Does particle moving in a straight line have Angular Momentum?

The answer to the title question is simple, ask yourself the question: Does the moving particle has the ability to create a rotation motion? If yes, then it has the angular momentum, otherwise: no. It all depends on the pivot you choose.  Let me show you by a gedankan experiment (The phrase Gedankan experiment, which means Thought Experiment, became popular after Einstein used in explaining his Relativity theory). 

Consider the figure below:

In the gedankan experiment, a bullet is approaching a wooden-block. The wooden-block is initially at rest that is free to rotate around a pivot. The bullet hits the wooden-block gets embedded into the log and the combined wooden log with embedded bullet starts rotating clockwise (CW)

Since it is rotating, it has angular momentum. Angular momentum of isolated system is always conserved. Where did this angular momentem come from? Initially there was no angular momentum of the block as it was at rest. So it could have only come from the moving bullet. 

But if the bullet was going to hit the pivot as shown in figure below:

Here the block would not rotate and therefore, in this situation, the bullet has no angular momentum. It all depends on the position of pivot.

Now, let me give you the answer. The magnitue of angular momentum of bullet is (also see the figure below): 

$$|\vec{L}| = mvb $$

where \( m \) is the mass of the bullet, \( v \)  is the initial velocity of the bullet, and \( b \) is the distance of the closest approach of the bullet to the pivot. It is also called as Impact Parameter.   Clearly, you can see that in the second case discussed above when the bullet is hitting on pivot the impact parameter is zero, and therefore the angular momentum is zero in second case.

This phrase became popular aftter Ernest Rutherford used when describing his model of the atom.

When dealing with non contact forces such as gravitational forces, or electromagnetic forces. The impact parameter should be considered as if there were no such forces, i.e., the bullet travelling in straight line and the distance of closest approach to pivot is the impact parameter.   

I will leave to reader to calculate the angular momentum,  based on mathematical formula  

$$\vec{L} = \vec{r} \times \vec{p} \quad or  \quad L = r_{\perp} \times p \quad or \quad  L = r \times p_{\perp} \quad =mvb \quad$$  

THE END