Perpendicular axis theorem — I_z = I_x + I_y — proof, restrictions: Question bank

Two words you must have crisp before starting:

Lamina / planar body = an idealised flat sheet where every mass point has $z = 0$ (zero thickness). Picture a coin flattened until it has no height at all.
Mutually perpendicular axes = the $x$ , $y$ , $z$ axes each meet at $9 0^{\circ}$ , all crossing at one common point $O$ , with $z$ poking straight out of the sheet.

True or false — justify

The theorem holds for a solid cube.

False — a cube has real extent in

z

, so the

z^{2}

terms in

I_{x}

and

I_{y}

never vanish, and the clean split

I_{z} = I_{x} + I_{y}

collapses. It works only for a

z = 0

lamina.

The theorem holds for a thin square plate of tiny but nonzero thickness.

Approximately true — a real thin plate has small

z

, so

z^{2} d m

terms are negligible; treated as a lamina it is essentially exact, but strictly it is an approximation, not the ideal law.

For a uniform disc,

I_{z} = I_{x} + I_{y}

and also

I_{x} = I_{y}

True — the disc is planar so the theorem applies, and by Symmetry arguments in MOI every diameter is equivalent, forcing

I_{x} = I_{y}

; hence

I_{z} = 2 I_{x}

The theorem requires

I_{x} = I_{y}

False — equal in-plane MOIs are only a symmetry convenience. A rectangular plate has

I_{x} \neq = I_{y}

yet still obeys

I_{z} = I_{x} + I_{y} = \frac{1}{12} M (a^{2} + b^{2})

If you tilt the in-plane axes by

3 0^{\circ}

(still perpendicular to each other, still through

O

), the theorem still gives

I_{z}

True — any two mutually perpendicular in-plane axes through the same point work; the Pythagoras split

x^{'2} + y^{'2} = r^{2}

is rotation-invariant, so the sum is unchanged.

The theorem lets you find

I

about an axis that lies outside the plane but is not the

z

-axis.

False — it only relates the three special axes (

x

y

in-plane and

z

out-of-plane) through one point; a slanted axis needs a different method entirely.

For a flat L-shaped bracket (non-symmetric lamina), the theorem still holds.

True — planarity, not symmetry, is the requirement; every mass point still has

z = 0

, so

I_{z} = I_{x} + I_{y}

regardless of the odd shape.

I_{z}

is always the largest of the three MOIs for a lamina.

True — since

I_{z} = I_{x} + I_{y}

and each of

I_{x}, I_{y}

is a non-negative integral of squares,

I_{z} \geq I_{x}

and

I_{z} \geq I_{y}

; it can only equal one of them if the other is zero.

Spot the error

"A solid sphere:

I_{x} = I_{y} = I_{z} = \frac{2}{5} M R^{2}

, so

I_{z} = I_{x} + I_{y} = \frac{4}{5} M R^{2}

. Contradiction — so MOI is undefined."

The error is applying the theorem to a 3-D body. There is no contradiction; the theorem simply does not apply, because a sphere is not planar.

I_{z} = \frac{2}{5} M R^{2}

is correct on its own.

"For a disc,

I_{z} = \frac{1}{2} M R^{2}

, so each diameter MOI is

\frac{1}{2} M R^{2}

as well."

Wrong — you must first split:

I_{z} = I_{x} + I_{y} = 2 I_{dia}

, so

I_{dia} = \frac{1}{4} M R^{2}

. Forgetting the factor of two doubles the answer.

"The three axes can be parallel and offset like in the parallel axis theorem."

No — that confuses the two theorems. Here the axes are mutually perpendicular and meet at one point; offsetting any of them breaks the Pythagoras relation

r_{⊥}^{2} = x^{2} + y^{2}

that the proof needs. Use Parallel axis theorem for offsets.

"For a ring, mass is all at radius

R

, so

I_{x} = I_{y} = I_{z} = M R^{2}

Wrong — only

I_{z} = M R^{2}

(all mass at distance

R

from the central perpendicular axis). About a diameter the mass sits at varying perpendicular distances, giving

I_{dia} = \frac{1}{2} M R^{2}

via

I_{z} = 2 I_{dia}

"Since the theorem uses

x^{2} + y^{2}

, it must be about the origin only."

