Ulearngo

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.

Theorem 6.11

The Symmetry Principle

If a region R is symmetric about a line l, then the centroid of R lies on l.

Let’s turn to more general laminas. Suppose we have a lamina bounded above by the graph of a continuous function \(f \left(\right. x \left.\right) ,\) 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 6.65 A 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 \(\left[\right. a , b \left]\right.\) and constructing rectangles.

For \(i = 0 , 1 , 2 ,\ldots , n ,\) let \(P = \left{\right. x_{i} \left.\right}\) be a regular partition of \(\left[\right. a , b \left]\right. .\) Recall that we can choose any point within the interval \(\left[\right. x_{i - 1} , x_{i} \left]\right.\) as our \(x_{i}^{\star} .\) In this case, we want \(x_{i}^{\star}\) to be the x-coordinate of the centroid of our rectangles. Thus, for \(i = 1 , 2 ,\ldots , n ,\) we select \(x_{i}^{\star} \in \left[\right. x_{i - 1} , x_{i} \left]\right.\) such that \(x_{i}^{\star}\) is the midpoint of the interval. That is, \(x_{i}^{\star} = \left(\right. x_{i - 1} + x_{i} \left.\right) / 2 .\) Now, for \(i = 1 , 2 ,\ldots , n ,\) construct a rectangle of height \(f \left(\right. x_{i}^{\star} \left.\right)\) on \(\left[\right. x_{i - 1} , x_{i} \left]\right. .\) The center of mass of this rectangle is \(\left(\right. x_{i}^{\star} , \left(\right. f \left(\right. x_{i}^{\star} \left.\right) \left.\right) / 2 \left.\right) ,\) 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 6.66 A representative rectangle of the lamina.

Next, we need to find the total mass of the rectangle. Let \(\rho\) represent the density of the lamina (note that \(\rho\) is a constant). In this case, \(\rho\) 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 \(\rho .\) Then, the mass of the rectangle is given by \(\rho f \left(\right. x_{i}^{\star} \left.\right) \Delta x .\)

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

\[m \approx \sum_{i = 1}^{n} \rho f \left(\right. x_{i}^{\star} \left.\right) \Delta x .\]

This is a Riemann sum. Taking the limit as \(n \rightarrow \infty\) gives the exact mass of the lamina:

\[m = \underset{n \rightarrow \infty}{\text{lim}} \sum_{i = 1}^{n} \rho f \left(\right. x_{i}^{\star} \left.\right) \Delta x = \rho \int_{a}^{b} f \left(\right. x \left.\right) d x .\]

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 \(\left(\right. x_{i}^{\star} , \left(\right. f \left(\right. x_{i}^{\star} \left.\right) \left.\right) / 2 \left.\right) .\) 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, \(\rho f \left(\right. x_{i}^{\star} \left.\right) \Delta x ,\) multiplied by the distance from the center of mass to the x-axis: \(\left(\right. f \left(\right. x_{i}^{\star} \left.\right) \left.\right) / 2 .\) Therefore, the moment with respect to the x-axis of the rectangle is \(\rho \left(\right. \left[\right. f \left(\right. x_{i}^{\star} \left.\right) \left]\right.^{2} / 2 \left.\right) \Delta x .\) 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 x-axis is

\[M_{x} = \underset{n \rightarrow \infty}{\text{lim}} \sum_{i = 1}^{n} \rho \frac{\left[\right. f \left(\right. x_{i}^{\star} \left.\right) \left]\right.^{2}}{2} \Delta x = \rho \int_{a}^{b} \frac{\left[\right. f \left(\right. x \left.\right) \left]\right.^{2}}{2} d x .\]

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 \(x_{i}^{\star} .\) Then the moment of the lamina with respect to the y-axis is given by

\[M_{y} = \underset{n \rightarrow \infty}{\text{lim}} \sum_{i = 1}^{n} \rho x_{i}^{\star} f \left(\right. x_{i}^{\star} \left.\right) \Delta x = \rho \int_{a}^{b} x f \left(\right. x \left.\right) d x .\]

We find the coordinates of the center of mass by dividing the moments by the total mass to give \(\overset{–}{x} = M_{y} / m \text{and} \overset{–}{y} = M_{x} / m .\) If we look closely at the expressions for \(M_{x} , M_{y} , \text{and} m ,\) we notice that the constant \(\rho\) cancels out when \(\overset{–}{x}\) and \(\overset{–}{y}\) are calculated.

