Class Notes for Fall, 2000

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

Phy 205, Week 9

16 Oct Do one of the homework problems in some detail.

For the special case of constant force, I had shown a special case of the work-energy theorem, W=DK. Now I want the much more useful version that applies for non-constant forces.

There are some problems that you just can't do with the tools we've covered so far. This next version of the work-energy theorem will greatly expand the possibilities.

The idea behind this is to do a sequence of approximations. The work-kinetic energy equation for constant force is (using "i" for initial and "f" for final)

If the force isn't a constant, you could get a rough approximation to the result by taking some middle point in the interval and pretending that the force is a constant, taking the value at that middle point. The approximate result would then be for the force function F_x(x),

This isn't exact of course, but you can improve on the approximation by dividing the interval into two pieces, and then approximating F_x by a constant in each piece. Call the division point x₁, then the two approximating equations are

Here I took two points in the middle of the two intervals: x^*₁ in the first interval and x^*₂ in the second.

Look at what happens when you add these two equations. Nothing special on the left, but on the right there's some cancellation, leaving only the difference of the final and the initial kinetic energies.

If you can understand the process of these two approximations using one interval and then using two intervals, you will have mastered a key mathematical concept. After doing two intervals, the extension to three, four or four million is just more of the same. Ideally, you take a limit as the length of each interval approaches zero (and the number of intervals approaches infinity). This defines the integral.

As the first example, I took a familiar problem, one that you can solve just as easily by other means. Throw an object straight up; how high will it go? Apply the work-kinetic energy theorem and the result comes out in a couple of lines.

The more interesting problem is one that you can't solve by the old methods: Throw the object straight up, but fast enough that it tries to escape from the Earth. In this case you better not assume that the Earth's gravitational field is uniform, because it drops off with distance.

18 Oct The gravitational pull of the Earth drops off with distance. In particular, it is described by the equation

Here, x is measured from the center of the Earth, not its surface. Notice that when x=R, you are at the surface of the Earth, and the force is expressed as F_x = - mg, the way you expect.

The actual gravitational field of the Earth is more complicated than this, because the Earth isn't a perfect sphere. The deviations from this equation would be important if you're trying to hit something such as the moon, but we don't need to worry about them.

If you fire something off from the surface of the Earth at a speed v₀ and let it coast, will it fall back or will it escape from the Earth altogether? Apply the work-energy theorem: The initial position will be x_i = R, so this is

Applied to this problem, the work integral becomes

Solve for this required initial speed to get

Use the numbers, g = 10 m/s² and R = 6400 km to get v₀ = 11.2 km/s.

The common way to evaluate an integral is to use the fundamental theorem of calculus. This says that if you find the anti-derivative of a function, evaluate it at the two endpoints, and subtract these two numbers, that is the integral of the original function between those two points.

What does that lead to when you apply it to the work-kinetic energy theorem? Let the anti-derivative of F_x be -U(x). Why the minus sign? If I don't put it here, it will just appear later in a more awkward way.