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 1 shows a point $P$ as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

Figure 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 $P{({\overline{x}},{\overline{y}})}$ of a lamina, we need to find the moment $M_x$ of the lamina about the $x$-axis and the moment $M_y$ about the $y$-axis. We also need to find the mass $m$ of the lamina. Then

${\large{\overline{x}} = {\frac{{M}_{y}}{m}} \ {\text{and}} \ {\overline{y}} = {\frac{{M}_{x}}{m}}}.$

Refer to Centers of Mass and Moments 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 ${\overline{x}} = {\frac{{M}_{y}}{m}} \ {\text{and}} \ {\overline{y}} = {\frac{{M}_{x}}{m}}$ give the centroid of the lamina.

Suppose that the lamina occupies a region $R$ in the $xy$-plane, and let ${\rho}{({{x},{y}})}$ be its density (in units of mass per unit area) at any point $(x, y)$. Hence, ${\rho}{({{x},{y}})} = {\displaystyle\lim_{{\Delta}{A}{\rightarrow}{0}}}{\frac{{\Delta}{m}}{{\Delta}{A}}}$, where ${\Delta}{m}$ and ${\Delta}{A}$ 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).

Figure 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.

Figure 3. Subdividing the lamina into tiny rectangles $R_{ij}$, each containing a sample point $(x_{ij}^{\ast},y_{ij}^{\ast})$.

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

Example: Finding the total mass of a lamina

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

Show Solution

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

Figure 4. A lamina in the $xy$-plane with density $\rho(x,y)=xy$.

Using the expression developed for mass, we see that

$\begin{aligned} m & = {\underset{R}{\displaystyle\iint}}{d}{m} = {\underset{R}{\displaystyle\iint}}{\rho}{({x},{y})}{dA} = {\displaystyle\int^{{x} = {3}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {3} - {x}}_{{y} = {0}}}{xy} \ {dy} \ {dx} = {\displaystyle\int^{{x} = {3}}_{{x} = {0}}} \ {\left [{x}{\frac{y^2}{2}}{\bigg\vert}^{{y} = {3} - {x}}_{{y} = {0}}\right ]}{dx} \\ & = {\displaystyle\int^{{x} = {3}}_{{x} = {0}}} \ \ {\frac{1}{2}}{x}{({3} - {x})}^{2}{dx} = {\left [{\frac{9x^2}{4}} - {{x}^{3}} + {\frac{x^4}{8}} \right]}{\bigg|}^{{x} = {3}}_{{x} = {0}} \\ & = {\frac{27}{8}}. \end{aligned}$

The computation is straightforward, giving the answer ${m} = {\frac{27}{8}}$ 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 $M_x$ about the $x$-axis for $R$ is the limit of the sums of moments of the regions ${R}_{ij}$ about the $x$-axis. Hence

$\large{{{M}_{x}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{{m}_{ij}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A} = \underset{R}{\displaystyle\iint}{y}{\rho}{({x},{y})}{dA}}.$

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

$\large{{{M}_{y}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{({x}^{*}_{ij})}{{m}_{ij}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{({y}^{*}_{ij})}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A} = \underset{R}{\displaystyle\iint}{x}{\rho}{({x},{y})}{dA}}.$

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 ${\overline{x}}$ and the $y$-coordinate by ${\overline{y}}$. Specifically,

$\large{{\overline{x}} = {\frac{{M}_{y}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint}{x}{\rho}{({x},{y})}{dA}}{\underset{R}{\displaystyle\iint}{\rho}{({x},{y})}{dA}}} \ {\text{and}} \ {\overline{y}} = {\frac{{M}_{x}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint}{y}{\rho}{({x},{y})}{dA}}{\underset{R}{\displaystyle\iint}{\rho}{({x},{y})}{dA}}}}.$

Example: finding the 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.

