Laminae With Holes and Cutouts

Why a Hole Changes the Centroid

A cutout removes material, so it changes both the mass of a lamina and the locations of its mass. The centroid of the remaining shape is generally not the centroid of the original shape: removing material on one side shifts the balance of the remaining mass away from that cutout.

The calculation is systematic. Find the mass and first moments of the intact lamina, then subtract the mass and first moments of each removed region. The removed material must be subtracted as a whole contribution—not merely by subtracting its area or its centroid coordinates.

Mass and First-Moment Notation

Let a uniform lamina occupy a planar region \(R\), with constant areal density \(\sigma\). If area is measured in square metres, then \(\sigma\) has units of kilograms per square metre. The mass element is \(dm=\sigma\,dA\). The mass \(m\), first moment \(Q_x\) about the \(x\)-axis, and first moment \(Q_y\) about the \(y\)-axis are

Thus \(Q_x\) weights each mass element by its vertical coordinate, while \(Q_y\) weights it by its horizontal coordinate. For \(m>0\), the centroid \((\bar{x},\bar{y})\) is given by

For a uniform lamina, the centroid of a region is also its center of mass. If a region has area \(A\) and centroid \((x_c,y_c)\), its mass is \(\sigma A\), and its first moments are \(Q_y=\sigma A x_c\) and \(Q_x=\sigma A y_c\).

The Subtraction Rule

Suppose the cutouts \(H_1,\ldots,H_n\) lie within \(R\) and do not overlap in area. Let the original region have area \(A_0\) and centroid \((x_0,y_0)\). Let cutout \(H_i\) have area \(A_i\) and centroid \((x_i,y_i)\). All the material that remains has the same density \(\sigma\). Then its mass and first moments are

The centroid follows from \(\bar{x}=Q_y/m\) and \(\bar{y}=Q_x/m\). These formulas require that the cutouts be contained in the original region, that they are counted without overlapping one another, and that the remaining material has uniform density. The remaining area, and hence the mass, must be positive.

Why Subtracting Moments Works

The rule follows directly from the additivity of area integrals. Write \(R\setminus H\) for the material left after removing one cutout \(H\). Except on boundaries, which have zero area, the indicator of \(R\setminus H\) is the indicator of \(R\) minus the indicator of \(H\). Integrating the constant density over the remaining region therefore subtracts the cutout’s mass. Integrating the density multiplied by \(x\) or \(y\) subtracts the corresponding first moment as well.

For multiple disjoint cutouts, apply the same subtraction to each one. This is the same mass-and-moment accounting used for components, except that each removed component has a negative contribution.

Worked Examples

A Common Error: Subtracting Area Instead of Mass

In the first example, a mistaken calculation might subtract the hole’s area-weighted horizontal coordinate directly from the plate’s mass moment:

This is not the remaining \(Q_y\). The value \(900\) is a mass moment, measured in \(\text{kg}\cdot\text{m}\), while \(4(3)=12\) is an area moment, measured in \(\text{m}^3\). The density must be included: the hole’s mass is \(3(4)=12\text{ kg}\), so its mass moment is \(12(3)=36\text{ kg}\cdot\text{m}\). The correct subtraction is \(900-36=864\text{ kg}\cdot\text{m}\).

Checks Before Accepting a Result

A reliable calculation should satisfy several basic checks. The remaining mass must be positive, and it must equal the density multiplied by the remaining area. Each centroid coordinate must be found using the matching first moment: \(\bar{x}\) uses \(Q_y\), while \(\bar{y}\) uses \(Q_x\). Finally, multiplying the proposed centroid coordinates by the net mass must recover the net moments.

The coordinates also provide a plausibility check. A cutout removes mass from its own location, so the centroid tends to shift away from it. This tendency is not a substitute for calculation, but it can reveal a sign error or an incorrect assignment of moments.