Calculating Centers of Mass and Moments of Inertia

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. [link] shows a point $P$

as the center of mass of a lamina. The lamina is perfectly balanced about its center of mass.

To find the coordinates of the center of mass $P(\stackrel{\text{−}}{x},\stackrel{\text{−}}{y})$

of a lamina, we need to find the moment $M_x$

of the lamina about the $x\text{-axis}$

and the moment $M_y$

about the $y\text{-axis}\text{.}$

We also need to find the mass $m$

of the lamina. Then

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 $\stackrel{\text{−}}{x}=\frac{M_y}{m}\text{and}\stackrel{\text{−}}{y}=\frac{M_x}{m}$

give the centroid of the lamina.

Suppose that the lamina occupies a region $R$

in the $xy\text{-plane},$

and let $\rho \left(x,y\right)$

be its density (in units of mass per unit area) at any point $\left(x,y\right).$

Hence, $\rho \left(x,y\right)=\underset{\text{Δ}A\to 0}{\text{lim}}\frac{\text{Δ}m}{\text{Δ}A},$

where $\text{Δ}m$

and $\text{Δ}A$

are the mass and area of a small rectangle containing the point $\left(x,y\right)$

and the limit is taken as the dimensions of the rectangle go to $0$

(see the following figure).

Just as before, we divide the region $R$

into tiny rectangles $R_{ij}$

with area $\text{Δ}A$

and choose $\left(x_{ij}^{*},y_{ij}^{*}\right)$

as sample points. Then the mass $m_{ij}$

of each $R_{ij}$

is equal to $\rho \left(x_{ij}^{*},y_{ij}^{*}\right)\text{Δ}A$

([link]). 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.

Hence, the mass of the lamina is

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

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 $x\text{-axis}$

for $R$

is the limit of the sums of moments of the regions $R_{ij}$

about the $x\text{-axis}\text{.}$

Hence

Similarly, the moment $M_y$

about the $y\text{-axis}$

for $R$

is the limit of the sums of moments of the regions $R_{ij}$

about the $y\text{-axis}\text{.}$

Hence

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 $\stackrel{\text{−}}{x}$

and the y-coordinate by $\stackrel{\text{−}}{y}.$

Specifically,

Once again, based on the comments at the end of [link], we have expressions for the centroid of a region on the plane:

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

referred to in the last three examples.

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 in Section $6.6.$

The moment of inertia of a particle of mass $m$

about an axis is $m{r}^2,$

where $r$

is the distance of the particle from the axis. We can see from [link] that the moment of inertia of the subrectangle $R_{ij}$

about the $x\text{-axis}$

is ${(y_{ij}^{*})}^2\rho (x_{ij}^{*},y_{ij}^{*})\text{Δ}A.$

Similarly, the moment of inertia of the subrectangle $R_{ij}$

about the $y\text{-axis}$

is ${(x_{ij}^{*})}^2\rho (x_{ij}^{*},y_{ij}^{*})\text{Δ}A.$

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 $x\text{-axis}$

for the region $R$

is the limit of the sum of moments of inertia of the regions $R_{ij}$

about the $x\text{-axis}\text{.}$

Hence

Similarly, the moment of inertia $I_y$

about the $y\text{-axis}$

for $R$

is the limit of the sum of moments of inertia of the regions $R_{ij}$

about the $y\text{-axis}\text{.}$

Hence

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

All these expressions can be written in polar coordinates by substituting $x=r\text{cos}\theta ,$

and $dA=rdrd\theta .$

For example, $I_0={\displaystyle \underset{R}{\iint }{r}^2\rho \left(r\text{cos}\theta ,r\text{sin}\theta \right)}dA.$

As mentioned earlier, the moment of inertia of a particle of mass $m$

about an axis is $m{r}^2$

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 $x\text{-axis,}$

the $y\text{-axis,}$

and the origin are

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.

Center of Mass and Moments of Inertia in Three Dimensions

All the expressions of double integrals discussed so far can be modified to become triple integrals.

