Rotating Reference Frames: Coriolis, Transverse, and Centrifugal Forces
We figure out the discrepancy between a stationary frame and a purely-rotating one โ and that's exactly where Coriolis, centrifugal, and transverse forces come from!
Alright, jumping right in!
Last time, the coordinate system wasn’t rotating. (We kinda quietly assumed that, lol.)
And we found the discrepancy between a stationary frame and a frame doing pure translational motion (just plain straight-line motion).
So now โ let’s find out what discrepancy shows up between a stationary frame and a frame that only rotates!!!!
Say we have a stationary frame $Oxyz$ and a purely-rotating frame $Ox'y'z'$.
(Since it’s only rotating, we can just say the two origins are the same!!)
And the rotation direction of $Ox'y'z'$ โ we can’t just say “clockwise lol” or anything dumb like that!!!!!
The direction of rotation is
$\hat{n}$
and let the angular velocity vector $\omega$ be
$$\overrightarrow{w} = \hat{n}w$$like so. heh heh heh heh
OK, two coordinate systems set up!!!! Now here โ
since the origin $O$ is the same, the position vector $\vec{r}$ and $\vec{r}'$ for some point $P$ are the same no matter which frame we look from.
$$\overrightarrow{r} = \overrightarrow{r}'$$Now, using this, we’ll do the sequential-differentiation thing like in the Galilean transformation, and find the discrepancy between $\vec{a}$ and $\vec{a}'$!
$$\overrightarrow{r} = \overrightarrow{r}' \\ x\hat{i} + y\hat{j} + z\hat{k} = x'\hat{i}' + y'\hat{j}' + z'\hat{k}'$$Now we differentiate both sides!!!
Here’s the catch.
$\hat{i}$
This guy stays the same~~ no matter how much time passes,
$\hat{i}$
but
$\hat{i}'$
is different before and after time passes!!! Because the frame is rotating!!!
That is,
$$\frac{d\hat{i}}{dt} = 0$$but
$$\frac{d\hat{i}'}{dt} \neq 0$$That’s the deal! So let’s differentiate both sides once.
$$\frac{d}{dt}\overrightarrow{r} = \frac{d}{dt}\overrightarrow{r}' \\ \frac{d}{dt}\left(x\hat{i} + y\hat{j} + z\hat{k}\right) = \frac{d}{dt}\left(x'\hat{i}' + y'\hat{j}' + z'\hat{k}'\right) \\ \frac{dx}{dt}\hat{i} + \frac{dy}{dt}\hat{j} + \frac{dz}{dt}\hat{k} = \overrightarrow{v} \quad (\text{LHS}) \\ \frac{dx'}{dt}\hat{i}' + \frac{dy'}{dt}\hat{j}' + \frac{dz'}{dt}\hat{k}' + x'\frac{d\hat{i}'}{dt} + y'\frac{d\hat{j}'}{dt} + z'\frac{d\hat{k}'}{dt} \\ = \overrightarrow{v}' + x'\frac{d\hat{i}'}{dt} + y'\frac{d\hat{j}'}{dt} + z'\frac{d\hat{k}'}{dt} \quad (\text{RHS}) \\ \text{So, } \overrightarrow{v} = \overrightarrow{v}' + x'\frac{d\hat{i}'}{dt} + y'\frac{d\hat{j}'}{dt} + z'\frac{d\hat{k}'}{dt}$$And now โ
$$\frac{d\hat{i}'}{dt}, \quad \frac{d\hat{j}'}{dt}, \quad \frac{d\hat{k}'}{dt}$$let’s figure out how the heck these guys (which aren’t zero) actually look!!!!
First, let’s find how $\hat{i}'$ changes over time $dt$.
Say it was the red one, and after time $dt$ it became the blue one.
Then the “change in $\hat{i}'$” is
$$\Delta\hat{i}' = \hat{i}' - \hat{i}'$$That’d be fine, right??
There โ
$\Delta\hat{i}'$
let’s use this to describe the $\vec{\omega}$ vector!!!
During time $dt$,
$\Delta\theta$
it rotated by that much, so the magnitude of $\Delta\hat{i}'$ is!!!!
$$|\Delta\hat{i}'| = ? \cdot \Delta\theta$$How do we define that question mark?
