6.6 Moments and Centers of Mass

In this section, we consider centers of mass (also called centroids, under certain conditions) and moments. The basic idea of the center of mass is the notion of a balancing point. Many of us have seen performers who spin plates on the ends of sticks. The performers try to keep several of them spinning without allowing any of them to drop. If we look at a single plate (without spinning it), there is a sweet spot on the plate where it balances perfectly on the stick. If we put the stick anywhere other than that sweet spot, the plate does not balance and it falls to the ground. (That is why performers spin the plates; the spin helps keep the plates from falling even if the stick is not exactly in the right place.) Mathematically, that sweet spot is called the center of mass of the plate.

In this section, we first examine these concepts in a one-dimensional context, then expand our development to consider centers of mass of two-dimensional regions and symmetry. Last, we use centroids to find the volume of certain solids by applying the theorem of Pappus.

Center of Mass and Moments

Let’s begin by looking at the center of mass in a one-dimensional context. Consider a long, thin wire or rod of negligible mass resting on a fulcrum, as shown in Figure 6.62(a). Now suppose we place objects having masses $m_{1}$ and $m_{2}$ at distances $d_{1}$ and $d_{2}$ from the fulcrum, respectively, as shown in Figure 6.62(b).

This figure has two images. The first image is a horizontal line on top of an equilateral triangle. It represents a rod on a fulcrum. The second image is the same as the first with two squares on the line. They are labeled msub1 and msub2. The distance from msub1 to the fulcrum is dsub1. The distance from msub2 to the fulcrum is dsub2. — Figure 6.62(a) A thin rod rests on a fulcrum. (b) Masses are placed on the rod.

The most common real-life example of a system like this is a playground seesaw, or teeter-totter, with children of different weights sitting at different distances from the center. On a seesaw, if one child sits at each end, the heavier child sinks down and the lighter child is lifted into the air. If the heavier child slides in toward the center, though, the seesaw balances. Applying this concept to the masses on the rod, we note that the masses balance each other if and only if $m_{1} d_{1} = m_{2} d_{2} .$

In the seesaw example, we balanced the system by moving the masses (children) with respect to the fulcrum. However, we are really interested in systems in which the masses are not allowed to move, and instead we balance the system by moving the fulcrum. Suppose we have two point masses, $m_{1}$ and $m_{2},$ located on a number line at points $x_{1}$ and $x_{2},$ respectively (Figure 6.63). The center of mass, is the point where the fulcrum should be placed to make the system balance.

This figure is an image of the x-axis. On the axis there is a point labeled x bar. Also on the axis there is a point xsub1 with a square above it. Inside of the square is the label msub1. There is also a point xsub2 on the axis. Above this point there is a square. Inside of the square is the label msub2. — Figure 6.63The center of mass $x -$ is the balance point of the system.

Thus, we have

m_{1} ∣ x - - x_{1} ∣ m_{1} (x - - x_{1}) m_{1} x - - m_{1} x_{1} x - (m_{1} + m_{2}) x - = = = = = m_{2} ∣ x_{2} - x - ∣ m_{2} (x_{2} - x -) m_{2} x_{2} - m_{2} x - m_{1} x_{1} + m_{2} x_{2} \frac{m _{1} x _{1} + m _{2} x _{2}}{m _{1} + m _{2}} .

The expression in the numerator, $m_{1} x_{1} + m_{2} x_{2},$ is called the first moment of the system with respect to the origin. If the context is clear, we often drop the word first and just refer to this expression as the moment of the system. The expression in the denominator, $m_{1} + m_{2},$ is the total mass of the system. Thus, the of the system is the point at which the total mass of the system could be concentrated without changing the moment.

This idea is not limited just to two point masses. In general, if n masses, $m_{1}, m_{2}, \dots, m_{n},$ are placed on a number line at points $x_{1}, x_{2}, \dots, x_{n},$ respectively, then the center of mass of the system is given by

x - = \frac{\sum _{i = 1}^{n} m _{i} x _{i}}{\sum _{i = 1}^{n} m _{i}} .

We apply this theorem in the following example.

We can generalize this concept to find the center of mass of a system of point masses in a plane. Let $m_{1}$ be a point mass located at point $(x_{1}, y_{1})$ in the plane. Then the moment $M_{x}$ of the mass with respect to the -axis is given by Similarly, the moment with respect to the -axis is given by Notice that the -coordinate of the point is used to calculate the moment with respect to the -axis, and vice versa. The reason is that the -coordinate gives the distance from the point mass to the -axis, and the -coordinate gives the distance to the -axis (see the following figure).

This figure has the x and y axes labeled. There is a point in the first quadrant at (xsub1, ysub1). This point is labeled msub1. — Figure 6.64Point mass $m_{1}$ is located at point $(x_{1}, y_{1})$ in the plane.

If we have several point masses in the xy-plane, we can use the moments with respect to the x- and y-axes to calculate the x- and y-coordinates of the center of mass of the system.

The next example demonstrates how to apply this theorem.

