Rotating Reference Frames: Coriolis, Transverse, and Centrifugal Forces [Classical Mechanics I Studied #7]

Let’s dive right in!

Previously, the coordinate system wasn’t rotating. (Actually we secretly assumed that, lol.) And we found the discrepancy between a coordinate system that stays still and one doing translational motion (just straight-line motion).

So now let’s find out what kind of discrepancy there is between a coordinate system that stays still and one that does only rotation!!!!

Let’s say we have a coordinate system Oxyz that stays still and a coordinate system Ox’y’z’ that only rotates. (Since it only rotates, we can say that the two origins are the same!!!)

And for Ox’y’z’s direction of rotation, we can’t just say “clockwise” lol!!!!! The direction of rotation is $\hat{n}$ and the angular velocity vector omega is $\vec{w} = \hat{n}w$ , let’s say hehehehe

Fixed axes x y z in black, rotating axes x’ y’ z’ in red, and the green rotation axis with the angular velocity vector w

Alright, we’ve established two different coordinate systems!!!! Now here, because point O (origin) is the same, we can get the result that, from either coordinate system, the position vectors r and r’ of some point P are the same: $\vec{r} = \vec{r}'$ .

Now with this, like we did with the Galilean transformation, we’ll take successive derivatives and then find the discrepancy between a and a'!

The position vector r equals r’, written out in components

So now we’ll take successive derivatives of the left and right sides!!! There’s this difference. $\hat{i}$ This one, even as time passes, is continuou~~sly $\hat{i}$ , whereas $\hat{i}'$ is different before and after as time passes!!! Because the coordinate system is spinning!!!

That is, $\frac{d\hat{i}}{dt} = 0$ whereas $\frac{d\hat{i}'}{dt} \neq 0$ is what we have! So let’s take the derivative of both sides once~

Differentiating the components of r = r’ on both sides, giving v = v’ plus the terms x’ di’/dt, y’ dj’/dt and z’ dk’/dt

And now, $\frac{d\hat{i}'}{dt}, \frac{d\hat{j}'}{dt}, \frac{d\hat{k}'}{dt}$ let’s find out how on earth we can describe these guys that aren’t zero!!!!

First, let’s figure out how i’ changes during time dt.

Vector diagram of the unit vector i’ before (red) and after (blue) a small rotation by delta theta about the axis n

Let’s say it was red, and after time dt it turned blue. Then ’the change in i’’ is $\Delta\hat{i}' = \hat{i}' - \hat{i}'$ , we can write it like this, right?? Over there, $\Delta\hat{i}'$ let’s describe the vector omega using this!!! During time dt, it rotated by $\Delta\theta$ , so the magnitude of delta-i-prime is!!!!

How should we define that question mark ? over there?

Defining the angle Φ between x’ and n, $|?| = |i'\sin\varphi|$ let’s say. Therefore,

And then let’s divide by Δt and perform the limit operation sending delta-t to 0!!!

Dividing by delta t and taking the limit gives di’/dt as the magnitude of i’ sin phi times d theta/dt

Hey, looks like we can make use of the cross product here?!?! $\hat{i}' \times \vec{w}$ We can write it like this!?!? Aaaaaaaah nope, let’s be careful about the direction!!!! So the conclusion is

Result: di’/dt equals w cross i’

this is what we get. That is ~~~ by the same principle for the others too,

The same principle: dj’/dt equals w cross j’

&&&&&

And dk’/dt equals w cross k’

Yes!!! So now let’s go back to the equation~!!! hehehe

The velocity: v equals v’ plus x’ di’/dt plus y’ dj’/dt plus z’ dk’/dt, and with the cross products v’ plus x’ (w cross i’) plus y’ (w cross j’) plus z’ (w cross k’)

Wow! Neat!!! We found the discrepancy in velocity between the two coordinate systems!! If we write that in derivative notation,

In derivative notation: the fixed-frame derivative of r equals the rotating-frame derivative of r’ plus w cross r’, and the same relation written as an operator on r

We found the con.nec.ting.link~~ between the fixed coordinate system and the rotating coordinate system!!

