I Coriolis force, real or just an illusion?


So does then huricans really rotate or just appear to rotate? Does wind particles in huricane feel centrifugal force because they move in curved path or maybe not because they actually go in straight line?

I never understand coriolis force 100%,is wind particles really rotate or this is just illusion becuase we see it from diffrent referent point?
 

A.T.

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Then you should try to understand it based on simpler examples:


Hurricanes, or fluid dynamics in general, are not the best starting point to learn about a simple mechanical principle, because they arise from a complex interaction of many forces.

We had this whole topic just recently:
https://www.physicsforums.com/threads/conceptual-question-about-the-coriolis-force-and-the-weather.959432/
I will ask just one simple question to your video..
If I "sit" inside this cannonball will I feel centrifugal force,(just like when drive car fast in corner) or not?

If I feel centrifugal force that mean that cannonball is curving,if not, that mean is going in straight line..
 

A.T.

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If I "sit" inside this cannonball will I feel centrifugal force,(just like when drive car fast in corner) or not?
In the car you feel the deformation from the interaction force by the car on you.

If I feel centrifugal force that mean that cannonball is curving,if not, that mean is going in straight line..
The ball is inertial here, so the passenger would not feel any forces from the walls.
 

russ_watters

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So does then huricans really rotate or just appear to rotate? Does wind particles in huricane feel centrifugal force because they move in curved path or maybe not because they actually go in straight line?
Yes, hurricanes really rotate, but it doesn't make sense to talk about centrifugal force because hurricanes are big and slow and the coriolis effect isn't really what causes the rotation anyway.
 
In the car you feel the deformation from the interaction force by the car on you.

The ball is inertial here, so the passenger would not feel any forces from the walls.
Ok that mean that cannonball is going in straight line and coriolis force is just optical illusion because of diffrent referent point of view..
 

PeroK

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Ok that mean that cannonball is going in straight line and coriolis force is just optical illusion because of diffrent referent point of view..
If you try to move the ball in a straight line in the rotating frame, then you have to deal with real forces. If for example you picked up the cannonball from where it landed and carried it back to the cannon. Then you'd have a real Coriolis force to deal with.
 
Why coriolis force do not exist on equator?

Why coriolis force exists when plane fly on same latitude?
 

A.T.

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boneh3ad

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I feel a discussion of so-called "fictitious forces" is relevant here. Coriolis and centrifugal forces both fall into this category in that they are not true forces but inertial effects. They only appear as forces in the equations when using a non-inertial frame of reference.
 

FactChecker

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I feel a discussion of so-called "fictitious forces" is relevant here. Coriolis and centrifugal forces both fall into this category in that they are not true forces but inertial effects. They only appear as forces in the equations when using a non-inertial frame of reference.
Good point. Suppose we call it the Coriolis effect. It certainly is the reason for the spin of a hurricane. It is easily understood by looking at the desired inertial path of particles being drawn North and South toward a hurricane by the low pressure. Likewise, the sustained low pressure can be understood by the desired inertial path of particles circling in a hurricane. All that together explains not only the spin of the hurricane, but also the stability of the spin in combination with the low pressure.
 

kuruman

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And here is an elaboration of the points made by @A.T. and @boneh3ad. Suppose you build a very tall tower of height ##h## on the equator somewhere in SE Asia. The figure shows the Earth in an inertial frame view looking down at the north pole; the arrow indicates the sense of the Earth's rotation. You stand at the base of the tower and ask a friend to climb to the top stick his arm out and release a red ball straight down.

Neglecting air resistance, where will you see the ball land on the Earth?
(A) At a point behind the base of the tower.
(B) At a point right at the base of the tower.
(C) At a point ahead of the base of the tower.

Polar View.png


The correct answer is (C).
Explanation: While the ball is in the air, the only force on it is the radial force of gravity; there is no tangential acceleration. Before the ball is released it has a higher tangential speed vball = Ω (RE + h) than the base of the tower vbase = Ω RE in the inertial frame (RE = Earth's radius). That higher speed is retained throughout the flight, so the ball will land ahead of the base of the tower. You, in the non-inertial frame, would expect the ball to land right at the base of the tower because you know that the only force acting on the ball is gravity which is along the tower. The fact that it does not, leads you to the conclusion that there must be some kind of horizontal force other than gravity acting on the ball. This is what @A.T. calls the Eötvös effect and @boneh3ad calls "inertial effects".
 

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vanhees71

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The expression "fictitious forces" is the most misleading expression in introductory physics. As the very examples from meteorology show the Coriolis force is everything else than fictitious but very real. It's of course depending on the interpretation, whether you count it to the accerlation with respect to an inertial frame expressed in coordinates defined in the non-inertial (rotating) frame or whether you bring it on the right-hand side of the equations of motion and reinterpret as additional force. I'd call this kind of forces in this more common 2nd interpretation "inertial forces" rather than "fictitious forces". Both interpretations are of course physically equivalent.
 
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The expression "fictitious forces" is the most misleading expression in introductory physics. As the very examples from meteorology show the Coriolis force is everything else than fictitious but very real.
Yes, it is real, but it is not a force. The term "inertial forces" doesn't include any hint that these "forces" are actually no forces. That makes it misleading as well.

