# What is the Coriolis force

1. Jan 3, 2005

### Reshma

What is Coriolis Force & what are its applications?

2. Jan 3, 2005

### quasar987

Coriolis force: I don't see a more effective way of describing it other than by the definition...

Suppose you have two coordinate sytems whose origin meet and one of them is rotating with respect to the other. The rotation can be anything, it doesn't have to be at contant in speed nor does it have to be around a fixed axis.

Let $\vec{\omega}$ denote that angular velocity (function) at which the rotating coordinate system rotates around the other and let $\vec{r}$ be the position vector of any given point in space. Since the origin of the two coordinate sytems meet, the vector position is the same in both coordinate systems (although the components of that vector will most likely be different depending on which coordinate system we express it).

A bit of vector analysis shows that the rate of change of the rate of change (acceleration) of the vector position, as "measured" in the fixed system, is related to the rate of change of the rate of change (acceleration) of the vector position as "measured" in the rotating system through the equation:

$$\frac{d^2\vec{r}}{dt^2}=\frac{d^{*2}\vec{r}}{dt^2} + \vec{\omega}\times(\vec{\omega}\times\vec{r})+2\vec{\omega}\times\frac{d^{*}\vec{r}}{dt}+\frac{d\vec{\omega}}{dt}\times\vec{r}$$

where the starred derivative denotes the derivative of the vector r as "measured" in the rotating system (it is different then the derivative measured in the fixed system).

We call this whole equation Coriolis' theorem and we call the third term on the RHS

$$2\vec{\omega}\times\frac{d^{*}\vec{r}}{dt}$$

the coriolis acceleration (because it has the dimensions of an acceleration [distance divided by time squared]).

If now, we multiply both sides of Coriolis' theorem by m and we suppose that the vector r refers to the position in space of a particle of mass m, then the left-hand side of Coriolis' theorem is (according to Newton's second law), the force on that particle. By rearanging the terms a little, we can write

$$m\frac{d^{*2}\vec{r}}{dt^2}= \vec{F} - m\vec{\omega}\times(\vec{\omega}\times\vec{r})-2m\vec{\omega}\times\frac{d^{*}\vec{r}}{dt}-m\frac{d\vec{\omega}}{dt}\times\vec{r}$$

and the term we called coriolis acceleration we now call coriolis force (because it now has the dimensions of a force [mass times distance divided by time squared]).

We see that the coriolis force is not a force at all but rather a "correction" to F=ma in a rotating coordinate system. In other words, Newton's second law does not hold in a rotating coordinate system in the sense that

$$m\frac{d^{*2}\vec{r}}{dt^2} \neq \vec{F}$$

but if we introduce the "ficticious forces"

$$-m\vec{\omega}\times(\vec{\omega}\times\vec{r})$$
$$-2m\vec{\omega}\times\frac{d^{*}\vec{r}}{dt}$$
and
$$-m\frac{d\vec{\omega}}{dt}\times\vec{r}$$

then it does "hold".

Its applications: It is often useful when studying a phenomenon, to adopt the point of view of a rotating coordinate system. Symon (pp.279) gives the exemple of the action of a cream separator, for which it is much more convenient to adopt a point of view in which the liquid is at rest and use the law of diffusion to study the diffusion of the cream towards the axis under the action of the centrigugal force field, than to try to study the motion from the point of view of a fixed observer watching the whirling liquid.

3. Jan 3, 2005

Staff Emeritus
The only thing I have to add to this is that "ficticious" is used in a special sense. Anyone who thinks Coriolis force is ficticious has never experienced a tornado or a hurricane. If you live on a rotating coordinate system, as we do, fiction becomes your reality.

4. Jan 3, 2005

### quasar987

Or has never ridden the bus .

5. Jan 4, 2005

### enigma

Staff Emeritus
The way I always pictured what the coriolis force "is" follows:

If you're on a merry go round, you're spinning about a certain center of rotation. You feel the force from your hands and feel holding you onto the rotating mass. That's the "fictional" centrifugal force you're feeling.

If you then take a step outward while still on the merry-go-round, your path is now different. You are following a path which has a "center" different from that of the MGR. Since you're getting pulled towards your "center" and the floor of the MGR is pulled towards it center, you end up falling. The difference in the two forces is the coriolis force.

6. Jan 4, 2005

### Andrew Mason

Think of the earth as thin disks rotatating about the earth axis. Then think of an object on the surface of a disk and plot the forces. There is the force of gravity directed toward the centre of the earth, mg. There is also the centripetal force directed toward the earth's axis along the plane of the disk.

The centripetal force can be broken down into a component tangential to the earth's surface and a component that is toward the earth's centre. Gravity supplies the radial component but cannot supply the tangential component. This means that the matter to which the object is attached must supply the tangential component. In the northern hemisphere this tangential component is in the northern direction (just work out the vectors). This means that north of the equator the earth is always tugging things north. For an object that is not rigidly connected to the earth (eg. water, air), the earth is always pulling away from them in a northern direction. This has the appearance that the object moves south but that is just an inertial effect. It is really that the earth surface is accelerating north from under the object.

The opposite occurs in the south. The earth pulls everything south and objects that are not fixed to the earth tend to move north.

Since the earth is spinning in a west to east direction, in the northern hemishere there is a clockwise rotation observed. In the sourthern hemisphere there is a counterclockwise rotation.

AM

7. Jan 4, 2005

### ceptimus

Assume the empire state building has straight sides and was built with each story aligned as perfectly vertical as possible. The Coriolis force explains why the building will curve slightly. It also allows you to work out how far to the south and east of the release point a dropped penny will land, on a windless day.

8. Jan 8, 2012

### Hsinha

What book is this? Can you tell me the full name? I need to find out a bit about coriolis based seperations.

9. Jan 8, 2012

### Staff: Mentor

10. Jan 8, 2012

### Staff: Mentor

And the book is even older! So old that it was one of my undergrad texts.

11. Jan 8, 2012

### Staff: Mentor

I've never used it myself, but one of my colleagues about twenty years ago used it in teaching our advanced mechanics course. He called it "fuchsia physics". Either Amazon's image doesn't have the color quite right (it looks more like red), or the publisher has changed it. I definitely remember it being a fuchsia sort of color.

Last edited: Jan 8, 2012
12. Jan 8, 2012

### Staff: Mentor

Just pulled it off the shelf. Yep, fuchsia it is.

13. Jan 8, 2012

### Hsinha

I realise now that the thread is so old. :P
Sorry if it caused inconvenience.

And Thanks a lot. :D