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

A quick vocabulary reminder before we start, so no symbol is unearned:

Definition The words we will keep using

Lamina / planar body — a body so thin we treat all its mass as lying in one flat plane. We put that plane as the $x y$ -plane, so ==every mass point has $z = 0$ ==.
Moment of inertia — the number $\int r_{⊥}^{2} d m$ that measures how hard a body is to angularly accelerate about a chosen axis (a mass far from the axis resists spinning far more than one close to it). See Moment of inertia — definition. Here $r_{⊥}$ = the shortest, straight-across distance from a tiny mass bit $d m$ to the axis.
$I_{z}$ — moment of inertia about the axis poking straight out of the plane (the perpendicular axis).
$I_{x}$ , $I_{y}$ — moments of inertia about two lines drawn on the face, at right angles to each other, both crossing the $z$ -axis at the same point (the two in-plane axes).

The picture below anchors every symbol above. Look at it before Q1: the teal and plum arrows are the two in-plane axes, the dark arrow is the perpendicular ( $z$ ) axis coming out of the sheet, and the dashed segment is $r_{⊥} = x^{2} + y^{2}$ — the very quantity we square inside $I_{z}$ .

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

Figure 1 shows why the $z$ -axis distance is $x^{2} + y^{2}$ : the mass bit, the origin, and its two coordinate projections form a right triangle, so Pythagoras gives the hypotenuse. Keep this triangle in mind — it is the whole theorem in one drawing.

Level 1 — Recognition

Can you spot when the theorem even applies, and read off the pieces?

Recall Solution Q1

The theorem's one requirement is $z = 0$ for every mass element — a flat lamina.

(a) CD disc → yes, essentially planar.
(b) solid ball → no, mass fills all three dimensions, $z \neq = 0$ .
(c) sheet of paper → yes, an ideal lamina.
(d) thick brick → no, real thickness in $z$ . Answer: (a) and (c).

Recall Solution Q2

All the ring's mass sits at distance $R$ from the $z$ -axis, so $I_{z} = M R^{2}$ . $I_{z} = (2) (0.5)^{2} = 2 \times 0.25 = 0.5 kg m^{2} .$

Recall Solution Q3

The distance to an axis uses the other two coordinates (Pythagoras in the plane perpendicular to that axis — exactly the right triangle drawn in Figure 1):

to $z$ -axis: $x^{2} + y^{2}$
to $x$ -axis: $y^{2} + z^{2} \to y^{2}$ (since $z = 0$ )
to $y$ -axis: $x^{2} + z^{2} \to x^{2}$ (since $z = 0$ ) These are exactly the integrands of $I_{z}$ , $I_{x}$ , $I_{y}$ .

Level 2 — Application

Plug the theorem in and get a number, usually with a symmetry shortcut.

Before Q4 — why symmetry lets us write $I_{x} = I_{y}$ . Figure 2 shows a uniform disc with two diameters drawn on it: a horizontal one and a vertical one. Because the disc is round and uniformly dense, rotating the whole picture by $9 0^{\circ}$ maps the horizontal diameter onto the vertical one and leaves the mass pattern unchanged. Two axes related by a symmetry that leaves the body identical must have the same moment of inertia — so $I_{x} = I_{y}$ . This is not a shortcut we take on faith; it is forced by the fact that $\int y^{2} d m$ and $\int x^{2} d m$ compute the same number once the picture is symmetric. (See also Symmetry arguments in MOI.)

Recall Solution Q4

Any two perpendicular diameters look identical (Figure 2: a $9 0^{\circ}$ turn swaps them and leaves the uniform disc unchanged), so $I_{x} = I_{y} = I_{dia}$ . $I_{z} = I_{x} + I_{y} = 2 I_{dia} \Rightarrow I_{dia} = \frac{1}{2} I_{z} = \frac{1}{4} M R^{2} .$ Numerically: $I_{dia} = \frac{1}{4} (4) (0.3)^{2} = \frac{1}{4} (4) (0.09) = 0.09 kg m^{2} .$

Recall Solution Q5

The ring has the same rotational symmetry as the disc (Figure 2), and uniform linear density, so $I_{x} = I_{y}$ . With $I_{z} = M R^{2}$ , $I_{dia} = \frac{1}{2} I_{z} = \frac{1}{2} M R^{2}$ . $I_{dia} = \frac{1}{2} (3) (0.4)^{2} = \frac{1}{2} (3) (0.16) = 0.24 kg m^{2} .$

Recall Solution Q6

Here $a \neq = b$ , so a $9 0^{\circ}$ turn does not map the plate onto itself — hence $I_{x} \neq = I_{y}$ . No symmetry shortcut, but the theorem still adds directly: $I_{z} = I_{x} + I_{y} = \frac{1}{12} M (a^{2} + b^{2}) .$ $I_{z} = \frac{1}{12} (5) (0. 6^{2} + 0. 4^{2}) = \frac{1}{12} (5) (0.36 + 0.16) = \frac{1}{12} (5) (0.52) = 0.2167 kg m^{2} .$

Level 3 — Analysis

Reason about which axis, reverse the theorem, or combine with another result.

