A particle moves along a closed trajectory in a central field of force where the particle's potential energy `U=kr^2` (k is a positive constant, r is the distance of the particle from the centre O of the field). Find the mass of the particle if its minimum distance from the point O equals `r_1` and its velocity at the point farthest from O equals `v_2`.

Explanation

If `r=` radial velocity of the particle then the total energy of the particle at any instant is
`1/2m r^2+(M^2)/(2mr^2)+kr^2=E` (1)
where the second term is the kinetic energy of angular motion about the centre O. Then the extreme value of r are determined by `r=0` and solving the resulting quardratic equation
`k(r^2)^2-Er^2+(M^2)/(2m)=0`
we get
`r^2=(E+-sqrt(E^2-(2M^2k)/(m)))/(2k)`
From this we see that
`E=k(r_1^2+r_2^2)`
where `r_1` is the minimum distance from O and `r_2` is the maximum distance. Then
`1/2mv_2^2+2kr_2^2=k(r_1^2+r_2^2)`
Hence, `m=(2kr^2)/(v_2^2)`
Note: Eq. (1) can be derived from the standard expression for kinetic energy and angular momentum in plane polar coordinates:
`T=1/2mr^2+1/2mr^2theta^2`
M=angular momentum `=mr^2overset(.)theta`