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

This page is the drill floor for the parent theorem. We take the one rule $I_{z} = I_{x} + I_{y}$ and throw every kind of problem at it — symmetric shapes, non-symmetric shapes, a degenerate flat rod, a real-world word problem, a trap that looks 3-D, and an exam twist that combines it with the Parallel axis theorem. Each worked example tells you which cell of the scenario matrix it covers.

The scenario matrix

#	Case class	What's special	Example that hits it
A	Planar, symmetric, common point	$I_{x} = I_{y}$ , split in half	Ex 1 (disc), Ex 2 (ring)
B	Planar, unequal in-plane axes	$I_{x} \neq = I_{y}$ , no shortcut	Ex 3 (rectangular plate)
C	Planar, solve backwards (find $I_{z}$ from two flats)	forecast-then-add	Ex 4 (annulus)
D	Degenerate input: a "plate" collapsed to a line	one in-plane $I = 0$	Ex 5 (thin rod as $b \to 0$ )
E	Trap: a 3-D body — theorem must be refused	$z \neq = 0$	Ex 6 (solid sphere)
F	Real-world word problem	translate words → axes	Ex 7 (spinning coin sign)
G	Exam twist: perpendicular then parallel axis	axis offset → two theorems	Ex 8 (disc, tangent perpendicular axis)
H	Limiting behaviour check	ratios & sanity limits	Ex 9 (square, and $a \to b$ collapse of Ex 3)

Everything below is built only on the $\int r^2\,dm$ definition, the symmetry argument, and results from Moment of inertia of standard bodies.

Worked Examples

Worked example 1. Disc about a diameter —

Cell A (planar, symmetric) A uniform disc has mass $M$ and radius $R$ . Its MOI about the central axis sticking straight out of its face is the known value $I_{z} = \frac{1}{2} M R^{2}$ . Find the MOI about a diameter (an in-plane axis through the centre).

Forecast: Guess before reading — is the answer bigger, smaller, or half of $I_{z}$ ? Why?

Name the axes. Put the disc in the $x y$ -plane, centre at the origin $O$ . Let the $z$ -axis poke out of the face; the $x$ - and $y$ -axes are two perpendicular diameters (red and yellow in the figure). Why this step? The theorem needs three mutually perpendicular axes through one point — we fix that point at the centre.
Use symmetry. Rotating the disc about $z$ by any angle leaves it looking identical, so no diameter is special: $I_{x} = I_{y}$ . Why this step? Symmetry replaces an integral. If the shape looks the same along $x$ and $y$ , their MOIs must be equal.
Apply the theorem. $I_{z} = I_{x} + I_{y} = 2 I_{x}$ . Why this step? This is the whole point of the theorem — the out-of-plane spin is the sum of the two in-plane spins.
Solve. $I_{x} = \frac{I _{z}}{2} = \frac{1}{2} \cdot \frac{1}{2} M R^{2} = \frac{1}{4} M R^{2}$ .

Verify: The diameter axis is "easier" to spin than the face axis, so $I_{x} < I_{z}$ — and indeed $\frac{1}{4} < \frac{1}{2}$ . ✓ Units: $[kg] [m^{2}]$ throughout. ✓

Worked example 2. Ring about a diameter —

Cell A (planar, symmetric, all mass at one radius) A thin ring (all mass on the rim) has mass $M$ , radius $R$ , and $I_{z} = M R^{2}$ . Find the MOI about a diameter.

Forecast: The ring has more of its mass far out than the disc does. Will the diameter MOI be a larger or smaller fraction of $I_{z}$ than the disc's $\frac{1}{2}$ ?

Symmetry. Same as the disc — every diameter looks alike, so $I_{x} = I_{y}$ . Why this step? Circular symmetry again forces the two in-plane axes to be equal.
Apply the theorem. $I_{z} = 2 I_{x} \Rightarrow I_{x} = \frac{I _{z}}{2} = \frac{1}{2} M R^{2}$ . Why this step? Identical algebra to Ex 1; only the input $I_{z}$ changed.

