3.3: Angular Momentum

Imagine the following experiment. You’ve got, somehow, a frictionless plane. (These frictionless planes are common in physics, but much more difficult to manufacture in the real world! If you wish, you can imagine it as an air hockey table, or a particularly smooth and slippery sheet of ice or teflon.) A hockey puck is sliding along the plane, where it hits a big clock hand, and sticks to the end of the clock hand. The other end of the clock hand is nailed into the ground, so that it’s not going anywhere. After the hockey puck hits the clock hand, the clock hand starts spinning around.

On first glance, you might think, wait! Momentum isn’t conserved here! The clock hand may be spinning around, but it’s no longer moving off in one direction, whereas before there was clearly momentum in the x-direction! However, remember that the clock hand is nailed into the ground. That means when the puck collides with the clock hand, the clock hand will push on that nail, which pushes on the ground, and effectively the whole earth is pushed off (very, very slowly!) to the right. Momentum is conserved, but you have to consider everything that’s interacting to keep track of all of it.

So, let’s do another experiment. Let’s collide the puck with a bar— and still have it stick— but not nail that bar to the ground. What happens now is that after the collision, the bar does move off to the right only not as fast. It’s still rotating, though.

Here, momentum is conserved. The combined system moves off to the right at a lower speed than the puck came in because it’s a more massive system. However, there’s also the rotation. There’s clearly some kinetic energy associated with that, as bits of the rod have motion about the center of the rod in addition to the bulk motion of the rod as a whole. It turns out that there is yet another quantity, called angular momentum, that is conserved in interactions. Be careful about the name! Angular momentum is not a “special kind” of momentum. It is a wholly different quantity, with different units, that is conserved separately. It bears some similarities to momentum, and thus the name is similar, but it is a different thing. You cannot mix momentum and angular momentum; again, remember that they are two different things.

Back to our example here: the question is, if there’s angular momentum afterwards in the rotation of the rod, what angular momentum is there before the collision? It must be there, if angular momentum is a conserved quantity!

In order to measure angular momentum, you must choose an axis to measure it about. How can angular momentum be a conserved quantity if you can choose any arbitrary axis you may ask? The answer is that angular momentum is conserved about any axis, as long as you stick with the same axis all the way through the problem.

If a particle is moving in a straight line directly towards or directly away from your chosen axis, then it has no angular momentum. However, if it’s motion is offset from the axis, even if it’s moving in a straight line, it still has angular momentum. To figure out the angular momentum, you multiply the lever arm by the momentum of the moving particle. The lever arm is the perpendicular distance from the axis to the line of motion of the particle. For example:

I’ve cleverly chosen my axis to be on line with the motion of the center of mass of the system after the collision. That means that the after the collision, the linear motion of the center of mass of the system makes no contribution to the angular momentum; all of the contribution comes from whatever the rotation is doing. Before, however, the linear momentum of the puck does contribute angular momentum. The lever arm is, as drawn, the perpendicular distance from the axis to the line of motion of the particle. If we call that perpendicular distance \(\ d\), then the angular momentum (for which we traditionally use the letter \(\ l\)) is:

\(\ l=d p=d m v\)

where \(\ m\) is the mass of the puck and \(\ v\) is the initial speed of the puck.¹

How do you figure out the angular momentum of a rotating object? The hard way to do it is to consider the object as a collection of a lot of little pieces of object. For each small piece of that object, you multiply the small mass of that piece by the speed of that piece resulting from the rotation by the lever arm from the axis to that piece. Add up what you get, and you have the object’s angular momentum. In practice, for most objects we’re able to define a single number that we call the moment of inertia, which takes care a of a lot of that for you. This is a quantity that adds up all of the bits of mass and the distances of those bits of mass from a specified axis of rotation for an object. It takes into account the mass part of momentum and the lever arms for all those little bits of the object. The angular momentum of an object rotating about a given axis is then:

\(\ l=I \omega\)

where \(\ I\) is the moment of inertia of that object about the axis and \(\ \omega\) is the angular speed of the rotation. To figure out angular speed, first figure how how long it takes for the object to make one complete rotation; call that the period \(\ T\). The angular speed is then:

\(\ \omega=\frac{2 \pi}{T}\)

\(\ \omega\) then has dimensionality of one over time; the SI unit for \(\ \omega\) is \(\ s^{-1}\).

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¹If you are familiar with vectors, in fact the real definition of angular momentum is \(\ \vec{l}=\vec{r} \times \vec{p}\), where \(\ \vec{r}\) is the displacement from the axis to the position of the moving particle and \(\ \vec{p}\) is the particle’s momentum.