Just teach newcomers not to confuse terms with definitions. That solves all problems with potentially misleading terms at once.
 

FactChecker

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The expression "fictitious forces" is the most misleading expression in introductory physics. As the very examples from meteorology show the Coriolis force is everything else than fictitious but very real. It's of course depending on the interpretation, whether you count it to the accerlation with respect to an inertial frame expressed in coordinates defined in the non-inertial (rotating) frame or whether you bring it on the right-hand side of the equations of motion and reinterpret as additional force. I'd call this kind of forces in this more common 2nd interpretation "inertial forces" rather than "fictitious forces". Both interpretations are of course physically equivalent.
IMHO, it is bad to say that there are forces acting on all stationary (in an inertial reference frame) objects just because one wants to use an accelerating/rotating reference frame. The forces would be dependent on the reference frame and there would be all sorts of confusing complications.
 

RPinPA

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This became very straightforward to me when I was doing ballistic calculations at one point in my life. If you launch an object in a sub-orbital path, where does it land? First you figure out the path it follows around the center of the earth and where that path again crosses the surface of the earth. Then you note that while the object was flying, the earth was moving under it, which just shifts the longitude of where it will land, not the latitude. Just doing ##\omega t## where ##\omega## is the rotational velocity of the earth gives you the amount of longitude that rotates under the object.

That's Coriolis "force" for artillery and missiles. The earth rotates under you while you're in the air. Simple as that.

Imagine firing something straight north (and up of course) from a place on the equator. Ignoring the rotation of the earth, it should land somewhere north of where you are. But while it's in the air, the surface of the earth is rotating from west to east. The point where you were aiming has moved to the right (east), and the point where it lands is to the left, on a point that was originally to the west of where you were aiming. Standing on the surface of the earth, it appears your object curved westward.
 

FactChecker

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This became very straightforward to me when I was doing ballistic calculations at one point in my life. If you launch an object in a sub-orbital path, where does it land? First you figure out the path it follows around the center of the earth and where that path again crosses the surface of the earth. Then you note that while the object was flying, the earth was moving under it, which just shifts the longitude of where it will land, not the latitude. Just doing ##\omega t## where ##\omega## is the rotational velocity of the earth gives you the amount of longitude that rotates under the object.

That's Coriolis "force" for artillery and missiles. The earth rotates under you while you're in the air. Simple as that.

Imagine firing something straight north (and up of course) from a place on the equator. Ignoring the rotation of the earth, it should land somewhere north of where you are. But while it's in the air, the surface of the earth is rotating from west to east. The point where you were aiming has moved to the right (east), and the point where it lands is to the left, on a point that was originally to the west of where you were aiming. Standing on the surface of the earth, it appears your object curved westward.
The projectile also has an initial velocity from being on the surface of a rotating earth. And it's path is changed by the aerodynamic forces of the atmosphere which is rotating with the earth. It can be complicated.
 

PeroK

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That's Coriolis "force" for artillery and missiles. The earth rotates under you while you're in the air. Simple as that.
Why doesn't that work for aircraft? You could take a helicopter from London to New York by hovering for 5 hours or so and New York would appear below you.

On the other hand, flying East would be impossible, as once you are in the air you could never catch up with the ground rotating under you.
 

kuruman

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On the other hand, flying East would be impossible, as once you are in the air you could never catch up with the ground rotating under you.
Not so impossible. You can still take a hovering helicopter from New York to London but it will take longer to get there. :smile:
 
That's Coriolis "force" for artillery and missiles. The earth rotates under you while you're in the air. Simple as that.

.
If it like you say,than you can hower in the air above africa untill south america will appear under you,this is nonsense!
Air,aircraft, everthing what is in earth gravity field moves with earth together,..
But I still dont understand why then projectil must do correction...
so I still do not understand what coriolis force really is ,it is confusing
 

A.T.

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so I still do not understand what coriolis force really is ,it is confusing
Try to understand what inertial forces are in general, and don't fixate on the Coriolis force.

In non-inertial frames we observe coordinate accelerations that aren't consistent with the basic Newton's 2nd Law. To extend the applicability of the 2nd Law to non-inertial frames we introduce these inertial force terms. You can call it a "math trick" if you want.

Also look up the difference between "coordinate acceleration" (frame dependent change in velocity) and "proper acceleration" (frame independent acceleration relative to free fall - what an accelerometer measures).
 
Try to understand what inertial forces are in general, and don't fixate on the Coriolis force.

In non-inertial frames we observe coordinate accelerations that aren't consistent with the basic Newton's 2nd Law. To extend the applicability of the 2nd Law to non-inertial frames we introduce these inertial force terms. You can call it a "math trick" if you want.

Also look up the difference between "coordinate acceleration" (frame dependent change in velocity) and "proper acceleration" (frame independent acceleration relative to free fall - what an accelerometer measures).
Inertial forces like acceleration,decelariotn or centrifugal force is very easy to understand and feel in real life.
Every day when drive car to fast in corner I feel centrifugal force,when brake I feel decelariton,when put full throttle I feel acceleration etc etc etc.

