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Fowles and Cassicy 4.1 - 4.4
Rules of vector algebra.
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.
The x and y forces are zero, so we get motion at a constant velocity: x = v0x t and y = v0y t. The motion in the z direction is of a body acted on by a constant force: z = -1/2 gt2 + v0z t. You know that the path of the projectile is a parabola, but how can we show this explicitly. We can regard the three expressions for x, y, and z as a function of time as a parameterization of the path -- if you plot the (x,y,z) points for a range of values of time, the points will form a curve. To show explicitly that the path is a parabola, we need to rewrite the equations to remove the parameter t, and have, for instance, a function z(x). This can be accomplished by noting that t = x/v0x, such that:
z(x) = x v0z/v0x - 1/2(g/(v0x)2) x2 and y(x) = (v0y/v0x) x
The equation for z is now clearly a parabola in terms of x. And if we use a coordinate that measures the distance from the origin in the x-y plane, instead of the x coordinate, then we get a similar equation for z, but in that coordinate.
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.
Let's use the same force discussed in section 4.1 and displayed in Fig. 4.1.2, F = -iby + jbx. And let's calculate the work required to move a particle from (x0,y0) to (x0+D x, y0+D y), but this time along the path y = y0 + (x - x0)2 D y/(D x)2. This is a parabolic path. Now we can parameterize everything in terms of the single variable (parameter) x. It is easy to do this for F, but what about ds? Well, ds is a small step along the tangent vector to the curve. What is the tangent vector to this curve? In general, if a curve is parameterized like (x(t), y(t)), then the tangent vector is (dx/dt, dy/dt). In our case, x=t, so the tangent vector is simply i + j dy/dx = i + j 2(x - x0) D y/(D x)2. A small step along this direction means a small change in our parameter x, so ds = idx + j 2(x - x0) D y/(D x)2dx. We can now do the dot product F× ds
F× ds = -bydx + 2bx(x - x0) D y/(D x)2dx = bdx[D y/(D x)2(2x(x-x0) - (x-x0)2) - y0]
= bdx[D y/(D x)2(x+x0)(x-x0) - y0] = bdx[D y/(D x)2(x2-x02) - y0]
This looks messy, but we can now do the integral. The limits of the integral are from x0 to x0 + D x, which corresponds to the values of x at the beginning and end of the path.
ò
F× ds = bò [D y/(D x)2(x2-x02) - y0]dx = b[D y/3(D x)2{(x0+D x)3 - x03} - x02D y/(D x) - y0D x]
Ó
11 February, 1999 R. Harr