Polar Moment Of Inertia

From Coordinate-Axis Moments to the Polar Moment

A planar lamina can rotate about different axes, and the moment of inertia depends on the axis chosen. The moments $I_x$ and $I_y$ measure how mass is distributed relative to the $x$- and $y$-axes in the plane. The polar moment of inertia about the origin measures the distribution relative to the axis through the origin and perpendicular to the plane.

For a small piece of mass, its contribution to a moment of inertia is its mass multiplied by the square of its distance from the axis. For the perpendicular axis through the origin, that distance is the distance from the point to the origin. This gives a direct link between the polar moment and the two coordinate-axis moments.

Definition: For a lamina occupying a planar region $R$, with nonnegative mass density $\rho(x,y)$, the polar moment of inertia about the origin is $$ J=\iint_R (x^2+y^2)\rho(x,y)\,dA. $$ Here $x^2+y^2$ is the squared distance from $(x,y)$ to the origin, and $\rho(x,y)\,dA$ is the mass of a small area element.

The definition assumes the integral exists and is finite. The density is mass per unit area, so if lengths are measured in metres and mass in kilograms, $J$ has units of kilogram-metres squared. A finite region with bounded density is a common setting in which the integral is finite.

Why the Polar Moment Is the Sum of Two Moments

The moment about the $x$-axis uses squared distance $y^2$, since a point $(x,y)$ is a distance $|y|$ from that axis. The moment about the $y$-axis uses squared distance $x^2$. Thus

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

Adding these integrals adds the two squared coordinate distances at each point. Since $x^2+y^2$ is the squared distance to the origin, this proves the relationship.

Theorem: For a planar lamina whose coordinate-axis moments are finite, the polar moment about the origin satisfies $$ J=I_x+I_y. $$

This identity is specific to the polar axis through the origin, perpendicular to the plane. It follows from adding the moments about the two axes in the plane; it does not require the lamina or its density to have any symmetry.

Writing the Moment in Polar Coordinates

In polar coordinates, $x=r\cos\theta$, $y=r\sin\theta$, and $x^2+y^2=r^2$. The area element is $dA=r\,dr\,d\theta$, where the factor $r$ is the Jacobian of the polar coordinate transformation. If the region has polar bounds $a\le r\le b$ and $\alpha\le\theta\le\beta$, then

J=\int_\alpha^\beta\int_a^b r^2\rho(r,\theta)\,r\,dr\,d\theta =\int_\alpha^\beta\int_a^b r^3\rho(r,\theta)\,dr\,d\theta.

The density in this expression is the original density written in polar coordinates: $\rho(r,\theta)=\rho(r\cos\theta,r\sin\theta)$. For regions whose polar bounds vary with angle, the radial limits must be replaced by the appropriate functions of $\theta$. The integrand still contains $r^3\rho(r,\theta)$.

Key formula: In polar coordinates, the squared distance contributes $r^2$, and the area element contributes another factor $r$. Therefore the polar moment integrand is $r^3\rho(r,\theta)$, not $r^2\rho(r,\theta)$.

Worked Example: A Uniform Quarter-Disk Sector

A quarter-disk of radius $2$ occupies $0\le r\le2$, $0\le\theta\le\pi/2$, with constant density $\rho(r,\theta)=\sigma$. Find its polar moment and verify it using $I_x+I_y$.

Use the polar moment integral:

\begin{aligned} J &=\int_0^{\pi/2}\int_0^2 \sigma r^3\,dr\,d\theta\\ &=\sigma\int_0^{\pi/2}\left[\frac{r^4}{4}\right]_0^2\,d\theta\\ &=\sigma\int_0^{\pi/2}\left(\frac{16}{4}-0\right)\,d\theta\\ &=\sigma\int_0^{\pi/2}4\,d\theta\\ &=4\sigma\left(\frac{\pi}{2}-0\right)\\ &=2\pi\sigma. \end{aligned}

For a check, calculate the two coordinate-axis moments. Since $y=r\sin\theta$ and $x=r\cos\theta$,

\begin{aligned} I_x &=\sigma\int_0^{\pi/2}\sin^2\theta\,d\theta \int_0^2 r^3\,dr\\ &=\sigma\left(\frac{\pi}{4}\right)(4) =\pi\sigma, \\[4pt] I_y &=\sigma\int_0^{\pi/2}\cos^2\theta\,d\theta \int_0^2 r^3\,dr\\ &=\sigma\left(\frac{\pi}{4}\right)(4) =\pi\sigma. \end{aligned}

The radial integral is $4$ in both calculations, and each angular integral is $\pi/4$. Therefore $I_x+I_y=\pi\sigma+\pi\sigma=2\pi\sigma=J$, as required. The result has units of density multiplied by length to the fourth power.

Radial Symmetry Simplifies the Integral

A lamina is rotationally symmetric about the origin when rotating it about the origin leaves both its region and its density unchanged. In polar coordinates, such a density depends only on $r$, so it can be written $\rho(r)$. If the region consists of complete circles, rings, or a union of full radial intervals, its angular range covers a full turn. The angle integral then contributes $2\pi$.

Radial symmetry formula: For a rotationally symmetric region with radial intervals $a\le r\le b$ and radial density $\rho(r)$, $$ J=2\pi\int_a^b r^3\rho(r)\,dr. $$ If there are several disjoint radial intervals, compute this integral on each interval and add the results.

The full-turn factor is not valid merely because the density depends on $r$. The region must also include every angle for the radial intervals being integrated. For example, an annular sector with radial density still requires integration over its actual angular range, not automatically over $0\le\theta\le2\pi$.

