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| Phy 205 page |
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| Oct 29 | Notes for Fall, 2000 |
8 Oct The equation F = m a could also have been written F = d( m v ) / dt. Does it matter? As long as the mass is a constant, so that dm/dt = 0, they are the same. If the mass is changing they can be very different. If you are trying to send a rocket into space, its mass is changing rapidly; these two versions will then give very different answers. Which is correct? the second one.
We'll do little if anything with changing masses, so the F=ma form is good enough for that.
There's a point that I've ignored about even this form of the equation: It says that the total force on a point mass divided by the size of that mass gives its acceleration. What I previously ignored is the fact that this is supposed to apply to point masses, and that's a sort of idealization or a limiting case of real masses. Is the Earth a point mass? Not if you're standing on it, but if you are describing the structure of the solar system it is reasonable to call it a point mass because its size makes no difference.
How do I get away with ignoring this? The answer is in a theorem about the motion of several point masses. If you understand how it works for two masses, the general case is just more terms of the same kind.
Here, I am being careful to distinguish between the forces on each mass as caused by the other mass and as caused by everything else. For example, the Earth and the Moon go around the Sun. The Moon is pulled on by the Earth and the Sun. The Earth is pulled on by the Moon and the Sun.
Now add the above equations. Remember Newton's third law? It says that if two objects act on each other then the forces are related by
Now I'll manipulate the above force equations. ADD them.
The first and third terms cancel. Now manipulate the right-hand side. The masses are constants, so they can go either inside or outside the differentiation. Bring them inside and then multiply by one.
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This last expression defines the center of mass. This says that the final equation is really quite simple:
The motion of the center of mass behaves as a point with just the external forces being applied. None of the complications of the internal forces matter for this purpose.
A demonstration using an off-center dumbbell, showing that the center of mass appears to follow a parabola when the whole thing is thrown. The center of mass of the Earth-Moon system is about a thousand kilometers below the Earth's surface. When you describe the motion of the Earth around the sun as being an ellipse (or approximately a circle) you are really describing this center of mass motion. The Earth wobbles around.
10 Oct Reprise the work leading to the concept of center of mass. Talk about the ambiguity in the word "average" and show different types of average:
and there are others. The sort of average that we have to contend with in mechanics is none of these. The center of mass is a weighted average in which the position of one of the masses counts for more if its mass is larger. Until about five years ago, the only planets known were the nine in our own solar system. Since then about 70 or 80 have been found, and more all the time. The method is based on this result about the center of mass. When a planet revolves around a star, it's not really revolving around the star, it (and the star) are revolving around their common center of mass. If the star is 500 times the mass of the planet, then the planet will have about 500 times the motion of the star, but the star's motion will be detectable (barely). This is what's been detected: the star's wobble. From this slight wobble, we can determine the orbit of the planet, its mass, and even if there is more than one such planet. Several multiple planet systems have been found, though so far this method isn't sensitive enough to detect a planet as small as Earth. That will have to wait until later.
The same sort of ideas that went into the relation between total external force and the acceleration of the center of mass can yield another result, conservation of momentum.
Suppose you have a system with no net external force. The only forces to deal with are the internal ones. In that case, the equations for the two masses are conveniently written as
This time, if you add the forces you get zero. Remember, I said that the net external force is zero, then all the internal forces pair off to add to zero also. This sum is
This in turn implies that the combination in parentheses is a constant.
This is "conservation of momentum." I discussed several applications of this result, including neutrinos and galaxies. Then I demonstrated it in a more mundane context with some colliding carts.
12 Oct Do one of the homework problems, taken from the old exam.
Center of mass. What do you do when you have many masses, not just two or three? You can do this by succesive approximations, dividing the object into pieces and approximating each piece by a point mass. This sounds crude, but it can give remarkable accurate results, and eventually leads to the concept of the integral. I showed a numerical example to see how it could work.
Derive the relation between force, displacemant and kinetic energy for the special case of constant force in one dimension. It's mostly a lot of routine algebra, eliminating the parameter, t, between two equations for position and velocity.
This is a special case because it's restricted to one dimension and to a constant force. I stated what the generalization is to three dimensions (though still constant force). Force is a vector. The change in position is a vector, not just x-x0. The generalization is
