Conservation of angular momentum of a bullet

[refrigerator]

I was thinking of a couple basic mechanics problems lately. What if you have a rod sitting in space at rest, and you shoot a bullet at its center. Linear momentum is conserved, and the problem is quite trivial.

Now what if the bullet hit the rod slightly off of the rod's CG? I think linear momentum is conserved so the CG velocity of the rod is the same as with the problem above. However, I think the rod should also rotate.

So my question is, is angular momentum conserved here? Initially it seems the rod and bullet have purely linear motion, so no angular momentum, but after collision there clearly would be angular momentum.

I have thought about this but can't seem to figure out where the error in my reasoning is. Can somebody show me what's going on, or at least try to give me a fresh perspective?

Thank you in advance,

Refrigerator

 

[D H]

Even a point mass can have angular momentum. Suppose you fire that bullet horizontally from a gun while standing upright. That bullet has non-zero angular momentum from the perspective of a reference frame with its origin at your feet.

When you are looking at angular momentum, you have to look at the total angular momentum of the system, not just that of spinning things like your rod.

 

[dauto]

The bullet has orbital angular momentum even though it has no spin angular momentum

 

[Khashishi]

Angular momentum is always conserved but it has a different value depending on where you put the origin. Even if the bullet is moving in a straight line, it still has angular momentum measured relative to a point that is displaced from the path of the bullet.

Usually you want to do this problem in the center of mass frame. If the bullet is hitting the end of the rod, the bullet flies some distance from the center of mass.