Moments Of Inertia For Uniform Rectangles And Disks

Why Shape and Axis Both Matter

A lamina’s moment of inertia measures how its mass is distributed relative to a chosen axis. Mass farther from the axis contributes more, because its contribution is proportional to the square of its distance. Thus, the same shape can have different moments of inertia about different axes.

For uniform rectangles and disks, symmetry identifies axes through the centroid, while direct integration gives the moments about those axes. The parallel axis theorem then relates those values to moments about other coordinate axes.

Definitions and Conventions

In this tutorial, a uniform lamina has constant mass density per unit area, denoted by \(\sigma\). If its total mass is \(M\) and its area is \(A\), then \(\sigma=M/A\). For a small area element \(dA\), its mass is \(dm=\sigma\,dA\).

Definition: The moment of inertia of a lamina about the \(x\)-axis is \(I_x=\int_R y^2\,dm\), where \(y\) is the perpendicular distance from the area element to that axis. The moment about the \(y\)-axis is \(I_y=\int_R x^2\,dm\). For a uniform lamina, these become \(I_x=\sigma\int_R y^2\,dA\) and \(I_y=\sigma\int_R x^2\,dA\).

The squared coordinate in each integrand is the distance to the axis: distance to the \(x\)-axis is \(|y|\), and distance to the \(y\)-axis is \(|x|\). The moments have units of mass times length squared, such as \(\text{kg}\cdot\text{m}^2\).

Uniform Rectangle: Deriving the Formulas

Take a rectangle of width \(a\) parallel to the \(x\)-axis and height \(b\) parallel to the \(y\)-axis, centered at the origin. Its bounds are \(-a/2\le x\le a/2\) and \(-b/2\le y\le b/2\). Its area is \(ab\), so its constant density is \(\sigma=M/(ab)\).

To find \(I_x\), integrate \(y^2\) over the rectangle. The integrand does not depend on \(x\), so the \(x\)-integration contributes the width \(a\). The remaining integral sums the squared distances from the \(x\)-axis across the height.

\begin{aligned} I_x &=\sigma\int_{-a/2}^{a/2}\int_{-b/2}^{b/2}y^2\,dy\,dx\\ &=\sigma a\left[\frac{y^3}{3}\right]_{-b/2}^{b/2}\\ &=\sigma a\left(\frac{b^3}{24}-\left(-\frac{b^3}{24}\right)\right)\\ &=\sigma a\frac{b^3}{12} =\frac{M}{ab}\frac{ab^3}{12} =\frac{Mb^2}{12}. \end{aligned}

For \(I_y\), integrate \(x^2\) instead. The \(y\)-integration contributes the height \(b\), and integrating across the width gives the corresponding result.

\begin{aligned} I_y &=\sigma\int_{-b/2}^{b/2}\int_{-a/2}^{a/2}x^2\,dx\,dy\\ &=\sigma b\left[\frac{x^3}{3}\right]_{-a/2}^{a/2}\\ &=\sigma b\left(\frac{a^3}{24}-\left(-\frac{a^3}{24}\right)\right)\\ &=\sigma b\frac{a^3}{12} =\frac{M}{ab}\frac{ba^3}{12} =\frac{Ma^2}{12}. \end{aligned}

Result: For a uniform rectangle of width \(a\), height \(b\), and mass \(M\), centered at the origin, \(I_x=Mb^2/12\) and \(I_y=Ma^2/12\). These are moments about the rectangle’s centroidal axes, which coincide with the coordinate axes in this placement.

The different dimensions in the two formulas are important. For \(I_x\), the relevant spread is the height \(b\), because vertical distance is measured from the \(x\)-axis. For \(I_y\), the relevant spread is the width \(a\).

Rectangles Not Centered at the Origin

Consider an axis-aligned rectangle with horizontal bounds \(x_0\le x\le x_1\) and vertical bounds \(y_0\le y\le y_1\). Its width is \(a=x_1-x_0\), its height is \(b=y_1-y_0\), and its centroid is at \((h,k)\), where \(h=(x_0+x_1)/2\) and \(k=(y_0+y_1)/2\). Symmetry about the lines \(x=h\) and \(y=k\) places the centroid there.

