Class Notes, Fall 2001

Week of	Aug 22	Sep 5	Oct 1	Nov 5	Phy 205 page
	Aug 27	Sep 10	Oct 8	Nov 12
		Sep 17	Oct 15	Nov 19
		Sep 24	Oct 22	Nov 26
			Oct 29		Notes for Fall, 2000

Phy 205, Week 13

12 Nov If I have a wheel that is pivoted on a fixed axis, I can get it rotating by wrapping a cord around it and pulling. This means that I do work on it, and that work appears as the kinetic energy (Iω²/2) of the wheel. If I pull the cord a distance L, this becomes the equation

Differentiate this with respect to time and you get

F0 d L d t = I 2 d ω2 d t

The chain rule allows you to do this derivative to obtain

I 2 d ω2 d t = I 2 d ω2 d ω d ω d t = I 2 2 ω d ω d t = I ω α

The quantity dL/dt is just the speed of the cord, v, this is related to the angular speed by v = R ω. Use that in the preceeding equation and you get

The quantity on the left side is called the torque. (Actually it's the magnitude of the torque, which is more generally defined and is a vector.)

I did another version of this, starting from F=ma, or rather, from F=dp/dt. As long as you do the same operation to the two sides of an equation you will get an equation. I manipulated this by taking the cross product with the displacement vector r. The text on page 309 does the same sort of calculation, only working in reverse. They start with the end result and work backwards -- it is probably easier to follow.

14 Nov Discuss a couple of problems by request. Then a qualitative demonstration of the vector nature of angular momentum and angular velocity.

Simple harmonic motion. A mass hanging on the end of a spring oscillates up and down. What is the mathematical description of the motion? I'll start with a simpler variation, where I don't have to include gravity among the forces. Let the spring move horizontally and suppose that the friction is negligible. This isn't realistic, but it's simpler, and when I do the other case it will be much easier to understand.

The force exerted by a spring on a mass attached to it is F_x = -k x, where the position x=0 is where the spring is relaxed. If this is the only force on a mass m, then F=ma becomes

The function x( t ) is what I want to determine. The methods we've used up to now won't work on this equation. (Actually they can be made to work, but it's a stretch.) If you decide to integrate with respect to t, that works for the left-hand side, giving you m v_x(t). You can't do it to the right side, because you don't know the function x(t). Whatever it is, it isn't a constant.

I can rearrange this a bit:

d2x( t ) d t2 = - k m x( t )

This says that whatever the function x(t) is, its second derivative with respect to t is a (negative) constant time x(t) itself. You know some functions that do that. Remember that the independent variable is t, not x. Sines, cosines, and exponentials are plausible candidates for this. The exponential function has the wrong sign for its second derivatibe though, so its eliminated.

d2 d t2 cos( t ) = - cos( t ),        and similarly        d2 d t2 sin( t ) = - sin( t ),

But this isn't quite enough, because the constant, k/m, isn't there. Also, the dimensions don't work because you can't take the cosine of a time and the output of a cosine isn't a length. Try again with a couple of constants thrown in

d2 d t2 A cos( ω0t ) = - A ω02 cos( ω0t ),

To find out if this is a solution to the equation (F=ma), substitute this x and this a_x and see if it works.

16 Nov There are various ways to write solutions to the harmonic oscillator equation.

The only difference between the graph of the sine and the graph of the cosine is in where you place the origin. The third form above includes the other two as special cases ( β = 0 and β = - p / 2 ).

Spend some time on the interpretation of this result. What does the parameter ω₀ mean, and what is its relation to the period of oscillation?

I was able to verify at least a part of this experimentally: the dependence on mass.

The next simple example of a harmonic oscillator is a pendulum. In order to find the equations that describe its motion, you can (1)~start from F=ma, (2)~use torque, (3)~use energy methods. I'll do the latter. Let L be the length of the cord and m the mass suspended on the end of the cord. With the angle θ(t) being measured from the vertical, the total energy is

Use the relation v = L ω here. [Watch out! there are two uses for the letter ω. One means d  θ / d t. The other is the number, the frequency of oscillation -- I tried to keep them a little clearer by calling the number ω₀ instead, but the text doesn't do this.

That the energy is a constant means that its derivative with respect to time is zero. To evaluate this derivative, use the chain rule a few times.

d cos θ d t =  d cos θ d θ  d θ d t = - sin θ ω

and

d ω2 d t =  d ω2 d ω  d ω d t = 2 ω  d2 θ d t