Multivariable-Calculus · Unit 15 · Video 3 · Interactive Practice

Moment of Inertia: Spinning a Disk About Its Center and Its Rim

IKey Formulas

Formula	Name	What it says
$\tfrac12 m v^2 = \tfrac12 m r^2 \omega^2$ , so $I_0 = m r^2$	Point mass	With $v = r\omega$ , the coefficient of $\tfrac12\omega^2$ plays the role mass plays in $\tfrac12 m v^2$
$I_0 = \iint_R r^2\,\delta\,dA$ , $\ r^2 = x^2 + y^2$	Plate about the origin	Each piece $\Delta m \approx \delta\,\Delta A$ counts with its squared distance; $KE = \tfrac12 I_0\,\omega^2$
$I_x = \iint_R y^2\,\delta\,dA$	Plate about the $x$ -axis	$\lvert y\rvert$ is the distance from a point to the $x$ -axis
$\int_0^{2\pi}\!\int_0^a r^3\,dr\,d\theta = \frac{\pi a^4}{2}$ $\int_{-\pi/2}^{\pi/2}\!\int_0^{2a\cos\theta} r^3\,dr\,d\theta = \frac32\pi a^4$	Uniform disk ( $\delta = 1$ )	About the center; about a rim point, with the origin moved to that point

Key Insight: The ratio is $\frac32\pi a^4 \div \frac{\pi a^4}{2} = 3$ . Moving the rotation point to the rim puts most of the disk farther away, and every piece counts with the square of its distance.

IISquared Distance, Piece by Piece

Each cell of the unit disk contributes $(\text{distance to the axis})^2 \times \Delta A$ ; the sums approach the moment of inertia.

Rotation axis Polar cells 4 rings × 32

💡 In $\iint_R r^2\,dA$ the distance $r$ varies over $0 \le r \le a$ ; replacing it by $a$ gives $a^2 \cdot \pi a^2 = \pi a^4$ , twice the true $\frac{\pi a^4}{2}$ .

IIIBounds for the Rim Disk

From a rim point, the ray at angle $\theta$ leaves the disk at $r = 2a\cos\theta$ , and only $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ meets it.

Angle θ 35°

IVQuiz Questions

Problem 1 · Kinetic Energy of a Point Mass

Given: A mass $m = 2$ kg moves on a circle of radius $r = 3$ m about the origin with angular velocity $\omega = 4$ rad/s — find its kinetic energy $\tfrac12 I_0\,\omega^2$ .

✅ Correct!

I_0 = mr^2 = 18

and

\tfrac12 \cdot 18 \cdot 4^2 = 144

J, the same as

\tfrac12 m v^2

with

v = r\omega = 12

m/s.

❌ Square the angular velocity.

\tfrac12 \cdot 18 \cdot 4 = 36

uses

\omega

, not

\omega^2

; the energy is

\tfrac12 I_0\,\omega^2

, just as it is

\tfrac12 m v^2

❌ The distance enters squared.

I_0 = m r = 6

gives

\tfrac12 \cdot 6 \cdot 16 = 48

. From

v = r\omega

\tfrac12 m v^2 = \tfrac12 m r^2 \omega^2

, so

I_0 = m r^2 = 18

❌ Keep the factor $\tfrac12$ .

I_0\,\omega^2 = 18 \cdot 16 = 288

is twice the kinetic energy

\tfrac12 I_0\,\omega^2

Show solution

The moment of inertia of a point mass about the origin is

I_0 = m r^2 = 2 \cdot 3^2 = 18 \ \text{kg}\cdot\text{m}^2.

The rotational kinetic energy is

KE = \tfrac12 I_0\,\omega^2 = \tfrac12 \cdot 18 \cdot 4^2 = 144 \ \text{J}.

Check with translation: $v = r\omega = 3 \cdot 4 = 12$ m/s and $\tfrac12 m v^2 = \tfrac12 \cdot 2 \cdot 144 = 144$ J.

Problem 2 · Disk About Its Center

Given: A uniform disk of radius $a = 2$ with density $\delta = 1$ , centered at the origin — find $I_0 = \iint_R r^2\,dA$ .

✅ Correct!

\frac{\pi a^4}{2} = \frac{16\pi}{2} = 8\pi

❌ The area element is $r\,dr\,d\theta$ .

\int_0^{2\pi}\!\int_0^2 r^2\,dr\,d\theta = \frac{16\pi}{3}

drops the factor

r

from

dA

; the integrand must be

r^2 \cdot r = r^3

❌ $r$ is not the radius $a$ . Plugging in

r = a

gives

a^2 \cdot \pi a^2 = 16\pi

. Here

r

is the distance from the center to a point of the disk; it ranges from

0

a

and is integrated.

❌ The outer integral is missing.

\int_0^2 r^3\,dr = 4

is only the inner integral; integrating the constant

4

over

0 \le \theta \le 2\pi

multiplies it by

2\pi

Show solution

In polar coordinates the disk is $0 \le r \le 2$ , $0 \le \theta \le 2\pi$ , and $dA = r\,dr\,d\theta$ :

I_0 = \int_0^{2\pi}\!\int_0^{2} r^2 \cdot r\,dr\,d\theta.

Inner integral: $\left[\frac{r^4}{4}\right]_0^2 = \frac{16}{4} = 4$ . It does not depend on $\theta$ , so

I_0 = 2\pi \cdot 4 = 8\pi,

which matches $\frac{\pi a^4}{2}$ with $a = 2$ .