We summarize these findings in the following theorem.

Theorem 6.12

Center of Mass of a Thin Plate in the xy-Plane

Let R denote a region bounded above by the graph of a continuous function \(f \left(\right. x \left.\right) ,\) below by the x-axis, and on the left and right by the lines \(x = a\) and \(x = b ,\) respectively. Let \(\rho\) denote the density of the associated lamina. Then we can make the following statements:

  1. The mass of the lamina is
    \[m = \rho \int_{a}^{b} f \left(\right. x \left.\right) d x .\]
  2. The moments \(M_{x}\) and \(M_{y}\) of the lamina with respect to the x- and y-axes, respectively, are
    \[M_{x} = \rho \int_{a}^{b} \frac{\left[\right. f \left(\right. x \left.\right) \left]\right.^{2}}{2} d x \text{and} M_{y} = \rho \int_{a}^{b} x f \left(\right. x \left.\right) d x .\]
  3. The coordinates of the center of mass \(\left(\right. \overset{–}{x} , \overset{–}{y} \left.\right)\) are
    \[\overset{–}{x} = \frac{M_{y}}{m} \text{and} \overset{–}{y} = \frac{M_{x}}{m} .\]

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

Example 6.31

Finding the Center of Mass of a Lamina

Let R be the region bounded above by the graph of the function \(f \left(\right. x \left.\right) = \sqrt{x}\) and below by the x-axis over the interval \(\left[\right. 0 , 4 \left]\right. .\) Find the centroid of the region.

Solution

The region is depicted in the following figure.

This figure is the graph of the curve f(x)=squareroot(x). It is an increasing curve in the first quadrant. Under the curve above the x-axis there is a shaded region. It starts at x=0 and is bounded to the right at x=4.
Figure 6.67 Finding the center of mass of a lamina.

Since we are only asked for the centroid of the region, rather than the mass or moments of the associated lamina, we know the density constant \(\rho\) cancels out of the calculations eventually. Therefore, for the sake of convenience, let’s assume \(\rho = 1 .\)

First, we need to calculate the total mass:

\[\begin{aligned} m & = \rho \int_{a}^{b} f \left(\right. x \left.\right) d x = \int_{0}^{4} \sqrt{x} d x \\ & = \left(\frac{2}{3} x^{3 / 2} \left|\right.\right)_{0}^{4} = \frac{2}{3} \left[\right. 8 - 0 \left]\right. = \frac{16}{3} . \end{aligned}\]

Next, we compute the moments:

\[\begin{aligned} M_{x} & = \rho \int_{a}^{b} \frac{\left[\right. f \left(\right. x \left.\right) \left]\right.^{2}}{2} d x \\ & = \int_{0}^{4} \frac{x}{2} d x = \left(\frac{1}{4} x^{2} \left|\right.\right)_{0}^{4} = 4 \end{aligned}\]

and

\[\begin{aligned} M_{y} & = \rho \int_{a}^{b} x f \left(\right. x \left.\right) d x \\ & = \int_{0}^{4} x \sqrt{x} d x = \int_{0}^{4} x^{3 / 2} d x \\ & = \left(\frac{2}{5} x^{5 / 2} \left|\right.\right)_{0}^{4} = \frac{2}{5} \left[\right. 32 - 0 \left]\right. = \frac{64}{5} . \end{aligned}\]

Thus, we have

\[\overset{–}{x} = \frac{M_{y}}{m} = \frac{64 / 5}{16 / 3} = \frac{64}{5} \cdot \frac{3}{16} = \frac{12}{5} \text{and} \overset{–}{y} = \frac{M_{x}}{m} = \frac{4}{16 / 3} = 4 \cdot \frac{3}{16} = \frac{3}{4} .\]

The centroid of the region is \(\left(\right. 12 / 5 , 3 / 4 \left.\right) .\)

Checkpoint 6.31

Let R be the region bounded above by the graph of the function \(f \left(\right. x \left.\right) = x^{2}\) and below by the x-axis over the interval \(\left[\right. 0 , 2 \left]\right. .\) Find the centroid of the region.

