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Moments Of Inertia For Nonuniform Laminae

Use density-weighted integrals to calculate moments of inertia and see how moving mass closer to or farther from an axis changes the result.

Intermediate 10 min read

What You'll Learn

  • Define moments of inertia about the coordinate axes and the origin for a nonuniform lamina.
  • Explain how density weights each area element according to its distance from an axis.
  • Set up and evaluate Cartesian moment integrals for coordinate-dependent densities.
  • Convert radial-density calculations to polar coordinates with the correct area element.
  • Compare equal masses distributed near and far from an axis.

Why Density Changes a Lamina’s Moment of Inertia

A moment of inertia measures how a lamina’s mass is distributed relative to an axis. Mass farther from the axis contributes more because its contribution is weighted by the square of its distance. If density varies across the lamina, equal-sized area elements need not contain equal masses, so the density must be included in the integral.

This weighting is useful in applications involving rotation: for the same angular acceleration, an object with more mass distributed far from the axis is harder to rotate. In this tutorial, we calculate moments for coordinate-dependent and radial densities, and compare distributions with equal total mass.

Definition for a Nonuniform Lamina

Let a lamina occupy a bounded planar region \(R\), with nonnegative mass density \(\rho(x,y)\). Density is mass per unit area. For a small area element \(dA\), its mass is approximately \(\rho(x,y)\,dA\). Assume the density is piecewise continuous on the region, so the integrals below are defined.

Definition: The moment of inertia about the \(x\)-axis, the moment of inertia about the \(y\)-axis, and the polar moment of inertia about the origin are $$ I_x=\iint_R y^2\rho(x,y)\,dA,\qquad I_y=\iint_R x^2\rho(x,y)\,dA,\qquad I_O=\iint_R (x^2+y^2)\rho(x,y)\,dA. $$

The perpendicular distance from \((x,y)\) to the \(x\)-axis is \(|y|\), and its squared distance is \(y^2\). The corresponding squared distance to the \(y\)-axis is \(x^2\). Distance to the origin has square \(x^2+y^2\). Each small mass is therefore multiplied by its squared distance from the axis in the relevant integral.

All three moments are nonnegative. If mass is measured in kilograms and distance in metres, a moment of inertia has units of \(\mathrm{kg}\,\mathrm{m}^2\). The total mass, by contrast, is \(M=\iint_R\rho(x,y)\,dA\), with units of kilograms.

Why the Integral Uses Density and Squared Distance

Divide the lamina into small pieces with areas \(\Delta A_i\), and choose a sample point \((x_i,y_i)\) in each piece. The mass of piece \(i\) is approximately \(\rho(x_i,y_i)\Delta A_i\). Its contribution to the moment about the \(x\)-axis is approximately \(y_i^2\rho(x_i,y_i)\Delta A_i\). Summing these contributions gives an approximation to the total moment.

For this approximation to converge to the area integral, the partition must become fine in every direction. In particular, the largest diameter of any piece must tend to zero; making only the largest piece area tend to zero is not enough.

Derivation: For partitions of \(R\) whose mesh (the maximum diameter of a piece) tends to zero, the Riemann sums for a piecewise continuous density converge to the integral of the corresponding integrand. Thus $$ \lim_{\text{mesh}\to 0} \sum_i y_i^2\rho(x_i,y_i)\Delta A_i = \iint_R y^2\rho(x,y)\,dA. $$ The same argument, using \(x_i^2\) or \(x_i^2+y_i^2\) as the squared distance, gives the other moment formulas.

The definition also yields a useful identity. Since \(x^2+y^2\) is the sum of the two squared coordinate distances, linearity of integration gives

$$ I_O = \iint_R (x^2+y^2)\rho\,dA = \iint_R x^2\rho\,dA+\iint_R y^2\rho\,dA = I_y+I_x. $$

Worked Example: A Density Depending on \(x\)

Worked Example: A Density Depending on \(x\)

Let \(R=[0,1]\times[0,2]\), with density \(\rho(x,y)=1+x\) kilograms per square metre. Find the mass and the moments about both coordinate axes.

