Frame of reference question: Car traveling at the equator

[jbriggs444]

How can you say that centrifugal force don't depend on velocity?

ω depend on velocity ,if you increase velocity ,you also increase ω..
ω =v/r

F=mω^2 r ... F=mv^2 /r ...these two formulas are same

In the frame that rotates with the object, the object IS NOT MOVING! Its velocity is zero.

Centrifugal force is found in a rotating frame and depends on the rotation rate of the frame, not on the velocity of any object measured against that frame.

Scroll back up a few posts to see your question: You yourself have noted that the object has zero velocity and expressed bewilderment that despite this, it is subject to centrifugal force.

 

[John Mcrain]

In the frame that rotates with the object, the object IS NOT MOVING! Its velocity is zero.

Centrifugal force is found in a rotating frame and depends on the rotation rate of the frame, not on the velocity of any object measured against that frame.

Scroll back up a few posts to see your question: You yourself have noted that the object has zero velocity and expressed bewilderment that despite this, it is subject to centrifugal force.

Rotation rate of frame compare to what?

 

[jbriggs444]

Rotation rate of frame compare to what?

Absolute rotation rate. Measured, for instance, by determining centrifugal force. Or Coriolis. There are also fancier ways like ring laser gyros.

Alternately, one might reference the rotation to the fixed stars and call it a sidereal rotation rate. It turns out that a frame in which the stars are fixed is also one that has zero centrifugal, zero Coriolis and zero Euler forces. So we can treat such a frame as a standard of zero rotation.

 

[John Mcrain]

Can we solve any problem/task with inertial frame,if does why than useing non inertial frames?

 

[jbriggs444]

Can we solve any problem/task with inertial frame,if does why than useing non inertial frames?

Any problem can be solved in any frame. Non inertial frames sometimes simplify matters. Doing weather prediction in the inertial frame would be positively painful because all of the land would be moving at hundreds of miles per hour. If you are trying to predict weather for Chicago, it helps if Chicago stays put.

 

[John Mcrain]

Any problem can be solved in any frame. Non inertial frames sometimes simplify matters. Doing weather prediction in the inertial frame would be positively painful because all of the land would be moving at hundreds of miles per hour. If you are trying to predict weather for Chicago, it helps if Chicago stays put.

Any live creature feel fictitious forces because our own reference frame moving with us all the time.
So when drive car in turn , is better to say that centrifugal force push our body out of turn ,then car seat push our body into turn(centripetal).

 

[jbriggs444]

Any live creature feel fictitious forces because our own reference frame moving with us all the time.
So when drive car in turn , is better to say that centrifugal force push our body out of turn ,then car seat push our body into turn(centripetal).

No live creatures feel fictitious forces. Fictitious forces come with reference frames. Reference frames are mathematical constructs. They have no physical effects.

Both descriptions of the situation are equally valid. Whether it is centrifugal force which keeps us in place in the face of a real physical centripetal force or whether the real physical centripetal force has the effect of accelerating us around a corner in lock-step with the car does not change the nature of the real physical force of the seat on our body.

What we feel are the strains on our body as it adjusts itself to the real physical force. Those real physical strains exist regardless of the reference frame we might use to predict or explain them.

 

[John Mcrain]

No live creatures feel fictitious forces. Fictitious forces come with reference frames. Reference frames are mathematical constructs. They have no physical effects.

Both descriptions of the situation are equally valid. Whether it is centrifugal force which keeps us in place in the face of a real physical centripetal force or whether the real physical centripetal force has the effect of accelerating us around a corner in lock-step with the car does not change the nature of the real physical force of the seat on our body.

What we feel are the strains on our body as it adjusts itself to the real physical force. Those real physical strains exist regardless of the reference frame we might use to predict or explain them.

But why you call centripetal force real physical force and centrifugal not,for men point of view?
They are both equaly real for men in car,just depend what decription you decide to use..

 

[jbriggs444]

But why you call centripetal force real physical force and centrifugal not,for men point of view?
They are both equaly real for men in car,just depend what decription you decide to use..

The centripetal force is [the horizontal component of] the force of the seat on your body. It is real. It is physical. It has a third law partner force: the force of your body on the seat. When we speak of a "centripetal force", we are not speaking about some special type of force. Instead, we are talking about the net force [from whatever source] that acts toward the center of a curved path. The centripetal force may, for example, be the force of road on tires, of seat on body, of centrifuge on test tube or of sling on stone.

The centrifugal force is the fictitious force associated with the adoption of a rotating frame. Its existence or not depends on the coordinate system that you have decided to use. It is fictitious. It has no third law partner force.

If you have adopted an inertial frame then the centrifugal force does not exist. Your body is accelerating around the corner in response to the only two forces there are: gravity and the force of the seat on your body. The net force is inward toward the center of your curved path. Your body accelerates. Newton's second law is upheld.

If you have adopted the rotating frame then the centrifugal force does exist. Your body is at rest and remains at rest throughout the exercise. There are three forces in play: gravity, the force of the seat on your body and centrifugal force. The net of these three forces acting on your body is zero. Your body remains at rest in its seat. Newton's second law is upheld.

 

[John Mcrain]

It has a third law partner force: the force of your body on the seat.

This force is called reactive centrifugal force.

From your logic gravity is also fictitous force.Is Earth in same strange sense accelrate "upward" in same manner like when elevator accelrate up weight scale show increase in reading?

 

[jbriggs444]

This force is called reactive centrifugal force.

From your logic gravity is also fictitous force.Is Earth in same strange sense accelrate "upward" in same manner like when elevator accelrate up weight scale show increase in reading?

Gravity is indeed a fictitious force. But you'll have to wait for General Relativity to make sense of that idea. The fact that you can transform your weight away by adopting a freely falling frame of reference is a key motivator for that theory.

However, that is for another day. We are here discussing classical Newtonian physics. In the Newtonian model, gravity is a real, physical, inertial force. The third law partner of the force of Earth's gravity on your body is the force of gravity from your body on the Earth.

The reactive centrifugal force you mention is not the centrifugal force that I mentioned.

 

[John Mcrain]

Gravity is indeed a fictitious force. But you'll have to wait for General Relativity to make sense of that idea.

In the Newtonian model, gravity is an real, physical, inertial force. The third law partner of the force of Earth's gravity on your body is the force of gravity from your body on the Earth.

How do you mean wait?Is general relativity proved?

 

[jbriggs444]

How do you mean wait?Is general relativity proved?

It is correct. It is just that you are better served learning Newtonian mechanics first before we start complicating matters. It does no good discussing curvature in four dimensional pseudo-Riemannian manifolds when you are hazy on how centrifugal force works.

 

[haruspex]

No live creatures feel fictitious forces.

As I mentioned before, I think most people would say they 'feel' centrifugal force. Of course, what is felt directly is the normal force from the wall or side of the car, but since no acceleration is observed the brain infers a force pushing the body against the wall.

 

[John Mcrain]

As I mentioned before, I think most people would say they 'feel' centrifugal force.

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...

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

 

[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.