Why Composite Laminae Need Mass Weighting
A lamina may be assembled from several pieces whose shapes, sizes, or densities differ. Each piece has its own mass and centroid. To locate the centroid of the whole object, we combine the pieces' contributions to the total mass and to the first moments.
The crucial point is that a centroid describes the distribution of mass. If two pieces have the same area but different densities, they do not have equal influence on the combined centroid. The correct weights are their masses, not their areas.
Assume that the lamina consists of a finite collection of components with nonoverlapping interiors. Components may touch along their boundaries. Each component has finite, positive mass, and its mass distribution has a well-defined centroid. The additivity of mass and first moments then gives a direct rule for the composite centroid.
Mass, First Moments, and Centroid Coordinates
For a lamina occupying a region \(R\) with areal density \(\rho(x,y)\), a small area element \(dA\) has mass \(dm=\rho(x,y)\,dA\). The total mass \(m\), first moment \(Q_x\) about the \(x\)-axis, and first moment \(Q_y\) about the \(y\)-axis are
The coordinate \(y\) is used in the moment about the \(x\)-axis because it measures the perpendicular distance to that axis; similarly, \(x\) appears in the moment about the \(y\)-axis. If \(m>0\), the centroid \((\bar{x},\bar{y})\) is defined by
For component \(i\), write its mass as \(m_i\), its centroid as \((\bar{x}_i,\bar{y}_i)\), and its first moments as \(Q_{x,i}\) and \(Q_{y,i}\). The centroid definition gives \(Q_{x,i}=m_i\bar{y}_i\) and \(Q_{y,i}=m_i\bar{x}_i\).
The Mass-Weighted Averaging Rule
Let the components be indexed by \(i=1,\ldots,n\). Since mass and first moments add across the components, the total mass is \(\sum_{i=1}^n m_i\). The total moments are \(\sum_{i=1}^n Q_{x,i}\) and \(\sum_{i=1}^n Q_{y,i}\). Substituting the component centroid relations into the centroid definition gives the rule:
This formula requires the components' masses and centroids to refer to the same coordinate axes. It also requires that each component's mass be counted exactly once. For ordinary laminae with nonnegative density, a positive total mass makes the composite centroid well-defined.
Why the Rule Works
The derivation is a consequence of additivity, not an extra assumption about how centroids behave. For example, the \(y\)-coordinate of the total centroid follows from adding the first moments about the \(x\)-axis:
The total mass is \(M=\sum_{i=1}^n m_i\), because the components have nonoverlapping interiors and each mass is included once.
The total first moment about the \(x\)-axis is \(Q_x=\sum_{i=1}^n Q_{x,i}\), by additivity of the integral over the component regions.
For every component, \(Q_{x,i}=m_i\bar{y}_i\). Therefore \(Q_x=\sum_{i=1}^n m_i\bar{y}_i\).
By definition, \(\bar{y}=Q_x/M\), so \(\bar{y}=(\sum_{i=1}^n m_i\bar{y}_i)/(\sum_{i=1}^n m_i)\). Applying the same argument to \(Q_y\) gives the formula for \(\bar{x}\).
Thus, each component contributes its mass multiplied by its distance-coordinate from the relevant axis. A heavier component contributes more to the total first moment than a lighter component at the same location.
Worked Examples
Worked Example: Two Rectangles With Different Densities
Rectangle A occupies \(0\le x\le2\), \(0\le y\le1\), and has uniform density \(3\) mass units per square unit. Rectangle B occupies \(2\le x\le3\), \(0\le y\le2\), and has uniform density \(2\) mass units per square unit. Find the centroid of their joined lamina.
Rectangle A has area \(2\cdot1=2\), so its mass is \(m_1=3\cdot2=6\) mass units. Its centroid is the midpoint of its sides, \((\bar{x}_1,\bar{y}_1)=(1,\frac12)\). Rectangle B has area \(1\cdot2=2\), so \(m_2=2\cdot2=4\) mass units. Its centroid is \((\bar{x}_2,\bar{y}_2)=(\frac52,1)\).
The total mass is \(M=6+4=10\) mass units. Weighting the \(x\)-coordinates by mass gives
For the \(y\)-coordinate,
The composite centroid is \((\frac85,\frac7{10})\) length units. To check, the component first moments about the \(y\)-axis sum to \(6(1)+4(\frac52)=16\) mass-length units, and those about the \(x\)-axis sum to \(6(\frac12)+4(1)=7\) mass-length units. Dividing by the total mass \(10\) recovers \(16/10=\frac85\) and \(7/10\), as required.
