Two skaters each of mass `50 kg`, approach each other along parallel paths separated by `3m`. They have equal ad opposite velocities of `10 m//s`. The first skater carries a long light pole, `3 m` long, and the second skater grabs the end of it as he passes (assume frictionless ice).
a. Described quantitatively the motion of the skaters after they are connected by the pole.
b. By pulling on the skaters reduce their distance to `1m`. What is their motion then?
c. Compare the `KEs` of the system in parts a. and b. where does the change come from?

Explanation

a. As the net linear momentum of the system (skater`+`pole) is zero, the centre of mass will be at rest before and after the collision.
The skaters and the pole will rotate around the cente of mass (at the midpoint of the pole).

Applying the conservation of angular moment about an axis through `C` and perpendicular to the plane of the figure
`mvl+mvl=Iomega`, where `I=2m(l)^(2)`
`omegamvl/I=v/l=20/3rad/s`
b. As the separation reduces to `2l'=1m`
`Iomega=I'omega`
Again applying conservation of angular mometum, we get
`omega^(')=(Iomega)/I^(')=(2ml^('2)omega)/(2ml^('2))=9omega=60rad//s`
Therefore, angular velocity increases
c. `(KE_(i))/(KE_(f))=(1/2I'omega^('2))/(1/2Iomega^(2))=(l'omega^('2))/(l^(2)omeg^(2))=9`
The kinetic energy increase because the skaters do positive work in pulling themselves towards the centre of pole.