Moment of Inertia Lab

Moment of inertia is how hard it is to change how fast something spins. Drag a ball on an arm, turn a 3D body in your hands, watch a double or triple integral add itself up one piece at a time, and see what happens when the axis is not through the center.

Drag the ball · drag any 3D view to turn it · click a dot to inspect it · every slider has a number box

I1.000kg·m² total

I_CM1.000spin about the center

Md²0.000from moving the axis

∫∫∫ doubleintegral this body needs

1 · What is moment of inertia?

start here

Moment of inertia, written I, measures how hard it is to change how fast something spins. It is the spinning version of mass.

Mass resists a push

Push a cart with a force F and it speeds up with acceleration a. A heavier cart speeds up less. It does not matter where the load sits in the cart.

$F = m a$

Moment of inertia resists a twist

Twist a wheel with a torque τ and its spin speeds up with angular acceleration α. Larger I, slower spin-up. Here it does matter where the mass sits.

$τ = I α$

Drag the ball, or focus the picture and use the arrow keys.

One ball on a light arm

The arm weighs nothing. You are looking straight down the axis. The gold square has side r, so its area is r². The ball’s size grows with its mass.

Distance from the axis r

Ball mass m

I = m × r²0.250 kg·m²

Twisting applies the same 1 N·m torque to your ball and to the faint starting ball (1 kg at 0.50 m).

Why distance counts twice (r × r)

Farther away means faster. In one turn a ball at distance r travels a circle of length 2πr. To speed up the spin at rate α, the ball must speed up at a = r·α. Twice as far needs twice the push: F = m·r·α.
A twist works through a lever. Torque is force × distance from the axis: τ = r·F = r·(m·r·α) = (m r²)·α.

Compare τ = (m r²)·α with F = m·a: m r² plays the role of mass. That is the moment of inertia of one small ball.

One ball

I = m r^{2}

Mass times distance squared.

Many balls

I = \sum_{i} m_{i} r_{i}^{2}

Every ball adds its own m r². Inertia simply adds up.

A solid body

I = \int r^{2} d m

Chop the body into tiny pieces and add them all. Section 3 shows this happening.

Words you will see

Axis: The fixed line the body spins around (copper in every picture).
r⊥: A piece’s distance from the axis, measured straight out at a right angle to it.
Torque τ: A twist: force times its distance from the axis, in N·m.
α: Angular acceleration: how fast the spin speeds up, in rad/s².
CM: Center of mass: the balance point, the average position of the mass.

Experiments

Body

Shape

Total mass M

Radius R

Uniform surface density. Disk thickness is neglected.

Rotation axis

Axis tilt β (degrees)

Move the axis off-center d

View

Camera

2 · Every dot is a piece of mass

Solid disk

small r⊥²Δmlarge r⊥²Δmrotation axisdot area ∝ its mass Δm

What this body tells you

One dot, explained ΔI = Δm × r⊥²

Inspect piece1 / 128

Mass Δm
Distance r⊥
Adds ΔI

Where does the inertia come from?

Table of the bands

Shares of mass and inertia by distance from the axis
r⊥ band (m)	Mass	Inertia

3 · Build the integral, one sum inside another

2D surface · double ∫∫

An integral sign ∫ means “add up”. You need one ∫ for each number it takes to say where a tiny piece is. Each ∫ is a sum, and the sums sit inside each other: first add the pieces along a ring, then add the rings into a slice, then stack the slices.

Speed

Piece size

Smaller pieces: the sum becomes the integral

Piece size	Pieces	Sum Σ r⊥²Δm	Off by

With infinitely many, infinitely small pieces, every Σ turns into an ∫ and the sum equals the exact integral. That limit is all an integral is.

How many sums? Count the numbers that locate a piece

Line · rod, hoop1 number (x, or angle φ)∫ single

Surface · disk, ring, plate2 numbers (x, y or r, φ)∫∫ double

Solid · cylinder, sphere3 numbers (x, y, z)∫∫∫ triple

“Thin” bodies are idealized: a hoop or disk is treated as having no thickness, which removes one dimension.

Every shape in this lab. Choose one to load it.
Shape	A piece is located by	Tiny mass dm	Integral	Symmetry shortcut

Why do textbooks often show fewer ∫ signs? Symmetry. About a disk’s center axis, every piece in a thin ring has the same distance r, so the sum around the ring is easy: a ring of radius r and width dr has mass σ·2πr·dr. That finishes the inner sum in your head. The integral is still double; the shortcut just does one sum for you. Watch the builder: each completed ring is exactly that shortcut.

