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## Homework Statement

Two ice skaters of mass ##m = 50\,\mathrm{kg}## each are moving towards each other frictionless on parallel paths with a distance of ##3\,\mathrm{m}##. They both have a velocity of ##v_o=10\,\frac{\mathrm m}{\mathrm s}##.

Skater 1 is holding a massless rod of lenght ##3\,\mathrm{m}## at one end perpendicular to his direction of movement. Skater 2 grabs the other end of the rod when the two have minimal distance of each other (again, ##3\,\mathrm{m}##). (At this point, they should start rotating.) Both skaters now pull themselves along the rod until their distance from each other is ##1\,\rm m##.

Hereafter, they both let go of the rod.

Compare the kinetic energies of the system before and after the encounter of the two skaters.

Prove that the difference is equal to the work the two have to perform while shifting towards each other.

## Homework Equations

Angular momentum: ##L = rp=rmv## (only in this case, because ##r\perp v##)

Kinetic energy: ##E_{\rm{kin}}=\frac12mv^2##

## The Attempt at a Solution

The angular momentum when the two start rotating is ##L=r(2m)v_0=1.5m\cdot(50kg\cdot2)\cdot10\frac ms=1500kg\frac m{s^2}##, and we now that ##L## is conserved. So let us call the skaters' velocities immediately before letting go of the rod ##v'## and the new (smaller) radius of their rotation ##r'##. This means $$r(2m)v=r'(2m)v'\Rightarrow v'=\frac{r}{r'}v=\frac{1.5m}{0.5m}\cdot10\frac ms=30\frac ms .$$

Calculating the kinetic energies before and after gives us a difference of ##\Delta E_{\rm{kin}}=40000J##, if I am not mistaken. But how should I proceed from here? I have no idea on how to calculate the work performed by the two ice skaters as I do not know the corresponding force with which I could calculate ##W=Fs=F(r-r')##.

Thanks in advance!