Taylor 4.1 {and part of 4.10} (today) and 4.2-4.3 (Wednesday).
Let's summarize. The motion of the center of mass is governed by the external forces acting on the system. Internal forces contribute to internal motion, but don't affect the motion of the CM. The time rate of change of the total angular momentum of a system equals the net external torque. The total angular momentum is composed of two components, the angular momentum due to rotations about the center of mass, and the angular momentum of the orbit of the center of mass.
The strategy for analyzing the motion of a system is to take a top down approach. Start with the motion of the center of mass of the system. Then look for rotations about the center of mass (a collective motion). Finally, consider the individual motions of particles relative to the center of mass.
Energy is another very useful constant, but even more abstract than momentum. Today we take conservation of energy as a fundamental precept, but it wasn't formulated until about 1880 by Helmholtz. To utilize conservation of energy, we must take many forms of energy into account: kinetic energy, potential energy, and thermal energy are of particular importance in mechanics. You will run into many other types in your study of physics: electromagnetic energy, chemical energy, nuclear energy, and still others.
We define the kinetic energy of a single particle as T=½mv², where m is the particle's mass and v is the magnitude of its velocity. The time derivative of the kinetic energy is
Multiply both sides by dt and note that vdt = dr to get a relation known as the Work-Energy Theorem
In this form, both sides of the equation can be integrated along the path followed by the particle. To integrate along a path from r1 to r2, first imagine dividing the path up into lots of small steps of length Δri, then adding up along the path to get
Take the limit as the Δr's go to zero, and the sum goes over to a line integral
A line integral is a one-dimensional integral performed over a path rather than a single coordinate like x. The normal way to calculate such an integral is to parameterize the path in terms of another variable, such as t, then integrate over that variable. Sometimes it is easy to break a single path into several segments, where each segment can be easily integrated and the results summed together.
Evaluate the work done
by the two-dimensional force F=(-yn, yxn) along the three paths joining the points P = (1,0) and Q = (0,1): (a) the path that goes from P to the origin, O = (0,0), and then to Q; (b) the path that goes from P to Q along a straight line (express y as a function of x and rewrite the integral as an integral over x); and (c) the path given parametrically by x = 1 - t³ and y = t².
For each case, we want to express the path in terms of a single variable to perform the integral, or part of the integral. Let's tackle the these one at a time.
(a) This is the easiest case. We can break the entire path into two segments, one running along x with y constant, and the other along y with x constant. When moving along x with y constant, dr = dx xhat, and when moving along y with x constant, dr = dy yhat. Then F⋅dr will reduce to Fxdx = -yndx when moving along x with y constant, and Fydy = yxndy when moving along y with x constant. These are integrated along the appropriate segment and summed to get the work,
(b) This integral is done by parameterizing the path in terms of one variable. The obvious way is to express y as a function of x, y=1-x. Then dy = -dx, so we can replace y and dy with x and dx and perform the integral as follows
(c) Use the parameterized x and y to write the force and the work integral in terms of t. Notice that x=1 and y=0 corresponds to t=0, and x=0 and y=1 occurs when t=1, and that dx = -3t²dt and dy = 2tdt.
The force in the integral for the work is the net force on a particle, Ftot = F1 + F2 +...+ Fn = ∑i=1n Fi. The work integral over the sum of forces can be changed to a sum over integrals of the forces:
So the total work done can be calculated by finding the work done by each force separately and summing.
If the total work done on an object is zero, then the kinetic energy is the same at the beginning and end of the path (but maybe not in between), and the magnitude of the velocity is the same. If the net force on an object is zero, then the work done along any path is zero, the kinetic energy is constant, and the speed of the particle (magnitude of the velocity) is constant.