Ok, I've been wondering about this question for many years and I don't know whether I just don't understand the basic concepts enough. If you have a rapidly-spinning neutron star that continues to collapse down to a singularity, what happens to the conservation of angular momentum from the spin? As the circumference of the neutron star approaches zero, what happens to the speed of the spin? Would it actually have to approach c? Would the energy get bled off into some other form?
When I clicked on this link I thought that there might be a 1kg mass in a completely elastic collision with something . Cool question though, sorry I don't know squat about this kinda stuff.
Well, neutron stars do not collapse to a singularity; they typically are about 20km in diameter, with masses on the order of the sun. Read about rotation here: http://en.wikipedia.org/wiki/Neutron_star#Rotation It suggests newborn (i.e. fastest) neutron stars rotate on the order of once per second, which with a circumference of a mere 64km, is quite slow.
A black hole can have angular momentum and charge. In fact, externally, once stabilized, a black hole is fully characterized by mass, charge, and angular momentum.
Cool. So, as the circumference of the black hole shrinks, the rate of spin increases, correct? If the circumference approaches zero, what happens to the rate of spin? Would this have any effect on a practical limit to how small a black hole could become?
Externally you get event horizon(s). Inside the event horizon you get a ring singularity. A point singularity only occurs for a non-rotating black hole. In GR, spacetime curvature (the metric) can incorporate angular momentum, so the complete black hole structure can be said to have angular momentum. Inside the horizon, nobody believes the idealized Kerr black hole describes what would really happen. This is outside currently understood physics (IMO).