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

What, Why, How

WHAT it gives you: a shortcut. Know two in-plane $I$ 's → get the perpendicular one for free (or vice versa, often by symmetry).

WHY restricted to flat bodies: it secretly assumes every particle has $z=0$ . A 3-D body breaks this.

HOW you use it: typically with symmetry. For a disc, $I_x = I_y$ (looks the same from any in-plane diameter), so $I_z = 2I_x \Rightarrow I_{\text{diameter}} = \tfrac12 I_z$ .

Derivation from scratch

Formula Step-by-step proof

Take a planar body in the $xy$ -plane. The three axes $x,y,z$ all pass through one common origin $O$ , with $z \perp$ plane.

Step 1 — Definition of moment of inertia. For a mass element $dm$ at position $(x,y,z)$ , the MOI about an axis is $\int (\text{perp. distance to axis})^2\,dm$ . Why this step? MOI is defined as $\sum m r_\perp^2$ ; we just write it as an integral for a continuous body.

Step 2 — Distances to each axis.

To the $z$ -axis: perpendicular distance $^2 = x^2 + y^2$ .
To the $x$ -axis: perpendicular distance $^2 = y^2 + z^2$ .
To the $y$ -axis: perpendicular distance $^2 = x^2 + z^2$ .

Why this step? The perpendicular distance from a point to an axis uses the other two coordinates (Pythagoras in the plane $\perp$ to that axis).

Step 3 — Write the three integrals. $I_z=\int(x^2+y^2)\,dm,\quad I_x=\int(y^2+z^2)\,dm,\quad I_y=\int(x^2+z^2)\,dm$

Step 4 — Apply the planar restriction $z=0$ . $I_x=\int y^2\,dm,\qquad I_y=\int x^2\,dm$ Why this step? This is the only place the "flat body" condition enters. No $z=0$ → no theorem.

Step 5 — Add $I_x$ and $I_y$ . $I_x+I_y=\int y^2\,dm+\int x^2\,dm=\int(x^2+y^2)\,dm=I_z \quad\blacksquare$ Why this step? Linearity of the integral lets us combine, and the sum is exactly the $I_z$ integrand.

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

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Recall Try before peeking

State the theorem and its one restriction.
Where exactly in the proof is the restriction used?
Derive disc-about-diameter from $I_z=\tfrac12MR^2$ .

Answers: $I_z=I_x+I_y$ for a planar body; restriction = body lies in a plane ( $z=0$ ). Used in Step 4 setting $z=0$ . Disc: $I_z=2I_x\Rightarrow I_x=\tfrac14MR^2$ .

Flashcards

State the perpendicular axis theorem.

I_z = I_x + I_y

, valid for a planar body, with

x,y,z

mutually perpendicular axes through a common point (

z\perp

plane).

What single condition does the proof require?

Every mass element has

z=0

(the body is a flat lamina).

At which step is the planar restriction used?

When setting

z=0

I_x=\int(y^2+z^2)dm

and

I_y=\int(x^2+z^2)dm

Why does the theorem fail for a solid sphere?

A sphere is 3-D (

z\neq0

); the

z^2

terms don't vanish, so the sum no longer equals

I_z

MOI of a disc about a diameter, given

I_z=\tfrac12MR^2

\tfrac14 MR^2

(since

I_z=2I_{\text{dia}}

MOI of a ring about a diameter, given

I_z=MR^2

\tfrac12 MR^2

For a rectangular plate,

I_z

in terms of sides?

I_z=\tfrac{1}{12}M(a^2+b^2)

Must the three axes intersect?

Yes — all three must pass through one common point.

Must

I_x

equal

I_y

for the theorem to hold?

No; symmetry is only a convenience. The theorem holds even when

I_x\neq I_y

Perpendicular distance² of a point

(x,y,0)

from the

z

-axis?

x^2+y^2

Recall Feynman: explain to a 12-year-old

Imagine a flat coin lying on a table. Spinning it like a fidget top (axis poking up through the middle) is the " $z$ spin." Now flipping it end-over-end about a line drawn across its face is an "in-the-plane spin." The rule says: the difficulty of the top-spin is just the two flipping difficulties added together — one for flipping left-right, one for flipping front-back. This only works because a coin is flat: every bit of metal sits on the table, none sticks up. For a thick ball, bits stick out in all directions, and the neat adding trick breaks.

Connections

Parallel axis theorem — the other shifting theorem; uses $I = I_{cm}+Md^2$ , works for any body but offset parallel axes.
Moment of inertia — definition — the $\int r^2\,dm$ this is all built on.
Moment of inertia of standard bodies — disc, ring, rectangular plate results used here.
Radius of gyration — alternative packaging of $I$ .
Symmetry arguments in MOI — why $I_x=I_y$ for round/square plates.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, perpendicular axis theorem sirf flat (planar) bodies ke liye hota hai — jaise coin, disc, ring, ya patli plate. Agar body table par flat padi hai (xy-plane mein), to uske har mass point ka $z=0$ hota hai. Isi ek baat ki wajah se magic hota hai: $z$ -axis (jo plane se bahar nikalti hai) ke baare mein moment of inertia exactly plane ke andar wali do perpendicular axes ( $x$ aur $y$ ) ke MOI ka sum ban jaata hai — $I_z = I_x + I_y$ .

Proof ka core simple hai. $z$ -axis se kisi point ki perpendicular distance $^2 = x^2+y^2$ . $x$ -axis se distance $^2 = y^2+z^2$ , aur $y$ -axis se $x^2+z^2$ . Ab kyunki body flat hai, $z=0$ daal do — to $I_x=\int y^2 dm$ aur $I_y=\int x^2 dm$ . Inko jod do: $I_x+I_y=\int(x^2+y^2)dm = I_z$ . Bas ho gaya. Yaad rakho — $z=0$ wali line hi pura theorem chala rahi hai.

Practical fayda: symmetry ke saath shortcut milta hai. Disc ka $I_z=\tfrac12 MR^2$ pata hai, aur diameter ke baare mein $I_x=I_y$ (symmetry). To $I_z=2I_x$ , matlab $I_x=\tfrac14 MR^2$ — bina koi integration kiye! Ring ke liye $I_z=MR^2$ se diameter MOI $=\tfrac12 MR^2$ .

Common galti: log isse solid sphere par laga dete hain — galat. Sphere 3-D hai, $z\neq0$ , isliye theorem fail. Doosri galti: teeno axes ek hi point par milni chahiye aur $x,y$ ek doosre ke perpendicular honi chahiye. Symmetry zaroori nahi (rectangle plate mein $I_x\neq I_y$ phir bhi theorem chalta hai), lekin planar hona zaroori hai.

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

What, Why, How

Derivation from scratch

Worked Examples

Common Mistakes (Steel-manned)

Active Recall

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Rotational Mechanics

Connections