Moments Of Inertia About Coordinate Axes

Why Moments of Inertia Depend on Distance

A lamina’s mass is spread across a region, rather than concentrated at a single point. When it rotates about an axis, mass farther from that axis contributes more strongly to its resistance to angular acceleration. The relevant distance is the perpendicular distance from each point of the lamina to the axis, and the contribution is proportional to the square of that distance.

For a point \((x,y)\), the perpendicular distance to the \(x\)-axis is \(|y|\), while the distance to the \(y\)-axis is \(|x|\). This difference determines which coordinate appears in each moment integral.

Definition: Let a lamina occupy a planar region \(R\) and have nonnegative areal density \(\rho(x,y)\), measured in mass per unit area. Its moments of inertia about the coordinate axes are

I_x=\iint_R y^2\rho(x,y)\,dA, \qquad I_y=\iint_R x^2\rho(x,y)\,dA.

Here \(I_x\) is the moment about the \(x\)-axis, and \(I_y\) is the moment about the \(y\)-axis. The integrals are assumed to exist and be finite.

The subscripts identify the axis of rotation, not the coordinate that is squared. To find the moment about the \(x\)-axis, use \(y^2\), because \(y\) measures distance from that axis. To find the moment about the \(y\)-axis, use \(x^2\).

Why the Integral Formula Works

For a small piece of area \(dA\) near \((x,y)\), its mass is approximately \(\rho(x,y)\,dA\). A small mass \(dm\), a perpendicular distance \(r\) from an axis, contributes \(r^2dm\) to the moment about that axis. Thus the small contributions are \(y^2\rho(x,y)\,dA\) about the \(x\)-axis and \(x^2\rho(x,y)\,dA\) about the \(y\)-axis.

To see how the whole-region formula follows, divide \(R\) into small pieces. Choose a point \((x_i,y_i)\) in each piece of area \(\Delta A_i\). The mass of that piece is approximately \(\rho(x_i,y_i)\Delta A_i\). Adding its contributions gives the approximations

I_x\approx\sum_i y_i^2\rho(x_i,y_i)\Delta A_i, \qquad I_y\approx\sum_i x_i^2\rho(x_i,y_i)\Delta A_i.

As the pieces become finer, these sums approach the double integrals of their respective integrands. That is the defining construction of the integrals above. The squares matter: doubling a piece’s distance from an axis multiplies its contribution by four, if its mass stays fixed.

Interpretation: A moment of inertia measures how strongly a mass distribution resists changes in rotation about a specified axis. It depends both on how much mass there is and on how far that mass lies from the axis. It is not determined by total mass alone.

Worked Examples

Worked Example: A Uniform Triangular Lamina

A triangular lamina occupies \(R=\{(x,y):0\leq x\leq 2,\ 0\leq y\leq x\}\), with coordinates measured in meters. Its constant density is \(3\text{ kg/m}^2\). Find \(I_x\) and \(I_y\).

For the moment about the \(x\)-axis, square the vertical distance \(y\):

\begin{aligned} I_x &=3\int_0^2\int_0^x y^2\,dy\,dx\\ &=3\int_0^2\left[\frac{y^3}{3}\right]_{0}^{x}\,dx\\ &=\int_0^2 x^3\,dx\\ &=\left[\frac{x^4}{4}\right]_{0}^{2}\\ &=4\text{ kg}\cdot\text{m}^2. \end{aligned}

For the moment about the \(y\)-axis, square the horizontal distance \(x\):

\begin{aligned} I_y &=3\int_0^2\int_0^x x^2\,dy\,dx\\ &=3\int_0^2 x^2(x-0)\,dx\\ &=3\int_0^2 x^3\,dx\\ &=3\left[\frac{x^4}{4}\right]_{0}^{2}\\ &=12\text{ kg}\cdot\text{m}^2. \end{aligned}

As a check on \(I_x\), reverse the order of integration. The same triangle has \(0\leq y\leq2\) and \(y\leq x\leq2\), so

I_x=3\int_0^2\int_y^2 y^2\,dx\,dy =3\int_0^2 y^2(2-y)\,dy =3\left(\frac{16}{3}-4\right) =4\text{ kg}\cdot\text{m}^2.

The reversed-order calculation agrees. Also, \(I_y>I_x\) is consistent with the shape: the triangle extends farther horizontally than vertically, placing more of its mass farther from the \(y\)-axis than from the \(x\)-axis.

Worked Example: A Uniform Rectangular Region

A lamina occupies \(0\leq x\leq1\text{ m}\), \(0\leq y\leq2\text{ m}\) and has density \(2\text{ kg/m}^2\). Evaluate its moments about both coordinate axes.

