We have looked at several examples in which objects were accelerating either linearly or centripetally and ended up feeling like gravity had been altered by the subtraction of the acceleration term from the gravitation vector.  In those cases we often asked "what is the direction that feels like downward", or "how strong would the effective force of gravity be".  Now we wish to consider one additional factor that may not at first seem to matter.  What if we let that effective gravitation actually cause an object to start falling?  What will the object's path be?  [br][br]The answer is that the path will simply be linear along the direction of [math]\vec{g}_{effective}[/math] if the accelerated reference frame is accelerating in a linear fashion like in the train example or the skateboarder going down a ramp.  However, if the object is accelerating centripetally on a rotating object like a Ferris wheel or on the earth itself, it will begin its fall in the direction of [math]\vec{g}_{effective}[/math], but will experience an additional acceleration once it has a velocity relative to the rotating reference frame.[br][br]This additional term is called the Coriolis acceleration and will only be present in rotating reference frames like earth itself, inside a car that is rounding a bend, or on rotating objects like merry-go-rounds and Ferris wheels, etc.

Weather patterns are influenced by the Coriolis force.  In fact, without the Coriolis force we can't have hurricanes.  You'll see in this chapter that near the equator we can't generate a Coriolis force to cause storms to rotate.  Consequently hurricanes cannot form within 5 degrees of latitude of the equator.  There has also never been such a storm that crossed the equator.  The circulation direction of hurricanes is opposite in the northern versus the southern hemispheres.  I hope that such facts seem obvious after our discussions in this chapter.[br][br]The article linked below is an interesting historical perspective of such Coriolis force studies, and as you read the article, you'll see that studies of the Coriolis effect were directly involved in proving that our planet is actually rotating on its axis - a notion that had been suggested since the 17th century, and was first proven in the 19th century by Gauss and Laplace and also by Foucault with his famous pendulum experiment.

The force that objects feel while on or inside a rotating object is called the Coriolis force.  It is often called a pseudo-force since it does not exist in a non-accelerating reference frame.  It is much better to call it an [b]apparent force[/b], because it appears in situations where objects move in a rotating reference frame like earth. In this sense it is just as real as any other forces on earth, and is just as measurable as friction or normal forces.[br][br][center][math]\vec{F}_{Coriolis}=-2m\vec{\omega}\times\vec{v}.[/math][/center]The associated acceleration of an object due to this force is just the force over the mass, or[br][br][center][math]\vec{a}_{Coriolis}=-2\vec{\omega}\times\vec{v} \\[br]\text{Since \vec{a}\times\vec{b}=-\vec{b}\times\vec{a}, we can write: } \\[br]\vec{a}_{Coriolis}=2\vec{v}\times\vec{\omega}.[/math][/center][br][br]In this equation, [math]\vec{\omega}[/math] is the angular velocity of the planet or rotating system, and [math]\vec{v}[/math] is the velocity of an object on (or in) the rotating object in which the coordinate system remains fixed with respect to the planet or rotating system. This means in the Coriolis equation if you are sitting still, your velocity is zero even if you are on a rotating planet. So this is the velocity we usually use in physics problems - one which assumes the earth is at rest.  If we're dealing with Coriolis effects due to the rotating earth, unless objects are moving with respect to the planet they will not feel this Coriolis force or undergo the associated acceleration.[br][br]On the other hand, if we are dealing with Coriolis forces we'd feel on a rotating merry-go-round, then we'd have to walk along the surface of the merry-go-round to feel the force. Just sitting on a rotating one (just like sitting still on rotating earth) reduces the velocity to a zero vector and eliminates the Coriolis force.

Adding this term completes our equation for effective gravitation. The way gravity acts ultimately depends on actual gravity, acceleration of an object (linear or centripetal) and motion relative to the rotating object. The completed expression that takes all of these things into account is[br][br][center][math]\vec{g}_{effective}=\vec{g}-\vec{a}+\vec{a}_{Coriolis}.[/math][/center]

[color=#1e84cc]PROBLEM: Suppose you are on a merry-go-round that is rotating such that its angular velocity is directed upward in the z-direction.  While sitting in the middle of the merry-go-round you slide an ice cube (assumed frictionless) radially outward.  How does the ice-cube move in your rotating reference frame?  [br]SOLUTION: While traveling radially outward, the Coriolis force on the ice cube would be to your right if you assume you are directing your gaze outward radially.  Thus the object does not just slide directly straight away from you but instead moves away from you while curving rightward along its path.  The acceleration rightward is just given by [math]\vec{a}_{Coriolis}=2\vec{v}\times\vec{\omega}.[/math]  If we assume, for instance that you slide the ice-cube at 0.5m/s away from you and the merry-go-round rotates at 3.0rad/s, then the acceleration will be 3.0m/s[sup]2[/sup] to the right... or quite substantial.[/color]

If you read the linked article, you saw that in 1668, Giovanni Borelli attempted to calculate the lateral deflection of a dropped object as it falls on a rotating planet in order to prove that the planet is in fact rotating.  His tower was 71m tall, and due to small effects like wind gusts, he was unable to prove the effect.  Later, however, Ferdinand Reich performed the same experiment in Freiburg, Saxony (the city in Germany where my brother went to the university which was founded in 1457).  Reich dropped a series of metal balls into a mine shaft that was 158.5m deep and measured the deflection.  Let's calculate what he should have expected to measure.[br][br]The first thing to discuss is how Reich could measure a deflection.  Previous discussions we've had about [math]\vec{g}_{effective}[/math] were focused on which way things would fall, or which way was "downward".  To measure a deflection we need to know which way is downward and then compare that direction with the way things actually fall.  If this sounds confusing, please read on. Downward will always be defined by using something like a plumb bob, and a plumb bob will hang in the direction of [math]\vec{g}_{effective}[/math] but will not experience a Coriolis force since the bob isn't moving.  Since the Coriolis force only exists for things that are moving in a rotating reference frame (like earth), it will affect falling objects like rocks. This means [b]the direction plumb bobs hang and the direction rocks fall is not the same direction!  [br][br][/b]Because of this, the Coriolis deflection is at least in principle measurable. 