Show Solution

Solution

Using the formulas we developed, we have

$\hspace{6cm}\large{\begin{align} \overline{x}&=\frac{M_y}m=\dfrac{\underset{R}{\displaystyle\iint}x\rho(x,y) \ dA}{\underset{R}{\displaystyle\iint}\rho(x,y) \ dA}=\frac{81/20}{27/8}=\frac65 \\ \overline{y}&=\frac{M_x}m=\dfrac{\underset{R}{\displaystyle\iint}y\rho(x,y)dA}{\underset{R}{\displaystyle\iint}\rho(x,y) \ dA}=\frac{81/20}{27/8}=\frac65 \end{align}}$

Therefore, the center of mass is the point $\left(\frac65,\frac65\right)$.

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,

$\begin{aligned} {{x}_{c}} = {\frac{{M}_{y}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint}{x} \ {dA}}{\underset{R}{\displaystyle\iint}{dA}}} = {\frac{9/2}{9/2}} = 1, \\ {{y}_{c}} = {\frac{{M}_{x}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint}{y} \ {dA}}{\underset{R}{\displaystyle\iint}{dA}}} = {\frac{9/2}{9/2}} = 1. \end{aligned}$

Notice that the center of mass ${\left ({\frac{6}{5}},{\frac{6}{5}} \right )}$ 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.

try it

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

Show Solution

$\overline{x}=\frac{M_y}m=\frac{81\pi/63}{9\pi/8}=\frac98$ and $\overline{y}=\frac{M_x}m\frac{81\pi/63}{9\pi/8}=\frac98$.

Once again, based on the comments at the end of Example “Finding the Center of Mass”, we have expressions for the centroid of a region on the plane:

$\large{{{x}_{c}} = {\frac{{M}_{y}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint}{x} \ {dA}}{\underset{R}{\displaystyle\iint}{dA}}} \ {\text{and}} \ {{y}_{c}} = {\frac{{M}_{x}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint}{y} \ {dA}}{\underset{R}{\displaystyle\iint}{dA}}}}.$

We should use these formulas and verify the centroid of the triangular region $R$ referred to in the last three examples.

Example: Finding Mass, Moments, and Center of Mass

Find the mass, moments, and the center of mass of the lamina of density ${\rho}{({x},{y})} = {x} + {y}$ occupying the region $R$ under the curve $y=x^{2}$ in the interval ${0} \leq {x} \leq {2}$ (see the following figure).

Figure 5. Locating the center of mass of a lamina $R$ with density $\rho(x,y)=x+y$.

Show Solution

First we compute the mass $m$. We need to describe the region between the graph of $y=x^{2}$ and the vertical lines $x=0$ and $x=2$:

$\begin{aligned} {m} & = {\underset{R}{\displaystyle\iint}}{dm} = {\underset{R}{\displaystyle\iint}}{\rho}{({x},{y})}{dA} = {\displaystyle\int^{{x} = {2}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{({x} + {y})}{dy} \ {dx} = {\displaystyle\int^{{x} = {2}}_{{x} = {0}}}{\left [{xy} + {\frac{{y}^{2}}{2}}{\bigg\vert}^{{y} = {{x}^{2}}}_{{y} = {0}} \right ]}{dx} \\ & = {\displaystyle\int^{{x} = {2}}_{{x} = {0}}} \ {\left [{{x}^{3}} + {\frac{{x}^{4}}{2}} \right ]}{dx} = {\left [ {\frac{{x}^{4}}{4}} + {\frac{{x}^{5}}{10}} \right ]}{\bigg\vert}^{{x} = {2}}_{{x} = {0}} = {\frac{36}{5}}. \end{aligned}$

Now compute the moments $M_x$ and $M_y$:

$\begin{aligned} {M}_{x} = & {\underset{R}{\displaystyle\iint}}{y}{\rho}{({x},{y})}{d}{A} = {\displaystyle\int^{{x} = {2}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{y}{({x} + {y})}{dy} \ {dx} = {\frac{80}{7}}, \\ {M}_{y} = & {\underset{R}{\displaystyle\iint}}{x}{\rho}{({x},{y})}{d}{A} = {\displaystyle\int^{{x} = {2}}_{{x} = {0}}} \ {\displaystyle\int^{{y} = {{x}^{2}}}_{{y} = {0}}}{x}{({x} + {y})}{dy} \ {dx} = {\frac{176}{15}}. \end{aligned}$