Problem 3 · Disk About a Rim Point

Given: A uniform disk of radius $a$ , $\delta = 1$ , placed with a point of its circle at the origin and its diameter along the positive $x$ -axis, so it lies to the right of the $y$ -axis — find the setup and the value of $I_0 = \iint_R r^2\,dA$ .

Which iterated integral equals I₀?

What is its value?

✅ Correct!

I_0 = 4a^4 \cdot \frac{3\pi}{8} = \frac32\pi a^4

, three times the

\frac{\pi a^4}{2}

about the center.

❌ Not every direction meets the disk. A ray pointing into the left half-plane never enters it, so

\theta

runs only from

-\frac{\pi}{2}

\frac{\pi}{2}

❌ The ray exits before $r = 2a$ . Only the ray along the diameter reaches

2a

; the ray at angle

\theta

leaves the disk at

r = 2a\cos\theta

, where the inscribed right angle sits.

❌ The area element is $r\,dr\,d\theta$ . The integrand

r^2

times

dA = r\,dr\,d\theta

gives

r^3\,dr\,d\theta

❌ Check the setup. Fix

\theta

, find where the ray enters and leaves the disk, then find which

\theta

hit it at all.

❌ That integrates over $0 \le \theta \le 2\pi$ .

4a^4\int_0^{2\pi}\cos^4\theta\,d\theta = 4a^4 \cdot \frac{3\pi}{4} = 3\pi a^4

covers the disk twice; the integral runs over

-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}

only.

❌ That is only half the interval.

4a^4 \cdot \frac{3\pi}{16} = \frac34\pi a^4

integrates

\cos^4\theta

from

0

\frac{\pi}{2}

. Since

\cos^4\theta

is even, the integral over

-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}

is twice that:

\frac{3\pi}{8}

❌ Raise the whole $2a\cos\theta$ to the fourth power.

\frac{(2a\cos\theta)^4}{4} = \frac{16a^4\cos^4\theta}{4} = 4a^4\cos^4\theta

; dropping the

16/4 = 4

gives

\frac38\pi a^4

❌ Check the evaluation. The inner integral is

4a^4\cos^4\theta

and

\int_{-\pi/2}^{\pi/2}\cos^4\theta\,d\theta = \frac{3\pi}{8}

Show solution

Bounds for $r$ . The ray at angle $\theta$ enters at $r = 0$ . Join its exit point to the far end $(2a, 0)$ of the diameter: the angle there is inscribed in a semicircle, so it is a right angle, and in this right triangle with hypotenuse $2a$ the side next to $\theta$ is $2a\cos\theta$ . So $0 \le r \le 2a\cos\theta$ .

Bounds for $\theta$ . Rays into the left half-plane miss the disk; the circle is tangent to the $y$ -axis at the origin, so $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$ .

I_0 = \int_{-\pi/2}^{\pi/2}\!\int_0^{2a\cos\theta} r^3\,dr\,d\theta.

Inner: $\left[\frac{r^4}{4}\right]_0^{2a\cos\theta} = \frac{16a^4\cos^4\theta}{4} = 4a^4\cos^4\theta$ .

Outer: from $\cos^4\theta = \frac38 + \frac12\cos 2\theta + \frac18\cos 4\theta$ , $\int_0^{\pi/2}\cos^4\theta\,d\theta = \frac{3\pi}{16}$ , and evenness doubles it:

I_0 = 4a^4 \cdot 2 \cdot \frac{3\pi}{16} = 4a^4 \cdot \frac{3\pi}{8} = \frac32\pi a^4.

Problem 4 · Disk About a Diameter

Given: A uniform disk of radius $a$ , $\delta = 1$ , centered at the origin, spun about the $x$ -axis — find $I_x = \iint_R y^2\,dA$ .

✅ Correct!

\int_0^{2\pi}\!\int_0^a r^3\sin^2\theta\,dr\,d\theta = \frac{a^4}{4} \cdot \pi = \frac{\pi a^4}{4}

, half of

I_0

about the center.

❌ That is the moment about the center. Spinning about the

x

-axis, a point's distance to the axis is

\lvert y\rvert

, not

r

; the integrand is

y^2 = r^2\sin^2\theta

❌ The area element is $r\,dr\,d\theta$ .

\int_0^{2\pi}\!\int_0^a r^2\sin^2\theta\,dr\,d\theta = \frac{\pi a^3}{3}

leaves out the factor

r

dA

❌ Square the distance.

\iint_R \lvert y\rvert\,dA = \frac{4a^3}{3}

uses the distance

\lvert y\rvert

itself; each piece counts with

y^2

Show solution

The distance from $(x, y)$ to the $x$ -axis is $\lvert y\rvert$ , so the integrand is $y^2 = r^2\sin^2\theta$ . With $dA = r\,dr\,d\theta$ :

I_x = \int_0^{2\pi}\!\int_0^a r^3\sin^2\theta\,dr\,d\theta = \frac{a^4}{4}\int_0^{2\pi}\sin^2\theta\,d\theta.

Using $\sin^2\theta = \frac12 - \frac12\cos 2\theta$ , $\int_0^{2\pi}\sin^2\theta\,d\theta = \pi$ , so

I_x = \frac{\pi a^4}{4}.

Solved: 0 / 4