4 · Same twist race

α = τ / I

Give your body and a reference body the same torque for the same time, both starting at rest. The one with the smaller I spins up faster.

Torque τ (N·m)

Reference body

Your body

Reference

Your body

I (kg·m²)
α (rad/s²)
ω (rad/s)
Angle (rad)

Reference

I (kg·m²)
α (rad/s²)
ω (rad/s)
Angle (rad)

Scrub time t (s)0.00 / 2.00 s

Spin speed over time slope = α

Ready. Both bodies start at rest.

Rigid bodies on fixed axes, no friction. The gold dot is the point farthest from the axis; with an off-center axis the dashed circle is the path of the center of mass. Drag either view to turn both.

5 · What if the axis isn’t through the center?

I = I_CM + M d²

Doors spin about their hinges, pendulums about their pivot, a bat about the hands. Spinning about an axis a distance d from the center of mass makes two motions happen at once:

① The center goes around

The center of mass travels around the axis on a circle of radius d. For this motion the body acts like one ball of mass M at distance d, which costs M d² (step 1).

② The body turns about its own center

At the same time it rotates about its own center, exactly as on the centered axis. That costs I_CM, the value of the integral.

Looking down the axis animated

Copper ⊙: the axis. Dashed circle: the path of the center. The arrow on the body shows that it also turns about its own center.

I for every axis position drag the dot

The curve is lowest at d = 0 and rises as a parabola. The gold bar is the extra M d².

Axis distance from the center d (m)

② Spin about own center I_CM
① Center going around M d²
Total I

The center is the easiest axis

M d² is never negative, so of all parallel axes the one through the CM has the smallest I.

Left or right doesn’t matter

d and −d give the same I, because d is squared.

Example: a rod about its end

ML²/12 + M(L/2)² = ML²/3: four times harder to spin than about its middle.

Why the two parts simply add, with nothing in between

Measure each piece from the CM: its position is r′, and the axis sits at d. Its squared distance to the axis is |r′ − d|² = r′² − 2 r′·d + d². Adding over all pieces: Σ m r′² = I_CM, Σ m d² = M d², and the middle term is −2 d·Σ m r′ = 0, because Σ m r′ = 0 is exactly what “center of mass” means. Nothing else is left.

The math for this body, step by step

Surface density

Add up every piece about z

Full integral, one ∫ per coordinate:

Using symmetry:

Turn the axis

Move the axis

What distance is actually squared?

The axis points along n = (sin β, 0, cos β) and is shifted by d(cos β, 0, −sin β). Distance is measured perpendicular to this line, not to a single point on it.

How to use

Drag the ball in step 1, any 3D view (it turns all of them), and the dot on the I(d) curve. Click a dot to inspect that piece.
Experiments load a body and a reference and say what to look for. They stay active while you follow them.
Build the integral: Play, or step one piece, one ring or one slice at a time. The colored boxes are the nested sums.
Same twist race: pin any body as the reference, change yours, and race them again.
Every slider has a number box for exact values.

Learn with this tool

Learning objectives

Explain moment of inertia as the spinning version of mass, and why distance enters squared.
Tell how many integrals a body needs, and read a double or triple integral as sums inside sums.
Use I = I_CM + M d² for an axis that does not pass through the center.

Prerequisites

Newton’s second law F = m a.
Area and volume of simple shapes.
Helpful, not required: what an integral means as an area under a curve.

Model

Uniform rigid bodies spin about a fixed axis. Every tiny piece contributes its mass times its squared perpendicular distance from the axis, and the moment of inertia is the sum: I = ∫ r⊥² dm. Under a constant torque τ from rest, α = τ/I, ω = αt and θ = ½αt².

M: total mass (kg)
r⊥: perpendicular distance from the axis (m)
λ, σ, ρ: mass per metre, per square metre, per cubic metre
I_CM: moment of inertia about a parallel axis through the center of mass
d: distance between the axis and the center of mass
β: axis tilt from z towards x

Exercise 1 — twice as far

In step 1, the ball sits 0.50 m from the axis. If you move it to 1.00 m, how many times larger is I?
Drag the ball to r = 1.00 m.
Read I and press Twist both.
Why does the ball at 1.00 m turn only a quarter as far in 2 s?

Show answer

4×. I = m r², so doubling r multiplies I by 2² = 4, and α = τ/I is four times smaller.

Exercise 2 — disk or hoop?