We conclude this section with an example of finding moments of inertia $I_x,I_y,$

and $I_z.$

Key Concepts

Finding the mass, center of mass, moments, and moments of inertia in double integrals:

• For a lamina $R$

  with a density function

  $\rho \left(x,y\right)$

  at any point

  $\left(x,y\right)$

  in the plane, the mass is

  $m={\displaystyle \underset{R}{\iint }\rho \left(x,y\right)dA}.$
• The moments about the $x\text{-axis}$

  and

  $y\text{-axis}$

  are

  ---

  $M_x={\displaystyle \underset{R}{\iint }y\rho \left(x,y\right)}dA\text{and}{M}_y={\displaystyle \underset{R}{\iint }x\rho \left(x,y\right)}dA.$
• The center of mass is given by $\stackrel{\text{−}}{x}=\frac{M_y}{m},\stackrel{\text{−}}{y}=\frac{M_x}{m}.$
• The center of mass becomes the centroid of the plane when the density is constant.
• The moments of inertia about the $x-\text{axis,}$ $y-\text{axis,}$

  and the origin are

  ---

  $I_x={\displaystyle \underset{R}{\iint }{y}^2\rho \left(x,y\right)dA,}{I}_y={\displaystyle \underset{R}{\iint }{x}^2\rho \left(x,y\right)dA,}\text{and}{I}_0=I_x+I_y={\displaystyle \underset{R}{\iint }\left(x^2+y^2\right)\rho \left(x,y\right)dA.}$

Finding the mass, center of mass, moments, and moments of inertia in triple integrals:

• For a solid object $Q$

  with a density function

  $\rho \left(x,y,z\right)$

  at any point

  $\left(x,y,z\right)$

  in space, the mass is

  $m={\displaystyle \underset{Q}{\iiint }\rho \left(x,y,z\right)dV}.$
• The moments about the $xy\text{-plane,}$

  the

  $xz\text{-plane,}$

  and the

  $yz\text{-plane}$

  are

  ---

  $M_{xy}={\displaystyle \underset{Q}{\iiint }z\rho \left(x,y,z\right)dV},M_{xz}={\displaystyle \underset{Q}{\iiint }y\rho \left(x,y,z\right)dV},M_{yz}={\displaystyle \underset{Q}{\iiint }x\rho \left(x,y,z\right)dV}.$
• The center of mass is given by $\stackrel{\text{−}}{x}=\frac{M_{yz}}{m},\stackrel{\text{−}}{y}=\frac{M_{xz}}{m},\stackrel{\text{−}}{z}=\frac{M_{xy}}{m}.$
• The center of mass becomes the centroid of the solid when the density is constant.
• The moments of inertia about the $yz\text{-plane,}$

  the

  $xz\text{-plane,}$

  and the

  $xy\text{-plane}$

  are

  ---

  $\begin{array}{}\\ I_x={\displaystyle \underset{Q}{\iiint }\left(y^2+z^2\right)\rho \left(x,y,z\right)dV,}{I}_y={\displaystyle \underset{Q}{\iiint }\left(x^2+z^2\right)\rho \left(x,y,z\right)dV,}\hfill \\ I_z={\displaystyle \underset{Q}{\iiint }\left(x^2+y^2\right)\rho \left(x,y,z\right)dV.}\hfill \end{array}$

Key Equations

• Mass of a lamina

  ---

  $m=\underset{k,l\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^l{m}_{ij}}}=\underset{k,l\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^l\rho (x_{ij}^{*},y_{ij}^{*})\text{Δ}A={\displaystyle \underset{R}{\iint }\rho (x,y)dA}}}$
• Moment about the x-axis

  ---

  $M_x=\underset{k,l\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^l\left(y_{ij}^{*}\right)m_{ij}}}=\underset{k,l\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^l\left(y_{ij}^{*}\right)\rho (x_{ij}^{*},y_{ij}^{*})\text{Δ}A={\displaystyle \underset{R}{\iint }y\rho (x,y)dA}}}$
• Moment about the y-axis

  ---

  $M_y=\underset{k,l\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^l\left(x_{ij}^{*}\right)m_{ij}}}=\underset{k,l\to \infty }{\text{lim}}{\displaystyle \sum _{i=1}^k{\displaystyle \sum _{j=1}^l\left(x_{ij}^{*}\right)\rho (x_{ij}^{*},y_{ij}^{*})\text{Δ}A={\displaystyle \underset{R}{\iint }x\rho (x,y)dA}}}$
• Center of mass of a lamina

  ---

  $\stackrel{\text{−}}{x}=\frac{M_y}{m}=\frac{{\displaystyle \underset{R}{\iint }x\rho \left(x,y\right)dA}}{{\displaystyle \underset{R}{\iint }\rho \left(x,y\right)dA}}$

  and

  $\stackrel{\text{−}}{y}=\frac{M_x}{m}=\frac{{\displaystyle \underset{R}{\iint }y\rho \left(x,y\right)dA}}{{\displaystyle \underset{R}{\iint }\rho \left(x,y\right)dA}}$