Defining the angle $\varphi$ between $x'$ and $\hat{n}$,
$$|?| = |i' \cdot \sin\varphi|$$let’s say.
So,
$$|\Delta\hat{i}'| = |i' \cdot \sin\varphi| \cdot \Delta\theta$$And then we do the limiting move โ divide by $\Delta t$ and send $\Delta t \to 0$!!!
$$\lim_{\Delta t \to 0}\frac{\Delta\hat{i}'}{\Delta t} = \frac{d\hat{i}'}{dt} = \lim_{\Delta t \to 0}|i' \cdot \sin\varphi| \cdot \frac{\Delta\theta}{\Delta t} = |i' \cdot \sin\varphi|\frac{d\theta}{dt} = |i' \cdot \sin\varphi|\dot{\theta} = |i' \cdot \sin\varphi|w$$Oh โ we can use the cross product here?!?!?!?!?
$$\hat{i}' \times \overrightarrow{w}$$We can write it like this, right?!?!?!?!!
Aaaaaaaaa wait no โ let’s be careful about direction!!!!
So the conclusion is
$$\frac{d\hat{i}'}{dt} = \overrightarrow{w} \times \hat{i}'$$That’s it. And~~~ by the same logic for the others,
$$\frac{d\hat{j}'}{dt} = \overrightarrow{w} \times \hat{j}'$$&&&&&
$$\frac{d\hat{k}'}{dt} = \overrightarrow{w} \times \hat{k}'$$Yes!!!
OK now let’s get back to the equations~!!! heh heh heh
$$\overrightarrow{v} = \overrightarrow{v}' + x'\frac{d\hat{i}'}{dt} + y'\frac{d\hat{j}'}{dt} + z'\frac{d\hat{k}'}{dt} \\ = \overrightarrow{v}' + x'\left(\overrightarrow{w} \times \hat{i}'\right) + y'\left(\overrightarrow{w} \times \hat{j}'\right) + z'\left(\overrightarrow{w} \times \hat{k}'\right) \\ = \overrightarrow{v}' + \overrightarrow{w} \times (x'\hat{i}' + y'\hat{j}' + z'\hat{k}') \\ = \overrightarrow{v}' + \overrightarrow{w} \times \overrightarrow{r}'$$Whoa! Beautiful!!! We found the velocity discrepancy between the two frames!!
If we write that in derivative notation,
$$\overrightarrow{v} = \overrightarrow{v}' + \overrightarrow{w} \times \overrightarrow{r}' \\ \left(\frac{d\overrightarrow{r}}{dt}\right)_{fixed} = \left(\frac{d\overrightarrow{r}'}{dt}\right)_{rot} + \overrightarrow{w} \times \overrightarrow{r}' \\ \left(\frac{d}{dt}\right)_{fixed} \cdot \overrightarrow{r} = \left\{\left(\frac{d}{dt}\right)_{rot} + \overrightarrow{w}\times\right\} \cdot \overrightarrow{r}' \\ = \left\{\left(\frac{d}{dt}\right)_{rot} + \overrightarrow{w}\times\right\} \cdot \overrightarrow{r} \quad \because \overrightarrow{r} = \overrightarrow{r}' \\ cf.)) \quad \left(\frac{d}{dt}\right)_{rot} : \text{ described w.r.t. time changes in the rotating frame} \\ \text{since it's expressed in } x'y'z' \text{ components, it's different from } \left(\frac{d}{dt}\right)_{fixed}$$We found the con-nect-ing lin-k~~ between the fixed frame and the rotating frame!!
$$\left(\frac{d}{dt}\right)_{fixed} = \left\{\left(\frac{d}{dt}\right)_{rot} + \overrightarrow{w}\times\right\}$$Let’s write this as an operator between fixed and rotating!!!!
Writing it as an operator just means we can apply it to any other arbitrary vector~~.
And as for “any other arbitrary vector,” you can think of it as: looking at something from the stationary frame, but viewed via the rotating frame!!!
Because we’re literally just plugging any vector in, lol.
