## Let’s explicitly take the Earth as the rotating frame. The angular frequency of rotation of the Earth is ## Let’s explicitly take the Earth as the rotating frame. The angular frequency of rotation of the Earth is ## We will assume that the inertial frame So and rotating frame S share the same origin, so the only motion of S relative to So is a rotation with angular velocity ****. For example, the common origin could be the center of the Earth. ## Now consider an arbitrary vector **Q**. The time rate of change of **Q** in frame So is related to its time rate of change in S by ## This says that even if **Q** is not changing length or direction in frame S, it is still changing in frame So by virtue of its rotation.
## Let’s now see what Newton’s second law looks like. Obviously, in frame So, it is just ## Let’s now see what Newton’s second law looks like. Obviously, in frame So, it is just ## where, as usual, **F** is the net force on the particle as identified in the inertial frame. ## We can now use our result from the previous slide, identifying **Q** and the position vector **r**, i.e. ## Differentiating a second time ## Now the vector **Q** is the entire quantity in square brackets, hence:
## We will use the dot notation for vectors in frame S, hence: ## We will use the dot notation for vectors in frame S, hence: ## Thus, Newton’s second law, which was ## is now ## Once again, we just have Newton’s second law, but this time with two inertial force terms on the right: ## We will look at each of these in more detail.
## You are already quite familiar with the centrifugal force (centripetal force in the inertial frame). You know it as *mv*2/*r*, but you may protest that this expression: looks nothing like that. ## You are already quite familiar with the centrifugal force (centripetal force in the inertial frame). You know it as *mv*2/*r*, but you may protest that this expression: looks nothing like that. ## But if you recall that *v* = *r*, and that we are using capital letters for quantities related to motion of one frame relative to another, then *v* = *r*. So you see that *mv*2/*r = m*2*r*, which does have the right variables. ## But what do we make of these complicated looking ## cross-products? First, let’s look at the magnitude: ## and since the cross-product of two vectors is ## perpendicular to both of the two vectors, the ## angle between **** **×** **r** and **** is **/2, so ## where ** is as shown in the figure. ## Now look carefully at the directions. **F**cf points ## radially outward from its circular path.
## Due to the centrifugal force, a plumb bob does not actually point in the direction to the center of the Earth except at the pole or equator. ## Due to the centrifugal force, a plumb bob does not actually point in the direction to the center of the Earth except at the pole or equator. ## Because a plumb bob will respond to both the force of gravity **F**grav downward, and **F**cf outward, the effective force is ## from which we can identify an effective gravity ## The centrifugal term is zero at the pole, and largest ## at the equator, where it is 2*R* = 0.034 m/s2. The ## tangential component of this effective gravity is ## gtang = 2*R *sin* *cos** and is maximum at ** = 45o. ## The maximum angular deviation is
## The Coriolis force may seem mysterious at first, but its origin is almost trivial. Before examining the equation, let me start by considering a missile at the equator. It’s speed in the inertial frame is just that of the Earth, ## The Coriolis force may seem mysterious at first, but its origin is almost trivial. Before examining the equation, let me start by considering a missile at the equator. It’s speed in the inertial frame is just that of the Earth, ## If we fire the missile straight north, heading for latitude ** = **/2 ** (** is called the colatitude), it will, of course, maintain its (the equator’s) sideways velocity, but it will be traveling over land for which the sideways ## velocity of the Earth is less: ## Because of that difference in speed, the rocket ## seems to drift to the east in the frame of Earth. ## This appears as an inertial force, the Coriolis ## force Note that it depends on the ## velocity of the moving object (the missile), and ## if you are used to finding the direction of ## **v** × **B** in E&M, note **F**cf ~ **v** × ****.
## The Coriolis force also acts in a flat geometry, such as movement on a rotating disk. ## The Coriolis force also acts in a flat geometry, such as movement on a rotating disk. ## The force is always perpendicular to the velocity, so the force ## and velocity have the relationships shown in the figure. ## Check the direction using the right-hand rule. ## You can see how the Coriolis force works in the flat geometry ## from the center of the rotating platform. From an inertial frame ## the puck will follow a straight line. However, from the rotating frame the path is a curve, as shown in the figures below.
## Let’s do a simple example that exercises the calculation of Coriolis force and introduces the useful technique of successive approximations. ## Let’s do a simple example that exercises the calculation of Coriolis force and introduces the useful technique of successive approximations. ## Consider an object at the surface of the Earth in free-fall with no other forces acting (i.e. no air resistance). We have to include both **F**cor and **F**cf. ## Since we are at the surface of the Earth, we can let *r = R*. Let’s also combine the first two terms as *m***g**, the effective gravity force. Then ## Since this does not depend on position **r**, only on its ## derivatives, we can move the origin to the surface of ## the Earth. Let’s now write the components of the ## vectors in east-north-up coordinates. ## Thus
## The equation of motion resolves to the three equations: ## The equation of motion resolves to the three equations: ## Let’s solve these using the method of successive approximations. First, write the equations for small (i.e. set to zero). This is the zeroeth-order approximation. This gives the equations you have solved in freshman Physics: ## These lead to the solutions ## where we have used initial conditions appropriate to dropping the object from rest from high *h*. We then plug these back into our original equations to get the first-order approximation: ## The last two equations have not changed, but the *x* equation has. Integrating twice, we get which shows that there will be an eastward deviation. We could repeat to get second-order approx., etc. ## An object dropped down a 100-m mine shaft at the equator will move east by:
## You may think that, since Coriolis force depends on the velocity, the force would be miniscule for something moving as slowly as a cloud in the atmosphere—and you would be right. However, a small force does not necessarily mean a small effect. The force on a slowly moving body can act over a long period of time (weeks in the case of a weather pattern), and so the effect can be important. ## You may think that, since Coriolis force depends on the velocity, the force would be miniscule for something moving as slowly as a cloud in the atmosphere—and you would be right. However, a small force does not necessarily mean a small effect. The force on a slowly moving body can act over a long period of time (weeks in the case of a weather pattern), and so the effect can be important. ## It is the Coriolis force that causes the cyclonic (counter-clockwise) weather patterns seen in the northern hemisphere. Consider a low pressure system that draws air from the North and South. The air moving south will be turned to the west, while the air moving north will be turned to the east. ## Although not directly due to Coriolis force, the flows from east and west also turn due to the turning of the north and south flows. ## To properly understand this, you have to consider the ## motions on a sphere, i.e. the north and south flows ## are not parallel to . ## What is the flow pattern like in the southern hemisphere?
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