The operator relation: the fixed-frame d/dt equals the rotating-frame d/dt plus w cross

Let’s use it as an Operator between fixed and rotating!!!! Using it as an operator means ~~ that it can be applied to any other arbitrary vector as well. You can think of it as: some arbitrary other vector, what it looks like in the fixed coordinate system when viewed from the rotating coordinate system, something like that!!! Since you can just stick any vector into it lol

Let’s put our velocity vector v in and find the acceleration vector.

The operator applied to v on both sides, then expanded step by step into the acceleration a, with the red term needing the operator treatment

That red part over there is a bit off. With d/dt(rot), you can’t just differentiate w expressed in coordinate-system components….. hehehe So we need to handle the omega vector using the operator!!! It’s not hard, let’s do it~

The operator applied to w itself, which gives the same result as w-dot

Because of this, we can write that as omega-dot!!! It’s not just casually differentiated!! Now let’s go to the final result!!!!!

The acceleration a collected into rotating-frame terms

Let’s multiply both sides by mass m~ heh

Multiplying by the mass m: m a = m a’ plus the extra terms, so F = F’ plus some discrepancies

Unlike when the coordinate system just does straight-line motion, a lot of fictitious force terms appear…….

One thing first: the terms above sit on the inertial side of $F = ma$ (they are what the inertial acceleration picks up in rotating coordinates). Move them over to the rotating frame’s side of Newton’s law and each one changes sign; those negatives are the fictitious forces we feel there: $-2m\vec\omega\times\vec v'$, $-m\dot{\vec\omega}\times\vec r'$ and $-m\vec\omega\times(\vec\omega\times\vec r')$.

Let’s tear them apart one by one~ First! $2m\vec{w} \times \vec{v}'$ with its minus sign is the Coriolis force.

How should we intuitively understand this?

A marble drops straight down onto a rotating disk while a stick figure outside the disk watches

You drop a marble on a disk. Someone looking from just the fixed coordinate system will see the marble drop straight from top to bottom, thud! right! But that dude was Spider-Man. Then, if he clings to the disk and watches the falling marble????

The same rotating disk with the observer riding on it and watching the marble fall

Probably a parabola??? is how it’ll look!! Try to imagine it well…. if you can’t imagine it well, you’d better go ride a Disco Pang Pang (a spinning Korean carnival ride)!!!! hehehe

$m\dot{\vec{w}} \times \vec{r}'$ With its minus sign, this term gives the transverse force. I’ll think about it like this

A stick figure standing on the rim of a disk

You’re standing like this, and if the disk is sud!!!!!!denly spun?? The person would lose their balance and fall over, right? That’s exactly that force. hehehehe

Oh right. And if $\dot{\vec{w}} = 0$ then naturally~ it doesn’t appear — you already knew that, right?

$m\vec{w} \times (\vec{w} \times \vec{r}')$ And well, with its minus sign this term is the familiar centrifugal force~~!! (Without the minus sign it points inward, toward the axis; the minus sign is what makes it point outward.)

Were you wondering why it didn’t come up???

The most familiar force to us!! hehehe The force pointing outward~~ Among amusement park rides, there’s one where you sit on the octopus legs and it spins you round and round, and there’s this thing like a tent that opens and closes — you feel it a lot when you ride that..?!! lol

We’ve felt transverse force and centrifugal force (strongly) with our bodies, but it seems like we haven’t (strongly) felt the Coriolis force….. hehe The reason is, well $\vec{v}' \neq 0$ only when this is the case does the force arise, and on something that’s spinning, well… it’s not easy for us to move, so I think we haven’t felt it strongly… hehe

We’ve now looked at “only rotation,” so now let’s make it fly while rotating!!!!! hehehe (like a slider, maybe? lol) Or you can imagine something like this

A sketch of a fire truck with its ladder raised and rotating

It’s sitting still and suddenly a fire truck rotates its ladder,,,, hehehe? If up until now it was that kind of situation, now the fire truck runs forward while doing that shit……. hehehe But this situation isn’t very hard hehe

You just do “discrepancy with the rotating coordinate system” + “discrepancy with the moving coordinate system” from the fixed coordinate system, and that’s it heheh

So the relationship between the force F seen by an observer in an inertial frame and the force F’ seen by someone living in a non-inertial frame is!!!

