One Framework for Several Physical Models
A rod, a thin sheet, and a three-dimensional solid have different geometries, but their mass and centroid calculations follow the same principle: add up small contributions of mass, then weight each contribution by its position to obtain a first moment. The key is choosing the correct expression for each small mass element.
For a rod, the element is linear; for a lamina, it is an area element; and for a solid described by slices, it is a thin volume slice. Composite objects require one more step: calculate the mass and moments of each part, then add them. A reliable solution keeps track of the geometry, density, units, and coordinate used in every integral.
The names of the planar moments can initially seem reversed: the moment about the \(x\)-axis weights mass by its \(y\)-coordinate, while the moment about the \(y\)-axis weights it by its \(x\)-coordinate. This convention follows the perpendicular distance to the relevant axis. For a one-dimensional coordinate calculation, it is often clearer to write \(Q_x=\int x\,dm\) and state explicitly that this is the first moment for the \(x\)-coordinate.
Why Mass and Moment Add
Suppose an object consists of separate parts indexed by \(i\), with masses \(m_i\) and centroid coordinates \((\bar{x}_i,\bar{y}_i)\). The total mass is \(\sum_i m_i\). The total first moment in the \(x\)-direction is \(\sum_i m_i\bar{x}_i\): each part contributes its own mass times its own centroid coordinate. Therefore, the centroid of the whole object is
The same reasoning applies when the object is divided into continuously many small elements. Each element contributes \(dm\) to mass and \(x\,dm\) to the first moment. Summing those contributions in the limit gives \(m=\int dm\) and \(Q_x=\int x\,dm\). This explains why the integral formula and the mass-weighted formula for composite objects are consistent.
Worked Example: A Rod With Variable Linear Density
Worked Example: A Rod With Variable Linear Density
A straight rod lies along the \(x\)-axis from \(x=0\) m to \(x=2\) m. Its linear density is \(\lambda(x)=1+x\) kg/m. Find its mass and centroid coordinate.
For a small length \(dx\), the mass is \(dm=\lambda(x)\,dx\). Thus the total mass is
The first moment for the \(x\)-coordinate is
Therefore,
Check: multiplying the mass by the proposed centroid gives \(4\text{ kg}\times(7/6)\text{ m}=14/3\text{ kg}\cdot\text{m}\), the calculated first moment. Also, \(0<7/6<2\), so the centroid lies within the rod. The units are consistent: a first moment has units of mass times length, and division by mass gives length.
Worked Example: A Lamina With Density Varying Across It
Worked Example: A Variable-Density Rectangular Lamina
A rectangular lamina occupies \(0\leq x\leq2\) m and \(0\leq y\leq1\) m. Its area density is \(\sigma(x,y)=1+x\) kg/m². Find its mass and centroid.
For an area element \(dA=dx\,dy\), the mass element is \(dm=\sigma\,dA\). Integrate first with respect to \(y\), then \(x\):
The moment about the \(y\)-axis weights mass by \(x\):
The moment about the \(x\)-axis weights mass by \(y\):
Divide each moment by the total mass:
Check: \(m\bar{x}=4(7/6)=14/3\text{ kg}\cdot\text{m}=Q_y\), and \(m\bar{y}=4(1/2)=2\text{ kg}\cdot\text{m}=Q_x\). The centroid satisfies \(0<7/6<2\) and \(0<1/2<1\), as required for a positive-density lamina contained in this rectangle. The mass units are kg because \(\text{kg}/\text{m}^2\) is multiplied by area in m².
Worked Example: Combining Regions With Different Densities
Worked Example: Two Rectangles With Different Densities
A lamina is made from two rectangles joined along an edge. The left rectangle occupies \(0\leq x\leq2\) m and \(0\leq y\leq1\) m, with density \(2\) kg/m². The right rectangle occupies \(2\leq x\leq3\) m and \(0\leq y\leq2\) m, with density \(1\) kg/m². Find the centroid of the combined lamina.
For the left rectangle, the area is \(2\text{ m}^2\), so its mass is \(2\text{ kg/m}^2\times2\text{ m}^2=4\) kg. Its geometric center is \((1,1/2)\) m. Since its density is constant, its centroid is also \((1,1/2)\) m. Its moments are \(Q_y=4(1)=4\text{ kg}\cdot\text{m}\) and \(Q_x=4(1/2)=2\text{ kg}\cdot\text{m}\).