Same 2 kg, same 1 m radius. Which spins up faster: a solid disk or a thin hoop?
Choose the experiment “Disk vs. hoop” and apply the twist.
Read both values of I and α.
Use “Where does the inertia come from?” to explain the difference.

Show answer

The disk: I = MR²/2 = 1 kg·m² versus MR² = 2 kg·m² for the hoop. All of the hoop’s mass is at the rim, the farthest place.

Exercise 3 — a double integral by hand

Choose “Build a plate” (4 × 4 pieces). How many strips will the outer sum add, and how many pieces are in each strip?
Press “Finish strip” four times.
Write down each strip’s total from the inner box.
Why do the outer strips contribute more than the middle ones?

Show answer

4 strips of 4 pieces. The outer strips sit farther from the axis, so their pieces have larger r⊥². Adding the four strip totals gives the full sum.

Exercise 4 — a rod about its end (limiting case)

A 2 kg, 2 m rod has I = ML²/12 about its middle. Predict I about one end.
Choose “Center vs. end”, or drag d to 1.00 m in step 5.
Read I_CM, M d² and the total.
Now set the tilt to 90°, so the axis runs along the rod. Why is I zero?

Show answer

0.667 + 2 × 1² = 2.667 kg·m² = ML²/3. Along its own line an ideal rod has every piece on the axis, so r⊥ = 0 for all of them; a real rod has a small nonzero I from its thickness.

Worked example: a solid cylinder about its axis as a triple integral

Locate a piece with three numbers: height z, radius r and angle φ. Its volume is r dr dφ dz, so dm = ρ r dr dφ dz with ρ = M/(πR²H).
Innermost sum, around a ring: every piece has the same r, so the angle sum is $\int_{0}^{2 π} d φ = 2 π$ .
Middle sum, rings into a slice: $\int_{0}^{R} r^{2} ρ 2 π r d r = ρ 2 π \frac{R^{4}}{4}$ .
Outer sum, stacking slices: $\int d z = H$ , so $I = ρ 2 π H \frac{R^{4}}{4} = \frac{M}{π R^{2} H} \cdot \frac{2 π H R^{4}}{4} = \frac{M R^{2}}{2}$ .
Check in the builder: choose the solid cylinder; the three nested boxes are exactly these three sums.

When the model fails

Thickness: “thin” rods, disks and hoops have zero thickness here; real ones have slightly larger I. Non-uniform bodies: density is uniform; a wheel with a heavy rim needs a density that changes with r. Non-rigid bodies: a skater pulling in their arms changes I while spinning. Free rotation: a body that is not held on a fixed axis can wobble; that needs the full inertia tensor.

References

OpenStax, University Physics Volume 1: 10.5 Calculating moments of inertia and 10.7 Newton’s second law for rotation.

Formula sheet: every shape, and where it comes from

I = ∫ r⊥² dm

Every formula below comes from the same recipe. 1. Cut the body into tiny pieces and write the mass of one piece, dm. 2. Write that piece’s distance from the axis, r⊥. 3. Put both into I = ∫ r⊥² dm and add up over the whole body. Use one ∫ for each number that locates a piece.

How do we find λ, σ and ρ?

For a uniform body, equal lengths, areas or volumes contain equal masses. The rule is density = total mass ÷ total length, area or volume occupied by the material. A tiny piece then has dm = density × the piece’s length, area or volume.

Along a line: λ (lambda)

Mass per metre, in kg/m.

λ = M / total length
dm = λ ds

ds is a tiny length. For a rod the total length is L; for a hoop it is the circumference 2πR.

Across a surface: σ (sigma)

Mass per square metre, in kg/m².

σ = M / total area
dm = σ dA

dA is a tiny area. A disk has area πR²; a plate has area LW. For a ring, subtract the hole: π(R² − a²).

Throughout a solid: ρ (rho)

Mass per cubic metre, in kg/m³.

ρ = M / total volume
dm = ρ dV

dV is a tiny volume. A cylinder has volume πR²H; a solid sphere has volume 4πR³/3.

Why divide? All the pieces must add back to the total mass: M = ∫ dm. For a uniform rod, M = ∫ λ dx = λL, so λ = M/L. For a hoop, M = λ(2πR), so λ = M/(2πR). A tiny arc has length ds = R dφ (φ in radians), giving dm = λR dφ = M/(2π) dφ.