Let’s plug in our velocity vector $\vec{v}$ and find the acceleration vector.
$$\text{Apply both sides of } \left(\frac{d}{dt}\right)_{fixed} = \left\{\left(\frac{d}{dt}\right)_{rot} + \overrightarrow{w}\times\right\} \text{ to } \overrightarrow{v} \\ \left(\frac{d}{dt}\right)_{fixed}\overrightarrow{v} = \left\{\left(\frac{d}{dt}\right)_{rot} + \overrightarrow{w}\times\right\}\overrightarrow{v} \\ \left(\frac{d\overrightarrow{v}}{dt}\right)_{fixed} = \left\{\left(\frac{d}{dt}\right)_{rot} + \overrightarrow{w}\times\right\}(\overrightarrow{v}' + \overrightarrow{w} \times \overrightarrow{r}') \quad \text{(reason for this swap: we need to differentiate with } \left(\frac{d}{dt}\right)_{rot} \text{, which is in } x'y'z' \text{ components)} \\ \overrightarrow{a} = \left(\frac{d}{dt}\right)_{rot}(\overrightarrow{v}' + \overrightarrow{w} \times \overrightarrow{r}') + \overrightarrow{w} \times (\overrightarrow{v}' + \overrightarrow{w} \times \overrightarrow{r}') \\ = \left(\frac{d\overrightarrow{v}'}{dt}\right)_{rot} + \left(\frac{d}{dt}\right)_{rot}(\overrightarrow{w} \times \overrightarrow{r}') + \overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ = \left(\frac{d\overrightarrow{v}'}{dt}\right)_{rot} + \left(\frac{d\overrightarrow{w}}{dt}\right)_{rot} \times \overrightarrow{r}' + \overrightarrow{w} \times \left(\frac{d\overrightarrow{r}'}{dt}\right)_{rot} + \overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}')$$That red bit there? Doesn’t quite make sense.
We can’t just go differentiating $\vec{w}$ (which is expressed in frame components) with $(d/dt)_{rot}$ all willy-nilly….. heh heh heh
So we need to verify that the omega vector behaves nicely under the operator!!!
Not hard, let’s just do it.
$$\left(\frac{d}{dt}\right)_{fixed}\overrightarrow{w} = \left\{\left(\frac{d}{dt}\right)_{rot} + \overrightarrow{w}\times\right\}\overrightarrow{w} \\ \left(\frac{d\overrightarrow{w}}{dt}\right)_{fixed} = \left(\frac{d\overrightarrow{w}}{dt}\right)_{rot} + \overrightarrow{w} \times \overrightarrow{w} \,(= 0) \\ \overrightarrow{\dot{w}} = \left(\frac{d\overrightarrow{w}}{dt}\right)_{rot}$$That’s why we get to write it as $\dot{\vec{w}}$!!
It’s not just blindly differentiated!!
OK now let’s hit the final result!!!!!
$$\overrightarrow{a} = \left(\frac{d\overrightarrow{v}'}{dt}\right)_{rot} + \left(\frac{d\overrightarrow{w}}{dt}\right)_{rot} \times \overrightarrow{r}' + \overrightarrow{w} \times \left(\frac{d\overrightarrow{r}'}{dt}\right)_{rot} + \overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ = \left(\frac{d\overrightarrow{v}'}{dt}\right)_{rot} + \overrightarrow{\dot{w}} \times \overrightarrow{r}' + \overrightarrow{w} \times \left(\frac{d\overrightarrow{r}'}{dt}\right)_{rot} + \overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ = \overrightarrow{a}' + \overrightarrow{\dot{w}} \times \overrightarrow{r}' + \overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ = \overrightarrow{a}' + \overrightarrow{\dot{w}} \times \overrightarrow{r}' + 2\overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}')$$Shall we multiply both sides by mass $m$? heh
$$m\overrightarrow{a} = m\overrightarrow{a}' + m\overrightarrow{\dot{w}} \times \overrightarrow{r}' + 2m\overrightarrow{w} \times \overrightarrow{v}' + m\overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ \overrightarrow{F} = \overrightarrow{F}' + m\overrightarrow{\dot{w}} \times \overrightarrow{r}' + 2m\overrightarrow{w} \times \overrightarrow{v}' + m\overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ \overrightarrow{F} = \overrightarrow{F}' + \text{ 'some discrepancies'}$$Unlike the case where the frame was just doing plain linear motion, a bunch of fictitious-force terms show up…….