Worked Example: A Radially Dense Annulus

An annulus has inner radius $1$ and outer radius $3$, and its density is $\rho(r)=kr$, where $k$ is a positive constant. Find its polar moment about the origin.

The annulus contains a full circle at every radius from $1$ to $3$, so use the radial symmetry formula:

\begin{aligned} J &=2\pi\int_1^3 r^3(kr)\,dr\\ &=2\pi k\int_1^3 r^4\,dr\\ &=2\pi k\left[\frac{r^5}{5}\right]_1^3\\ &=2\pi k\left(\frac{3^5}{5}-\frac{1^5}{5}\right)\\ &=2\pi k\left(\frac{243-1}{5}\right)\\ &=\frac{484\pi k}{5}. \end{aligned}

The antiderivative checks because $\frac{d}{dr}(r^5/5)=r^4$. Also, $J$ is positive: $k>0$, and the integrated quantity $r^4$ is positive throughout the annulus. The $r$ in the density makes the radial integrand $kr^4$, in addition to the factor $2\pi$ from the full angular range.

Worked Example: A Disk with Density Decreasing Outward

A disk of radius $a$ has density $\rho(r)=\rho_0(1-r^2/a^2)$, where $a>0$ and $\rho_0>0$. Find its polar moment about the origin.

The density depends only on $r$, and the region includes every angle for $0\le r\le a$. Apply the radial symmetry formula:

\begin{aligned} J &=2\pi\int_0^a r^3\rho_0\left(1-\frac{r^2}{a^2}\right)\,dr\\ &=2\pi\rho_0\int_0^a\left(r^3-\frac{r^5}{a^2}\right)\,dr\\ &=2\pi\rho_0\left[\frac{r^4}{4}-\frac{r^6}{6a^2}\right]_0^a\\ &=2\pi\rho_0\left(\frac{a^4}{4}-\frac{a^6}{6a^2}\right)\\ &=2\pi\rho_0\left(\frac{a^4}{4}-\frac{a^4}{6}\right)\\ &=2\pi\rho_0\left(\frac{a^4}{12}\right)\\ &=\frac{\pi\rho_0a^4}{6}. \end{aligned}

Differentiating the antiderivative gives $r^3-r^5/a^2$, the radial expression being integrated. At $r=0$ the density is $\rho_0$; at $r=a$ it is $0$. It remains nonnegative throughout the disk, consistent with a nonnegative mass density and a positive moment.

A Common Error: Omitting the Area Factor

A frequent mistake is to replace $x^2+y^2$ by $r^2$ but then use $dA=dr\,d\theta$. That omits the Jacobian factor $r$, so the resulting integral does not represent the area integral.

Common Mistake: Forgetting the Polar Jacobian

For the uniform quarter-disk of radius $2$, the incorrect calculation would be

J_{\text{wrong}} =\sigma\int_0^{\pi/2}\int_0^2 r^2\,dr\,d\theta =\sigma\left(\frac{\pi}{2}\right)\left[\frac{r^3}{3}\right]_0^2 =\frac{4\pi\sigma}{3}.

This disagrees with the correctly computed value $2\pi\sigma$. The error is that the Cartesian area element transforms as $dA=r\,dr\,d\theta$, not $dr\,d\theta$. Including that factor gives

J =\sigma\int_0^{\pi/2}\int_0^2 r^2\underbrace{r}_{\text{area factor}}\,dr\,d\theta =\sigma\left(\frac{\pi}{2}\right)\left[\frac{r^4}{4}\right]_0^2 =2\pi\sigma.

The corrected answer also agrees with the independent check $I_x+I_y=\pi\sigma+\pi\sigma=2\pi\sigma$.

Using the Polar Moment Correctly

The polar moment combines the two coordinate-axis moments into one measure of how mass is distributed away from the origin. Mass farther from the origin contributes more strongly because its squared distance is larger. This explains the $r^2$ in the definition; the additional factor $r$ in polar coordinates comes from the area element, not from the distance.

Before evaluating a polar moment, check three things: that the axis is the perpendicular axis through the origin, that the bounds describe the entire region being integrated, and that the density is expressed in the same coordinates as the integrand. Use the full-turn simplification only when both the density and region have the needed rotational symmetry. These checks help distinguish the general formula from its especially convenient radial form.

Check Your Understanding

Use the definitions and formulas above to answer these questions.

Why does the polar moment about the origin equal $I_x+I_y$?
What factor from the polar area element changes $r^2\rho(r,\theta)$ into the polar-coordinate integrand?
For a full annulus with radial density $\rho(r)$ on $a\le r\le b$, write the formula for $J$.
If the density depends only on $r$ but the region is only a sector, why can the angular factor not automatically be set to $2\pi$?
For a full disk of radius $a$ with constant density $\rho_0$, set up the polar integral for $J$, without evaluating it.

What You'll Learn

From Coordinate-Axis Moments to the Polar Moment

Why the Polar Moment Is the Sum of Two Moments

Writing the Moment in Polar Coordinates

Worked Example: A Uniform Quarter-Disk Sector

Worked Example: A Uniform Quarter-Disk Sector

Radial Symmetry Simplifies the Integral

Worked Example: A Radially Dense Annulus

Worked Example: A Radially Dense Annulus

Worked Example: A Disk with Density Decreasing Outward

Worked Example: A Disk with Density Decreasing Outward

A Common Error: Omitting the Area Factor

Common Mistake: Forgetting the Polar Jacobian

Using the Polar Moment Correctly

Check Your Understanding

Previous

Next