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First mid-term exam next Tuesday, 9 February, 11:45-1:10
The exam will be on the topics we covered in chapters 2 and 3. More precisely, 2.1 - 2.3 and 3.1 - 3.3.
I strongly recommend that you review homework problems, and pay a little extra attention to the homework due week after next since those problems are on topics in chapter 3.
Fowles and Cassidy 4.1 - 4.4
The rules for vector algebra, and the (time) derivative of a vector.
Suppose we have a reference frame (coordinate system), and in that frame we know the position of a particle is:
r = ix + jy + kz
If x, y, and z are functions of time, then we know the position as a function of time, and r as a function of time. We find the velocity as a function of time by taking the time derivative of r, v = dr/dt = i[xdot] + j[ydot] + k[zdot] . The speed v is the absolute value of the velocity and is given by v = |v| = Sqrt([xdot]2 + [ydot]2 + [zdot]2).
The acceleration is given by the time derivative of the velocity, a = dv/dt .
In 3 dimensions we will rewrite Newton's 2nd and 3rd laws in vector form:
F = dp/dt = ma
F12 = -F21
Each of these stands for 3 component equations, equality of the x, y, and z components. This compactness is the reason that we use vector notation! If the equation of motion can be separated into its 3 components giving 3 equations of the types studied previously (i.e. force is constant, depends only on time, or each component depends only on the corresponding component of position) then the equations can be solved with the techniques used previously. Only now there are 2 or 3 equations to solve instead of only 1! The simplest example of this case is the motion of a projectile in uniform gravity and without air resistance.
A projectile is fired from the origin at t = 0 with velocity v0 = (v0x, v0y, v0z). Describe its motion.
First, choose the coordinate system. Let's take z to be the up (vertical) direction with x and y parallel to the ground. The equation of motion is: ma = -mgk, which represents the three scalar equations:
max = 0
may = 0
maz = -mg
This set of equations are "separated", meaning that a given variable appears in only one of the equations. The variables here are x, y, and z. The time t is a parameter, and m and g are constants. We know how to solve each of these separately, therefore we can assemble the separate solutions to solve the 3-dimensional problem.
So, you can apply what you learned about one-dimensional problems to solve 3-dimensional problems that are "separable". These include problems where the force is only a function of time, or a function of position, but with the x component a function of x only, the y component a function of y only, and the z component a function of z only. Of course, this is a very small subset of all the possible situations one might encounter, so we will push on to more general and powerful methods.
The first is to invoke the ideas of work and energy. In 1-dimension, work is defined as: W = ò F dx. That is, the work required to move from x0 to x1 is the integral of the force over that distance. The generalization to 3-dimensions is similar, the work required to move from r0 to r1 is the integral of the force over the path taken from r0 to r1. This sounds easy enough, but in the beginning, the part "over the path taken from r0 to r1" causes some confusion. This means that we must step along the actual path the object moves along, and for each step we sum up the force on the object times the distance moved in that step. Then, we take the limit that the step sizes go to zero, and we get the integral:
W = ò F× ds
Pay careful attention to this integral, it is not your standard integral. This is called a line integral. To compute this integral you must first define the path that the object follows. Then you parameterize the path in terms of a single variable (parameter). Then you can express ds in terms of this parameter, perform the dot product, and integrate. An example is clearly in order.
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1 February, 1999 R. Harr