Not quite — the theorem holds about any common point through which all three perpendicular axes pass; you are free to choose

O

wherever you like, as long as the three axes still intersect there.

I_{x} = \int (y^{2} + z^{2}) d m

and for a lamina that is

\int (y^{2} + 0) d m

, but I'll keep

z

just in case."

Keeping

z

is the error for a lamina — the whole theorem rests on setting

z = 0

. Retaining

z^{2}

means you are no longer describing a planar body and the split fails.

Why questions

Why does the restriction enter exactly at one step of the proof?

Because only

I_{x}

and

I_{y}

carry

z^{2}

terms; setting

z = 0

is the single move that kills them, and it happens once when writing

I_{x} = \int y^{2} d m

I_{y} = \int x^{2} d m

Why must the two in-plane axes be perpendicular to each other?

Because the identity

x^{2} + y^{2} = r^{2}

(perpendicular distance

^{2}

from the

z

-axis) is Pythagoras, which only holds for orthogonal coordinate directions; non-perpendicular axes give cross-terms that ruin the clean sum.

Why does the theorem "come for free" — no integration for the diameter of a disc?

Because symmetry hands you

I_{x} = I_{y}

, and the theorem ties them to a known

I_{z}

; two equations, one unknown, so algebra replaces the integral. See Moment of inertia of standard bodies.

Why is

I_{z}

the sum and not the average of

I_{x}

and

I_{y}

Because the

z

-axis distance

^{2}

literally equals

x^{2} + y^{2}

— a sum of the very squares that build

I_{y}

and

I_{x}

— so integrating gives an addition, never an average.

Why does the theorem never depend on the shape's outline?

Because the derivation uses only the coordinate identity

r^{2} = x^{2} + y^{2}

valid at every point with

z = 0

; the boundary of the lamina never appears, so any flat shape qualifies.

Why can the same theorem be read "backwards" to get

I_{z}

from two in-plane MOIs?

Because

I_{z} = I_{x} + I_{y}

is an equality, not a one-way recipe; knowing any two of the three determines the third.

Edge cases

A thin straight rod lying along the

x

-axis (a degenerate lamina). What is

I_{x}

I_{x} = \int y^{2} d m = 0

because every point has

y = 0

; then

I_{z} = I_{y}

, matching that spinning a rod about its own length costs nothing.

A point mass at the origin. What do the three MOIs read?

All zero — with

x = y = z = 0

the perpendicular distance to every axis is zero, so

I_{x} = I_{y} = I_{z} = 0

and the sum

0 = 0 + 0

holds trivially.

Two point masses on the

x

-axis, symmetric about

O

. Does the theorem hold?

Yes — it is still a lamina (

z = 0

). Here

I_{x} = 0

(points on the

x

-axis) and

I_{z} = I_{y}

, so all mass "spin difficulty" about

z

comes entirely from the

y

-axis term.

A lamina where all mass lies on the

y

-axis. Relate the three MOIs.

Then

x = 0

for all mass, so

I_{y} = \int x^{2} d m = 0

and

I_{z} = I_{x}

; the out-of-plane and the in-line-perpendicular MOIs coincide.

Limiting a thin plate's thickness

t \to 0

— what happens to the error in

I_{x} = \int y^{2} d m

The dropped

\int z^{2} d m

term scales like

t^{2}

, so as

t \to 0

the error vanishes quadratically and the lamina result becomes exact.

A lamina bent slightly out of the plane (a shallow dome). Does the theorem still apply?

No — once points acquire nonzero

z

, the

z^{2}

terms revive and

I_{z} = I_{x} + I_{y}

fails; the body must be genuinely flat, not merely thin-and-curved.

Connections

Parallel axis theorem — the sibling theorem for offset parallel axes; contrast the "Plus" vs "Plus a distance term".
Symmetry arguments in MOI — why $I_{x} = I_{y}$ for discs and rings, and why symmetry is optional here.
Radius of gyration — recasts these MOIs as an equivalent single distance.
Moment of inertia — definition — the $\int r^{2} d m$ everything above rests on.