I am not a physicist (math/comp-sci) but I understand that Angular Momentum is supposed to be conserved. I find this confusing because there seems to be many simple, common cases where a restrained, revolving body, when released seems to lose it's angular momentum.
For example, let's say that a hammer thrower is revolving a 10kg hammer around herself at a 1 meter distance from their combined center of mass every second. Taking the hammer as a point-mass, then before release, the hammer and the thrower each have an angular momentum of
$l\omega = (mr^2)*(2\Pi) = (10*1*1)*(2*3.14..) =~ 62.8 kg\text{-}m^2/s$
Now, after the hammer is released, the thrower still has her same angular momentum (also 62.8), but the hammer no longer seems to have any.
Yes, I know that the hammer is still rotating (spinning) as it did when it was revolving around the thrower, so that angular momentum seems to be conserved, but the angular momentum from revolution (orbiting) seems to be gone. AFAIK, it's not transferred to it's spin angular momentum (observation seems to bear this out).
Nor is it transferred to the thrower. Besides having no mechanism for this (she let go of it), if it did, we would have noticed hammer throwers being knocked over by having their angular momentum suddenly doubled.
So where does it go? Or is it not actually conserved in this case?
Best Answer
Let us see a similar example: two people on skates going with some velocity towards each other both a bit on left off their common center, and in the moment of the closest approach, they just catch each other by right arms and they start to rotate.
In fact they have (as one system) the same angular momentum all the time.
When you have a projectile that aims a bit off center towards the target, the angular momentum of the system is nonzero. I think you find it under name collision parameter, usually $b$. If collision parameter is zero, angular momentum is zero.
The hammer and thrower is just time reversed situation. Angular momentum conserved. He might dissipate some his remaining bits of energy into the ground.
To better constraint the problem: imagine that nothing else exists in universe, only the hammer and the thrower. Forget any rotations of hammer or thrower. For eternity the system of hammer+thrower will keep the total angular momentum. Once you remove your thrower from the system, it is another excersise.
Small remark: The angular momentum is not partially there and there, the complete system has it.
The same picture is in first pages of nuclear physics textbooks, particle $a$ going to nucleus $C$ a little bit off axis. The distance between center of $C$ and axis of flight is that $b$. And you have defined the angular momentum of the system $=b \cdot p$ for any situation, any force between $a$ and $C$, any time before or after. Notice, that in an empty universe, the thrower cannot properly throw a hammer starting from rest (due to the conservation).