Check the units: (kg/m) × m = kg, just as (kg/m²) × m² = kg and (kg/m³) × m³ = kg. Moving or tilting the rotation axis changes r⊥, but keeps the body’s density and dm the same. If density varies with position, use its local value inside the integral; dividing total mass by total size gives only the average.

M is the total mass. R, L, W and H are the body’s radius, length, width and height; a is the ring’s inner radius. φ and θ are angles in radians. “Load” sets the lab to that body and axis with M = 2 kg, R = 1 m, L = 2 m, W = 1 m and H = 2 m, so you can check the number in the stats bar.

Moments of inertia of uniform bodies, derived from I = ∫ r⊥² dm
Body and axis	I	What you plug in	The integral, worked out	Try it
Point massat distance rno ∫	$I = m r^{2}$	All the mass m sits at one distance r, so there is nothing to add up.	$I = m r^{2}$	not in the lab
Thin rod, length Lthrough its center, perpendicular to it∫ single	$I = \frac{M L^{2}}{12}$	Length = L ⇒ λ = M/L (kg/m).Piece length dx ⇒ dm = λ dx; r⊥ = \|x\|.	$\int_{- \frac{L}{2}}^{\frac{L}{2}} x^{2} \frac{M}{L} d x = \frac{M}{L} \cdot \frac{L^{3}}{12} = \frac{M L^{2}}{12}$
Thin rod, length Lthrough one end, perpendicular to it∫ single	$I = \frac{M L^{2}}{3}$	Length = L ⇒ λ = M/L (kg/m).Piece length dx ⇒ dm = λ dx; r⊥ = x, with x from 0 to L.	$\int_{0}^{L} x^{2} \frac{M}{L} d x = \frac{M}{L} \cdot \frac{L^{3}}{3} = \frac{M L^{2}}{3}$
Thin hoop, radius Rthrough its center, perpendicular to its plane∫ single	$I = M R^{2}$	Circumference = 2πR ⇒ λ = M/(2πR) (kg/m).Arc length R dφ ⇒ dm = λ R dφ; every piece has r⊥ = R.	$\int_{0}^{2 π} R^{2} \frac{M}{2 π} d φ = M R^{2}$
Thin hoop, radius Ralong a diameter∫ single	$I = \frac{M R^{2}}{2}$	Circumference = 2πR ⇒ λ = M/(2πR) (kg/m).Arc length R dφ ⇒ dm = λ R dφ; r⊥ = R \|sin φ\|.	$\int_{0}^{2 π} R^{2} \sin^{2} φ \frac{M}{2 π} d φ = \frac{M R^{2}}{2}$
Solid disk, radius Rthrough its center, perpendicular to its face∫∫ double	$I = \frac{M R^{2}}{2}$	Area = πR² ⇒ σ = M/(πR²) (kg/m²).Patch area r dr dφ ⇒ dm = σ r dr dφ; r⊥ = r.	$\int_{0}^{2 π} \int_{0}^{R} r^{2} σ r d r d φ = 2 π σ \frac{R^{4}}{4} = \frac{M R^{2}}{2}$
Solid disk, radius Ralong a diameter∫∫ double	$I = \frac{M R^{2}}{4}$	Area = πR² ⇒ σ = M/(πR²) (kg/m²).Patch area r dr dφ ⇒ dm = σ r dr dφ; r⊥ = r \|sin φ\|.	$\int_{0}^{2 π} \int_{0}^{R} r^{2} \sin^{2} φ σ r d r d φ = π σ \frac{R^{4}}{4} = \frac{M R^{2}}{4}$
Ring (annulus), radii a to Rthrough its center, perpendicular to its face∫∫ double	$I = \frac{M (R^{2} + a^{2})}{2}$	Material area = πR² − πa² ⇒ σ = M/[π(R² − a²)] (kg/m²).Patch area r dr dφ ⇒ dm = σ r dr dφ; r⊥ = r, from a to R. For a = R, use the thin-hoop model above.	$\int_{0}^{2 π} \int_{a}^{R} r^{2} σ r d r d φ = \frac{π σ (R^{4} - a^{4})}{2} = \frac{M (R^{2} + a^{2})}{2}$
