This week you have learned, or will learn, in physics and chemistry classes about centers of mass and moments of inertia, their meaning and importance. Now we will study how to use calculus to compute these things. In engineering class you will be doing parallel computations in AutoCad.
Temporary home of the graphics for the summary:
An everyday object contains something like 1025 particles; we can't keep track of them individually. Instead, we have to treat the body as consisting of a smooth distribution of "stuff".
We do that by mentally cutting up space into small cubes. (READ ME) (PDF version)
It's time for an example! The concept of a double or triple integral is (we hope) intuitively clear for an arbitrary region, but so far we have covered the technique of evaluating them only for rectangular and circular regions.
Suppose that the density in the rectangular block
0 < x < 1, 0 < y < 1, 0 < z < 2
is [rho](x,y,z) = 1 + 2y kg/m3. (The coordinates are in meters.)
Problem 1: What is the mass of the block? (Try to answer the question yourself before looking at the worked-out example.) (PDF version)
Before continuing with the example, we have to return to the general definitions. What do we mean by the center of mass of a solid body? Remember that we have to think of the cloud of particles as just "stuff" (a continuum of matter). Therefore, we replace the sums over particles in the particulate definition by integrals over the density function. <--(READ TOP) (PDF version)
Special case: If [rho] is a constant (the material is homogeneous), then it cancels from these formulas. The total mass equals [rho] times the volume of E, etc.: (READ BOTTOM). (PDF version) The vector with these components is called the centroid of the region E.
Notice that the centroid is a purely geometrical concept -- it does not depend at all on what sort of matter occupies the region. Nevertheless, it is also equal to the center of mass (a physical quantity) when the matter is homogeneous.
Problem 2: Find the center of mass of our block. (We will do this as a class exercise.)
Back to the general theory one more time: What are the moments of inertia of a solid body? It's the same general idea as before: replace the sum over particles by a triple integral. (PDF version)
Doing this for the rectangular block is both harder and less useful than this:
Problem 3: Find Iz for the cylinder
-2 < z < 2, 0 < r = (x2 + y2)1/2 < 3
with constant density [rho] throughout.
Here are two versions of the solution. (PDF version)
Temporary home of the graphics for the solution of Problem 3:
Notice that the center of mass and moment of inertia depend on where E is, relative to the coordinate axes. Unlike the volume and the total mass, they are not intrinsic properties of the body E. (However, in elementary physics one is primarily interested in the moment of inertia about an axis of symmetry of the body, since in that case the equation L = I [omega] applies (L = angular momentum, [omega] = angular velocity).)
Now we consider special cases where all the matter lies (approximately) in the x-y plane. Such problems (which have important practical applications) are relatively easy to solve, because we need to integrate over only two dimensions, not three.
(READ ME). (PDF version).
Example (PDF version)
Temporary home of the graphics for this example:
Challenge: Try finding ycm for this example by slicing the region horizontally instead of vertically. Be prepared to encounter an integral even harder than the one we just did.
Remark: The two-dimensional moment formulas turn up in engineering in the theory of bending of beams. (Of course, this application has nothing to do with the original physical application to a flat plate spinning in space.) The moment of inertia of the cross section of the beam about an axis in the plane of the cross section is related to the behavior of the beam under bending in the corresponding transverse direction. The moment of inertia of the cross section about the axis perpendicular to the plane is related to the behavior of the beam under twisting. We refer to later courses on engineering mechanics for the details.
Now you are ready to study these topics in more detail by reading Stewart, Sections 13.7, 13.5, 13.3, and 8.4. (The backwards order is actually the most natural one, conceptually! Read them in whatever order you find them easiest to understand (and as many times as necessary).)
You may have noticed that so far we have carefully avoided giving either a name or a symbol to the numerators in the center-of-mass formulas. The reason is that there are several unfortunate inconsistencies in the terminology and notation of this subject. We now have the obligation to warn you about these.