For \(I_x\), integrate \(y^2\) over the region:

\begin{aligned} I_x &=2\int_0^{1}\int_0^{2} y^2\,dy\,dx\\ &=2\int_0^{1}\left[\frac{y^3}{3}\right]_0^2\,dx\\ &=2\int_0^{1}\frac{8}{3}\,dx\\ &=\frac{16}{3}\text{ kg}\cdot\text{m}^2. \end{aligned}

For \(I_y\), integrate \(x^2\):

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

Check the integrations by separating the factors on this rectangular region: \(I_x=2(1)(8/3)=16/3\), while \(I_y=2(2)(1/3)=4/3\), with units of \(\text{kg}\cdot\text{m}^2\) in both cases. The longer vertical extent makes the moment about the \(x\)-axis larger, as expected.

Worked Example: Density Varying Across a Square

Let \(R\) be the square \(0\leq x\leq L\), \(0\leq y\leq L\), where \(L=1\text{ m}\). Suppose the density is \(\rho(x,y)=\kappa(1+y/L)\), with \(\kappa=2\text{ kg/m}^2\). The density increases with height. Find the two moments.

First calculate \(I_x\). Since the density depends on \(y\), retain its factor inside the \(y\)-integral:

\begin{aligned} I_x &=\kappa\int_0^L\int_0^L y^2\left(1+\frac{y}{L}\right)\,dy\,dx\\ &=\kappa L\int_0^L\left(y^2+\frac{y^3}{L}\right)\,dy\\ &=\kappa L\left(\frac{L^3}{3}+\frac{L^4}{4L}\right)\\ &=\frac{7}{12}\kappa L^4\\ &=\frac{7}{6}\text{ kg}\cdot\text{m}^2. \end{aligned}

Now calculate \(I_y\). The factor \(x^2\) depends only on \(x\), while the density factor depends only on \(y\), so the integrations separate:

\begin{aligned} I_y &=\kappa\left(\int_0^L x^2\,dx\right) \left(\int_0^L\left(1+\frac{y}{L}\right)\,dy\right)\\ &=\kappa\left(\frac{L^3}{3}\right)\left(L+\frac{L}{2}\right)\\ &=\frac{1}{2}\kappa L^4\\ &=1\text{ kg}\cdot\text{m}^2. \end{aligned}

For a check, integrating the density alone gives the mass \(m=\kappa\int_0^L\int_0^L(1+y/L)\,dy\,dx=(3/2)\kappa L^2=3\text{ kg}\). Both moments are positive and have the correct units. Since the density increases with \(y\), more mass is farther from the \(x\)-axis than for a uniform square with the same overall scale; the larger value of \(I_x\) is consistent with that distribution.

Units and Conditions

If density is measured in \(\text{kg/m}^2\), then \(dA\) is measured in \(\text{m}^2\), and \(x^2\) and \(y^2\) are measured in \(\text{m}^2\). Therefore each moment has units

\frac{\text{kg}}{\text{m}^2}\cdot\text{m}^2\cdot\text{m}^2 =\text{kg}\cdot\text{m}^2.

More generally, the units are mass times length squared. A result with units of mass alone, or mass times length, signals that a factor has been omitted or misapplied. For a physical lamina, the density should be nonnegative and integrable over the region, and the coordinate-squared integrals must be finite for the moments to be finite.

Each moment is nonnegative because \(x^2\), \(y^2\), and \(\rho\) are nonnegative. A moment is zero only if the mass is concentrated entirely on the corresponding axis (or the lamina has zero mass). For an ordinary lamina occupying a region with positive area, both moments are positive.

A Common Mix-Up

A frequent error is to use the coordinate matching the axis name, rather than the coordinate measuring distance from that axis. For the triangular lamina above, an incorrect setup for the moment about the \(x\)-axis would be

I_x\ \text{(incorrect)} =3\int_0^2\int_0^x x^2\,dy\,dx =12\text{ kg}\cdot\text{m}^2.

This calculation actually squares the distance to the \(y\)-axis, so it computes \(I_y\), not \(I_x\). The correction is to use \(y^2\) for distance to the \(x\)-axis:

I_x =3\int_0^2\int_0^x y^2\,dy\,dx =4\text{ kg}\cdot\text{m}^2.

Before integrating, name the axis and identify the perpendicular distance to it. This simple check helps prevent both swapped coordinates and incorrect interpretations of the result.

Check Your Understanding

Answer these questions using the definitions and calculations above.

Which coordinate is squared when calculating the moment about the \(x\)-axis, and why?
For a point at \((3,4)\), what squared distance contributes to its moment about the \(y\)-axis?
What are the units of a planar moment of inertia if density is measured in \(\text{kg/m}^2\) and coordinates in meters?
In the triangular example, why does reversing the order of integration provide a useful check on \(I_x\)?
If the same small mass is moved to twice its original distance from an axis, how does its contribution to the moment change?