Why Locate the Center of a Lamina?
A lamina is a thin, planar object whose mass is distributed over a region \(D\) in the \(xy\)-plane. Knowing its total mass does not tell us where that mass is concentrated. A single point that captures important information about the distribution is its center of mass.
For a lamina with uniform density, the center of mass depends only on the shape of the region. If the density varies, however, more heavily weighted parts of the region influence the center more strongly. This distinction separates the center of mass from the geometric centroid, which averages over area without weighting by density.
Mass and First Moments
Let \(\rho(x,y)\) be the mass density of the lamina at \((x,y)\), measured as mass per unit area. Assume \(D\) is measurable, \(\rho\) is nonnegative and integrable over \(D\), and the total mass is finite and positive. The mass is
The first moments about the coordinate axes are
The coordinate in each integrand is the signed perpendicular distance to the corresponding axis: \(y\) for the \(x\)-axis and \(x\) for the \(y\)-axis. Thus, material below the \(x\)-axis contributes negatively to \(M_x\), and material left of the \(y\)-axis contributes negatively to \(M_y\).
The absolute-moment conditions matter. Finite mass alone does not guarantee finite first moments: mass far from an axis can have a divergent moment even when the total mass is finite. Requiring the absolute integrals to be finite ensures that the displayed moments are finite numbers, so the center coordinates are well-defined finite coordinates.
Why These Coordinates Represent the Mass Distribution
The center of mass is the point at which the entire mass could be represented, for purposes of first moments, by a single point mass of size \(m\). A point mass \(m\) at \((a,b)\) has first moment \(mb\) about the \(x\)-axis and \(ma\) about the \(y\)-axis. Therefore, matching the lamina’s first moments requires \(mb=M_x\) and \(ma=M_y\).
This characterization also explains the coordinate order: a point’s moment about the \(y\)-axis uses its \(x\)-coordinate, whereas its moment about the \(x\)-axis uses its \(y\)-coordinate.
Worked Example: A Uniform Rectangular Lamina
Worked Example: A Uniform Rectangular Lamina
Let \(D=[0,4]\times[0,2]\), measured in metres, and let the constant density be \(\rho=3\ \mathrm{kg/m^2}\). Find the center of mass.
First, calculate the mass:
The first moment about the \(y\)-axis is
The first moment about the \(x\)-axis is
Therefore,
Check by substituting these coordinates into the point-mass first moments: \(m\bar{x}=24(2)=48\ \mathrm{kg\,m}=M_y\), and \(m\bar{y}=24(1)=24\ \mathrm{kg\,m}=M_x\). The calculated point therefore matches both moments.
Density-Weighted Center of Mass
When density varies, each part of the region contributes according to its mass, not just its area. The factors \(\rho(x,y)\) in the mass and moment integrals perform this weighting. A region with more mass on one side can have its center of mass pulled toward that side, even if the region’s shape itself is balanced.
Worked Example: A Lamina with Density Increasing in the \(x\)-Direction
Let \(D=[0,2]\times[0,1]\), in metres, and define \(\rho(x,y)=(1\ \mathrm{kg/m^2})(1+x/(1\ \mathrm{m}))\). Find the center of mass. In the calculations below, \(x\) and \(y\) are coordinates measured in metres.
The mass is
For \(M_y\), the inner integral is \(\int_0^2 (x+x^2/(1\ \mathrm{m}))\,dx =2\ \mathrm{m^2}+\frac{8}{3}\ \mathrm{m^2} =\frac{14}{3}\ \mathrm{m^2}\). Integrating over the \(1\)-metre \(y\)-interval gives
For \(M_x\), the \(y\)-integral is \(\int_0^1 y\,dy=\frac12\ \mathrm{m^2}\), and the remaining \(x\)-integral of \(1+x/(1\ \mathrm{m})\) is \(4\ \mathrm{m}\). Hence
It follows that
Check: \(m\bar{x}=4(7/6)=14/3\ \mathrm{kg\,m}=M_y\), and \(m\bar{y}=4(1/2)=2\ \mathrm{kg\,m}=M_x\). The center is shifted toward larger \(x\), where the density is greater.
