Frame of reference question: Car traveling at the equator

[jbriggs444]

Of course they feel centrifugal force.
In sharp left turn your head wonts to go the right,so your left muscle in neck must contract to keep head in stright position.
So from human perspective centrifugal force exist..

If you want imitate force on your head during left turn,you can put head strap which pull you to the right...

View attachment 270706

for same reason F1 drivers lean head into turn,so head gravity helps them fighting against centrifugal force..

And yet, with a flick of pencil on paper you are using the inertial frame in which centrifugal force does not exist. So what you are feeling, though real, must not now be the centrifugal force.

 

[PeterDonis]

In sharp left turn your head wonts to go the right,so your left muscle in neck must contract to keep head in stright position.

So the force you feel is the force from your neck muscles. And also from the car, pushing on you to move with it in the left turn.

What you do not feel as a force is "your head wants to go to the right". And that is what "centrifugal force" is--it's "your head wants to go to the right" in the non-inertial frame in which the car making the left turn, and you making the turn with it, are at rest.

 

[italicus]

@haruspex

Given the minus sign, it’s opposite.
I don't know why it is given as −2mΩ×v instead of 2mv×Ω. Maybe there's some standard about writing the rotation vector first in cross products.
Source https://www.physicsforums.com/threa...on-car-traveling-at-the-equator.994297/page-2

Sorry, I am not yet able to quote.
I think the reason for the minus sign is that every inertial force is opposite to the acceleration of the NON inertial r.f. that causes it. Give a look at chapter 10 of this book by David Morin:

https://scholar.harvard.edu/files/david-morin/files/cmchap0.pdf

as a simple example (not for you, of course, but for students) let us take a bus, which is traveling at constant speed ( I should say : constant v, supposing the Earth to be an inertial r.f. for some seconds ) ; on the floor of the bus there is a bead, traveling with the bus. If the bus suddenly accelerates , the bead slips backward, and people sitting inside ( pushed against their chair) say that the bead, of a mass m, has been accelerated by a force opposite to the acceleration :

F_t = - m a

no friction force considered here.

All signs are a matter of convention, but we normally say that, in a rotating reference frame , the centrifugal (fictitious) force is given by :

$\vec F_c= - \vec\omega\times(\vec\omega\times\vec r)$

applying the right-hand rule for cross product .

With the same rule, we find the Coriolis force to be : $\vec F_ (cor) = -2m \vec\omega\times\vec v$

in which $\vec v$ is the velocity of particle (P,m) in the rotating r.f.

@JohnMcrain

when you travel at the equator from East to West, Coriolis force pushes you against earth. The opposite when you travel West to East.

There is something wrong with Latex , I am not able to fix. The main scope for my posts is to learn writing, mainly Latex, with which I have no practice. By the way, could you please tell me how to load images ?
How do you find my English ? It is poor , isn’t it?

Thank you for attention.

 

[Chestermiller]

What's the speed limit in the country where this car is traveling? Sounds like they better watch out for a speed trap.

 

[italicus]

What's the speed limit in the country where this car is traveling? Sounds like they better watch out for a speed trap.

You are right. This is only a theoretical exercise. I’ve made some simple calculations. Gravitational force at the equator is $\ mg$ , and the fictitious centrifugal force, in the rotating Earth ref. frame, is $\ m \omega^2R \approx \frac {mg}{300}$ . Coriolis force intensity, is given by $\ 2m\omega v$ . So, if you want this intensity comparable with that of the centrifugal force , you should have :

$$2\omega v \approx\omega^{2} R \rightarrow v \approx \frac{ \omega R}{2}$$

replacing numerical values :

$$v = \ 0.5 * (7.29*10^{-5} rad/s) * (6.37*10^{6}m) \approx 232 m/s = 835 km/h $$

This is the speed of a plane, not of a car.

 

[jbriggs444]

You are right. This is only a theoretical exercise. I’ve made some simple calculations. Gravitational force at the equator is $mg$ .

To a good approximation, yes. But "g" denotes the local apparent acceleration of gravity. It already contains a deduction for centrifugal force.

and the fictitious centrifugal force, in the rotating Earth ref. frame, is $\ m \omega^2R \approx \frac {mg}{300}$ . Coriolis force intensity, is given by $\ 2m\omega v$.

Yes. Let us work this out for the case of someone rotating at $-\omega$ as measured against a rotating frame which is itself rotating at $+\omega$.

The object's velocity measured against the rotating frame is $-\omega R$. The Coriolis force $mv\times R$ will then be given by $-2\omega^2R$.

