Just a simple question. I can see that for this to work I need: T_{rot} = 1/5 ma^{2}(thetaDOT + phiDOT)^{2} Just can't work out what phi has to do with rotational kinetic energy. I would have thought it would need to be simply the same thing but without the phiDOT term.
[itex]\theta[/itex] is the angular position of an identified point on the smaller sphere, measured with respect to the line joining the centers of the two spheres, right? Note that not only does [itex]\theta[/itex] change with time, but so does the orientation of that joining line, so you need to take that into account as well. That's where the [tex]\dot\phi[/tex] comes from.