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Deriving the Center of Mass of a Lamina

Use torque balance to derive the lamina center-of-mass formulas, track the signs of the first moments, and verify the formulas in worked examples.

Intermediate 9 min read

What You'll Learn

  • State the mass and first-moment conditions needed for finite center-of-mass coordinates
  • Derive the coordinate formulas by matching torque components
  • Explain why the moment about the y-axis determines the x-coordinate, and vice versa
  • Evaluate mass and first moments for laminae using double integrals
  • Keep signed coordinates distinct from unsigned distances to the axes
  • Check a proposed center of mass by comparing its torques with the distributed weight

Why Torque Determines the Center of Mass

A lamina has mass spread across a two-dimensional region. Its weight is therefore distributed across that region rather than concentrated at one point. The center of mass is the point at which the lamina's total weight can be treated as acting while producing the same torque as the distributed weight.

To derive its coordinates, imagine the lamina lying in the \(xy\)-plane, with gravity acting in the negative \(z\)-direction. This arrangement makes the torque components about the \(x\)- and \(y\)-axes directly comparable with the lamina's first moments about those axes. The central feature is that the torque about one axis depends on the signed coordinate parallel to the other axis.

Mass and First Moments

Let \(R\) be the region occupied by a lamina, and let \(\rho(x,y)\) be its areal mass density. Assume \(\rho\ge 0\), and define the mass element by \(dm=\rho(x,y)\,dA\). The total mass and the first moments about the coordinate axes are

$$ M=\iint_R \rho(x,y)\,dA,\qquad M_y=\iint_R x\,\rho(x,y)\,dA,\qquad M_x=\iint_R y\,\rho(x,y)\,dA. $$

The subscripts name the axes about which the moments are taken. The first moment about the \(y\)-axis uses the \(x\)-coordinate; the first moment about the \(x\)-axis uses the \(y\)-coordinate. These are signed coordinates. A point with negative \(x\), for example, contributes negatively to \(M_y\).

Definition: If the lamina has finite positive mass and both first moments are finite, its center of mass is the point \((\bar{x},\bar{y})\) whose weight has the same total force and torque as the lamina's distributed weight.

For a precise finiteness condition, it is sufficient to assume

$$ 0<M<\infty \qquad\text{and}\qquad \iint_R (|x|+|y|)\rho(x,y)\,dA<\infty. $$

This condition guarantees that both signed first moments are finite. A bounded region with finite mass also ensures this condition, since the coordinates are bounded there. Finite mass alone is not enough for an unbounded region: mass can be finite even when a first moment diverges.

Deriving the Coordinate Formulas

A mass element \(dm\) at \((x,y,0)\) has weight \(d\mathbf{F}=(0,0,-g\,dm)\), where \(g>0\) is the gravitational acceleration. The torque of this force about the origin is the position vector crossed with the force. In components, it is \((-gy\,dm,\;gx\,dm,\;0)\). Thus the torque about the \(x\)-axis depends on \(y\), while the torque about the \(y\)-axis depends on \(x\).

Derivation: Replace the distributed weight by a single weight \(Mg\) acting at \((\bar{x},\bar{y},0)\). Matching the torque components gives the center-of-mass formulas.
1
Find the distributed torque.
Integrating the torque components of all mass elements gives \(\tau_x=-g\iint_R y\,dm=-gM_x\) and \(\tau_y=g\iint_R x\,dm=gM_y\).
2
Find the torque of the equivalent point mass.
A point mass \(M\) at \((\bar{x},\bar{y},0)\) has weight \((0,0,-Mg)\), so its torque components are \(\tau_x=-Mg\bar{y}\) and \(\tau_y=Mg\bar{x}\).
3
Match corresponding components.
Since \(g>0\), equating the \(x\)-components gives \(M\bar{y}=M_x\), and equating the \(y\)-components gives \(M\bar{x}=M_y\). Since \(M>0\), division by \(M\) is valid.

Therefore, the coordinates are

$$ \bar{x}=\frac{M_y}{M}, \qquad \bar{y}=\frac{M_x}{M}. $$

The cross-axis pairing follows from torque geometry, not from a naming convention alone. The \(x\)-coordinate is the signed lever arm for torque about the \(y\)-axis, so \(M_y\) determines \(\bar{x}\). The \(y\)-coordinate is the signed lever arm for torque about the \(x\)-axis, so \(M_x\) determines \(\bar{y}\). The unsigned distances to those axes are \(|x|\) and \(|y|\), but torque components must retain their signs.

Theorem: If \(0<M<\infty\) and \(\iint_R (|x|+|y|)\rho(x,y)\,dA<\infty\), then the center of mass is finite and is given by \(\bar{x}=M_y/M\) and \(\bar{y}=M_x/M\). These coordinates produce the same total weight and torque as the lamina under uniform gravity.

Worked Examples

Worked Example: A Triangular Lamina with Variable Density

Let \(R\) be the triangle \(0\le x\le 1,\;0\le y\le x\), with density \(\rho(x,y)=2+x\). Find its center of mass.