Geometric Centroid
The geometric centroid is the area-based average position of a region. It does not use a varying density. If \(D\) has finite positive area \(A=\iint_D 1\,dA\), and its first absolute area moments are finite, its coordinates are
The finiteness conditions are \(\iint_D |x|\,dA<\infty\) and \(\iint_D |y|\,dA<\infty\). Finite area by itself is not enough to guarantee finite coordinate averages, because a small amount of area could extend arbitrarily far from an axis.
If the density is a positive constant \(\rho_0\), then \(m=\rho_0 A\), \(M_y=\rho_0\iint_D x\,dA\), and \(M_x=\rho_0\iint_D y\,dA\). Dividing the moments by the mass cancels the constant density, so the center of mass equals the geometric centroid. For nonconstant density, the two points need not coincide.
Worked Example: Comparing the Centroid and Center of Mass
Use the region and varying density from the previous example. Its area is \(A=(2\ \mathrm{m})(1\ \mathrm{m})=2\ \mathrm{m^2}\). The area integrals are
Thus the geometric centroid is
The density-weighted center of mass was \((7/6,1/2)\ \mathrm{m}\), while the centroid is \((1,1/2)\ \mathrm{m}\). The \(x\)-coordinates differ because density increases with \(x\). The \(y\)-coordinates happen to agree in this example. Check the centroid by multiplying its coordinates by the area: \(A x_c=2(1)=2\ \mathrm{m^3}=\iint_D x\,dA\), and \(A y_c=2(1/2)=1\ \mathrm{m^3}=\iint_D y\,dA\).
Signed Moments and a Common Mistake
First moments are signed quantities, not integrals of unsigned distances. For example, if a lamina lies entirely to the left of the \(y\)-axis, then \(x<0\) throughout its interior and \(M_y\) is negative for nonnegative density. This leads to a negative \(\bar{x}\), placing the center of mass on the correct side of the axis.
Worked Example: A Lamina Left of the \(y\)-Axis
Take \(D=[-2,0]\times[0,1]\), in metres, with constant density \(1\ \mathrm{kg/m^2}\). The mass is
The signed first moments are
Therefore, \(\bar{x}=M_y/m=-1\ \mathrm{m}\) and \(\bar{y}=M_x/m=\frac12\ \mathrm{m}\). Check: \(m\bar{x}=2(-1)=-2\ \mathrm{kg\,m}=M_y\), and \(m\bar{y}=2(1/2)=1\ \mathrm{kg\,m}=M_x\).
A frequent error is to swap the moments when assigning coordinates. In the first rectangular example, the incorrect assignment \(\bar{x}=M_x/m\), \(\bar{y}=M_y/m\) gives \((1,2)\ \mathrm{m}\) instead of \((2,1)\ \mathrm{m}\). It fails the moment check: a point mass of \(24\ \mathrm{kg}\) at \((1,2)\ \mathrm{m}\) would have \(M_y=24\ \mathrm{kg\,m}\), not the lamina’s \(48\ \mathrm{kg\,m}\). The correction is to pair \(\bar{x}\) with \(M_y\), and \(\bar{y}\) with \(M_x\).
Using the Formulas Reliably
For a density-weighted center of mass, first compute the mass and the two first moments using the same region and density. Then divide each first moment by the mass, remembering the cross-pairing of coordinates and moments. Check units: mass has units of mass, first moments have units of mass times length, and the resulting coordinates have units of length.
For a geometric centroid, replace the density-weighted integrals by area integrals and divide by the area. Do not use the area centroid in place of the center of mass when density varies. In either calculation, verify that the required mass or area is finite and positive and that the relevant absolute first moments are finite before interpreting the quotients as finite coordinates.
Check Your Understanding
Use the definitions and checks in this tutorial to answer the following questions.
- Which first moment determines \(\bar{x}\), and which determines \(\bar{y}\)?
- Why must first absolute moments be finite in addition to finite positive mass?
- For a region left of the \(y\)-axis with positive density, what sign do you expect for \(M_y\), and why?
- When does the center of mass necessarily agree with the geometric centroid?
- For \(D=[0,1]\times[0,1]\) and \(\rho(x,y)=1+x\), describe which direction the center of mass shifts relative to the centroid and explain why.