The Physics Philes, lesson 60: Momentum, It’s Not Just for Lines Anymore!
This will be a relatively quick post this week because of…laziness. But still! Physics!
Today, we start a more in depth discussion of angular momentum. Before we get too deep, we need to start shallow. Let’s wade in.
A loooooong time ago, we had a little discussion on momentum. The analog to the momentum of a particle is angular momentum, which is a vector quantity. (We use a capital L with a little arrow over it.) The relationship of angular momentum to linear momentum is the same as the relationship of torque to force; that is:
So, for a particle with any particular mass, velocity, momentum, and position, we can define the angular momentum of that particle as the cross product of the position vector and the momentum vector:
Because the position vector r is the position of relative to the origin, the value of angular momentum depends on the choice of origin.
Alrighty, here’s a crappy graph thing I tried to draw in order to illustrate what’s next. I’m so sorry you have to look at it.
Like I said, sorry. But let’s see if we can make sense of it. We have a particle moving on an xy-plane, and we have labeled the position vector r and the momentum vector p. The angular momentum vector L is perpendicular to the xy-plane. If we use the right hand rule, we know that the direction is along the +z-axis and the magnitude is
The l is the perpendicular distance from the line of vector v to O.
Fine, but what if a force acts on the particle? When that happens, the velocity and momentum of the particle are likely to change, right? That means that the angular momentum of the particle could also change. So what then, science?
Don’t worry. Science and math have teamed up to give you some tools to handle that particular situation. Together, we can show that the rate of change of angular momentum is equal to the torque of the net force.
Say whaaaaaa? It’s true. *cracks knuckles* Let’s get to it. First, we need to take the time derivative of the definition of angular momentum:
Whoa! Big equation. Don’t worry. We’re going to shrink it down. See that first term? It’s actually zero because it contains the vector product of vector v with itself. In the second term, we can replace ma with the net for F. So what we end up with is:
Muuuuuch easier. What does it mean? It means that the rate of change of angular momentum equals the torque of the net force acting on the particle. Boom! Science.
That’s all from me today. Drawing crappy pictures and using simple LaTeX really takes it out of me. Never fear! Next week I’ll be back with a little something about the angular momentum of rigid bodies. I know you won’t want to miss it.
Featured image credit: Marcy Kellar