The centroidal axes are parallel to the coordinate axes but pass through \((h,k)\). The parallel axis theorem says that a lamina of mass \(M\), with centroidal moment \(I_{\mathrm{cent}}\), has moment \(I=I_{\mathrm{cent}}+Md^2\) about a parallel axis at distance \(d\) from the centroidal axis. It applies here because the compared axes are parallel and the centroidal moments are finite. Consequently, the coordinate-axis moments are as follows.

I_x=M\left(\frac{b^2}{12}+k^2\right), \qquad I_y=M\left(\frac{a^2}{12}+h^2\right).

For the \(x\)-axis, the parallel-axis distance is \(|k|\), the vertical distance from the centroid to that axis; for the \(y\)-axis, it is \(|h|\). Squaring makes the sign of either coordinate irrelevant.

Worked Example: A Rectangle in the First Quadrant

A uniform rectangle occupies \(0\le x\le6\) and \(0\le y\le4\), with total mass \(12\) kg. Find its moments about the coordinate axes and its centroidal moments.

The area is \(6\cdot4=24\ \text{m}^2\), so \(\sigma=12/24=0.5\ \text{kg}/\text{m}^2\). Directly integrating distance squared from the \(x\)-axis gives

\begin{aligned} I_x &=0.5\int_0^6\int_0^4y^2\,dy\,dx\\ &=0.5(6)\left[\frac{y^3}{3}\right]_0^4\\ &=3\left(\frac{64}{3}-0\right) =64\ \text{kg}\cdot\text{m}^2. \end{aligned}

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

\begin{aligned} I_y &=0.5\int_0^4\int_0^6x^2\,dx\,dy\\ &=0.5(4)\left[\frac{x^3}{3}\right]_0^6\\ &=2\left(\frac{216}{3}-0\right) =144\ \text{kg}\cdot\text{m}^2. \end{aligned}

The centroid is \((3,2)\). The centroidal moments are \(I_{x,\mathrm{cent}}=12(4^2)/12=16\ \text{kg}\cdot\text{m}^2\) and \(I_{y,\mathrm{cent}}=12(6^2)/12=36\ \text{kg}\cdot\text{m}^2\). Check with the parallel axis theorem: \(16+12(2^2)=64\) and \(36+12(3^2)=144\), agreeing with the direct integrals.

Uniform Disk: Derivation in Polar Coordinates

Let a uniform disk of radius \(R\) and mass \(M\) be centered at the origin. Its area is \(\pi R^2\), so its density is \(\sigma=M/(\pi R^2)\). A disk is symmetric about both coordinate axes and about every line through its center; its centroid is at the origin.

Use polar coordinates \(x=r\cos\theta\), \(y=r\sin\theta\), with area element \(dA=r\,dr\,d\theta\). The coordinate-axis moment integrands become \(y^2=r^2\sin^2\theta\) and \(x^2=r^2\cos^2\theta\). Thus, for the \(x\)-axis,

\begin{aligned} I_x &=\sigma\int_0^{2\pi}\int_0^R(r\sin\theta)^2r\,dr\,d\theta\\ &=\sigma\left(\int_0^Rr^3\,dr\right) \left(\int_0^{2\pi}\sin^2\theta\,d\theta\right)\\ &=\sigma\left(\frac{R^4}{4}\right)(\pi) =\frac{MR^2}{4}. \end{aligned}

The radial integral is \(R^4/4\), and the angular integral is \(\pi\), since \(\sin^2\theta=(1-\cos2\theta)/2\). For \(I_y\), the angular integral of \(\cos^2\theta\) over the same full revolution is also \(\pi\), so \(I_y=MR^2/4\) as well.

Result: For a uniform disk centered at the origin, \(I_x=I_y=MR^2/4\). Both coordinate axes pass through the disk’s centroid, so these are also its centroidal moments about axes in the plane.

