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.

 

[PJC01]

I can confirm that conservation of angular momentum applies in this situation. Angular momentum is a fundamental principle in physics that states that the total angular momentum of a system remains constant unless acted upon by an external torque. In this case, the system consists of the rod and the bullet, and the external torque is the force of the bullet hitting the rod off-center.

Initially, the system has no angular momentum because both the rod and the bullet are moving in a straight line. However, when the bullet hits the rod off-center, it exerts a torque on the rod, causing it to rotate. This rotation results in the creation of angular momentum.

According to the principle of conservation of angular momentum, the total angular momentum of the system before and after the collision must be equal. This means that the initial angular momentum of zero must be balanced by the final angular momentum created by the rotation of the rod. Therefore, angular momentum is indeed conserved in this scenario.

To better understand this concept, it may be helpful to consider the equation for angular momentum: L = Iω, where L is angular momentum, I is moment of inertia (a measure of an object's resistance to rotation), and ω is angular velocity. In this case, the moment of inertia of the rod would change as it rotates, but the angular momentum would remain constant.

I hope this helps to clarify your understanding of conservation of angular momentum in this scenario. It is a fundamental principle in physics and applies to a wide range of systems and situations. If you have any further questions, please do not hesitate to ask.