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Lets say you have a free particle in a rotating frame of reference with constant angular velocity $\mathbf{\omega}$. By free, I mean there are no real forces on it. Lets call the moving system "primed".

The equation of the motion is $m\mathbf{a} = \mathbf{F} + \mathbf{F_{fic}}$ but $F = 0$ and the only fictitious forces acting on this particle are the centrifugal force and the coriolis force.

Just to double checked, 1) there is no azimuthal (or transverse) force since there is no angular acceleration. Correct?

2) And ofcourse there is no translation force since the origin of the primed and nonprimed systems are the same? Right?

Okay so lets assume I am right (which I am since I am following an example from a text :P) So we have $$m\mathbf{a} = −m \mathbf{\omega} × (\mathbf{\omega} × \mathbf{r'}) − 2m \mathbf{\omega} × \mathbf{v}'$$

Lets making the rotating axis fixed with the z axis so $\mathbf{w} = \omega \cdot \mathbf{e_z}$

$$\omega = \omega \mathbf{e_z} $$ $$\mathbf{r} = x \mathbf{e_x} + y \mathbf{e_y} + z \mathbf{e_z} $$ and similarly v is pretty much r, but x = x dot, and such. Don't know how to write x dot with latex.

So my main problem is that for example lets say we are calculating the x component of everything. So then in the equation the coriolis force has a cross product between $e_z$ and $e_x$. This is $e_y$. I get that, but how do I know what sign is it? How come $e_z$ and $e_y$ gives me A NEGATIVE $e_x$?

I am on page 11 of this pdf,

I am also confused about how $e_x$ crossed with $e_z$ gives you "y dot" and not "x dot" Look at the pdf (specifically the coriolis force term).

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Most of this question is unnecessary. You could delete the stuff at top and just say you're looking at the equation for Coriolis force and confused about the sign of cross products - it would save a lot of effort for people reading the question. Also, to put a dot over a character, type "\dot{x}" for $\dot{x}$ and "\ddot{x}" for $\ddot{x}$. – Mark Eichenlaub Mar 11 '11 at 5:16

Some rules for the cross product are:

$$ A \times (\beta B + \gamma C) = \beta (A \times B) + \gamma (A \times C)$$

$$ (e_x \times e_y) \cdot e_z = 1$$

The equation

$$ (B \times C)\cdot A = d$$

is antisymmetric with respect to transpositions, meaning that if you take any two of those vectors and switch them, you multiply $d$ by $-1$.

These rules uniquely define the cross product. You can derive other rules from them, for example that the cross product of any vector with itself is zero or more generally that the magnitude of the cross product is the product of the magnitudes times the sine of the angle between the vectors.

Here is the derivation of the specific result you requested:

$$ (e_x \times e_y) \cdot e_z = 1$$

which is by assumption. Use rule 3 to switch $e_x$ and $e_z$.

$$ (e_z \times e_y) \cdot e_x = -1$$

This shows that the $x$-component of $e_z \times e_y$ is -1. It remains to show that the $y$ and $z$-components are zero.

$$ (e_z \times e_y) \cdot e_y = c_y$$

where $c_y$ is the unknown $y$-component of $e_z \times e_y$. Use rule 3 to switch the two appearances of $e_y$.

$$ (e_z \times e_y) \cdot e_y = -c_y$$

The two expressions are the same, so

$$c_y = -c_y$$

and $c_y = 0$, so the $y$-component of $e_z \times e_y$ must be zero. Similarly for the $z$-component. Hence

$$e_z \times e_y = -e_x$$

In general,

$$\omega \times v = - v \times \omega$$

For the Coriolis force, this equation is telling you that the direction of the Coriolis force depends on the velocity. If you turn around and go the other way, the Coriolis force switches directions. The Coriolis force pushes things clockwise or counter-clockwise at a certain place, rather than always left or always right. Hurricanes in the Northern hemisphere all rotate the same way, and the would continue to even if they moved different directions.

As for your last question, you can derive from the above that $e_x \times e_z = - e_y$, not $\dot{y}$.

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I will answer the question of why $\dot{\hat{e}_x} = \hat{e}_y \omega$ when the rotation is $\vec{\omega}=\hat{e}_z \omega$

At any instant the components of $e_x$ and $e_y$ are

$$\hat{e}_{x}=\begin{bmatrix}\cos\theta & \sin\theta & 0\end{bmatrix}$$

$$\hat{e}_{y}=\begin{bmatrix}\mbox{-}\sin\theta & \cos\theta & 0\end{bmatrix}$$

where $\theta$ is the rotation angle. Using the chain rule

$$\dot{\hat{e}}_{x}=\begin{bmatrix}\omega\sin\theta & \mbox{-}\omega\cos\theta & 0\end{bmatrix} = \hat{e}_y\,\omega$$

$$\dot{\hat{e}}_{y}=\begin{bmatrix}\mbox{-}\omega\cos\theta & \mbox{-}\omega\sin\theta & 0\end{bmatrix} = \mbox{-}\hat{e}_x\,\omega$$

The general rule is the for any rotating constant vector $\vec{A}$ riding on a rotating frame (or on a rigid body) with angular velocity $\vec{\omega}$ the rate of change is given by

$$ \dot{\vec{A}} = \vec{\omega}\times\vec{A} $$

and in your case

$$\dot{\hat{e}}_{x} = \left( \hat{e}_z \omega \right) \times \hat{e}_{x} = \hat{e}_{y} \omega $$

I hope this helps.

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