Euler's Equations of Motion for a Rigid Body

I feel like I’ve been cursing Euler’s name non-stop since I started university. lol. Seriously, man.

He just did too much. T_T

I literally wrote a whole report on this guy once — lol lol lol lol lol.

Apparently he went blind from doing too much research or something.

But.

Even after losing his eyesight, he just kept researching….. ha….. that’s insane…… heh.

Anyway. Let’s get going.

Up through the last post, we studied the principal axis. What’s a principal axis again?

When you slap a coordinate axis onto some object, if you just grab any random direction, the product-of-inertia terms stick around in the inertia tensor and the math gets gross. But if you grab the principal axis — snap!!! — those product-of-inertia terms vanish, and then

$$\vec{L} = \mathbf{I}\vec{\omega}, \qquad T = \tfrac{1}{2}\vec{\omega}\cdot\mathbf{I}\vec{\omega}$$

these equations get absurdly clean.

So by now we’ve picked up some tools — tensors, principal axes, the whole kit!! lol

Now we’re going to tackle the more general thing: full-on 3D rotation of a rigid body.

Hmm, how do I put this. It’s like skewering a sweet potato, spinning the skewer round and round, and then chucking it. lol. (Okay, terrible analogy, I know.)

That’s the vibe though — the rotation axis itself is moving now. heh heh heh.

So now two coordinate systems show up.

A fixed one (the inertial frame) and one that rotates along with the spinning object (the rotating frame).

OK so —

The frame that just sits still watching the spinning-flying sweet potato? That’s the inertial frame.

The frame that’s glued onto the spinning-flying sweet potato and goes along for the ride? Rotating frame. (Like, imagine you’ve climbed on and you’re riding the sweet potato? More or less?)

The relation that connects the two is something we learned waaay back:

$$\left( \frac{d\vec{r}}{dt} \right)_{\text{fixed}} = \left( \frac{d\vec{r'}}{dt} \right)_{\text{rot}} + \vec{\omega} \times \vec{r'}$$

We saw this one before!

For now though, we spin the sweet potato but we don’t throw it.

And let’s look at angular momentum:

$$\left( \frac{d\vec{L}}{dt} \right)_{\text{fixed}} = \left( \frac{d\vec{L}}{dt} \right)_{\text{rot}} + \vec{\omega} \times \vec{L}$$

(Here, $\vec{L}$ and $\vec{L'}$ are the same vector, right? Why?? Because it’s not flying off anywhere!!)

So one frame is just sitting there, one frame shares the same origin O but is spinning along with the sweet potato, and the bridge between them is exactly that equation. heh heh heh heh.

(You can basically read $\vec{\omega} \times \vec{L}$ as a kind of inertial force. We learned that inertial forces — fictitious forces — pop out whenever you switch between frames, right???)

Now look at the left side of that equation. That’s …….. just torque, right? Torque??? (And we’re gonna say the coordinate axes are the principal axes.)

$$\left( \frac{d\vec{L}}{dt} \right)_{\text{fixed}} = \vec{N} = \begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix}$$

And the right side is!!!!!!!!!!

$$\left( \frac{d\vec{L}}{dt} \right)_{\text{rot}} = \begin{pmatrix} \dot{\omega}_1 I_1 \\ \dot{\omega}_2 I_2 \\ \dot{\omega}_3 I_3 \end{pmatrix}$$$$\vec{\omega} \times \vec{L} = \begin{vmatrix} \hat{e}_1 & \hat{e}_2 & \hat{e}_3 \\ \omega_1 & \omega_2 & \omega_3 \\ \omega_1 I_1 & \omega_2 I_2 & \omega_3 I_3 \end{vmatrix}$$$$= \hat{e}_1(\omega_2 \omega_3(I_3 - I_2)) + \hat{e}_2(\omega_1 \omega_3(I_1 - I_3)) + \hat{e}_3(\omega_1 \omega_2(I_2 - I_1))$$$$= \begin{pmatrix} \omega_2 \omega_3(I_3 - I_2) \\ \omega_1 \omega_3(I_1 - I_3) \\ \omega_1 \omega_2(I_2 - I_1) \end{pmatrix}$$

Take the left side, take the right side, mash them together — zzapooooong-ppiri-ppara-ppoppop-ppoppoppong —

$$\begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} = \begin{pmatrix} \dot{\omega}_1 I_1 \\ \dot{\omega}_2 I_2 \\ \dot{\omega}_3 I_3 \end{pmatrix} - \begin{pmatrix} \omega_2 \omega_3(I_3 - I_2) \\ \omega_1 \omega_3(I_1 - I_3) \\ \omega_1 \omega_2(I_2 - I_1) \end{pmatrix}$$

And that right there is Euler’s equations of motion for a rigid body — written out component by component along the rigid body’s principal axes.

OK let’s try one. Say we’ve got an object spinning at a constant angular velocity.

(Constant $\omega$, so angular acceleration is 0.)

Then Euler’s equation becomes

$$\begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} \omega_2 \omega_3(I_2 - I_3) \\ \omega_1 \omega_3(I_3 - I_1) \\ \omega_1 \omega_2(I_1 - I_2) \end{pmatrix}$$

Now let’s throw in one more condition: the rotation axis is along the 1-axis — which is one of the principal axes.

That means $\omega_1 = \omega$, $\omega_2 = 0$, $\omega_3 = 0$.

Then the right side alllll goes to 0?

$$\begin{pmatrix} N_1 \\ N_2 \\ N_3 \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} + \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$$

$N_1 = 0$, $N_2 = 0$, $N_3 = 0$.

All zero.

What does that even mean….

$\vec{N} = 0$ — there’s no torque at all.

Earlier I mentioned that when $\vec{\omega}$ points along a principal axis, $\vec{L} \parallel \vec{\omega}$ (they line up) and this is called dynamic balance — and now, with Euler’s equations, we’ve got it nailed down, snap! The torque is zero.

Ahhh~~ that’s why this kind of rotation comes out stable.

(This last bit isn’t in the book, by the way — I just went on this tangent ‘cause I was curious……. T_T)

---

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