F equals F’ plus some discrepancies, written out as the sum of the extra force terms and the translation term m A0

Let’s solve a few problems and wrap up!!

Example 5.2.1. A wheel of radius b moves forward along the ground at constant velocity V0. Find the acceleration of a point on the wheel’s circumference with respect to the ground.

A wheel rolling along the ground at speed V0 with the rotating axes x’ y’ z’ attached to its centre

First, setting up the axes — nobody has any complaints about taking the O’x’y’z’ axis like that, right? Now, what we need to find is

A circle with a red point on its rim and its acceleration vector a

finding this, right? hehehe

Let’s call the position vector of that red point $\vec{r}'$ ..lol And since we said the radius of that little circle is b, $\vec{r}' = b\hat{i}'$ it would be this.

Now if we’re looking at the red point from the rotating blue axis, how would it look??

At least it’s not going to be “it looks like it’s rotating.”! hehehe $\vec{v}' = 0$ That’s it. Then naturally $\vec{a}' = 0$ this as well hehe

Alright alright alright alright, and the omega vector is??

Now, since we’re being asked for the acceleration a with respect to the ground, let’s use that con.nec.ting.link operator!!! If we write it all~~ out

it would be this, but let’s apply v’=0 and a’=0. Then

But since it’s also doing uniform circular motion, dot(w) = 0!!!

With w-dot zero, a equals w cross (w cross r’), worked out for the wheel

Done~~~

Example 5.3.1. A bug is crawling outward at v’ along the spoke of a wheel that rotates at constant angular velocity w about a vertical axis. Find the force acting on this bug~~~

You should spray insecticide, but before that~ lol Looking at the problem, it seems like it’s asking how much centrifugal force, Coriolis force, and transverse force it experiences~~~ the problem is asking to figure that out.

First, let’s interpret the problem

A bicycle wheel with a bug crawling outward along a spoke with speed v’

This is a bicycle wheel. A little bug has stuck onto the wheel.

Ugh!!! hate them the most

The bug is crawling outward at v’, and in order to get rid of the bug, I spin the wheel at a constant angular velocity w.

How much force does the little bug receive, and will it fall off as a result???

heh (easy to understand the problem right, hehe)

The wheel with the rotating axes x’ and y’ and the wheel labelled

Let’s set the rotating axis like that and call the bug’s position vector $\vec{r}'$ ! $\dot{\vec{r}}' = \vec{v}'$ — it’s constant!!! Therefore $\vec{a}' = 0$ . @@@@@

And when the wheel rotates, the angular velocity is $\vec{w} = w\hat{k}'$ , let’s say this.

<Hmm, it seems like we could also add a situation where the wheel is rolling forward,,,hehehe? Weird>

Alright so then

The force F written out for the bug, ending in 2mwv’ j’ plus m w squared times minus i’

A force diagram with the blue force F as the sum of the red components 2mwv’ and m w squared

It was receiving a force F equal to the blue part!!!!! Let’s find the magnitude of this force F once hehehe

Aha~ If this little bug doesn’t fall off, we need to make w bigger so it receives a bigger force!!!!

And if this guy stays still, v’=0 and the force it receives gets drastically reduced…. Still, it depends on r’ times w squared, but it seems like we’d have to spin it crazy hard for the kid to fall off — yeah right, my ass, insecticide is just the answer lol

hahahahaha, this is fun. Well then, see you next time~~~!! lol

My “General Mechanics Vol. I that I studied” is almost coming to an end! hehehehe

Heads-up. The figures here are the original notes, slips and all. A few things to watch for:

The first line of the velocity picture has a doubled equals sign: it reads $\vec v = \vec v' + \vec\omega\times\vec r'$.
In the rolling-wheel picture the reason in brackets should be $\vec v = \vec\omega\times\vec r$ (so $\omega = V_0/b$), not $\vec\omega = \vec r\times\vec v$.
In the last line of the force picture the $r'$ is missing: the centripetal term is $m\omega^2 r'(-\hat i')$, as the magnitude $|\vec F|$ that follows uses.

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