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 \(f \left(\right. x \left.\right) ,\) 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, \(g \left(\right. x \left.\right) ,\) 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 6.68 A region between two functions.

Again, we partition the interval \(\left[\right. a , b \left]\right.\) 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 6.69 A representative rectangle of the region between two functions.

Note that the centroid of this rectangle is \(\left(\right. x_{i}^{\star} , \left(\right. f \left(\right. x_{i}^{\star} \left.\right) + g \left(\right. x_{i}^{\star} \left.\right) \left.\right) / 2 \left.\right) .\) 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 y-axis, the height of each rectangle is given by \(f \left(\right. x_{i}^{\star} \left.\right) - g \left(\right. x_{i}^{\star} \left.\right) ,\) which leads to the expression \(f \left(\right. x \left.\right) - g \left(\right. x \left.\right)\) 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, \(\rho \left[\right. f \left(\right. x_{i}^{\star} \left.\right) - g \left(\right. x_{i}^{\star} \left.\right) \left]\right. \Delta x ,\) by the distance of the centroid from the x-axis, \(\left(\right. f \left(\right. x_{i}^{\star} \left.\right) + g \left(\right. x_{i}^{\star} \left.\right) \left.\right) / 2 ,\) which gives \(\rho \left(\right. 1 / 2 \left.\right) \left{\right. \left[\right. f \left(\right. x_{i}^{\star} \left.\right) \left]\right.^{2} - \left[\right. g \left(\right. x_{i}^{\star} \left.\right) \left]\right.^{2} \left.\right} \Delta x .\) Summarizing these findings, we arrive at the following theorem.

Theorem 6.13

Center of Mass of a Lamina Bounded by Two Functions

Let R denote a region bounded above by the graph of a continuous function \(f \left(\right. x \left.\right) ,\) below by the graph of the continuous function \(g \left(\right. x \left.\right) ,\) and on the left and right by the lines \(x = a\) and \(x = b ,\) respectively. Let \(\rho\) denote the density of the associated lamina. Then we can make the following statements:

  1. The mass of the lamina is
    \[m = \rho \int_{a}^{b} \left[\right. f \left(\right. x \left.\right) - g \left(\right. x \left.\right) \left]\right. d x .\]
  2. The moments \(M_{x}\) and \(M_{y}\) of the lamina with respect to the x- and y-axes, respectively, are
    \[M_{x} = \rho \int_{a}^{b} \frac{1}{2} \left(\right. \left[\right. f \left(\right. x \left.\right) \left]\right.^{2} - \left[\right. g \left(\right. x \left.\right) \left]\right.^{2} \left.\right) d x \text{and} M_{y} = \rho \int_{a}^{b} x \left[\right. f \left(\right. x \left.\right) - g \left(\right. x \left.\right) \left]\right. d x .\]
  3. The coordinates of the center of mass \(\left(\right. \overset{–}{x} , \overset{–}{y} \left.\right)\) are
    \[\overset{–}{x} = \frac{M_{y}}{m} \text{and} \overset{–}{y} = \frac{M_{x}}{m} .\]

We illustrate this theorem in the following example.

Example 6.32

Finding the Centroid of a Region Bounded by Two Functions

Let R be the region bounded above by the graph of the function \(f \left(\right. x \left.\right) = 1 - x^{2}\) and below by the graph of the function \(g \left(\right. x \left.\right) = x - 1 .\) Find the centroid of the region.