Rectangular plate, L × Wthrough its center, perpendicular to its face∫∫ double	$I = \frac{M (L^{2} + W^{2})}{12}$	Area = LW ⇒ σ = M/(LW) (kg/m²).Patch area dx dy ⇒ dm = σ dx dy; r⊥² = x² + y².	$\int_{- \frac{L}{2}}^{\frac{L}{2}} \int_{- \frac{W}{2}}^{\frac{W}{2}} (x^{2} + y^{2}) σ d y d x = \frac{M (L^{2} + W^{2})}{12}$
Rectangular plate, L × Walong its center line, parallel to L∫∫ double	$I = \frac{M W^{2}}{12}$	Area = LW ⇒ σ = M/(LW) (kg/m²).Patch area dx dy ⇒ dm = σ dx dy; r⊥ = \|y\|.	$\int_{- \frac{L}{2}}^{\frac{L}{2}} \int_{- \frac{W}{2}}^{\frac{W}{2}} y^{2} σ d y d x = \frac{M W^{2}}{12}$
Thin-walled tube, radius Rits own axis∫∫ double	$I = M R^{2}$	Wall area = circumference × height = 2πRH ⇒ σ = M/(2πRH) (kg/m²).Wall patch area R dφ dz ⇒ dm = σ R dφ dz; every piece has r⊥ = R. The tube has no end caps.	$\int_{0}^{H} \int_{0}^{2 π} R^{2} σ R d φ d z = R^{2} \cdot (2 π R H σ) = M R^{2}$	not in the lab
Solid cylinder, R and Hits own axis∫∫∫ triple	$I = \frac{M R^{2}}{2}$	Volume = πR²H ⇒ ρ = M/(πR²H) (kg/m³).Piece volume r dr dφ dz ⇒ dm = ρ r dr dφ dz; r⊥ = r.	$\int_{0}^{H} \int_{0}^{2 π} \int_{0}^{R} r^{2} ρ r d r d φ d z = ρ \cdot 2 π \cdot H \cdot \frac{R^{4}}{4} = \frac{M R^{2}}{2}$
Solid cylinder, R and Hthrough its center, across its length∫∫∫ triple	$I = \frac{M (3 R^{2} + H^{2})}{12}$	Volume = πR²H ⇒ ρ = M/(πR²H) (kg/m³).Piece volume r dr dφ dz ⇒ dm = ρ r dr dφ dz; r⊥² = y² + z² = r² sin² φ + z².	$\int \int \int (r^{2} \sin^{2} φ + z^{2}) ρ r d r d φ d z = \frac{M R^{2}}{4} + \frac{M H^{2}}{12}$
Thin spherical shell, radius Rany diameter∫∫ double	$I = \frac{2 M R^{2}}{3}$	Surface area = 4πR² ⇒ σ = M/(4πR²) (kg/m²).Patch area R² sin θ dθ dφ ⇒ dm = σ R² sin θ dθ dφ; r⊥ = R sin θ.	$\int_{0}^{2 π} \int_{0}^{π} R^{2} \sin^{2} θ σ R^{2} \sin θ d θ d φ = \frac{2 M R^{2}}{3}$	not in the lab
Solid sphere, radius Rany diameter∫∫∫ triple	$I = \frac{2 M R^{2}}{5}$	Volume = 4πR³/3 ⇒ ρ = M/(4πR³/3) = 3M/(4πR³) (kg/m³).Piece volume r² sin θ dr dθ dφ ⇒ dm = ρ r² sin θ dr dθ dφ; r⊥ = r sin θ.	$\int_{0}^{2 π} \int_{0}^{π} \int_{0}^{R} r^{2} \sin^{2} θ ρ r^{2} \sin θ d r d θ d φ = ρ \cdot 2 π \cdot \frac{4}{3} \cdot \frac{R^{5}}{5} = \frac{2 M R^{2}}{5}$
Any bodyaxis moved a distance d from the center of mass (parallel-axis theorem)theorem	$I = I_{CM} + M d^{2}$	Measure pieces from the CM: r⊥ = r′ − d, and ∫ r′ dm = 0 at the CM	$\int {(r′ - d)}^{2} d m = \int {r′}^{2} d m - 2 d \int r′ d m + d^{2} \int d m = I_{CM} + M d^{2}$
Any flat bodyaxis perpendicular to its plane (perpendicular-axis theorem)theorem	$I = I_{z} = I_{x} + I_{y}$	For a piece in the plane, its distance to z is r⊥² = x² + y²	$\int (x^{2} + y^{2}) d m = \int y^{2} d m + \int x^{2} d m$	not in the lab

Spot the pattern. Every result is M × (a length)² × a number. The number depends only on where the mass sits: 1 when all of it is at the rim (hoop, tube), 1/2 for a filled disk or cylinder, 2/3 for a hollow sphere, 2/5 for a solid one. The more mass near the axis, the smaller the number.