At first glance, one might look at this and think "that's twice as much force as is needed to balance centrifugal force. So half that fast and he's canceled centrifugal".

Whoah there. Do the accounting more carefully. You are either jumping frames or missing a term.

In the rotating frame, he is rotating about the center at $-\omega$. This means that he has centripetal acceleration. We do not just need enough Coriolis to cancel centrifugal force. We need enough Coriolis to cancel centrifugal and explain the centripetal acceleration. [Arguably that's why there's a 2 in the formula for Coriolis]

So, if you want this intensity comparable with that of the centrifugal force , you should have :

$$2\omega v \approx\omega^{2} R \rightarrow v \approx \frac{ \omega R}{2}$$

Correcting the error described above, the result should simply be $\omega R$.

This should be no surprise. If you want to cancel centrifugal force, you have to be standing still in the inertial frame.

 

[italicus]

To a good approximation, yes. But "g" denotes the local apparent acceleration of gravity. It already contains a deduction for centrifugal force.

Sorry, I mean that “g" is simply the intensity of attractive gravitational force divided by mass, that is :

$g = GM/R^{2}$

not the apparent local acceleration, already corrected for the centrifugal acceleration, which has a maximum value at the equator . When considered in the rotating r.f. , on which we live, it is obvious that you have to sum vectorially the a.m. $\vec g$ with the apparent centrifugal acceleration. The resulting acceleration gives you the local vertical.

Yes. Let us work this out for the case of someone rotating at −ω as measured against a rotating frame which is itself rotating at +ω.
The object's velocity measured against the rotating frame is −ωR. The Coriolis force mv×R will then be given by −2ω^2R.

Why do you say : “the Coriolis force mvxR ...“? This fictitious force, which is to be considered in the rotating frame only, is given by a cross product , and a factor -2m :

$\vec F_{cor} = -2m\vec\omega\times\vec v$

in particular, factor “-2” comes out by accurately deriving velocity vectors against the rotating r.f. , please refer to the book by Morin, already cited.
$\vec F_{cor}$ is perpendicular to the plane which contains $\vec\omega$ and $\vec v$ , of course. Its modulus , is given in our case by : $F_{cor} = 2m\omega^{2}R$ , right.

At first glance, one might look at this and think "that's twice as much force as is needed to balance centrifugal force. So half that fast and he's canceled centrifugal".

Whoah there. Do the accounting more carefully. You are either jumping frames or missing a term.

No, that’s not my reasoning here, it would be wrong. May be I have not been clear here, apologises for that.
I am just comparing the intensity of centrifugal acceleration (let’s take off mass) $\omega^{2}R$ with the intensity of Coriolis acceleration $A_{cor} = 2\omega^{2}R$ , which I can write $2\omega v$ , isn’t it ?
So factor 2 is correct in this contest.

But now, let’s see an example regarding what you said. Let’s take a rotating carousel which, as seen from the above by a inertial observer, rotates counterclockwise, and put a bead on it , at a certain distance R from the center. Suppose friction is completely absent here, so that the bead keeps its “fixed” position relative to the external inertial observer; but for an observer sitting at the center of platform and rotating with it, the bead seems to describe a circumference clockwise , with tangential speed v =ωR .

Taking into account the directions of vector $\vec\omega$ (oriented from platform toward the inertial observer above it, because platform rotates counterclockwise) and of vector $\vec v$ , oriented clockwise, and also the “-“ sign, vector $\vec F_{c}$ points toward the center, while centrifugal force, also considered by the NON inertial observer, points outward. Because the first ( put m=1 for simplicity) has module $F_{cor} = 2\omega^{2}R$ and the second is $F_{centrifugal} = \omega^{2}R$ , the resultant is a force directed toward the center , that is centripetal, of module $F_{centripetal} = \omega^{2}R$ .
So, the non inertial observer thinks he has found the centripetal force that, in his r.f. , causes the bead to move in circular uniform motion of radius R , and tangential speed v = ωR .

But this is actually one more demonstration that fictitious forces are...fictitious! They don’t exist, neither in inertial frames nor in non inertial frames. No real force acts on the bead of the example , exception made for vertical equilibrated forces. So all effects that we attribute to inertial fictitious forces are, in reality, due
due to the “bad” reference frame used to describe motion.

Anyway, we live on a Non-inertial r.f. , the Earth, so we are obliged, in a certain sense, to include fictitious forces in our description of motion on Earth, if we can’t ignore, at least locally and for a short time, the non-inertiality of Earth.

Sorry for my bad English, and thank you for attention.