Solution

The region is depicted in the following figure.

This figure is a graph. It has two curves. They are labeled f(x)=1-x^2 and g(x)=x-1. In between the curves is a shaded region. The shaded region is bounded to the left by x=a and to the right by x=b.
Figure 6.70 Finding the centroid of a region between two curves.

The graphs of the functions intersect at \(\left(\right. −2 , −3 \left.\right)\) and \(\left(\right. 1 , 0 \left.\right) ,\) so we integrate from −2 to 1. Once again, for the sake of convenience, assume \(\rho = 1 .\)

First, we need to calculate the total mass:

\[\begin{aligned} m & = \rho \int_{a}^{b} \left[\right. f \left(\right. x \left.\right) - g \left(\right. x \left.\right) \left]\right. d x \\ & = \int_{−2}^{1} \left[\right. 1 - x^{2} - \left(\right. x - 1 \left.\right) \left]\right. d x = \int_{−2}^{1} \left(\right. 2 - x^{2} - x \left.\right) d x \\ & = \left(\left[\right. 2 x - \frac{1}{3} x^{3} - \frac{1}{2} x^{2} \left]\right. \left|\right.\right)_{−2}^{1} = \left[\right. 2 - \frac{1}{3} - \frac{1}{2} \left]\right. - \left[\right. −4 + \frac{8}{3} - 2 \left]\right. = \frac{9}{2} . \end{aligned}\]

Next, we compute the moments:

\[\begin{aligned} M_{x} & = \rho \int_{a}^{b} \frac{1}{2} \left(\right. \left[\right. f \left(\right. x \left.\right) \left]\right.^{2} - \left[\right. g \left(\right. x \left.\right) \left]\right.^{2} \left.\right) d x \\ & = \frac{1}{2} \int_{−2}^{1} \left(\right. \left(\right. 1 - x^{2} \left.\right)^{2} - \left(\right. x - 1 \left.\right)^{2} \left.\right) d x = \frac{1}{2} \int_{−2}^{1} \left(\right. x^{4} - 3 x^{2} + 2 x \left.\right) d x \\ & = \frac{1}{2} \left(\left[\right. \frac{x^{5}}{5} - x^{3} + x^{2} \left]\right. \left|\right.\right)_{−2}^{1} = - \frac{27}{10} \end{aligned}\]

and

\[\begin{aligned} M_{y} & = \rho \int_{a}^{b} x \left[\right. f \left(\right. x \left.\right) - g \left(\right. x \left.\right) \left]\right. d x \\ & = \int_{−2}^{1} x \left[\right. \left(\right. 1 - x^{2} \left.\right) - \left(\right. x - 1 \left.\right) \left]\right. d x = \int_{−2}^{1} x \left[\right. 2 - x^{2} - x \left]\right. d x \\ = \int_{−2}^{1} \left(\right. 2 x - x^{3} - x^{2} \left.\right) d x & \\ & = \left(\left[\right. x^{2} - \frac{x^{4}}{4} - \frac{x^{3}}{3} \left]\right. \left|\right.\right)_{−2}^{1} = - \frac{9}{4} . \end{aligned}\]

Therefore, we have

\[\overset{–}{x} = \frac{M_{y}}{m} = - \frac{9}{4} \cdot \frac{2}{9} = - \frac{1}{2} \text{and} \overset{–}{y} = \frac{M_{x}}{m} = - \frac{27}{10} \cdot \frac{2}{9} = - \frac{3}{5} .\]

The centroid of the region is \(\left(\right. − \left(\right. 1 / 2 \left.\right) , − \left(\right. 3 / 5 \left.\right) \left.\right) .\)

Checkpoint 6.32

Let R be the region bounded above by the graph of the function \(f \left(\right. x \left.\right) = 6 - x^{2}\) and below by the graph of the function \(g \left(\right. x \left.\right) = 3 - 2 x .\) Find the centroid of the region.

This lesson is part of:

Applications of Integration

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