Verify: The fraction $I_{x} / I_{z} = \frac{1}{2}$ is the same as the disc's. That's expected: the theorem's halving comes from symmetry, not from where the mass sits. ✓

Worked example 3. Rectangular plate —

Cell B (planar, $I_{x} \neq = I_{y}$ , no shortcut) A uniform rectangular plate has side $a$ along $x$ , side $b$ along $y$ , mass $M$ . The standard in-plane results (through the centre) are $I_{x} = \frac{1}{12} M b^{2}$ and $I_{y} = \frac{1}{12} M a^{2}$ . Find $I_{z}$ about the central perpendicular axis.

Forecast: With $a \neq = b$ the two axes are different. Can we still just add them? Or does the theorem need $I_{x} = I_{y}$ ?

Check applicability. The plate is flat, all axes through the centre — theorem applies. Symmetry is not needed. Why this step? The proof only used $z = 0$ , never $I_{x} = I_{y}$ . Symmetry was a convenience in Ex 1–2, not a requirement.
Add directly. $I_{z} = I_{x} + I_{y} = \frac{1}{12} M b^{2} + \frac{1}{12} M a^{2} = \frac{1}{12} M (a^{2} + b^{2})$ . Why this step? Linearity of $\int$ lets us combine the two integrals; the theorem does the rest.

Verify: Set $a = b$ (a square) → $I_{z} = \frac{1}{12} M (2 a^{2}) = \frac{1}{6} M a^{2}$ , the known square-plate result. ✓ Dimensions: mass × length². ✓

Worked example 4. Annulus about a diameter —

Cell C (solve backwards, symmetric) A flat washer (annulus) has inner radius $R_{1}$ , outer radius $R_{2}$ , mass $M$ . Its perpendicular-axis MOI is $I_{z} = \frac{1}{2} M (R_{1}^{2} + R_{2}^{2})$ . Find its MOI about a diameter.

Forecast: Which known quantity is the "known", and which is the target? Which do we divide by 2?

Symmetry. Circular symmetry again gives $I_{x} = I_{y}$ . Why this step? An annulus is rotationally symmetric about $z$ , so all diameters are equivalent.
Halve $I_{z}$ . $I_{z} = 2 I_{dia} \Rightarrow I_{dia} = \frac{I _{z}}{2} = \frac{1}{4} M (R_{1}^{2} + R_{2}^{2})$ . Why this step? Exactly the disc logic; the annulus is just a disc with a hole, still planar and symmetric.

Verify: Let $R_{1} \to 0$ (hole shrinks to nothing → solid disc): $I_{dia} \to \frac{1}{4} M R_{2}^{2}$ , matching Ex 1. ✓ Let $R_{1} \to R_{2}$ (thin ring): $I_{dia} \to \frac{1}{4} M (2 R_{2}^{2}) = \frac{1}{2} M R_{2}^{2}$ , matching Ex 2. ✓ Two independent limits both land on earlier answers — strong evidence.

Worked example 5. Degenerate input — the plate collapsed to a rod —

Cell D Take the rectangular plate of Ex 3 and shrink side $b \to 0$ : the plate becomes a thin rod of length $a$ lying along the $x$ -axis. Predict its perpendicular-axis MOI $I_{z}$ and its MOI about its own length ( $I_{x}$ ).

Forecast: A rod has no width. What should the MOI about its own axis ( $x$ ) equal? Zero, or something small?

Kill the $b$ terms. From Ex 3, $I_{x} = \frac{1}{12} M b^{2} \to 0$ as $b \to 0$ . Why this step? $I_{x} = \int y^{2} d m$ ; when every point sits on the $x$ -axis, $y = 0$ , so the integral vanishes. Spinning a line about the line itself is free.
Keep the $a$ term. $I_{y} = \frac{1}{12} M a^{2}$ survives (mass is still spread along $x$ ). Why this step? $I_{y} = \int x^{2} d m$ has nonzero $x$ ; that's the real "flip end-over-end" difficulty.
Add. $I_{z} = I_{x} + I_{y} = 0 + \frac{1}{12} M a^{2} = \frac{1}{12} M a^{2}$ . Why this step? The theorem still holds; one of its two ingredients simply went to zero.