First compute the mass by integrating across the region:

$$ M=\int_0^1\int_0^x(2+x)\,dy\,dx =\int_0^1(2+x)x\,dx =\int_0^1(2x+x^2)\,dx =1+\frac13=\frac43. $$

For the first moment about the \(y\)-axis, multiply the density by the signed coordinate \(x\):

$$ M_y=\int_0^1\int_0^x x(2+x)\,dy\,dx =\int_0^1x^2(2+x)\,dx =\int_0^1(2x^2+x^3)\,dx =\frac23+\frac14=\frac{11}{12}. $$

For the first moment about the \(x\)-axis, multiply by \(y\):

$$ M_x=\int_0^1\int_0^x y(2+x)\,dy\,dx =\int_0^1\frac{(2+x)x^2}{2}\,dx =\int_0^1\left(x^2+\frac{x^3}{2}\right)\,dx =\frac13+\frac18=\frac{11}{24}. $$

Hence

$$ \bar{x}=\frac{11/12}{4/3}=\frac{11}{16}, \qquad \bar{y}=\frac{11/24}{4/3}=\frac{11}{32}. $$

Check by comparing torques. The distributed torques are \(\tau_x=-gM_x=-11g/24\) and \(\tau_y=gM_y=11g/12\). At the proposed point, they are \(-gM\bar{y}=-(4/3)g(11/32)=-11g/24\) and \(gM\bar{x}=(4/3)g(11/16)=11g/12\). Both components agree.

Worked Example: A Lamina Above the \(x\)-Axis

Let \(R=[0,1]\times[1,2]\) and \(\rho(x,y)=1+x\). Find the center of mass.

The mass is

$$ M=\int_0^1\int_1^2(1+x)\,dy\,dx =\int_0^1(1+x)\,dx =1+\frac12=\frac32. $$

The first moment about the \(y\)-axis is

$$ M_y=\int_0^1\int_1^2x(1+x)\,dy\,dx =\int_0^1(x+x^2)\,dx =\frac12+\frac13=\frac56. $$

The first moment about the \(x\)-axis is

$$ M_x=\int_0^1\int_1^2y(1+x)\,dy\,dx =\left(\int_1^2y\,dy\right)\left(\int_0^1(1+x)\,dx\right) =\frac32\cdot\frac32=\frac94. $$

Thus

$$ \bar{x}=\frac{5/6}{3/2}=\frac59, \qquad \bar{y}=\frac{9/4}{3/2}=\frac32. $$

For the torque check, the distributed components are \(\tau_x=-9g/4\) and \(\tau_y=5g/6\). The point-mass components are \(-g(3/2)(3/2)=-9g/4\) and \(g(3/2)(5/9)=5g/6\). The values match.

Worked Example: A Lamina with Negative \(x\)-Coordinates

Let \(R\) be the triangular region \(-1\le x\le 0,\;0\le y\le -x\), with constant density \(\rho=2\). Its negative \(x\)-coordinates make the role of signed moments explicit.

The mass is

$$ M=\int_{-1}^0\int_0^{-x}2\,dy\,dx =2\int_{-1}^0(-x)\,dx =2\left[\!-\frac{x^2}{2}\right]_{-1}^0=1. $$

The first moment about the \(y\)-axis is

$$ M_y=\int_{-1}^0\int_0^{-x}2x\,dy\,dx =2\int_{-1}^0x(-x)\,dx =-2\int_{-1}^0x^2\,dx =-\frac23. $$

The first moment about the \(x\)-axis is

$$ M_x=\int_{-1}^0\int_0^{-x}2y\,dy\,dx =\int_{-1}^0(-x)^2\,dx =\int_{-1}^0x^2\,dx =\frac13. $$

Therefore \(\bar{x}=-2/3\) and \(\bar{y}=1/3\). The torque components of the lamina are \(\tau_x=-g/3\) and \(\tau_y=-2g/3\). At the computed point they are \(-g(1)(1/3)=-g/3\) and \(g(1)(-2/3)=-2g/3\), as required.

A Common Sign Error

In the last example, replacing \(x\) by the unsigned distance \(|x|\) when computing \(M_y\) would give \(M_y=2/3\), and hence the incorrect coordinate \(\bar{x}=2/3\). That calculation uses distance from the \(y\)-axis, but it loses the sign of the torque. The actual points lie to the left of the \(y\)-axis, so their contributions to the \(y\)-component of torque are negative. The correct first moment is \(\int_R x\,dm=-2/3\), giving \(\bar{x}=-2/3\). Indeed, the incorrect positive coordinate would produce torque \(+2g/3\), whereas the distributed lamina's torque is \(-2g/3\).

Takeaway: Use \(x\) and \(y\), not \(|x|\) and \(|y|\), in the first-moment integrals. The signed coordinates preserve the directions of the torque components.

Why the Conditions and Checks Matter

The ratios \(M_y/M\) and \(M_x/M\) are meaningful as finite coordinates only when the mass is positive and finite and the first moments are finite. If a first moment diverges, there is no finite torque component to match, even if the total mass is finite. This is why the hypotheses include absolute integrability of the coordinates against the density, rather than only finite mass.

In an application, calculating the three integrals is not the end of the reasoning. Check that the mass is positive, that the first moments have the expected signs, and that both torque components agree when the proposed center is substituted back. The units also provide a useful check: \(M\) has units of mass, each first moment has units of mass times length, and dividing a first moment by mass gives a coordinate with units of length.

Check Your Understanding

Use the definitions and torque balance to answer these questions.

  1. Why does \(M_y\), rather than \(M_x\), determine \(\bar{x}\)?
  2. Write the finiteness condition that guarantees both first moments exist as finite signed quantities.
  3. If all mass lies at negative \(x\)-coordinates, what sign should \(M_y\) have, assuming the mass is positive?
  4. For a proposed center of mass, what two torque components should be checked?
  5. What goes wrong if \(|x|\) is substituted for \(x\) in the first moment about the \(y\)-axis?

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