Recall Solution Q7

Reverse the theorem: $I_{z} = 2 I_{dia} = 2 (0.05) = 0.1 kg m^{2}$ . Since $I_{z} = \frac{1}{2} M R^{2}$ : $R = \frac{2 I _{z}}{M} = \frac{2 ( 0.1 )}{2} = 0.1 \approx 0.316 m .$

Recall Solution Q8

A square is mapped onto itself by a $9 0^{\circ}$ turn (Figure 2 logic applied to a square), so both in-plane central axes are equal, $I_{x} = I_{y}$ : $I_{x} = \frac{1}{2} I_{z} = \frac{1}{2} \cdot \frac{1}{6} M a^{2} = \frac{1}{12} M a^{2} .$ This matches the direct plate result $\frac{1}{12} M a^{2}$ . ✓ Numerically: $I_{x} = \frac{1}{12} (1) (0.2)^{2} = \frac{0.04}{12} = 0.00333 kg m^{2} .$

Recall Solution Q9

Ring about a diameter: $I_{dia} = \frac{1}{2} M R^{2}$ . Setting $M k^{2} = \frac{1}{2} M R^{2}$ : $k = \frac{R}{2} .$ $k = \frac{0.5}{2} \approx 0.354 m .$

Level 4 — Synthesis

Combine the perpendicular theorem with the Parallel axis theorem or with integration.

Recall Solution Q10

Two steps, two theorems: 1. Perpendicular axis theorem → central diameter MOI: $I_{dia} = \frac{1}{4} M R^{2}$ . 2. Parallel axis theorem → shift that in-plane axis out by $d = R$ to the tangent: $I_{tan} = I_{dia} + M R^{2} = \frac{1}{4} M R^{2} + M R^{2} = \frac{5}{4} M R^{2} .$ Numerically: $I_{tan} = \frac{5}{4} (2) (0.3)^{2} = \frac{5}{4} (2) (0.09) = 0.225 kg m^{2} .$

Recall Solution Q11

Start from $I_{z} = \frac{1}{2} M R^{2}$ (central perpendicular), then parallel-axis shift by $d = R$ : $I_{z, tan} = \frac{1}{2} M R^{2} + M R^{2} = \frac{3}{2} M R^{2} .$ $= \frac{3}{2} (2) (0.09) = 0.27 kg m^{2} .$ Compare Q10 vs Q11: shifting the in-plane axis gave $\frac{5}{4} M R^{2}$ , the perpendicular axis gave $\frac{3}{2} M R^{2}$ — different because the central starting values ( $\frac{1}{4}$ vs $\frac{1}{2}$ ) differ.

Recall Solution Q12

1. Central perpendicular: $I_{z, cm} = \frac{1}{12} M (a^{2} + b^{2})$ . 2. Parallel-axis shift to a corner, distance from centre $d = \frac{1}{2} a^{2} + b^{2}$ , so $d^{2} = \frac{1}{4} (a^{2} + b^{2})$ : $I_{z, corner} = \frac{1}{12} M (a^{2} + b^{2}) + M \cdot \frac{1}{4} (a^{2} + b^{2}) = M (a^{2} + b^{2}) (\frac{1}{12} + \frac{1}{4}) = \frac{1}{3} M (a^{2} + b^{2}) .$ Numerically: $a^{2} + b^{2} = 0.64 + 0.36 = 1.0$ , so $I_{z, corner} = \frac{1}{3} (6) (1.0) = 2.0 kg m^{2} .$

Level 5 — Mastery

Prove, generalise, or catch a subtle failure.

Recall Solution Q13

From $I_{x} = \int y^{2} d m$ and $I_{y} = \int x^{2} d m$ , both integrands are squares → non-negative → $I_{x} \geq 0$ , $I_{y} \geq 0$ . Then $I_{z} = I_{x} + I_{y} \geq I_{x}$ (since we add a non-negative $I_{y}$ ), and likewise $I_{z} \geq I_{y}$ . ∎ Equality $I_{z} = I_{x}$ requires $I_{y} = \int x^{2} d m = 0$ , i.e. all mass has $x = 0$ — the lamina collapses onto the $y$ -axis (a thin rod along $y$ ). Symmetrically $I_{z} = I_{y}$ means a rod along $x$ . So a genuine 2-D lamina always has $I_{z}$ strictly larger than each in-plane MOI.

Recall Solution Q14

For a sphere the true $I_{z} = \frac{2}{5} M R^{2} = 0.4$ (with $M = R = 1$ ). The student's sum gives $\frac{4}{5} M R^{2} = 0.8$ . These disagree: $0.8 \neq = 0.4$ . Where it breaks: the proof step $I_{x} = \int (y^{2} + z^{2}) d m \to \int y^{2} d m$ used $z = 0$ . A sphere fills all three dimensions, so $z \neq = 0$ and the $\int z^{2} d m$ terms do not vanish. Keeping them: $I_{x} + I_{y} = \int (x^{2} + 2 z^{2} + y^{2}) d m = I_{z} + 2 \int z^{2} d m > I_{z} .$ So for a 3-D body $I_{x} + I_{y}$ overcounts $I_{z}$ by exactly $2 \int z^{2} d m$ . The theorem is planar-only.

Score yourself

Recall What each level tested

L1 — recognise when $z = 0$ actually holds.
L2 — direct plug-in, with and without symmetry.
L3 — reverse the theorem, bring in radius of gyration.
L4 — chain it with the Parallel axis theorem.
L5 — prove an inequality and diagnose the failure mode. If any L4/L5 slipped, revisit the parent proof Step 4 — that single $z = 0$ line is the hinge of everything.