^5$, here $r$ is distance from center, then kinetic energy of particle is proportional to

• A. $r^4$
• B. $r^{3/2}$
• C. $r^5$
• D. $r^{5/2}$

Answer 1

The correct option is C $r^5$

$U=Vm=-km{r}^5$
$F=-\frac{dU}{dr}=+5km{r}^4$
$F=F_c=\frac{m{v}^2}{r}$
$5km{r}^4=\frac{m{v}^2}{r}$
$\frac{m{v}^2}{2}=\frac{5}{2}km{r}^5$
K.E =$\frac{5}{2}km{r}^5$