First calculate the mass. The density does not depend on \(y\), so

$$ M = \int_0^1\int_0^2(1+x)\,dy\,dx = \int_0^1 2(1+x)\,dx = 2\left[x+\frac{x^2}{2}\right]_0^1 = 3\text{ kg}. $$

For the \(x\)-axis, the squared distance is \(y^2\):

$$ \begin{aligned} I_x &=\int_0^1\int_0^2 y^2(1+x)\,dy\,dx\\ &=\left(\int_0^1(1+x)\,dx\right) \left(\int_0^2y^2\,dy\right)\\ &=\left[ x+\frac{x^2}{2}\right]_0^1 \left[\frac{y^3}{3}\right]_0^2\\ &=\frac{3}{2}\cdot\frac{8}{3} =4\text{ kg}\,\mathrm{m}^2. \end{aligned} $$

For the \(y\)-axis, use \(x^2\):

$$ \begin{aligned} I_y &=\int_0^1\int_0^2 x^2(1+x)\,dy\,dx\\ &=2\int_0^1(x^2+x^3)\,dx\\ &=2\left[\frac{x^3}{3}+\frac{x^4}{4}\right]_0^1\\ &=2\left(\frac13+\frac14\right) =\frac76\text{ kg}\,\mathrm{m}^2. \end{aligned} $$

As a check, the polar moment must be the sum of these two moments: \(I_O=4+\frac76=\frac{31}{6}\text{ kg}\,\mathrm{m}^2\). Directly, integrating \((x^2+y^2)(1+x)\) over the rectangle gives the same sum. The larger \(I_x\) reflects the rectangle’s extent in the \(y\)-direction, which is the distance relevant to that axis.

Radial Density and Polar Coordinates

If the region is naturally described using distance \(r\) from the origin and angle \(\theta\), polar coordinates can simplify the integral. Recall that \(x=r\cos\theta\), \(y=r\sin\theta\), and \(dA=r\,dr\,d\theta\). The extra factor \(r\) in the area element is essential.

For a radial density \(\rho(x,y)=g(r)\), the squared distance to the origin is \(r^2\), so the polar moment on a disk of radius \(a\) is

$$ I_O = \int_0^{2\pi}\int_0^a r^2g(r)\,r\,dr\,d\theta = 2\pi\int_0^a r^3g(r)\,dr. $$

When the region is a disk and the density depends only on \(r\), symmetry also gives \(I_x=I_y\). This can be verified in the integral: \(y^2=r^2\sin^2\theta\), \(x^2=r^2\cos^2\theta\), and the integrals of \(\sin^2\theta\) and \(\cos^2\theta\) over a full turn are both \(\pi\).

Worked Example: A Disk with Increasing Radial Density

Consider the unit disk \(0\le r\le1\) with density \(\rho=1+r^2\) kilograms per square metre. Find its mass and moments of inertia.

The mass is

$$ \begin{aligned} M &=\int_0^{2\pi}\int_0^1(1+r^2)r\,dr\,d\theta\\ &=2\pi\left[\frac{r^2}{2}+\frac{r^4}{4}\right]_0^1 =2\pi\left(\frac12+\frac14\right) =\frac{3\pi}{2}\text{ kg}. \end{aligned} $$

The polar moment is

$$ \begin{aligned} I_O &=2\pi\int_0^1r^3(1+r^2)\,dr\\ &=2\pi\left[\frac{r^4}{4}+\frac{r^6}{6}\right]_0^1\\ &=2\pi\left(\frac14+\frac16\right) =\frac{5\pi}{6}\text{ kg}\,\mathrm{m}^2. \end{aligned} $$

The density and disk are symmetric about both axes, so \(I_x=I_y\). Since \(I_O=I_x+I_y\), each is half of \(I_O\): \(I_x=I_y=\frac{5\pi}{12}\text{ kg}\,\mathrm{m}^2\). To check one directly,

$$ I_x = \left(\int_0^{2\pi}\sin^2\theta\,d\theta\right) \left(\int_0^1r^3(1+r^2)\,dr\right) = \pi\left(\frac14+\frac16\right) = \frac{5\pi}{12}\text{ kg}\,\mathrm{m}^2. $$

Because the density increases with \(r\), more of the mass is placed toward the outer part of the disk than in a uniform disk of the same area. The factor \(r^2\) in the polar moment makes that outer mass contribute strongly.

