Perpendicular axis theorem — I_z = I_x + I_y — proof, restrictions: Visual walkthrough

Step 1 — What is a "flat body," and where do we put it?

WHAT we did: placed the object so every particle lives on one flat sheet.

WHY: the whole theorem lives or dies on one fact — no part of the object sticks up off the table. We build that in from line one so we never forget it.

PICTURE: Look at the figure. The black outline is the object. The red dot is one tiny mass piece we will track — call its mass $d m$ ("a crumb of mass"). It sits on the sheet, so its height above the table is $0$ .

Figure — Perpendicular axis theorem — I_z = I_x + I_y — proof, restrictions

Hold onto that $z = 0$ . It is the single seed the whole proof grows from.

Step 2 — Plant three axes through one shared pin

WHAT we did: fixed one common point and three perpendicular directions there.

WHY one shared pin: in the next steps we will add three integrals term by term. Adding only works if all three measurements start from the same place. Slide any axis off $O$ and the algebra collapses (we prove this in Step 8).

PICTURE: Two black axes lie flat; the red $z$ -axis rises out of the page. This red axis is the "spin like a fidget-top" axis — the odd one out that the theorem will express using its two flat teammates.

See also Symmetry arguments in MOI for why the choice of $x, y$ directions is often free.

Step 3 — What does it cost to spin? Define moment of inertia

WHAT we did: wrote the definition of $I$ for a continuous body.

WHY the square, and why this tool? We need a single number saying "how stubborn is this object to get spinning?" Speed at radius $r_{⊥}$ grows like $r_{⊥}$ , and energy grows like speed-squared — so difficulty must weight each crumb by $r_{⊥}^{2}$ , not $r_{⊥}$ . That is the why behind the exponent.

PICTURE: The red segment is $r_{⊥}$ — the shortest reach from crumb to axis. The crumb's cost is (length of red segment) $^{2} \times$ (its mass).

Step 4 — Read off the perpendicular distance to each axis

Now the key geometric move: for each axis, what is $r_{⊥}$ for our crumb at $(x, y, z)$ ?

PICTURE: The red right triangle shows the distance to the $z$ -axis. The crumb is $x$ to the right and $y$ forward; the straight-line reach across the floor is the red hypotenuse, and Pythagoras gives its square as $x^{2} + y^{2}$ .

WHY this step: these three little Pythagoras facts are the only geometry the proof needs. Everything after is bookkeeping.

Step 5 — Write the three moments of inertia as integrals

WHAT we did: plug each $r_{⊥}^{2}$ from Step 4 into the definition from Step 3.

WHY: now the three "spin-difficulties" are all written from the same crumb coordinates $(x, y, z)$ . Because they share coordinates, we can compare and combine them.

PICTURE: Three panels, one per axis, each showing the same crumb and the distance that axis "sees." Notice how each integrand is a sum of two squared coordinates.

Step 6 — Flatten it: the single moment where $z = 0$

Here is the heartbeat of the whole proof. From Step 1, every crumb has $z = 0$ . Feed that into $I_{x}$ and $I_{y}$ :

WHAT we did: deleted the two $z^{2}$ terms.

WHY this is the step: this is the only line where "flat" is used. Remove it and $I_{x}, I_{y}$ keep their $z^{2}$ baggage — and the clean sum below never appears. Remember this the next time someone tries the theorem on a solid ball (Step 7).

PICTURE: The two $z^{2}$ boxes get crossed out in red; the survivors $y^{2}$ (for $I_{x}$ ) and $x^{2}$ (for $I_{y}$ ) glow.

Step 7 — Add them and watch $I_{z}$ appear

WHAT we did: add the two survivors. Integrals add term by term (this is called linearity — the integral of a sum is the sum of the integrals).

WHY it works: $x^{2} + y^{2}$ is precisely the $z$ -axis distance-square from Step 4. The two flat spins carry exactly the two pieces the top-spin needs.

PICTURE: Two black bricks labelled $\int y^{2} d m$ and $\int x^{2} d m$ slide together into one red brick labelled $I_{z}$ . That is the theorem, drawn.

Step 8 — Edge & degenerate cases (never hit a surprise)

Every case the reader might meet, shown here.

PICTURE: Four mini-panels — ball (fails), offset axis (fails), skew axes (fails), thin plate (≈ works). The failing configurations are marked red.

The one-picture summary

Everything above, compressed: crumb at $(x, y, 0)$ → three Pythagoras distances → $z = 0$ kills the $z^{2}$ terms → the two flat integrals reassemble into the up-axis integral.

Recall Feynman: tell the whole walkthrough to a friend

Put a coin flat on the table. Pick one speck of metal on it. Because the coin is flat, that speck has no height — it sits on the table. Now ask three questions: how hard is it to spin the coin like a top (axis poking up), how hard to flip it front-to-back, how hard to flip it left-to-right? Each "hardness" adds up (distance-from-axis)² over every speck. The distance to the up-axis, by Pythagoras across the tabletop, is $x^{2} + y^{2}$ . The flip-hardnesses want to include a height term $z^{2}$ too — but every speck has zero height, so those terms disappear. What's left: one flip counts $y^{2}$ , the other counts $x^{2}$ . Add them and you get exactly $x^{2} + y^{2}$ — the top-spin hardness. So top = front-flip + side-flip. And it only works because the coin is flat: on a fat ball the specks stick up, the height terms survive, and the tidy adding breaks.

Connections

Same idea in Hinglish
Parallel axis theorem — the offset cousin ( $I = I_{c m} + M d^{2}$ ), works for any body.
Moment of inertia — definition — the $\int r^{2} d m$ underneath every step.
Moment of inertia of standard bodies — disc, ring, plate results this unlocks.
Radius of gyration — repackages any $I$ as $I = M k^{2}$ .
Symmetry arguments in MOI — why $I_{x} = I_{y}$ so often, letting you halve $I_{z}$ .