Loading [MathJax]/jax/element/mml/optable/MathOperators.js
Skip to main content
Library homepage
 

Text Color

Text Size

 

Margin Size

 

Font Type

Enable Dyslexic Font
Mathematics LibreTexts

7.6: Calculating Centers of Mass and Moments of Inertia

This page is a draft and is under active development. 

( \newcommand{\kernel}{\mathrm{null}\,}\)

We have already discussed a few applications of multiple integrals, such as finding areas, volumes, and the average value of a function over a bounded region. In this section we develop computational techniques for finding the center of mass and moments of inertia of several types of physical objects, using double integrals for a lamina (flat plate) and triple integrals for a three-dimensional object with variable density. The density is usually considered to be a constant number when the lamina or the object is homogeneous; that is, the object has uniform density.

Center of Mass in Two Dimensions

The center of mass is also known as the center of gravity if the object is in a uniform gravitational field. If the object has uniform density, the center of mass is the geometric center of the object, which is called the centroid. Figure 7.6.1 shows a point P as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

alt
Figure 7.6.1: A lamina is perfectly balanced on a spindle if the lamina’s center of mass sits on the spindle.

To find the coordinates of the center of mass P(ˉx,ˉy) of a lamina, we need to find the moment Mx of the lamina about the x-axis and the moment My about the y-axis. We also need to find the mass m of the lamina. Then

ˉx=Mym

and

ˉy=Mxm.

Refer to Moments and Centers of Mass for the definitions and the methods of single integration to find the center of mass of a one-dimensional object (for example, a thin rod). We are going to use a similar idea here except that the object is a two-dimensional lamina and we use a double integral.

If we allow a constant density function, then ˉx=Mym and ˉy=Mxm give the centroid of the lamina.

Suppose that the lamina occupies a region R in the xy-plane and let ρ(x,y) be its density (in units of mass per unit area) at any point (x,y). Hence,

ρ(x,y)=limΔA0ΔmΔA

where Δm and ΔA are the mass and area of a small rectangle containing the point (x,y) and the limit is taken as the dimensions of the rectangle go to 0 (see the following figure).

alt
Figure 7.6.2: The density of a lamina at a point is the limit of its mass per area in a small rectangle about the point as the area goes to zero.

Just as before, we divide the region R into tiny rectangles Rij with area ΔA and choose (xij,yij) as sample points. Then the mass mij of each Rij is equal to ρ(xij,yij)ΔA (Figure 7.6.2). Let k and l be the number of subintervals in x and y respectively. Also, note that the shape might not always be rectangular but the limit works anyway, as seen in previous sections.

alt
Figure 7.6.3: Subdividing the lamina into tiny rectangles Rij each containing a sample point (xij,yij).

Hence, the mass of the lamina is

m=limk,lki=1lj=1mij=limk,lki=1lj=1ρ(xij,yij)ΔA=

Let’s see an example now of finding the total mass of a triangular lamina.

Example \PageIndex{1}: Finding the Total Mass of a Lamina

Consider a triangular lamina R with vertices (0,0), \, (0,3), \, (3,0) and with density \rho(x,y) = xy \, kg/m^2. Find the total mass.

Solution

A sketch of the region R is always helpful, as shown in the following figure.

15.6.1.png
Figure \PageIndex{4}: A lamina in the xy-plane with density \rho (x,y) = xy.

Using the expression developed for mass, we see that

m = \iint_R \, dm = \iint_R \rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} xy \, dy \, dx = \int_{x=0}^{x=3} \left[ \left. x \dfrac{y^2}{2} \right|_{y=0}^{y=3} \right] \, dx = \int_{x=0}^{x=3} \dfrac{1}{2} x (3 - x)^2 dx = \left.\left[ \dfrac{9x^2}{4} - x^3 + \dfrac{x^4}{8} \right]\right|_{x=0}^{x=3} = \dfrac{27}{8}.

The computation is straightforward, giving the answer m = \dfrac{27}{8} \, kg.

Exercise \PageIndex{1}

Consider the same region R as in the previous example, and use the density function \rho (x,y) = \sqrt{xy}. Find the total mass.

