Moment of Inertia
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 5.6: “Calculating Centers of Mass and Moments of Inertia” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/5-6-calculating-centers-of-mass-and-moments-of-inertia |
| Textbook used in class | Stewart, Calculus, Section 15.5: “Applications of Double Integrals” (Examples 5, 6) |
Quick Reference
Moments of inertia: $$I_x = \iint_D y^2\,\rho(x,y)\,dA \quad \text{(about the }x\text{-axis)},$$ $$I_y = \iint_D x^2\,\rho(x,y)\,dA \quad \text{(about the }y\text{-axis)},$$ $$I_0 = I_x + I_y = \iint_D (x^2+y^2)\,\rho\,dA \quad \text{(polar moment, about origin)}.$$
Radius of gyration: $\tilde x = \sqrt{I_y/m}$, $\tilde y = \sqrt{I_x/m}$.
Motivation
The moment of inertia (also called the second moment) measures resistance to rotation: it appears in the formula for rotational kinetic energy $KE = \frac{1}{2}I\omega^2$ and in Newton’s second law for rotation $\tau = I\alpha$. A mass concentrated far from the axis contributes more to $I$ than the same mass near the axis, because it takes more force to accelerate it rotationally.
The formula uses $y^2$ (distance from the $x$-axis squared) or $x^2$ (distance from the $y$-axis squared), because angular effects depend on the lever arm.
Key Concepts
1. Difference from First Moments
First moments ($M_x, M_y$) use the distance to the axis ($y$ or $x$). Second moments (moments of inertia) use the squared distance ($y^2$ or $x^2$). The extra power of the distance reflects that rotation, unlike balance, depends on the square of the lever arm.
2. Polar Moment
$I_0 = I_x + I_y$ is the moment of inertia about the origin (or the $z$-axis for a 2D plate). It measures resistance to rotation about the $z$-axis and equals $\iint (x^2+y^2)\rho\,dA$, which is well suited to polar coordinates.
Worked Example
Find the moment of inertia about the $y$-axis for the unit disk $x^2+y^2 \leq 1$ with uniform density $\rho = 1$. (Stewart 15.5, Example 6.)
$$I_y = \iint_D x^2\,dA.$$
In polar: $x = r\cos\theta$, $x^2 = r^2\cos^2\theta$.
$$I_y = \int_0^{2\pi}\int_0^1 r^2\cos^2\theta\cdot r\,dr\,d\theta = \int_0^{2\pi}\cos^2\theta\,d\theta\cdot\int_0^1 r^3\,dr.$$
$$\int_0^{2\pi}\cos^2\theta\,d\theta = \pi, \qquad \int_0^1 r^3\,dr = \frac{1}{4}.$$
$$I_y = \pi\cdot\frac{1}{4} = \frac{\pi}{4}.$$
By symmetry, $I_x = \pi/4$ as well. $I_0 = \pi/2$.
The mass is $m = \pi(1)^2 = \pi$ (area times density). Radius of gyration: $\tilde x = \sqrt{I_y/m} = \sqrt{1/4} = 1/2$.
$I_x$ involves $x^2$ (because the subscript is $x$). The moment of inertia about the $x$-axis is $I_x = \iint y^2\rho\,dA$ -- it uses $y^2$ (the squared distance from the $x$-axis), not $x^2$. The subscript indicates the AXIS of rotation; the integrand uses the PERPENDICULAR distance to that axis. This is the same naming issue as for $M_x$ and $M_y$.
Leveled Practice
Problem 1. Find $I_0$ for the unit square $[0,1]\times[0,1]$ with uniform density $\rho = 1$.
Show answer
$I_0 = \iint(x^2+y^2)\,dA = \int_0^1\int_0^1(x^2+y^2)\,dy\,dx$.
$= \int_0^1\left[x^2 y + \frac{y^3}{3}\right]_0^1 dx = \int_0^1\left(x^2+\frac{1}{3}\right)dx = \left[\frac{x^3}{3}+\frac{x}{3}\right]_0^1 = \frac{2}{3}$.