Angular Momentum of a System of Particles

Last time we wrapped up linear momentum. This time — rotation!! Spinny spinny~

For a single particle rotating about some axis, angular momentum is $\vec{L} = \vec{r} \times m\vec{v}$ (bold = vector, you know the drill).

So for a system of particles, the total angular momentum is just the sum of each little guy’s:

$$\vec{L} = \sum \vec{r}_i \times m_i \vec{v}_i$$

Quick sanity check — remember for translational motion, what was force $F$? Yeah: $F = \dot{\vec{p}}$.

Same deal for rotation. The time derivative of angular momentum is the “twisty force” — torque, $\vec{N}$.

And just like before, if there’s no external torque kicking the system around, $\vec{N} = 0 \Rightarrow \dot{\vec{L}} = 0$, so $\vec{L}$ is constant.

Boom — that’s conservation of angular momentum.

OK but the real fun starts now.

The thing is, we’re dealing with a system of particles here, not one lonely point. And a system has these sneaky little quirks that a single particle doesn’t — quirks that tweak the equations juuust enough to matter.

So today’s mission: how do we describe a system of particles centered on the center of mass?

Let’s go.

We’re going to split the position vector $\vec{r}_i$ (origin $O$ to particle $i$) into two pieces — one from the origin to the center of mass, and one from the center of mass to the particle. All at once~~~~, for every particle.

(And since velocity is just the derivative of position, same split, same story, just swap the letter $r$ for $v$. No extra work.)

OK, with that setup, let’s redo the angular momentum sum using the new variables. The whole point is to see the system from the center of mass.

$$\begin{aligned} \vec{L} &= \sum \vec{r}_i \times m_i \vec{v}_i \\ &= \sum \left(\vec{r}_{cm} + \vec{\bar{r}}_i\right) \times m_i \left(\vec{v}_{cm} + \vec{\bar{v}}_i\right) \\ &= \sum \vec{r}_{cm} \times m_i \vec{v}_{cm} + {\color{red}\sum \vec{r}_{cm} \times m_i \vec{\bar{v}}_i} + {\color{red}\sum \vec{\bar{r}}_i \times m_i \vec{v}_{cm}} + \sum \vec{\bar{r}}_i \times m_i \vec{\bar{v}}_i \end{aligned}$$$$\begin{aligned} &= \vec{r}_{cm} \times \left\{\sum m_i\right\} \vec{v}_{cm} + {\color{red}\vec{r}_{cm} \times \sum m_i \vec{\bar{v}}_i} + {\color{red}\left(\sum m_i \vec{\bar{r}}_i\right) \times \vec{v}_{cm}} + \sum \vec{\bar{r}}_i \times m_i \vec{\bar{v}}_i \\ &= \vec{r}_{cm} \times m\vec{v}_{cm} + \sum \vec{\bar{r}}_i \times m_i \vec{\bar{v}}_i \end{aligned}$$

And look what pops out. The front term is the system — the whole system — treated like one big lump sitting at the center of mass, swinging around the origin. That’s the orbital piece.

The back term is each individual particle moving relative to the center of mass. That’s the spin piece.

Now — the red terms. Those are zero. Why? Because once you hop onto the center of mass and look around, the sum of position vectors and the sum of velocity vectors (weighted by mass) as seen from there is… zero. Obvious once you picture it, right?

Try to actually visualize it. Riding on the center of mass. Everything balanced around you.

If you want the math, here’s why that term is zero:

$$\begin{aligned} \sum m_i \vec{\bar{r}}_i &= \sum m_i\left(\vec{r}_i - \vec{r}_{cm}\right) = \sum m_i \vec{r}_i - \left(\sum m_i\right)\vec{r}_{cm} \\ &= \sum m_i \left\{\frac{\sum m_i \vec{r}_i}{\sum m_i}\right\} - m\vec{r}_{cm} = \left(\sum m_i\right)\vec{r}_{cm} - m\vec{r}_{cm} = 0 \end{aligned}$$

Cool. So the math checks out, but seriously — whenever you can, just hop on the center of mass and go.

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OK so for angular momentum, we tore the position vector apart and poked at each piece’s meaning. Which made me wonder — why didn’t we bother doing that for linear momentum?

So I tried it:

$$\begin{aligned} \vec{p} &= \sum m_i \vec{v}_i = \sum m_i\left(\vec{v}_{cm} + \vec{\bar{v}}_i\right) \\ &= \left(\sum m_i\right)\vec{v}_{cm} + \left(\sum m_i \vec{\bar{v}}_i\right) \\ &= m\vec{v}_{cm} \end{aligned}$$

Aaah — got it.

The split just collapses back to the same thing. That’s why nobody bothers. Or, if you trust your intuition, you could’ve just said “yeah, sum of the shards = center-of-mass answer” and moved on.

I’m putting the derivation here in case someone actually wanted to see it.

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Alright, we’ve covered the vector quantities. Now let’s hit the scalar headliner — energy.

For the total (kinetic) energy of a system of particles, the obvious move is to add up each each each each each each each particle’s energy. Then relate that mess to the center of mass.

$$\begin{aligned} T &= \frac{1}{2}\sum m_i v_i^2 \\ &= \sum \frac{1}{2} m_i \left(\vec{v}_{cm} + \vec{\bar{v}}_i\right)^2 \\ &= \sum \frac{1}{2} m_i v_{cm}^2 + {\color{red}\sum m_i \vec{v}_{cm} \cdot \vec{\bar{v}}_i} + {\color{orange}\sum \frac{1}{2} m_i \bar{v}_i^2} \\ &= \frac{1}{2} m v_{cm}^2 + {\color{orange}\sum \frac{1}{2} m_i \bar{v}_i^2} \end{aligned}$$

And there it is.

Notice — unlike with linear momentum, there’s an extra orange term hanging off the back.

That orange term is exactly the thing that makes a system of particles different from a single particle. That’s the whole difference showing up, right there.

And here’s why that orange term matters:

Imagine this picture — two weights flying off from the center, thud thud thud thud thud, in opposite directions.

Hop onto the center of mass. You’re sitting there. Watching.

Now — is the energy of this system as seen from your seat on the center of mass the same as the energy seen from the outside?

That’s not worth half a penny of thinking about — obviously not.

And the orange term is what’s telling us that. It’s accounting for all the motion you can’t see from the center of mass but is absolutely there. heh heh.

I’ll leave it here.

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Originally written in Korean on my Naver blog (2015-01). Translated to English for gdpark.blog.