<p>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?</p>

CC BY-SA 4.0
2022-02-12T13:36:52.210Z
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.
Daddy Kropotkin
CC BY-SA 4.0
2022-02-12T17:58:47.480Z
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 https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html
PM 2Ring
CC BY-SA 4.0
2022-02-13T22:41:03.103Z
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.
zephyr
<black-hole><astrophysics>
Why don’t black holes have an infinite rotational velocity?
Rose
# Why don’t black holes have an infinite rotational velocity?

(Asked by: username_0 on 2022-02-12)

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?

**Tags:** black-hole, astrophysics

> **Comments:**
> **username_1**: 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.  
> **username_2**: 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 https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/black_gravity.html  
> **username_3**: 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.  

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