Chapter 8
Integral Calculus and Its Uses
8.2 Two-Dimensional Centers of Mass
8.2.4 Calculation of \(\bar{x}\)
Now that we know how to find both the mass \(m\) and the horizontal moment \(M_y\), the final step in finding the \(x\)-coordinate \(\bar{x}\) is to divide the two:
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We hesitate to dignify this equation with the name "formula," because it is not something you should think of as a universal solution to center-of-mass problems, and you certainly don't want to just plug something into it and start simplifying. Rather, you should think of it as a guide to the process of finding \(\bar{x}\), more or less as we outlined the steps. Notice that \(\theta\) and \(\delta\) cancel out in this formula - because we took them to be constant, which meant that they factored out of the sums and the integrals.
The equation
has a compelling logic that should not be overlooked - and that will serve as a guide to finding other center of mass coordinates, e.g., \(\bar{y}\).
First, locating the center of mass of a cardboard cutout should be a purely geometric (as opposed to physical) problem, in the sense that it should depend only on the geometry of the shape. Hence the disappearance of \(\theta\) and \(\delta\). Alternatively, we can declare both \(\theta\) and \(\delta\) to be \(1\); that is, we can set the unit of density to be whatever the density of our shape is and the unit of distance (in the third dimension) to be whatever the thickness is. Clearly, these declarations should have nothing to do with the location of the two-dimensional center of mass. With either of these interpretations, area and mass become the same number, and the moment \(M_y\) is identified with the integral
Second, we can read
in a less elegant but more meaningful way:
Thus we accumulate mass (area) for the denominator by adding up height \(\times\) width for infinitely many rectangles, each with infinitesimal area. We accumulate the total moment for the numerator by multiplying each small rectangle by its distance from the \(y\)-axis and adding up the results.
There is still another way to read this formula: The average \(x\) has to be a weighted average, because the distribution of mass (area) is not uniform. In the numerator, we can think of the height and width factors \([h(x) \Delta x]\) as weighting the typical \(x\) factor. Then all the weighted \(x\)'s are added up - and to find an average, the sum of weighted \(x\)'s is divided by the sum of the weights.