Worked Example: Equal Mass Near and Far from an Axis

Worked Example: Comparing Two Densities

On the unit square \(R=[0,1]\times[0,1]\), compare the densities \(\rho_f(x,y)=2y\) and \(\rho_n(x,y)=2(1-y)\). The first places more mass at larger \(y\), farther from the \(x\)-axis; the second places more mass near that axis. Find the mass and moments about each coordinate axis for both distributions.

For the first density,

$$ M_f=\int_0^1\int_0^1 2y\,dx\,dy =\int_0^1 2y\,dy =[y^2]_0^1=1\text{ kg}. $$

Its moments are

$$ \begin{aligned} I_{x,f} &=\int_0^1\int_0^1 2y^3\,dx\,dy =2\left[\frac{y^4}{4}\right]_0^1 =\frac12\text{ kg}\,\mathrm{m}^2,\\ I_{y,f} &=\int_0^1\int_0^1 2x^2y\,dy\,dx =\int_0^1x^2\,dx =\left[\frac{x^3}{3}\right]_0^1 =\frac13\text{ kg}\,\mathrm{m}^2. \end{aligned} $$

For the second density,

$$ M_n=\int_0^1\int_0^1 2(1-y)\,dx\,dy =\int_0^1 2(1-y)\,dy =[2y-y^2]_0^1=1\text{ kg}. $$

Its moments are

$$ \begin{aligned} I_{x,n} &=\int_0^1 2y^2(1-y)\,dy\\ &=2\left[\frac{y^3}{3}-\frac{y^4}{4}\right]_0^1 =2\left(\frac13-\frac14\right) =\frac16\text{ kg}\,\mathrm{m}^2,\\ I_{y,n} &=\int_0^1x^2\,dx\int_0^1 2(1-y)\,dy =\frac13\cdot1 =\frac13\text{ kg}\,\mathrm{m}^2. \end{aligned} $$

Both masses are \(1\) kg, but the distribution farther from the \(x\)-axis has three times the \(x\)-axis moment: \(\frac12\) rather than \(\frac16\). Their \(y\)-axis moments agree because both densities depend only on \(y\), and both have the same total mass over each vertical strip. As a check, the polar moments are \(I_{O,f}=\frac12+\frac13=\frac56\) and \(I_{O,n}=\frac16+\frac13=\frac12\), each equal to the sum of the corresponding axis moments.

A Common Pitfall: Omitting the Density

A frequent error is to use the geometric moment integral for a uniform lamina, such as \(\iint_R y^2\,dA\), even when the density varies. That calculation treats every unit of area as if it had the same mass. For a nonuniform lamina, each contribution must be weighted by \(\rho(x,y)\).

For example, for the unit square with density \(\rho=2y\), omitting the density would give \(\int_0^1\int_0^1 y^2\,dx\,dy=\frac13\). The correct moment about the \(x\)-axis is \(\int_0^1\int_0^1 y^2(2y)\,dx\,dy=\frac12\). The former is the geometric integral for unit density, not the moment for the stated density.

Key takeaway: For a nonuniform lamina, multiply the squared distance to the chosen axis by the local density before integrating. In polar coordinates, also include the area factor \(r\) in \(dA=r\,dr\,d\theta\).

Check Your Understanding

Use the definitions and methods above to answer these questions.

  1. For a lamina with density \(\rho(x,y)\), which squared distance belongs in the integrand for the moment about the \(y\)-axis?
  2. Why does a polar integral over a disk include an extra factor of \(r\) in its area element?
  3. If \(I_x=2\) and \(I_y=5\) in consistent units, what is \(I_O\)?
  4. For equal total masses, why does placing more mass farther from the \(x\)-axis tend to increase \(I_x\)?
  5. What condition on the partitions is used to justify convergence of the Riemann sums for a moment integral?

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