Answer

\dfrac{9\pi}{8} \, kg

Now that we have established the expression for mass, we have the tools we need for calculating moments and centers of mass. The moment M_z about the x-axis for R is the limit of the sums of moments of the regions R_{ij} about the x-axis. Hence

M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R y\rho (x,y) dA

Similarly, the moment M_y about the y-axis for R is the limit of the sums of moments of the regions R_{ij} about the y-axis. Hence

M_x = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (x_{ij}^*)m_{ij} = \lim_{k,l \rightarrow \infty} \sum_{i=1}^k \sum_{j=1}^l (y_{ij}^*) \rho(x_{ij}^*,y_{ij}^*) \Delta A = \iint_R x\rho (x,y) dA

Example \PageIndex{2}: Finding Moments

Consider the same triangular lamina R with vertices (0,0), \, (0,3), \, (3,0) and with density \rho (x,y) = xy. Find the moments M_x and M_y.

Solution

Use double integrals for each moment and compute their values:

M_x = \iint_R y\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x y^2 \, dy \, dx = \dfrac{81}{20},

M_y = \iint_R x\rho (x,y) dA = \int_{x=0}^{x=3} \int_{y=0}^{y=3-x} x^2 y \, dy \, dx = \dfrac{81}{20},

The computation is quite straightforward.

Exercise \PageIndex{2}

Consider the same lamina R as above and use the density function \rho (x,y) = \sqrt{xy}. Find the moments M_x and M_y.

Answer

M_x = \dfrac{81\pi}{64} and M_y = \dfrac{81\pi}{64}

Finally we are ready to restate the expressions for the center of mass in terms of integrals. We denote the x-coordinate of the center of mass by \bar{x} and the y-coordinate by \bar{y}. Specifically,

\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA}

and

\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA}

Example \PageIndex{3}: center of mass

Again consider the same triangular region R with vertices (0,0), \, (0,3), \, (3,0) and with density function \rho (x,y) = xy. Find the center of mass.

Solution

Using the formulas we developed, we have

\bar{x} = \dfrac{M_y}{m} = \dfrac{\iint_R x\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5},

\bar{y} = \dfrac{M_x}{m} = \dfrac{\iint_R y\rho (x,y) dA}{\iint_R \rho (x,y)dA} = \dfrac{81/20}{27/8} = \dfrac{6}{5}.

Therefore, the center of mass is the point \left(\dfrac{6}{5},\dfrac{6}{5}\right).

Analysis

If we choose the density \rho(x,y) instead to be uniform throughout the region (i.e., constant), such as the value 1 (any constant will do), then we can compute the centroid,

x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1,

y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA} = \dfrac{9/2}{9/2} = 1.

Notice that the center of mass \left(\dfrac{6}{5},\dfrac{6}{5}\right) is not exactly the same as the centroid (1,1) of the triangular region. This is due to the variable density of R If the density is constant, then we just use \rho(x,y) = c (constant). This value cancels out from the formulas, so for a constant density, the center of mass coincides with the centroid of the lamina.

Exercise \PageIndex{3}

Again use the same region R as above and use the density function \rho (x,y) = \sqrt{xy}. Find the center of mass.

Answer

\bar{x} = \dfrac{M_y}{m} = \dfrac{81\pi/64}{9\pi/8} = \dfrac{9}{8} and \bar{y} = \dfrac{M_x}{m} = \dfrac{81\pi}{9\pi/8} = \dfrac{0}{8}.

Once again, based on the comments at the end of Example \PageIndex{3}, we have expressions for the centroid of a region on the plane:

x_c = \dfrac{M_y}{m} = \dfrac{\iint_R x \, dA}{\iint_R dA} \, \text{and} \, y_c = \dfrac{M_x}{m} = \dfrac{\iint_R y \, dA}{\iint_R dA}.

We should use these formulas and verify the centroid of the triangular region


This page titled 7.6: Calculating Centers of Mass and Moments of Inertia is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Pamini Thangarajah.

Support Center

How can we help?