Center of Mass of Thin Plates

So far we have looked at systems of point masses on a line and in a plane. Now, instead of having the mass of a system concentrated at discrete points, we want to look at systems in which the mass of the system is distributed continuously across a thin sheet of material. For our purposes, we assume the sheet is thin enough that it can be treated as if it is two-dimensional. Such a sheet is called a lamina. Next we develop techniques to find the center of mass of a lamina. In this section, we also assume the density of the lamina is constant.

Laminas are often represented by a two-dimensional region in a plane. The geometric center of such a region is called its centroid. Since we have assumed the density of the lamina is constant, the center of mass of the lamina depends only on the shape of the corresponding region in the plane; it does not depend on the density. In this case, the center of mass of the lamina corresponds to the centroid of the delineated region in the plane. As with systems of point masses, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes.

We first consider a lamina in the shape of a rectangle. Recall that the center of mass of a lamina is the point where the lamina balances. For a rectangle, that point is both the horizontal and vertical center of the rectangle. Based on this understanding, it is clear that the center of mass of a rectangular lamina is the point where the diagonals intersect, which is a result of the symmetry principle, and it is stated here without proof.

Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function $f (x),$ below by the x-axis, and on the left and right by the lines $x = a$ and $x = b,$ respectively, as shown in the following figure.

This image is a graph of y=f(x). It is in the first quadrant. Under the curve is a shaded region labeled “R”. The shaded region is bounded to the left at x=a and to the right at x=b. — Figure 6.65A region in the plane representing a lamina.

As with systems of point masses, to find the center of mass of the lamina, we need to find the total mass of the lamina, as well as the moments of the lamina with respect to the x- and y-axes. As we have done many times before, we approximate these quantities by partitioning the interval $[a, b]$ and constructing rectangles.

For $i = 0, 1, 2, \dots, n,$ let $P = {x_{i}}$ be a regular partition of $[a, b] .$ Recall that we can choose any point within the interval $[x_{i - 1}, x_{i}]$ as our In this case, we want to be the -coordinate of the centroid of our rectangles. Thus, for we select such that is the midpoint of the interval. That is, Now, for construct a rectangle of height on The center of mass of this rectangle is as shown in the following figure.

This figure is a graph of the curve labeled f(x). It is in the first quadrant. Under the curve and above the x-axis there is a vertical shaded rectangle. the height of the rectangle is labeled f(xsubi). Also, xsubi = xsubi – 1 + xsubi/2. — Figure 6.66A representative rectangle of the lamina.

Next, we need to find the total mass of the rectangle. Let $ρ$ represent the density of the lamina (note that $ρ$ is a constant). In this case, $ρ$ is expressed in terms of mass per unit area. Thus, to find the total mass of the rectangle, we multiply the area of the rectangle by $ρ .$ Then, the mass of the rectangle is given by $ρ f (x_{i}^{*}) Δ x .$

To get the approximate mass of the lamina, we add the masses of all the rectangles to get

m \approx i = 1 \sum n ρ f (x_{i}^{*}) Δ x .

This is a Riemann sum. Taking the limit as $n \to \infty$ gives the exact mass of the lamina:

m = n \to \infty lim i = 1 \sum n ρ f (x_{i}^{*}) Δ x = ρ \int_{a}^{b} f (x) d x .

Next, we calculate the moment of the lamina with respect to the x-axis. Returning to the representative rectangle, recall its center of mass is $(x_{i}^{*}, (f (x_{i}^{*})) / 2) .$ Recall also that treating the rectangle as if it is a point mass located at the center of mass does not change the moment. Thus, the moment of the rectangle with respect to the x-axis is given by the mass of the rectangle, $ρ f (x_{i}^{*}) Δ x,$ multiplied by the distance from the center of mass to the -axis: Therefore, the moment with respect to the -axis of the rectangle is Adding the moments of the rectangles and taking the limit of the resulting Riemann sum, we see that the moment of the lamina with respect to the -axis is

M_{x} = n \to \infty lim i = 1 \sum n ρ \frac{[ f ( x _{i}^{*} )] ^{2}}{2} Δ x = ρ \int_{a}^{b} \frac{[ f ( x )] ^{2}}{2} d x .

We derive the moment with respect to the y-axis similarly, noting that the distance from the center of mass of the rectangle to the y-axis is $x_{i}^{*} .$ Then the moment of the lamina with respect to the y-axis is given by

M_{y} = n \to \infty lim i = 1 \sum n ρ x_{i}^{*} f (x_{i}^{*}) Δ x = ρ \int_{a}^{b} x f (x) d x .

We find the coordinates of the center of mass by dividing the moments by the total mass to give $x - = M_{y} / m and y - = M_{x} / m .$ If we look closely at the expressions for we notice that the constant cancels out when and are calculated.

We summarize these findings in the following theorem.

In the next example, we use this theorem to find the center of mass of a lamina.

We can adapt this approach to find centroids of more complex regions as well. Suppose our region is bounded above by the graph of a continuous function $f (x),$ as before, but now, instead of having the lower bound for the region be the x-axis, suppose the region is bounded below by the graph of a second continuous function, $g (x),$ as shown in the following figure.

This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded region labeled “R”. The shaded region is bounded to the left by x=a and to the right by x=b. — Figure 6.68A region between two functions.