Let’s go through them one by one.
First!
$$2m\overrightarrow{w} \times \overrightarrow{v}'$$is called the Coriolis force.
OK, how should we picture this intuitively?
You drop a marble onto a rotating disk.
Someone watching from the fixed frame just sees the marble go plop! โ straight down, right!
But say that someone was Spider-Man. And he climbs onto the disk and hangs on while it spins, watching the falling marble????
It’d probably look like a parabola???, right??!!
Try to picture it well…. if you can’t picture it well, you’d better go ride one of those spinning teacup rides!!!! heh heh heh
$$m\overrightarrow{\dot{w}} \times \overrightarrow{r}'$$This term is called the transverse force (a.k.a. the Euler force),
and we can think about it like this:
You’re standing like that โ and suddenly!!!!!!! the disk starts spinning?? You’d lose balance and topple over, right?
That’s exactly that force. heh heh heh heh
Oh right โ and if
$$\overrightarrow{\dot{w}} = 0$$then naturally~ this guy doesn’t show up. You knew that already, right?
$$m\overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}')$$And this term โ well, this is the centrifugal force we all know and love~~!!
Were you wondering why it hadn’t shown up yet??? The most familiar force of them all!! heh heh heh The one that pushes you outward~~
You know those amusement park rides where you sit on the octopus arms and spin round and round โ
or those tent-things that open and close while spinning, you really feel it on those..?!! lol
The transverse force and centrifugal force are forces we’ve (very much) felt in our bodies,
but the Coriolis force is one we don’t really feel in everyday life….. heh heh
The reason is โ well, it only shows up when
$$\overrightarrow{v}' \neq 0$$and on top of something that’s already spinning, it’s not exactly easy to move around, sooo, yeah… we don’t really get to feel it… heh heh
So we’ve been looking at “pure rotation,”
but now let’s make it rotate while flying!!!!! heh heh heh (kinda like a slider? lol)
Or maybe picture this:
A fire truck is parked, standing still, and suddenly it swings its ladder around,,,, like that…. heh heh heh? If that’s what we had so far, then now โ
the fire truck is also driving forward while doing that nonsense……. heh heh heh
But this situation isn’t really that hard, heh heh.
Just take “the discrepancy from the rotating frame” + “the discrepancy from the moving frame” relative to the fixed frame, and that’s it, heh heh heh.
So the relationship between the force $\vec{F}$ as seen by an inertial-frame observer and the force $\vec{F}'$ as seen by someone living in the non-inertial frame is!!!
$$\overrightarrow{F} = \overrightarrow{F}' + \text{ 'some discrepancies'} \\ = \overrightarrow{F}' + m\overrightarrow{\dot{w}} \times \overrightarrow{r}' + 2m\overrightarrow{w} \times \overrightarrow{v}' + m\overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') + m\overrightarrow{A}_0$$Let’s run a couple of problems and wrap up!!
Example 5.2.1
A wheel of radius $b$ rolls along the ground at constant velocity $V_0$. Find the acceleration, relative to the ground, of a point on the rim of the wheel.
First, setting up axes โ nobody’s going to object to setting up the $O'x'y'z'$ axis like that, right?
OK then โ what we want to find is
That’s what we’re after, right? heh heh heh
Let the position vector of that red dot be
$\overrightarrow{r}'$
like that.. lol
And since we said the radius of the circle is $b$,
$$\overrightarrow{r}' = b\hat{i}'$$that’s gotta be it, heh.
Now if we’re looking at the red dot from the rotating blue axis, what does it look like??
At the very least, “it looks like it’s spinning” โ that’s not it! heh heh heh
$$\overrightarrow{v}' = 0$$And then naturally
$$\overrightarrow{a}' = 0$$that’s gotta be it, heh heh.