Worked Example: A Triangle Joined to a Rectangle
A right triangular lamina has vertices \((0,0)\), \((3,0)\), and \((0,3)\), with uniform density \(2\) mass units per square unit. A rectangle occupies \(3\le x\le5\), \(0\le y\le2\), with uniform density \(\frac32\) mass units per square unit. Find the centroid of the composite lamina.
The triangle has area \(\frac12(3)(3)=\frac92\) square units, so its mass is \(m_1=2(\frac92)=9\) mass units. Its centroid is \((1,1)\): the centroid of a uniform triangle is the average of its three vertices, and here \(((0+3+0)/3,(0+0+3)/3)=(1,1)\).
The rectangle has area \((5-3)(2-0)=4\) square units, so its mass is \(m_2=\frac32(4)=6\) mass units. Its centroid is its midpoint, \((4,1)\). Thus the total mass is \(M=9+6=15\) mass units, and
The centroid is \((\frac{11}{5},1)\) length units. For a check, the total first moment about the \(y\)-axis is \(9(1)+6(4)=33\) mass-length units, and the total first moment about the \(x\)-axis is \(9(1)+6(1)=15\) mass-length units. Dividing each by \(15\) gives \(\bar{x}=\frac{33}{15}=\frac{11}{5}\) and \(\bar{y}=\frac{15}{15}=1\).
Worked Example: A Component With Variable Density
A rectangular lamina occupies \(0\le x\le1\), \(0\le y\le2\), with density \(\rho(x,y)=1+x\) mass units per square unit. It is joined along an edge to a uniform rectangle occupying \(1\le x\le3\), \(0\le y\le1\), with density \(2\) mass units per square unit. Find the composite centroid.
First find the variable-density component's mass and moments. Integrating over its rectangular region gives
Its first moment about the \(y\)-axis is
Its first moment about the \(x\)-axis is
Therefore its centroid is \((\bar{x}_1,\bar{y}_1)=(Q_{y,1}/m_1,Q_{x,1}/m_1)=(\frac59,1)\). The uniform rectangle has area \(2\), mass \(m_2=2(2)=4\), and centroid \((2,\frac12)\).
The composite mass is \(M=3+4=7\) mass units. Its centroid coordinates are
The result is \((\frac{29}{21},\frac57)\) length units. As a check, the summed first moments are \(Q_y=\frac53+4(2)=\frac{29}{3}\) and \(Q_x=3+4(\frac12)=5\), in mass-length units. Dividing by \(M=7\) gives \(\bar{x}=\frac{29}{21}\) and \(\bar{y}=\frac57\), matching the calculation.
Common Pitfall: Weighting by Area Instead of Mass
In the first example, both rectangles have area \(2\), but their densities are \(3\) and \(2\), so their masses are \(6\) and \(4\). If one incorrectly weights the component centroids by area alone, the equal areas lead to the unweighted midpoint:
This is wrong for the specified densities. For instance, the proposed \(x\)-coordinate would imply a first moment about the \(y\)-axis of \(10(1.75)=17.5\) mass-length units. But the actual summed first moment is \(6(1)+4(2.5)=16\) mass-length units. The correct coordinate is therefore \(16/10=1.6\), not \(1.75\). Area can replace mass as a weight only when the components have the same uniform density.
Interpreting and Checking the Result
The mass-weighted formula is a weighted average: its weights are \(m_i/M\), which are positive and sum to \(1\). Thus, for example, \(\bar{x}=\sum_i(m_i/M)\bar{x}_i\). The result lies between the smallest and largest component \(x\)-coordinates, and the same is true for \(y\). This provides a useful plausibility check, although the coordinate bounds alone do not determine the centroid.
Dimensional consistency is another check. Each mass multiplied by a centroid coordinate has units of mass times length; dividing the summed first moment by total mass leaves units of length. A centroid coordinate should not have units of area, density, or mass.
When component masses and centroids are known, the composite calculation avoids setting up a new integral over the entire shape. If a component's density varies, its own mass and centroid must first reflect that density; once those quantities are correct, the same mass-weighted rule applies.
Check Your Understanding
Use mass, rather than area, as the weight in each composite-centroid calculation.
- Two components have masses \(2\) and \(6\), with centroids \((0,1)\) and \((3,4)\). Find the composite centroid.
- Why do first moments add when a lamina is divided into components with nonoverlapping interiors?
- Two equal-area pieces have different uniform densities. Is their composite centroid necessarily the average of their two centroids? Explain.
- A component has mass \(5\) and centroid \((2,-1)\). What are its first moments about the \(x\)-axis and the \(y\)-axis?
- What units should a composite centroid coordinate have if mass is measured in kilograms and length in metres?