But this coriolis force I do not know how to explain myself on easy intuitive way...


In wikipedia write that stone fall down from 50m tower on equator will fall down 7.7mm east,becuase of coriolis.
Why if everything on earth is moving together,this stone is part of earth rotating system so how can fall down 7.7mm east?

From this example,you can make conclusion that you can hover with helicopter and earth will be rotate under you...!!!!!
 

A.T.

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Inertial forces like acceleration,decelariotn or centrifugal force is very easy to understand and feel in real life. Every day when drive car to fast in corner I feel centrifugal force,when brake I feel decelariton,when put full throttle I feel acceleration etc etc etc.
You are obviously misunderstanding what inertial forces are, which is likely the reason you cannot incorporate the Coriolis force into that understanding:

You don't "feel" inertial forces, because that sensation is frame-independent, while inertial forces "exist" only when doing the analysis using some frames of reference. What you "feel" are just the interaction forces, that the car exerts on you. These cause frame-independent proper acceleration and deformation of your body.

But you can observe the frame-dependent coordinate accelerations relative to the non-inertial car. That is the only way you can actually "perceive" the effect of inertial forces.

But this coriolis force I do not know how to explain myself on easy intuitive way...
It's not made for intuition, but for calculations. In the rotating frame you introduce an inertial force to make Newton's 2nd Law work. That inertial force has:
- a position dependent part (called Centrifugal force)
- a velocity dependent part (called Coriolis force)
As you see, the above classification is purely mathematical, and handy for doing analysis. It has nothing to do with making it intuitive.

In wikipedia write that stone fall down from 50m tower on equator will fall down 7.7mm east,becuase of coriolis.
Why if everything on earth is moving together,this stone is part of earth rotating system so how can fall down 7.7mm east?

From this example,you can make conclusion that you can hover with helicopter and earth will be rotate under you...!!!!!
Introducing the atmosphere and objects which strongly interact with it (helicopter) makes things unnecessary complicated. You should try to understand simple examples in vacuum first. Preferably in a plane, not on the curved surface of the Earth.
 

vanhees71

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The most simple way to understand the motion in a rotating frame of reference is, of course, Lagrangian mechanics. Take a frame, which rotates around the ##x_3##-axis with respect to an inertial frame. Let the unprimed vector components be the inertial and the primed the rotating coordinates of a point particle. Then for rotation with constant angular velocity ##\omega## we have
$$\vec{x} =\hat{D}_3(\omega t) \vec{x}'=\begin{pmatrix} x_1' \cos(\omega t) - x_2' \sin(\omega t) \\ x_1' \sin(\omega t) + x_2' \cos \omega t\\x_3' \end{pmatrix}.$$
The kinetic energy reads
$$T=\frac{m}{2} \left [\dot{\vec{x}}^{\prime 2} + 2 \dot{\vec{x}}' \cdot (\vec{\omega} \times \vec{x}') +(\vec{\omega} \times \vec{x}')^2, \quad \vec{\omega}=\begin{pmatrix}0 \\ 0 \\ \omega \end{pmatrix}. \right] $$
Then let's write the potential of the forces in terms for the rotating coordinates, and we have
$$L=T-V.$$
The generalized momenta read
$$\vec{p}'=\frac{\partial T}{\partial \dot{\vec{x}}'} = m \dot{\vec{x}}'+m \vec{\omega} \times \vec{x}'$$
and thus the equations of motion
$$\dot{\vec{p}}'=m \ddot{\vec{x}}' + m \vec{\omega} \times \vec{x}'=\frac{\partial L}{\partial \vec{x}'}=-m \vec{\omega} \times \dot{\vec{x}}'-m \vec{\omega} \times (\vec{\omega} \times \vec{x}')-\frac{\partial V}{\partial \vec{x}'}.$$
This you can now order in two ways:

(a) bring ##\vec{F}'=-\vec{\nabla}' V## to the right-hand side. Then you get
$$m \ddot{\vec{x}}' + 2 m \vec{\omega} \times \vec{x}' + m \vec{\omega} \times (\vec{\omega} \times \vec{x}') = \vec{F}'.$$
On the left-hand side you have the accelaration wrt. the inertial frame, expressed in terms of rotating coordinates. In this interpretation, seen from the inertial frame, there are no inertial forces.

Now you can reinterpret the equation of motion from the point of view of the rotating observer by bringing ##m \ddot{\vec{x}}'## to the left side:
$$m \ddot{\vec{x}}'=-2 m \vec{\omega} \times \vec{x}' - m \vec{\omega} \times (\vec{\omega} \times \vec{x}') + \vec{F}'.$$
Then the observer interpretes this as if he were in an inertial frame. From this point of view despite the external force ##\vec{F}'## there are two types of inertial forces, the Coriolis and centrifugal forces. There's no reason to call them "fictitious", because they are definitely there on the right-hand side for the rotating observer.

Note, however, what we "feel" in a rotating situation (e.g., on a merry-go-around) are the constraint forces of the material acting on us to keep us in circular motion!
 

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