Verify: $\frac{1}{12} M a^{2}$ is the textbook MOI of a thin rod about its centre — the theorem reproduces it as a limit. ✓ And $I_{z} = I_{y}$ here because the degenerate rod's out-of-plane spin is its end-over-end spin. ✓

Worked example 6. The 3-D trap — refuse the theorem —

Cell E A student writes: "For a solid sphere of mass $M$ , radius $R$ , all three axes through the centre are equal, $I_{x} = I_{y} = I_{z} = \frac{2}{5} M R^{2}$ . So $I_{z} = I_{x} + I_{y}$ gives $\frac{2}{5} M R^{2} = \frac{4}{5} M R^{2}$ ." Find the error.

Forecast: Where is the hidden assumption of the theorem, and does a sphere satisfy it?

Recall the one condition. The proof set $z = 0$ for every mass element. A sphere has mass at all $z$ from $- R$ to $+ R$ . Why this step? This is the single line in the proof (Step 4 in the parent) where "flat" was used. Break it and everything after is invalid.
Show the numbers can't reconcile. $I_{x} + I_{y} = \frac{4}{5} M R^{2} \neq = \frac{2}{5} M R^{2} = I_{z}$ . Why this step? The contradiction is the proof that the theorem was misapplied — it is not a rule for solids.

Verify: The general 3-D identity is $I_{x} + I_{y} + I_{z} = 2 \int (x^{2} + y^{2} + z^{2}) d m$ ; only when $z = 0$ everywhere does this collapse to $I_{z} = I_{x} + I_{y}$ . For the sphere $z \neq = 0$ , so the correct statement is $I_{x} + I_{y} + I_{z} = 3 \cdot \frac{2}{5} M R^{2}$ , not $I_{z} = I_{x} + I_{y}$ . ✓ The answer is: the theorem does not apply — the sphere is not planar.

Worked example 7. Real-world word problem —

Cell F A coin (uniform disc, mass $M = 8.0 g$ , radius $R = 1.2 cm$ ) sits flat on a table. You (a) spin it like a fidget top about the vertical line through its centre, and (b) flip it end-over-end about a horizontal line drawn across its face through the centre. Which spin has the larger MOI, and by what factor? Give both numeric values.

Forecast: Vertical top-spin is the $z$ -axis; the flip is a diameter. From Ex 1, guess the factor before computing.

Translate words to axes. Top-spin = perpendicular axis → $I_{z} = \frac{1}{2} M R^{2}$ . Flip = diameter → $I_{dia} = \frac{1}{4} M R^{2}$ (Ex 1). Why this step? The whole difficulty of word problems is naming the axis; once named, the formulas are automatic.
Numbers. With $M = 8.0 \times 1 0^{- 3} kg$ , $R = 1.2 \times 1 0^{- 2} m$ : $I_{z} = \frac{1}{2} (8.0 \times 1 0^{- 3}) (1.2 \times 1 0^{- 2})^{2} = 5.76 \times 1 0^{- 7} kg m^{2}$ $I_{dia} = \frac{1}{4} (8.0 \times 1 0^{- 3}) (1.2 \times 1 0^{- 2})^{2} = 2.88 \times 1 0^{- 7} kg m^{2}$ Why this step? Plug into the two formulas; consistent SI units give SI answers.
Factor. $I_{z} / I_{dia} = 2$ . Why this step? The theorem predicts exactly this ratio for any symmetric disc — a clean sanity check.

Verify: $I_{z} = I_{dia} + I_{dia} = 2.88 \times 1 0^{- 7} \times 2 = 5.76 \times 1 0^{- 7}$ . ✓ Top-spin is twice as hard — matching the everyday feeling that flipping a coin edge-over-edge is easier than getting a fast top-spin. ✓

Worked example 8. Exam twist: perpendicular

then parallel — Cell G Find the MOI of a disc (mass $M$ , radius $R$ ) about an axis perpendicular to its plane but passing through a point on its rim (a tangent-point perpendicular axis).

