Question

A particle moves along a closed trajectory in a central field of force where the particle's potential energy $U=k{r}^2$ ($k$ is a positive constant, $r$ is the distance of the particle from the centre $O$ of the field). If the mass of the particle for minimum distance from the point $O$ equals $r_1$ and its velocity at the point farthest from $O$ equals $v_2$ is $m=\frac{xk{r}_1^2}{v_2^2}$. Find $x$.

Answer 1

If $\dot{r}=$ radial velocity of the particle then the total energy of the particle at any instant is
$\frac{1}{2}m{\dot{r}}^2+\frac{M^2}{2m{r}^2}+k{r}^2=E$ (1)
where the second term is the kinetic energy of angular motion about the centre $O$. Then the extreme values of $r$ are determined by $\dot{r}=0$ and solving the resulting quadratic equation
$k{\left(r^2\right)}^2-E{r}^2+\frac{M^2}{2m}=0$
we get
$r^2=\frac{E\pm \sqrt{E^2-\frac{2M^{2}k}{m}}}{2K}$
From this we see that
$E=k(r_1^2+r_2^2)$ (2)
where $r_1$ is the minimum distance from $O$ and $r_2$ is the maximum distance. Then
$\frac{1}{2}m{v}^2+2k{r}_2^2=k(r_1^2+r_2^2)$
Hence, $m=\frac{2k{r}_1^2}{v_2^2}$
$T=\frac{1}{2}m{\dot{r}}^2+\frac{1}{2}m{r}^2{\dot{\theta }}^2$
$M=$ angular momentum $=m{r}^2\dot{\theta }$