How do I calculate Coriolis force? For figuring Coriolis force, I've seen this equation:
$$
\vec a_{cor} = -2\vec \omega \times \vec v
$$
So, say that I have an angular velocity of 5 rpms, and a velocity of 10 m/s going toward the axis of rotation. How do I go about solving this equation? I studied vector math for about a week in high school, and never quite understood it. Please explain like I am 5, or an English major (full disclosure: BA in English Literature).
Note: I don't want you to solve it for me, just explain how to solve it, or link me somewhere that does. Thanks!
 A: Remember that $\vec{\omega}$ is a vector: its direction is given by the right hand rule: curl the fingers of your right hand in the direction of the rotation and see where your thumb points. The magnitude of this vector is the rotation rate in radians per second (IOW, you'll need to convert the 5 rpm datum to radians per second - remember there are $2\pi$ radians in a full rotation ($360^\circ$) and of course 60 seconds in a minute.)
Now you should realize that $\vec{\omega}$ points along the axis of rotation, so since $\vec{v}$ points towards the axis of rotation, the two vectors are perpendicular.
The rest follows from the definition of the cross product: the magnitude of the cross product is the product of the magnitudes of the individual vectors times the sine of the angle between them (but since they are perpendicular, that's $\sin{90^\circ} = 1$). The direction of the cross product is again given by the right hand rule: put the two vectors on a common tail, point the straightened fingers of your right hand in the direction of the first vector, but make sure that when you curl your fingers, they point in the direction of the second vector; the direction of the cross product is perpendicular to the two vectors in the direction of your thumb.
And don't forget that you have to multiply the resulting cross product by $-2$, that is double its magnitude and reverse its direction (because of the minus sign).
A: First of all, you have to define the parameters of your problem. Let's use cylindrical coordinates, using the base $(\mathbf{e}_z, \mathbf{e}_r, \mathbf{e}_\theta)$, where $\mathbf{e}_z$ is the axis of rotation.There is an inertial frame of reference $\mathcal{R}_0$, and another one $\mathcal{R}$ which rotates with and angular velocity $\mathbf{\omega} = \omega \mathbf{e}_z$ around $\mathcal{R}_0$, where $\omega = 5\,\textrm{rpm}$. Now, we are studying an object moving with a velocity $\mathbf{v} = -v\mathbf{e}_r$ where $v =10\,\textrm{m/s}$.
Now, you have to use the formula giving the Coriolis acceleration: $\mathbf{a}_c = -2\mathbf{\omega}\times\mathbf{v}$. To find the acceleration, you must find the cross product of two vectors. First of all, you can use the bilinearity of the cross product, ie. the fact that it behaves like a normal multiplication, except that it is not symmetric:
$$(\omega \mathbf{e}_z) \times (-v\mathbf{e}_r) = -\omega v (\mathbf{e}_z \times\mathbf{e}_r)$$
Finally, the cross product of two base vectors can be found easily: use the right-hand rule. Here, $\mathbf{e}_z$ is represented by your middle finger, $\mathbf{e}_z$ by your thumb, so their cross product (in this order) gives your forefinger, that is $\mathbf{e}_\theta$.