Again, we partition the interval $[a, b]$ and construct rectangles. A representative rectangle is shown in the following figure.

This figure is a graph of the first quadrant. It has two curves. They are labeled f(x) and g(x). f(x) is above g(x). In between the curves is a shaded rectangle. — Figure 6.69A representative rectangle of the region between two functions.

Note that the centroid of this rectangle is $(x_{i}^{*}, (f (x_{i}^{*}) + g (x_{i}^{*})) / 2) .$ We won’t go through all the details of the Riemann sum development, but let’s look at some of the key steps. In the development of the formulas for the mass of the lamina and the moment with respect to the -axis, the height of each rectangle is given by which leads to the expression in the integrands.

In the development of the formula for the moment with respect to the x-axis, the moment of each rectangle is found by multiplying the area of the rectangle, $ρ [f (x_{i}^{*}) - g (x_{i}^{*})] Δ x,$ by the distance of the centroid from the x-axis, $(f (x_{i}^{*}) + g (x_{i}^{*})) / 2,$ which gives Summarizing these findings, we arrive at the following theorem.

We illustrate this theorem in the following example.

The Symmetry Principle

We stated the symmetry principle earlier, when we were looking at the centroid of a rectangle. The symmetry principle can be a great help when finding centroids of regions that are symmetric. Consider the following example.

Student ProjectThe Grand Canyon Skywalk

The Grand Canyon Skywalk opened to the public on March 28, 2007. This engineering marvel is a horseshoe-shaped observation platform suspended 4000 ft above the Colorado River on the West Rim of the Grand Canyon. Its crystal-clear glass floor allows stunning views of the canyon below (see the following figure).

Figure 6.72The Grand Canyon Skywalk offers magnificent views of the canyon. (credit: 10da_ralta, Wikimedia Commons)

The Skywalk is a cantilever design, meaning that the observation platform extends over the rim of the canyon, with no visible means of support below it. Despite the lack of visible support posts or struts, cantilever structures are engineered to be very stable and the Skywalk is no exception. The observation platform is attached firmly to support posts that extend 46 ft down into bedrock. The structure was built to withstand 100-mph winds and an 8.0-magnitude earthquake within 50 mi, and is capable of supporting more than 70,000,000 lb.

One factor affecting the stability of the Skywalk is the center of gravity of the structure. We are going to calculate the center of gravity of the Skywalk, and examine how the center of gravity changes when tourists walk out onto the observation platform.

The observation platform is U-shaped. The legs of the U are 10 ft wide and begin on land, under the visitors’ center, 48 ft from the edge of the canyon. The platform extends 70 ft over the edge of the canyon.

To calculate the center of mass of the structure, we treat it as a lamina and use a two-dimensional region in the xy-plane to represent the platform. We begin by dividing the region into three subregions so we can consider each subregion separately. The first region, denoted $R_{1},$ consists of the curved part of the U. We model as a semicircular annulus, with inner radius 25 ft and outer radius 35 ft, centered at the origin (see the following figure).

Theorem of Pappus

This section ends with a discussion of the theorem of Pappus for volume, which allows us to find the volume of particular kinds of solids by using the centroid. (There is also a theorem of Pappus for surface area, but it is much less useful than the theorem for volume.)

Proof

We can prove the case when the region is bounded above by the graph of a function $f (x)$ and below by the graph of a function $g (x)$ over an interval $[a, b],$ and for which the axis of revolution is the y-axis. In this case, the area of the region is $A = \int_{a}^{b} [f (x) - g (x)] d x .$ Since the axis of rotation is the -axis, the distance traveled by the centroid of the region depends only on the -coordinate of the centroid, which is

x - = \frac{M _{y}}{m},

where

m = ρ \int_{a}^{b} [f (x) - g (x)] d x and M_{y} = ρ \int_{a}^{b} x [f (x) - g (x)] d x .

Then,

d = 2 π \frac{ρ \int _{a}^{b} x [ f ( x ) - g ( x )] d x}{ρ \int _{a}^{b} [ f ( x ) - g ( x )] d x}

and thus

d \cdot A = 2 π \int_{a}^{b} x [f (x) - g (x)] d x .

However, using the method of cylindrical shells, we have

V = 2 π \int_{a}^{b} x [f (x) - g (x)] d x .

So,

V = d \cdot A

and the proof is complete.

□

Key Concepts

Mathematically, the center of mass of a system is the point at which the total mass of the system could be concentrated without changing the moment. Loosely speaking, the center of mass can be thought of as the balancing point of the system.
For point masses distributed along a number line, the moment of the system with respect to the origin is $M = \sum_{i = 1}^{n} m_{i} x_{i} .$ For point masses distributed in a plane, the moments of the system with respect to the - and -axes, respectively, are and respectively.

Key Equations

Mass of a lamina	$m = ρ \int_{a}^{b} f (x) d x$
Moments of a lamina	$M_{x} = ρ \int_{a}^{b} \frac{[ f ( x )] ^{2}}{2} d x and M_{y} = ρ \int_{a}^{b} x f (x) d x$