Warp wrote:
> Warp <war### [at] tagpovrayorg> wrote:
>>   Besides, we can just forget the cable: Simply shoot the projectile and
>> that's it. With the correct amount of speed it will stop the object from
>> rotating. Where did the angular momentum go?
> 
>   I thought about this and became to a conclusion. You could have explained
> it if you knew it instead of just saying "you are wrong" without any
> explanation.
So now you're saying that *I* don't understand the topic.  Very, very rich.
> 
>   The answer is that the object-projectile system still has the angular
> momentum. If we calculate the angular momentum of this system after the
> firing, ie. the how the system is oriented with regard to the center of
> mass of the system and the distance between the two objects, we will
> probably get an angular momentum equivalent to the original one.
Sounds like you are coming around to the correct view that total angular 
momentum is conserved.
> 
>   The same is probably true for two approaching objects which collide.
> Even though each object by itself didn't have any angular momentum, the
> two-object system did. The entire two-object system is actually rotating
> around the center of mass of the two objects (even though they two objects
> are travelling almost rectilinearly; this is because they are not travelling
> along the same line in space). When they collide and stick to each other,
> the "speed of rotation" they had just before they collided will be kept. The
> angular momentum will be unmodified. Only if the two objects were travelling
> exactly on the same line in space will there be no rotation because there's
> no angular momentum.
> 
Yes.  Angular momentum is conserved.

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