Finally, evaluate the center of mass,

$\begin{aligned} {\overline{x}} = & {\frac{{M}_{y}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint} \ {x}{\rho}{({x},{y})}{d}{A}}{\underset{R}{\displaystyle\iint} \ {\rho}{({x},{y})}{d}{A}}} = {\frac{176/15}{36/5}} = {\frac{44}{27}}, \\ {\overline{y}} = & {\frac{{M}_{x}}{m}} = {\dfrac{\underset{R}{\displaystyle\iint} \ {y}{\rho}{({x},{y})}{d}{A}}{\underset{R}{\displaystyle\iint} \ {\rho}{({x},{y})}{d}{A}}} = {\frac{80/7}{36/5}} = {\frac{100}{63}}. \end{aligned}$

Hence the center of mass is ${({\overline{x}},{\overline{y}})} = {\left ({\frac{44}{27}},{\frac{100}{63}} \right )}.$

try it

Calculate the mass, moments, and the center of mass of the region between the curves ${y} = {x}$ and ${y} = {{x}^{2}}$ with the density function ${\rho}{({x},{y})} = {x}$ in the interval ${0} \leq {x} \leq {1}$.

Show Solution

$\overline{x}=\frac{M_y}m=\frac{1/20}{1/12}=\frac35$ and $\overline{y}=\frac{M_x}m\frac{1/24}{1/12}=\frac12$.

Watch the following video to see the worked solution to the above Try It

You can view the transcript for “CP 5.36” here (opens in new window).

Example: finding a centroid

Find the centroid of the region under the curve ${y} = {{e}^{x}}$ over the interval ${1} \leq {x} \leq {3}$ (see the following figure).

Figure 6. Finding a centroid of a region below the curve $y=e^{x}$

Show Solution

To compute the centroid, we assume that the density function is constant and hence it cancels out:

$\hspace{4cm}\begin{align} x_c&=\frac{M_y}m=\dfrac{\underset{R}{\displaystyle\iint}x \ dA}{\underset{R}{\displaystyle\iint}dA}\text{ and }y_c=\frac{M_x}m=\dfrac{\underset{R}{\displaystyle\iint}y \ dA}{\underset{R}{\displaystyle\iint}dA}, \\ x_c&=\frac{M_y}m=\dfrac{\underset{R}{\displaystyle\iint}x \ dA}{\underset{R}{\displaystyle\iint}dA}=\dfrac{\displaystyle\int^{x=3}_{x=1}\displaystyle\int^{y=e^x}_{y=0}x \ dy \ dx}{\displaystyle\int^{x=3}_{x=1}\displaystyle\int^{y=e^x}_{y=0}dy \ dx}=\dfrac{\displaystyle\int^{x=3}_{x=1}xe^x{dx}}{\displaystyle\int^{x=3}_{x=1}e^x{dx}}=\frac{2e^3}{e^3-e}=\frac{2e^2}{e^2-1}, \\ y_c&=\frac{M_x}m=\dfrac{\underset{R}{\displaystyle\iint}y \ dA}{\underset{R}{\displaystyle\iint}dA}=\dfrac{\displaystyle\int_{x=1}^{x=3}\displaystyle\int_{y=0}^{y=e^x}y \ dy \ dx}{\displaystyle\int_{x=1}^{x=3}\displaystyle\int_{y=0}^{y=e^x}dy \ dx}=\dfrac{\displaystyle\int_{x=1}^{x=3}\frac{e^{2x}}2dx}{\displaystyle\int_{x=1}^{x=3}e^xdx}=\frac{\frac14e^2(e^4-1)}{e(e^2-1)}=\frac14e(e^2+1). \end{align}$

Thus the centroid of the region is

$\large{{({{x}_{c}},{{y}_{c}})} = {\left ( {\frac{{2}{e}^{2}}{{e}^{2} - {1}}, {\frac{1}{4}}{e} \ {\left ( {e}^{2} + {1} \right )}} \right )}}.$

try it

Calculate the centroid of the region between the curves $y=x$ and ${y} = {\sqrt{x}}$ with uniform density in the interval ${0} \leq {x} \leq {1}$.

