We know that blackholes retain the properties of their progenitor, so if we think of the total angular momentum of a blackhole’s progenitor star now in an infinitesimal point (0 radius), should it’s angular velocity not be infinitely large?

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    $\begingroup$ A collapsing star does not collapse to an infinitesimal point. A collapsing star that forms into a black hole (i.e. if it's massive enough) forms a Kerr black hole, the physical singularity of which has a ring structure. The spin of a Kerr black hole has an upper limit, where if its spin exceeds that limit then it forms a naked singularity. $\endgroup$ Commented Feb 12, 2022 at 13:36
  • $\begingroup$ Bear in mind that once stuff crosses the event horizon it can no longer have an effect on what's outside the event horizon. See math.ucr.edu/home/baez/physics/Relativity/BlackHoles/… $\endgroup$
    – PM 2Ring
    Commented Feb 12, 2022 at 17:58
  • $\begingroup$ The short answer is that you don't consider the rotational velocity of the singularity. With black holes we can only ever say anything about the event horizon so at best, you can talk about the rotational velocity of said event horizon. This means that you compare the initial angular momentum of the star to the angular momentum of the event horizon, not the singularity. Anything inside the event horizon is beyond our knowledge. $\endgroup$
    – zephyr
    Commented Feb 13, 2022 at 22:41


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