To derive the expected lateral displacement of the dropped steel balls in the mine shaft, we need kinematics. From the right hand rule, we should first determine the direction of the anticipated deflection. The velocity of a dropping steel ball is in toward earth's center (or locally downward). The vector for earth's angular velocity is along its rotation axis from south to north. The cross product of these two vectors is directed eastward. I will use for this derivation, the x-direction as eastward and the y-direction as downward (into the mine shaft). Thus the Coriolis acceleration is eastward with a magnitude given by:[br][br][center][math]a_{Coriolis,x}=2v_y\Omega \cos\Phi, \\[br]\text{where we use in this case symbols \Omega and \Phi to} \\[br]\text{denote earth's angular velocity and the latitude as} \\[br]\text{used in the linked article. $v_y$ is downward velocity.}[br][/math][/center]

We start by doing the kinematics in the downward (y) direction because without [math]v_y[/math] we can't figure out the Coriolis acceleration. It is assumed that the air drag is negligible as Gauss and Laplace had done. The depth of the shaft will be denoted h, and the drop time [math]t_{drop}.[/math] The kinematics then gives:[br][br][center][math][br]\Delta v_y=\int_0^t a_y \; dt \\[br]\Delta v_y=\int_0^t g \; dt \\[br]\Delta v_y=gt \\[br]v_y=v_{y,0}+\Delta v_y \\[br]\text{Since the balls are dropped from rest, $v_{y,0}=0$.} \\[br]v_y=gt \\[br]\Delta r_y = \int_0^t v_y \; dt \\[br]\Delta r_y = \int_0^t gt \; dt \\[br]\Delta r_y = \frac{gt^2}{2} \\[br]\text{The vertical displacement is just the depth of the mine shaft.}\\[br]h = \frac{gt_{drop}^2}{2} \\[br]t_{drop}= \sqrt{\frac{2h}{g}}. \\[br]\text{} \\[br]\text{The Coriolis deflection in the x-direction (eastward) is calculated as follows: } \\[br]\Delta v_x = \int_0^t a_{Coriolis,x} \; dt \\[br]\Delta v_x = \int_0^t 2v_y\Omega\cos\Phi \; dt \\[br]\Delta v_x = \int_0^t 2gt\Omega\cos\Phi \; dt \\[br]\Delta v_x = gt^2\Omega\cos\Phi \\[br]v_x = v_{x,0}+\Delta v_x \\[br]v_x = gt^2\Omega\cos\Phi \\[br]\Delta r_x = \int_0^t gt^2\Omega\cos\Phi \; dt \\[br]\Delta r_x = \frac{gt^3\Omega\cos\Phi}{3} \\[br]\text{evaluated at $t=t_{drop}=\sqrt{2h/g}$, we get the expected eastward deflection: } \\[br]\Delta r_x = \frac{2\Omega\cos\Phi}{3}\sqrt{\frac{2h^3}{g}} \\[br]\text{Plugging in values and using h=158.5m, $\Phi=48^o$ and $\Omega=\tfrac{2\pi \: rad}{86,400s}$ gives:}\\[br]\Delta r_x = 2.9cm. [br][/math][/center][br][br]How to group terms in the last line of algebra is a matter of taste, but the way shown above matches the article that's linked. To reproduce the algebra leading to the last step it is useful to note that [math](\sqrt{x})^3=x\sqrt{x}.[/math]

[color=#1e84cc]The Coriolis force or acceleration term will impact any object that has a velocity relative to a rotating reference frame. Our home planet is such a rotating reference frame. Even ordinary things are affected by the Coriolis force. Here I will show how the west bank of a river like the Mississippi is measurably higher than the east bank. [br][br]Suppose a river flows southward at a speed [math]v[/math] at a latitude [math]\theta[/math] and that the river is [math]w[/math] meters wide. Let's start by asking in which direction the Coriolis force should act. Using the right hand rule you should determine that it is directly westward. The angle between the river's current and the earth's rotation axis is [math]\pi-\theta.[/math] Therefore the Coriolis acceleration's magnitude is [math]a_{Cor}=2v\omega\sin(\pi-\theta)=2v\omega\sin(\theta).[/math] If we assume a river flows at 3m/s (typical) and has a width of 1000m (many of the world's rivers are much wider than this), and is at latitude of 40 degrees north, let's see how much higher the west bank is. To solve this, we will consider the gravitation vector to act downward and the Coriolis acceleration to act westward. Thus the tilt of the water surface is given by [math]\tan(\phi)=a_{Cor}/g = 2.8\times 10^{-5}rad.[/math] So the west bank is [math]h=1000m\left(2.8\times10^{-5}rad\right)=2.8cm[/math] higher. That is measurable by a surveyor, and is also a factor affecting erosion on river banks and the way the course of a river changes over time. [/color]