| Week of | Aug 23
| Sep 6
| Oct 2
| Nov 6
| Phy 205 page |
| Aug 28 | Sep 11 | Oct 9 | Nov 13 | ||
| Sep 18 | Oct 16 | Nov 20 | Notes for Fall, 1999 | ||
| Sep 25 | Oct 23 | Nov 27 | |||
| Oct 30 |
13 Nov Discuss something about the exam coming up this evening.
Show again how to manipulate F = dp/dt by taking a cross product with r. The result is t = dL/dt, relating torque to the time-derivative of angular momentum.
What is angular momentum for two masses rotating about a common axis. This is the same system that I used previously to find the kinetic energy of a rotating rigid body. Before, I found the moment of inertia appearing as part of the result, and the same sort of thing happens here too. The system is a dumbbell rotating about an axis perpendicular to the bar. I computed
and used the fact that when everything is rotating at w, the speed of the first mass is r1w and of the second is r2w.
Put this all together, using the definition of the cross product, and find the total angular momentum. It is
where the direction of the w vector is along the axis of rotation. The moment of inertia appears again.
15 Nov Harmonic motion.
The example that I can most easily show consists of a mass hanging on
the end of a spring. If this were the only example it wouldn't be a very
interesting topic, but the same basic ideas appear in so many places
that it practically becomes a subject of its own.
A pendulum
The vibrations of a piece of quartz in a wristwatch
The oscillations of currents in an electrical circuit
The molecular vibrations that enable a microwave oven to function
The motion of a globular cluster of 100,000 stars above and below the
plane of our galaxy
Some people have even had the idea of applying this subject (developed
into Fourier Analysis) to the stock market, but not all ideas are good ones.
For the simplest example, the force applied by a spring when you extend it is pretty nearly proportional to the distance by which it's elongated. If x is the displacement of the spring (one dimensional) then the force it applies obeys Fx = - k x. If the mass m is attached to the end of the spring (neglecting all friction) then
The methods that we've used to date for solving these equations won't work. In the past, I've just taken an anti-derivative or two (integrated) and then figured out what the arbitrary constants are by using some initial conditions. If you try that here it fails.
In this example though you can just about guess the solution. The equation says that (ignoring constants) the second derivative of a function x(t) is proportional to minus the function itself. You know some functions with that property. The exponential, et, remains the same if you differentiate it once or twice, so it's almost right. The minus sign gets in the way though. Sines and cosines however give you a minus sign when you differentiate them twice, so they are likely candidates for solutions.
Two difficulties with saying x(t) = cos(t): The dimensions of x are length and the output of the cosine is dimensionless. Also, you can't take the cosine of a time, only an angle. You can patch this up however by putting in some parameters.
Substitute this into the equation to be solved and see if it works. You find that it does provided that the value of w is chosen correctly. The same thing works for the sine instead of the cosine, so there are two solutions. In the text he writes the solution as
and that includes both of these as special cases. Take d = 0. Take d = - p/2.
17 Nov Briefly go through the verification of the solution to the harmonic oscillator differential equation. Use the terminology for some of the constants that show up: amplitude, frequency, period.
The period is related to the frequency by T = 2p / w = 2p( m / k )1 / 2. This implies that the period of the mass oscillating on a spring varies as the square root of the mass. Double the mass and the period goes up by a factor of square root of two. (There's a correction here for the mass of the spring, but I'll ignore that.) Do the experiment and the result is pretty close to the prediction.
The pendulum: Set up the t = I a equation, and it is not quite the same as the original equation for a mass on a spring. The approximation that the angles are small however gives a much easier result. For small angles in radians, the sine of the angle is very nearly equal to the angle itself. With the approximation that you can use the angle instead of its sine, the equation to be solved is the same one as before.
