LECTURE NOTES FOR PHYSICS 307: CLASSICAL MECHANICS
TEXT: THORNTON AND MARION'S CLASSICAL MECHANICS OF PARTICLES AND SYSTEMS
SOME USEFUL REFERENCE STUFF: Greek alphabet, metric prefixes, conversion factors
ASIGNMENTS: (Subject to change: check back often.)
HW #11 (Extra credit) Due Friday, Dec. 12: Problems 10.7, 12.
WEEK 15: GO TO LECTURE 25, 26
LECTURE 25: SECTION 10.1-3 (Return to top.)
CHAPTER 10: NONINERTIAL REFERENCE FRAMES
“How can we dance when our earth is turning? / "How do we sleep while our beds are burning?" -- Beds
are Burning by Midnight Oil, 1986
SECTION 2: ROTATION COORDINATE SYSTEMS
There is an awful lot of mathematics here. Read the book and try to follow it. The most important part of this is to
realize that if you take the time derivative of some quantity in a fixed coordinate system, calculating the time
derivative of the same quantity in a rotating coordinate system leads to an extra "omega cross" term. In general,


 dQ 
 dQ 
 

 
  Q
 dt 



 f  dt  r
where the subscript corresponds to the reference frame in which we are measuring, fixed or rotating, and ω is the
angular velocity with which the rotating frame is rotating. Applying this to velocities, we get
 



v f  vtransl  v r    r
where vf is the velocity as measured in the fixed system, and vr is the velocity measured in the rotating system,
and vtransl is the velocity with which the rotating frame is translating -- its linear velocity relative to the fixed frame.
Why would we be interested in this? Because Newton's laws are valid in inertial reference frames, not in rotating
reference frames. And so, if we wish to measure forces in the rotating reference frame --- remember: we live on a
large rotating reference frame --- we need to come up with the way of relating those forces to the inertial forces,
that is, the forces as measured outside that rotating frame.
LECTURE 26: SECTION 10.4 (Return to top.)
SECTION 3: CENTRIFUGAL AND CORIOLIS FORCES

When we start with Newton's second law,

 F  ma , we must apply it in a fixed coordinate system. This is
exactly how the mathematical derivation in this section begins.
We differentiate twice with respect to time in the fixed reference frame. We then switch reference frames to get
the effective force felt in the rotating frame. This expression consists of five terms


 
  
 

Feff  F  mR  m  r  m    r   2m  vr
From left to right, these are...
1. All actual forces, F --- weight, normal, tension, friction, etc.
2. The translational acceleration of the reference frame (If you are in an moving elevator, for example,
the elevator's acceleration)
3. Rotational acceleration. Imagine you are in an amusement park ride as it begins to speed up.
4. Centrifugal force (This is the one that "pushes" you towards the outside of the car as you around a
curve.)
5. Coriolis force (We'll talk about this in much more detail because it's not one we are directly familiar
with, and yet it has a lot of very interesting consequences.)
SECTION 4: MOTION RELATIVE TO THE EARTH
We will take the expression that we've just developed for the effective force in a rotating reference frame, and
focus in on a particular case: motion relative to the rotating earth. (In Problem 10.1 we compare this acceleration
to that due to the Earth's motion about the Sun.)
Two interesting effects to note. First of all, the gravitational force pulling all objects towards the center of earth can
be lumped together with a centrifugal term to produce the effective gravitational accelerations, g. What is
interesting about this is, first, that the centrifugal term produces a bit of a bulge in the Earth's equator. Second,
this "effective gravitational force" does not point directly towards the center of the earth. If we could very precisely
follow the direction that a pendulum bob points, we would end up not in the center of the earth, but at a point in
the plane of the equator. We can still say that the pendulum points straight down: we just have to revise our
definition of "down".

  
 
g  g0      R


where R is the position (relative to the center of the Earth) of the object experiencing this apparent force.
The second interesting effect is the Coriolis force. This is a force that depends on the velocity of an object which
is travelling on the surface of the earth. Problem 8 is a very interesting problem because it considers a missile
going straight up in the air in the Northern Hemisphere, and then falling straight down. One can show that the
average Coriolis force going up exactly cancels the average Coriolis force coming down, and yet the missile does
not land where it was launched. Why is this? (Consider the equations of motion going up and down.)

 
FCor  2m  vr
Another interesting consequence of the Coriolis force is something called Foucault's pendulum. This is usually a
very long pendulum which is allowed to swing back and forth, usually with some magnetic device at the bottom of
the swing to replace the energy the pendulum loses to friction. If the pendulum is released from perfect rest, the
plane in which it swings will rotate. If it begins swinging North to South, eventually it will swing East to West, and
so on.
The cause of this rotation is the Coriolis force, and, just as in Problem 8, the average Coriolis force is zero. I'll
have a handout in class to show how it is that the two halves don't cancel. By the way, Foucault's pendulum has a
period of
T  24hr / sin 
where λ is the angle of latitude. Convince yourself, using this equation, that a Foucault pendulum on the North
Pole has a 24 hour period and that a Foucault pendulum at the equator has an infinite period - that is, it doesn't
rotate.
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