For the right rectangle, the area is \(1\text{ m}\times2\text{ m}=2\text{ m}^2\), so its mass is \(1\text{ kg/m}^2\times2\text{ m}^2=2\) kg. Its center, and hence its centroid, is \((5/2,1)\) m. Its moments are \(Q_y=2(5/2)=5\text{ kg}\cdot\text{m}\) and \(Q_x=2(1)=2\text{ kg}\cdot\text{m}\).
Add the two masses and the corresponding moments:
The combined centroid is therefore
Check: \(6(3/2)=9\text{ kg}\cdot\text{m}\) and \(6(2/3)=4\text{ kg}\cdot\text{m}\), reproducing both total moments. The centroid lies within the overall bounding rectangle \(0\leq x\leq3\), \(0\leq y\leq2\). Its \(x\)-coordinate is between the component centroids \(1\) m and \(5/2\) m, and its \(y\)-coordinate is between \(1/2\) m and \(1\) m, consistent with the masses and positions.
Worked Example: A Solid Described by Cross-Sectional Slices
Worked Example: A Solid With Varying Area and Density
A solid extends from \(x=0\) m to \(x=2\) m. The cross-sectional area perpendicular to the \(x\)-axis is \(A(x)=1+x\) m², and the density is \(\rho(x)=1+x\) kg/m³. Find the mass and the centroid coordinate along \(x\).
A slice of thickness \(dx\) has volume \(dV=A(x)\,dx\), so its mass is \(dm=\rho(x)A(x)\,dx\). The mass is
The slice at position \(x\) contributes \(x\,dm\) to the first moment for the \(x\)-coordinate:
Thus,
Check: \((26/3)(17/13)=34/3\text{ kg}\cdot\text{m}\), which matches the calculated moment. Also, \(0<17/13<2\), so the coordinate lies within the solid’s extent along the \(x\)-axis. The density times area has units \((\text{kg}/\text{m}^3)(\text{m}^2)=\text{kg}/\text{m}\); integrating with respect to \(x\) gives kilograms.
This slice calculation determines the \(x\)-coordinate because each slice is at a known \(x\). To find transverse coordinates, such as \(\bar{y}\) or \(\bar{z}\), one would also need the locations of the cross-sectional centroids and their mass contributions.
A Common Mistake: Using the Geometric Center for Unequal Density
The geometric center of a region is not automatically its centroid of mass. The shortcut works for a uniform-density rectangle because every area element has the same mass per unit area. If density varies, regions with greater density contribute more strongly to both mass and moments.
For the variable-density rectangle in the earlier example, a tempting but incorrect calculation is to use its geometric center, \((1,1/2)\) m, as the centroid. That would predict \(Q_y=m\bar{x}=4(1)=4\text{ kg}\cdot\text{m}\). But direct integration gives \(Q_y=14/3\text{ kg}\cdot\text{m}\), not \(4\text{ kg}\cdot\text{m}\). The corrected coordinate is \(\bar{x}=(14/3)/4=7/6\) m. The larger density at larger \(x\) shifts the centroid to the right.
A Reliable Setup for Mixed Problems
Use linear density \(\lambda\) for a rod, area density \(\sigma\) for a lamina, and volume density \(\rho\) for a solid. State the coordinate or axes whose centroid is required.
Use \(dm=\lambda\,dx\), \(dm=\sigma\,dA\), or \(dm=\rho\,dV\). For slices, \(dV=A(x)\,dx\) when slices are perpendicular to the \(x\)-axis.
Multiply each mass element by the coordinate that measures its contribution to the desired first moment. For an \(x\)-coordinate, this is \(x\,dm\).
For a composite object, add all component masses and all corresponding moments. Divide the total moment by the total mass, not by a component mass.
Check that units reduce to length, that \(m\bar{x}\) reproduces the calculated first moment, and that the centroid is physically consistent with the object and its density.
These checks do not replace the integral setup, but they catch many errors before a result is accepted. In particular, dimensions help distinguish a density from a mass, while the moment balance checks whether the final coordinate agrees with the integral that produced it.
Check Your Understanding
For each question, identify which mass and moment quantities are needed before evaluating any integrals.
- A rod lies on \(0\leq x\leq L\) with linear density \(\lambda(x)\). Write the integrals for its mass and \(x\)-coordinate of its centroid.
- For a lamina with area density \(\sigma(x,y)\), which coordinate weights the mass in the moment used to find \(\bar{x}\)?
- Two components have masses \(m_1,m_2\) and centroid coordinates \(x_1,x_2\). Write the total \(x\)-coordinate of their combined centroid.
- A solid has slice area \(A(x)\) and density \(\rho(x)\). Write the mass element for a slice of thickness \(dx\), including its units.
- Why is the geometric center generally not sufficient to find the centroid of a variable-density lamina?