OK OK OK OK, and what about the omega vector??????????
$$w = w\hat{k}' = \frac{V_0}{b}\hat{k}' \quad \left(\because w = \overrightarrow{r} \times \overrightarrow{v}\right)$$Now then โ we’re asked for the acceleration $\vec{a}$ relative to the ground, so let’s use that con-nect-ing lin-k operator!!!
Plug everything in~~~ and apply,
$$\overrightarrow{a} = \overrightarrow{a}' + \overrightarrow{\dot{w}} \times \overrightarrow{r}' + 2\overrightarrow{w} \times \overrightarrow{v}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}')$$that’s what we get, but apply $\vec{v}'=0$ and $\vec{a}'=0$.
Then
$$\overrightarrow{a} = \overrightarrow{\dot{w}} \times \overrightarrow{r}' + \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}')$$But also, since it’s uniform circular motion, $\dot{\vec{w}} = 0$!!!
$$\therefore \overrightarrow{a} = \overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ = \frac{V_0}{b}\hat{k}' \times \left(\frac{V_0}{b}\hat{k}' \times b\hat{i}'\right) \\ = \frac{V_0}{b}\hat{k}' \times \left(V_0\hat{j}'\right) = \frac{V_0^2}{b}\left(-\hat{i}'\right)$$Done~~~
Example 5.3.1
A bug is crawling outward at $v'$ along the spoke of a wheel that’s rotating at constant angular velocity $w$ about a vertical axis.
Find the force acting on this bug~~~
We’d love to just spray bug spray, but first~ lololol.
Reading the problem straight, it’s basically: how much centrifugal force, Coriolis force, and transverse force is this thing getting hit with~~~
Seems like that’s what we’re being asked to figure out.
First, let’s interpret the setup:
This is a bicycle wheel. A nasty little bug has crawled onto the wheel.
Ugh!!! I hate bugs the most.
The bug is crawling outward at $v'$, and I, wanting to fling the bug off,
if I spin the wheel at constant angular velocity $w$, how much force does the dang bug experience โ
and will it actually fly off???heh heh (the problem itself is easy enough, lol)
Setting up the rotating axis like that, let the bug’s position be the position vector
$\overrightarrow{r}'$
like so!
$$\overrightarrow{\dot{r}}' = \overrightarrow{v}'$$It’s constant!!! Therefore
$$\overrightarrow{a}' = 0$$@@@@@
And the angular velocity when the wheel rotates is
$$\overrightarrow{w} = w\hat{k}'$$let’s say.
<Hmm, feels like we could also throw in the case where the wheel is rolling forward.,,, heh heh heh? Weird.>
OK then,
$$\overrightarrow{F} = \overrightarrow{F}' + m\overrightarrow{\dot{w}} \times \overrightarrow{r}' + 2m\overrightarrow{w} \times \overrightarrow{v}' + m\overrightarrow{w} \times (\overrightarrow{w} \times \overrightarrow{r}') \\ \text{what we can kill: } \dot{w}=0 \text{ and } \overrightarrow{a}'=0 \\ = 2mwv'(\hat{k}' \times \hat{i}') + mw\hat{k}' \times \left(wr'(\hat{k}' \times \hat{i}')\right) \\ = 2mwv'\hat{j}' + mw\hat{k}' \times wr'\hat{j}' \\ = 2mwv'\hat{j}' + mw^2(-\hat{i}')$$
So it’s getting hit with force $\vec{F}$ equal to the blue arrow!!!!!
Shall we find the magnitude of this $\vec{F}$? heh heh heh
$$|\overrightarrow{F}| = \sqrt{(2mwv')^2 + (mw^2r')^2}$$Aha~ if this little bug isn’t falling off, we need to crank up $w$ so it gets hit harder!!!!
And if the bug were just sitting still, $v'=0$, so the force would drop way off….
Still, $r'$ rides on $w^2$, so we’d have to spin like crazy for it to actually fly off.
Whatever, bug spray is just the answer, lol.
Lol that was fun.
OK, see you next time~~~!! lol
Classical Mechanics I (the one I studied) is almost wrapping up! heh heh heh heh
Originally written in Korean on my Naver blog (2015-06). Translated to English for gdpark.blog.