Euler Angles (Part 2)

Picking up right where we left off…

Out of nowhere, this equation just appears:

$$\overrightharpoonup{w} \quad=\quad \overrightharpoonup{\dot{\theta}} \quad+\quad \overrightharpoonup{\dot{\varphi}} \quad+\quad \overrightharpoonup{\dot{\psi}}$$

And the prof goes: “Don’t try to understand this one~~~~”

Wait what.

I HATE memorizing things!!!!

Fine. Fine. Just gonna swallow it and move on T_T T_T T_T T_T

Whatever!!

But OK — at least I think I get what each piece means:

$$\dot{\psi} \quad:$$

How fast is the top spinning?!

$$\dot{\theta} \quad:$$

How fast is the top’s rotation axis tilting away from the vertical (or tipping back toward it).

Basically — how fast is the top losing its balance and falling over??

$$\dot{\varphi} \quad:$$

This one’s how fast the rotation axis itself swings around!!

I took a video for this:

The pen is the top’s rotation axis. The axis! The axis! THE AXIS!!!! Think of it as the 3-axis, or the z’-axis, yeah?

So $\dot{\varphi}$ is how fast that pen swings around like that, right???

OK now let’s figure out the directions of these vectors.

$$\overrightharpoonup{\dot{\theta}} \quad:$$

Direction?? The x’-axis, right~~~?

$$\overrightharpoonup{\dot{\varphi}} \quad:$$

Direction??? Straight up out of the floor — the z-axis!! (or straight down into it, take your pick)

$$\overrightharpoonup{\dot{\psi}} \quad:$$

Direction’s gotta be the same as the z’(3)-axis~~~~~~~~~~~~~~~~~~~~~~~???????

So $\psi$ is, as expected, the spin. Spin spin.

Which means — let’s strip the spin out and look at what’s left!! What you get with $\psi$ peeled off isn’t $\omega$ anymore, it’s $\omega'$:

$$\overrightharpoonup{w'} \quad=\quad \overrightharpoonup{\dot{\theta}} \quad+\quad \overrightharpoonup{\dot{\varphi}}$$

Like that.

And the relationship between $\omega$ and $\omega'$ is painfully obvious but~~~~

$$\overrightharpoonup{w} \quad=\quad \overrightharpoonup{w'} \quad+\quad \overrightharpoonup{\dot{\psi}}$$

Cool, done~~~ Now let’s express these vectors in terms of the unit vectors $(\hat{e}_1, \hat{e}_2, \hat{e}_3)$ of the O123 axes.

You can totally get there by projecting step by step — so just follow along, projection by projection!!

Here we go:

$$\overrightharpoonup{\dot{\varphi}} \quad=\quad \dot{\varphi}\cos\theta\,\hat{z}' \quad+\quad \dot{\varphi}\sin\theta\,\hat{y}'\\ \overrightharpoonup{\dot{\theta}} \quad=\quad \dot{\theta}\,\hat{x}'\\ \overrightharpoonup{\dot{\psi}} \quad=\quad \dot{\psi}\,\hat{z}'$$$$\overrightharpoonup{w} \quad=\quad \overrightharpoonup{\dot{\theta}} \quad+\quad \overrightharpoonup{\dot{\varphi}} \quad+\quad \overrightharpoonup{\dot{\psi}} \quad=\quad \dot{\varphi}\cos\theta\,\hat{z}' \quad+\quad \dot{\varphi}\sin\theta\,\hat{y}' \quad+\quad \dot{\theta}\,\hat{x}' \quad+\quad \dot{\psi}\,\hat{z}'\\ \quad=\quad \dot{\theta}\,\hat{x}' \quad+\quad \dot{\varphi}\sin\theta\,\hat{y}' \quad+\quad (\dot{\varphi}\cos\theta + \dot{\psi})\,\hat{z}'$$

Now dot both sides with $(\hat{e}_1, \hat{e}_2, \hat{e}_3)$:

$$\overrightharpoonup{w}\cdot(\hat{e}_1, \hat{e}_2, \hat{e}_3) \quad=\quad \left\{ \dot{\theta}\,\hat{x}' \quad+\quad \dot{\varphi}\sin\theta\,\hat{y}' \quad+\quad (\dot{\varphi}\cos\theta + \dot{\psi})\,\hat{z}' \right\} \cdot (\hat{e}_1, \hat{e}_2, \hat{e}_3) \\ \overrightharpoonup{w} \quad=\quad \left( \dot{\theta}\cos\psi \right)\hat{e}_1 \quad+\quad \left( \dot{\varphi}\sin\theta\cos\!\left(\tfrac{\pi}{2}-\psi\right) \right)\hat{e}_1\\ \quad+\quad \left( -\dot{\theta}\cos\!\left(\tfrac{\pi}{2}-\psi\right) \right)\hat{e}_2 \quad+\quad \left( \dot{\varphi}\sin\theta\cos\psi \right)\hat{e}_2 \quad+\quad \left( \dot{\varphi}\cos\theta + \dot{\psi} \right)\hat{e}_3\\ \quad=\quad \left( \dot{\theta}\cos\psi + \dot{\varphi}\sin\theta\sin\psi \right)\hat{e}_1 \quad+\quad (\dot{\varphi}\sin\theta\cos\psi - \dot{\theta}\sin\psi)\hat{e}_2 \quad+\quad (\dot{\varphi}\cos\theta + \dot{\psi})\hat{e}_3$$

And BOOM — we successfully ripped the $\omega$ vector to shreds and rewrote it in the unit vectors of the O123 axes~~~~~

Why did we do all that???

Because the principal axes are precious… heh.

Once you’ve got this much under your belt, the rest of the book — precession rate, top motion, all that — is easy to follow along.

So I’m just gonna boldly skip posting it.

Classical mechanics is so brutally hard T_T T_T T_T — I made a deal with myself to only post the parts where the concept is the hard bit!

Still, let’s keep grinding together!!!!

Everyone, fighting!!!!

Honestly kinda sad to end the classical mechanics posts here… T_T T_T But —

I never took Classical Mechanics 1, you know???

Catch you again in spring ‘15, lol lol lol lol lol

Oh and — thanks to this whole study-by-posting strategy, I came in 1st in every physics major class this semester

and pulled an A+ across the board! lol lol lol

Yessss!

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