The equality of the two moments follows from the disk’s symmetry: rotating the disk through a right angle exchanges the \(x\)- and \(y\)-axes without changing the mass distribution.

Worked Example: A Disk Centered at the Origin

A uniform disk has radius \(3\) m and mass \(18\) kg. Find its moments about the coordinate axes.

Its density is \(\sigma=18/(9\pi)=2/\pi\ \text{kg}/\text{m}^2\). Using the polar integral for \(I_x\),

\begin{aligned} I_x &=\frac{2}{\pi} \left(\int_0^3r^3\,dr\right) \left(\int_0^{2\pi}\sin^2\theta\,d\theta\right)\\ &=\frac{2}{\pi}\left(\frac{3^4}{4}\right)(\pi) =\frac{2}{\pi}\left(\frac{81}{4}\right)(\pi) =\frac{81}{2}\ \text{kg}\cdot\text{m}^2. \end{aligned}

For \(I_y\), replacing \(\sin^2\theta\) by \(\cos^2\theta\) leaves the angular integral equal to \(\pi\), so \(I_y=81/2\ \text{kg}\cdot\text{m}^2\). As a check, the formula \(MR^2/4\) gives \(18(9)/4=162/4=81/2\) for each axis.

Using Symmetry for a Translated Disk

If the center of a disk is at \((h,k)\), symmetry places its centroid there, so the centroidal axes are the lines \(x=h\) and \(y=k\). The disk’s centroidal moment about either in-plane axis is \(MR^2/4\). Applying the parallel axis theorem gives the moments about the coordinate axes.

I_x=M\left(\frac{R^2}{4}+k^2\right), \qquad I_y=M\left(\frac{R^2}{4}+h^2\right).

Worked Example: A Disk Away from the Origin

A uniform disk of radius \(2\) m and mass \(8\) kg has center \((2,-1)\). Find its coordinate-axis moments.

Symmetry gives the centroid at \((2,-1)\). The centroidal moment about either axis through the center is \(MR^2/4=8(2^2)/4=8\ \text{kg}\cdot\text{m}^2\). The distance from the centroid to the \(x\)-axis is \(1\) m, so \(I_x=8+8(1^2)=16\ \text{kg}\cdot\text{m}^2\). Its distance to the \(y\)-axis is \(2\) m, so \(I_y=8+8(2^2)=40\ \text{kg}\cdot\text{m}^2\).

Checking by the same formulas gives \(8(2^2/4+(-1)^2)=8(1+1)=16\) and \(8(2^2/4+2^2)=8(1+4)=40\), respectively.

A Common Mix-Up: Which Coordinate Is Squared?

A frequent error is to use \(x^2\) for the moment about the \(x\)-axis. That uses the distance to the \(y\)-axis instead. For the \(6\)-by-\(4\) rectangle of mass \(12\) kg above, incorrectly inserting the width into the \(x\)-axis formula gives \(12(6^2)/12=36\ \text{kg}\cdot\text{m}^2\). The correct moment is \(12(4^2)/12=16\ \text{kg}\cdot\text{m}^2\) about the centroidal \(x\)-axis, because its perpendicular distances are measured vertically and therefore depend on the height.

Keep the distance interpretation in view: \(I_x\) uses \(y^2\), while \(I_y\) uses \(x^2\). Also identify the axis before using a centroidal formula: a shape’s dimensions alone do not determine its moment about an axis displaced from its center.

Check Your Understanding

Use the definitions and results above to answer these questions.

For a uniform rectangle of width \(a\), height \(b\), and mass \(M\), which dimension appears in its centroidal moment about the \(x\)-axis?
Why does a disk centered at the origin have equal moments about the \(x\)- and \(y\)-axes?
A rectangle has centroid \((h,k)\). What distance is used in the parallel axis theorem when converting its centroidal \(x\)-axis moment to its moment about the coordinate \(x\)-axis?
What is the density of a uniform disk of mass \(M\) and radius \(R\)?
For a uniform disk of radius \(R\), what are its centroidal moments about axes through its center and parallel to the coordinate axes?