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3.7: Moments and Centers of Mass

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This section shows how to calculate the masses and moments of two- and three- dimensional objects in Cartesian (x,y,z) coordinates.

Mass

We saw before that the double integral over a region of the constant function 1 measures the area of the region. If the region has uniform density 1, then the mass is the density times the area which equals the area. What if the density is not constant. Suppose that the density is given by the continuous function

Density=ρ(x,y).

In this case we can cut the region into tiny rectangles where the density is approximately constant. The area of mass rectangle is given by

Mass=(Density)(Area)=(ρ(x,y))(ΔxΔy)

You probably know where this is going. If we add all to masses together and take the limit as the rectangle size goes to zero, we get a double integral.

Definition: Mass of a Two-Dimensional lamina

Let ρ(x,y) be the density of a lamina (flat sheet) R at the point (x,y). Then the total mass of the lamina is the double integral

Masslamina=ρ(x,y)dydx

or written as an integral over an area (A):

Masslamina=baρdA

Example 3.7.1

A rectangular metal sheet with 2<x<5 and 0<y<3 has density function

ρ(x,y)=x+y.

Set up the double integral that gives the mass of the metal sheet.

Solution

We just have to evaluate the integral in Equation ???

5230(x+y)dydx.

Extending this to three-dimensional solids requires redefining ρ(x,y,z) to be the density (mass per unit volume) of an object occupying a region D in space. The integral over D gives us the mass of the object. To see why, imagine partitioning the object into n mass elements. And when summing these mass elements up, it is the total mass.

M=lim

The integral of \rho (x,y,z) gives us the mass of the object.

Definition: Mass of a Three-Dimensional Solid

Let \rho(x,y,z) be the density of a solid R at the point (x,y,s). Then the total mass of the solid is the triple integral

\text{Mass}_{\text{solid}} = \iiint \rho (x,y,z)\, dy\,dx, dz \label{solid}

or written as an integral over an volume (V):

\text{Mass}_{\text{solid}}=\iiint_{a}^{b}\rho\, dV\nonumber

Moments and Center of Mass

The moments about an axis are defined by the product of the mass times the distance from the axis.

M_x=(\text{Mass}(y))\nonumber

M_y=(\text{Mass}(x)) \nonumber

If we have a region R with density function \rho (x,y), then we do the usual thing. We cut the region into small rectangles for which the density is constant and add up the moments of each of these rectangles. Then take the limit as the rectangle size approaches zero. This will give us the total moment.

Definition: Moments of Mass and Center of Mass

Suppose that \rho (x,y) is a continuous density function on a lamina R. Then the moments of mass are

M_x = \int_0^1\int_0^2 k(x^2+y^2) y\, dy\, dx\nonumber

and

M_y = \int_0^1\int_0^2 k(x^2+y^2) x\, dy \,dx\nonumber

and if M is the mass of the lamina, then the center of mass is

(\bar{x},\bar{y})=\left ( \dfrac{M_y}{M},\dfrac{M_x}{M}\right ).\nonumber

Example \PageIndex{2}

Set up the integrals that give the center of mass of the rectangle with vertices (0,0), (1,0), (1,1), and (0,1) and density function proportional to the square of the distance from the origin. Use a calculator or computer to evaluate these integrals.

Solution

The mass is given by

M = \int_0^1\int_0^2 k(x^2+y^2)\, dy\, dx = \dfrac{2k}{3}.\nonumber \nonumber

The moments are given by (definition 2a):

M_x = \int_0^1\int_0^2 k(x^2+y^2) y\, dy\, dx\nonumber \nonumber

and

M_y = \int_0^1\int_0^2 k(x^2+y^2) x\, dy\, dx.\nonumber \nonumber

These evaluate to

M_x = \dfrac{5k}{12}\nonumber \nonumber

and

M_y = \dfrac{5k}{12}.\nonumber \nonumber

It should not be a surprise that the moments are equal since there is complete symmetry with respect to x and y. Finally, we divide to get

(x,y) = (\dfrac{5}{8},\dfrac{5}{8}).\nonumber \nonumber

This tells us that the metal plate will balance perfectly if we place a pin at (\frac{5}{8},\frac{5}{8}).

Moments of Inertia

We often call M_x and M_y the first moments. They have first powers of y and x in their definitions and help find the center of mass. We define the moments of inertia (or second moments) by introducing squares of y and x in their definitions. The moments of inertia help us find the kinetic energy in rotational motion. Below is the definition

Definition: Moments of Inertia

Suppose that \rho (x,y) is a continuous density function on a lamina R. Then the moments of inertia are

I_x = \iint_R \rho(x,y) y^2 \, dy\, dx\nonumber

I_y = \iint_R \rho(x,y) x^2 \, dy\, dx.\nonumber

Exercise \PageIndex{1}

Find the moments of inertia for the square metal plate in Example \PageIndex{2}.

First Moment

The first moment of a 3-D solid region D about a coordinate plane is defined as the triple integral over D of the distance from a point (x,y,z) in D to the plane multiplied by the density of the solid at that point. First moments about the coordinate planes:

M(yz)=\iiint_{a}^{b}\delta x\, dV\nonumber

M(xz)=\iiint_{a}^{b}\delta y\,dV\nonumber

M(xy)=\iiint_{a}^{b}\delta z \,dV\nonumber

The first moment about the y-axis is the double integral over the region R forming the 2-D plate of the distance from the axis multiplied by the density.

M(y)=\iint_{a}^{b}\delta x\; dV\nonumber

M(x)=\iint_{a}^{b}\delta y\; dV\nonumber

Center of Mass

We defined center of mass located in \bar{x}, \bar{y}, \bar{z}. Then it is found from the first moments:

\bar{x} =\dfrac{M(y)}{M}\nonumber

\bar{y} =\dfrac{M(x)}{M}.\nonumber

Contributors and Attributions

  • Shengqiao Luo (UCD)
  • Integrated by Justin Marshall.


3.7: Moments and Centers of Mass is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by LibreTexts.

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