Show Solution

$\overline{x}=\frac{M_y}m=\frac{1/15}{1/6}=\frac25$ and $\overline{y}=\frac{M_x}m\frac{1/12}{1/6}=\frac12$.

Moments of Inertia

For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition of moments and centers of mass in Section 6.6 of Volume 1. The moment of inertia of a particle of mass $m$ about an axis is $mr^{2}$ where $r$ is the distance of the particle from the axis. We can see from Figure 3 that the moment of inertia of the subrectangle ${R}_{ij}$ about the $x$-axis is ${({y}^{*}_{ij})^{2}}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A}$. Similarly, the moment of inertia of the subrectangle ${R}_{ij}$ about the $y$-axis is ${({x}^{*}_{ij})^{2}}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A}$. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis.

The moment of inertia $I_x$ about the $x$-axis for the region $R$ is the limit of the sum of moments of inertia of the regions ${R}_{ij}$ about the $x$-axis. Hence

$\large{{{I}_{x}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {y}^{*}_{ij} \right )}^{2}{{m}_{ij}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {y}^{*}_{ij} \right )}^{2}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A} = {\underset{R}{\displaystyle\iint}}{{y}^{2}}{\rho}{({x},{y})}{dA}}.$

Similarly, the moment of inertia $I_y$ about the $y$-axis for $R$ is the limit of the sum of moments of inertia of the regions ${R}_{ij}$ about the $y$-axis. Hence

$\large{{{I}_{y}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {x}^{*}_{ij} \right )}^{2}{{m}_{ij}} = {\displaystyle\lim_{{k},{l}{\rightarrow}{\infty}}}{\displaystyle\sum^{k}_{{i} = {1}}}{\displaystyle\sum^{l}_{{j} = {1}}}{\left ( {x}^{*}_{ij} \right )}^{2}{\rho}{({{x}^{*}_{ij}},{{y}^{*}_{ij}})}{\Delta}{A} = {\underset{R}{\displaystyle\iint}}{{x}^{2}}{\rho}{({x},{y})}{dA}}.$

Sometimes, we need to find the moment of inertia of an object about the origin, which is known as the polar moment of inertia. We denote this by $I_0$ and obtain it by adding the moments of inertia $I_x$ and $I_y$. Hence

$\large{{{I}_{0}} = {{I}_{x}} + {{I}_{y}} = {\underset{R}{\displaystyle\iint}}{({x^{2}} + {y^{2}})}{\rho}{({x},{y})}{dA}}.$

All these expressions can be written in polar coordinates by substituting ${x} = {r} \ {\cos} \ {\theta}, {y} = {r} \ {\sin} \ {\theta},$ and ${dA} = {r} \ {dr} \ {d}{\theta}$. For example, ${{I}_{0}} = {\underset{R}{\displaystyle\iint}}{{r}^{2}}{\rho}{({{r}{\cos}{\theta}},{{r}{\sin}{\theta}})}{dA}.$

Watch the following video to see the worked solution to the above Try It

As mentioned earlier, the moment of inertia of a particle of mass $m$ about an axis is $mr^{2}$ where $r$ is the distance of the particle from the axis, also known as the radius of gyration.

Hence the radii of gyration with respect to the $x$-axis, the $y$-axis, and the origin are

$\large{{{R}_{x}} = {\sqrt{\frac{{I}_{x}}{m}}}, {{R}_{y}} = {\sqrt{\frac{{I}_{y}}{m}}}, \ {\text{and}} \ {{R}_{0}} = {\sqrt{\frac{{I}_{0}}{m}}}},$

respectively. In each case, the radius of gyration tells us how far (perpendicular distance) from the axis of rotation the entire mass of an object might be concentrated. The moments of an object are useful for finding information on the balance and torque of the object about an axis, but radii of gyration are used to describe the distribution of mass around its centroidal axis. There are many applications in engineering and physics. Sometimes it is necessary to find the radius of gyration, as in the next example.