Forecast: The perpendicular axis theorem alone gives centre values. This axis is offset. Which extra theorem do we need?

Get the centre perpendicular MOI. $I_{z}^{centre} = \frac{1}{2} M R^{2}$ (given/standard). This came from the perpendicular-axis family. Why this step? Parallel axis needs the MOI about the centre-of-mass axis parallel to the target — that's the central perpendicular axis.
Shift with the parallel axis theorem. The rim axis is parallel to the centre axis, offset by $d = R$ : $I = I_{z}^{centre} + M d^{2} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2} .$ Why this step? The perpendicular-axis theorem only relates axes through one common point; to move to a different point we must use the Parallel axis theorem. Never use perpendicular-axis to shift an axis sideways.

Verify: $\frac{3}{2} M R^{2} > \frac{1}{2} M R^{2}$ — moving the axis away from the centre always increases MOI, as $+ M d^{2} \geq 0$ guarantees. ✓ Units: mass × length². ✓ This is the classic exam trap: perpendicular-axis for changing axis direction at a point, parallel-axis for changing axis position.

Worked example 9. Limiting behaviour — square and the

$a \to b$ collapse — Cell H (i) For a square plate of side $a$ , mass $M$ , use Ex 3 to get $I_{z}$ and the diameter (in-plane centre) MOI. (ii) Confirm the rectangular result of Ex 3 collapses correctly to the square as $a \to b$ .

Forecast: For a square, is $I_{x} = I_{y}$ ? What does that force the diameter MOI to be as a fraction of $I_{z}$ ?

Square from Ex 3. Set $b = a$ : $I_{z} = \frac{1}{12} M (a^{2} + a^{2}) = \frac{1}{6} M a^{2}$ . Why this step? The square is the symmetric special case of the rectangle, so its result must fall out by substitution.
In-plane axis by symmetry. A square looks the same along $x$ and $y$ , so $I_{x} = I_{y}$ , and $I_{x} = \frac{I _{z}}{2} = \frac{1}{12} M a^{2}$ . Why this step? Symmetry returns for the square (unlike the general rectangle), letting us halve $I_{z}$ .
Collapse check. In Ex 3, $I_{x} - I_{y} = \frac{1}{12} M (b^{2} - a^{2}) \to 0$ as $a \to b$ — the two axes merge into one value. Why this step? A continuous limit test: the unequal-axis rectangle must smoothly become the equal-axis square.

Verify: $\frac{1}{12} M a^{2} \times 2 = \frac{1}{6} M a^{2} = I_{z}$ . ✓ Both in-plane axes equal $\frac{1}{12} M a^{2}$ ; their sum is exactly the perpendicular value. ✓

Active Recall

Recall Which theorem, which cell?

For each, name the cell (A–H) and the tool needed.

Disc about a diameter ::: Cell A — perpendicular axis + symmetry (halve $I_{z}$ ).
Rectangular plate $I_{z}$ from two unequal in-plane axes ::: Cell B — perpendicular axis, no symmetry.
Disc about a rim perpendicular axis ::: Cell G — perpendicular value then parallel axis theorem.
"It works for a sphere too" ::: Cell E — refuse; sphere is 3-D, $z \neq = 0$ .
Thin rod MOI from a collapsing plate ::: Cell D — degenerate limit, one in-plane MOI $\to 0$ .

Connections

Parent (Hinglish) — the proof these examples exercise.
Parallel axis theorem — needed in Ex 8 to shift the axis off-centre.
Moment of inertia — definition — the $\int r^{2} d m$ every answer rests on.
Moment of inertia of standard bodies — disc, ring, plate, annulus, rod base values.
Symmetry arguments in MOI — the $I_{x} = I_{y}$ trick used in Cells A, C, H.
Radius of gyration — a related way to repackage these MOIs.