Question

A particle of mass m moves under the action of a central force whose potential is given by $V\left(r\right)=K{r}^3,(K>0)$
(i) For what energy and angular momentum will the orbit be a circle of radius a about the origin?
(ii) What is the period of this circular motion?
(iii) If the particle be slightly disturbed from this circular motion, what will be the period of small radial oscillations about r=a?

Answer 1

Given data $v=k{r}^3$

d.w.r to 'r' m both sides

$dv=3k{r}^2$

Now $\Delta V=\frac{\Delta u}{m}\Rightarrow \frac{dv}{m}=dv=3k{r}^{2}dr$

[$u=PE$ of mass m]

$\frac{du}{dr}=3km{r}^2$

$F=\frac{-du}{dr}=|F|=3km{r}^2$

$\Rightarrow 2km{r}^2=\frac{m{v}^2}{r}$

(1) For circular motion $F=F_{centripetal}$

$\therefore v=\sqrt{3k{r}^3}$

for $r=a\therefore v=\sqrt{3k{a}^3}$

For circular motion

total energy $\Rightarrow E=K+U$

$=\frac{1}{2}m{v}^2+m{v}_r$

$=\frac{1}{2}m\left(3k{a}^3\right)+m\left(k{a}^3\right)$

$E=\frac{5}{3}mk{a}^3$

Angular momentum = $mvr=m\sqrt{3k{a}^3}$

$L=\left(m{a}^2\right)\sqrt{3ka}$

(2) Time for circular motion

$T=\frac{2\pi r}{v}=\frac{2\pi \left(a\right)}{\sqrt{3k{a}^3}}=\frac{2\pi }{\sqrt{3ka}}$