Perpendicular Axis Theorem: Statement, Proof, Examples

The perpendicular axis theorem states that for a plane lamina, the moment of inertia about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia about two mutually perpendicular axes lying in that plane and meeting the perpendicular axis at the same point: Iz = Ix + Iy. The one restriction that decides every problem you will meet is this: the theorem holds only for plane laminae, that is, flat two-dimensional bodies of negligible thickness. It does not hold for a sphere, a cylinder, a cone or any other solid. The parallel axis theorem, by contrast, works for every body.

Statement of the theorem of perpendicular axis

Take a flat lamina of any shape lying in the xy-plane. Choose any point O on the lamina. Draw the x and y axes in the plane of the lamina through O, at right angles to each other. Draw the z axis through the same point O, perpendicular to the lamina. Then:

I_z = I_x + I_y

Three conditions have to hold together, and dropping any one of them makes the result wrong:

The body must be a plane lamina. All of its mass lies in one plane. A thin sheet, a disc, a ring, a flat plate, a machine-cut section.
The x and y axes must be perpendicular to each other and must both lie in the plane of the lamina.
All three axes must be concurrent. They meet at one common point O. That point does not have to be the centre of mass. It can be a corner, a rim point, anywhere on or off the body, as long as the three axes pass through it.

Notice what the theorem does not require: the lamina need not be symmetric, uniform in shape, or of constant density. Any flat body works.

Proof of the perpendicular axis theorem

Let the lamina lie in the xy-plane, so that every mass element has z = 0. Consider a small element of mass dm at the point (x, y, 0).

Moment of inertia about the x-axis is the sum of dm times the square of the distance from the x-axis. That distance squared is y² + z², and z = 0, so:

I_x = ∫(y² + z²) dm = ∫y² dm

Similarly, about the y-axis:

I_y = ∫(x² + z²) dm = ∫x² dm

About the z-axis, the perpendicular distance of the element from the axis is r, where r is its distance from O measured in the plane. By Pythagoras, r² = x² + y². So:

I_z = ∫r² dm = ∫(x² + y²) dm = ∫x² dm + ∫y² dm

The two integrals on the right are exactly I_y and I_x. Therefore:

I_z = I_x + I_y

The whole proof turns on one step: r² = x² + y² with no z² term, and that is true only because z = 0 for every element of a lamina.

Why it applies only to plane laminae

Run the same algebra for a solid body, where elements have a non-zero z:

I_x + I_y = ∫(y² + z²) dm + ∫(x² + z²) dm = ∫(x² + y²) dm + 2∫z² dm

So for any body, I_x + I_y = I_z + 2∫z² dm. The leftover term 2∫z² dm is the price of thickness. It vanishes only when every element sits at z = 0, which is the definition of a lamina. For a solid body the sum I_x + I_y is always larger than I_z.

Two quick demonstrations that the theorem breaks for solids:

Solid sphere. By symmetry I_x = I_y = I_z = 2MR²/5. If the theorem applied, we would need 2MR²/5 = 4MR²/5, which is false.
Solid cylinder of radius R and length L about its own axis: I_z = MR²/2, while I_x = I_y = M(3R² + L²)/12. Their sum is M(3R² + L²)/6 = MR²/2 + ML²/6, and the extra ML²/6 is exactly 2∫z² dm. Let L go to zero and the cylinder becomes a disc, the extra term disappears, and the theorem is recovered.

Student tip: if the question says “solid”, “sphere”, “cone” or gives a thickness that matters, the perpendicular axis theorem is not the tool. If it says “thin”, “lamina”, “plate”, “sheet”, “disc” or “ring”, it is.

Applications: deriving standard results

Circular disc

For a uniform disc of mass M and radius R, the moment of inertia about the axis through the centre and perpendicular to the disc is I_z = MR²/2. A diameter is an in-plane axis through the centre, and by symmetry every diameter gives the same value, so I_x = I_y.

I_z = I_x + I_y = 2I_x, so I_diameter = MR²/4.

That is the standard book result, obtained in one line instead of a fresh integration.

Thin circular ring

Every element of a ring is at distance R from the central perpendicular axis, so I_z = MR² directly. Applying the theorem with I_x = I_y:

I_diameter = MR²/2

Annular disc (washer)

For an annulus of inner radius R₁ and outer radius R₂, I_z = M(R₁² + R₂²)/2. Symmetry again gives I_x = I_y, so each diameter gives:

I_diameter = M(R₁² + R₂²)/4

Set R₁ = 0 and you get the disc; set R₁ = R₂ = R and you get the ring. A good check on any annulus formula.

Rectangular lamina

Take a rectangular plate of mass M with side b along x and side d along y, axes through the centroid. Standard results give I_x = Md²/12 and I_y = Mb²/12. The theorem then gives the perpendicular axis through the centre straight away:

I_z = M(b² + d²)/12

Here the two in-plane values are different, so this case shows the theorem does not need any symmetry.

Square lamina

For a square of side a, I_x = I_y = Ma²/12, so I_z = Ma²/6.

The square hides a neat consequence. Because I_x + I_y must equal the same I_z for any pair of perpendicular in-plane axes through the centre, and symmetry forces the two to be equal for the square, the moment of inertia of a square lamina about any axis in its plane through the centre, including both diagonals, is Ma²/12. This is a favourite one-mark question.

Standard moments of inertia for plane laminae

Lamina (mass M)	In-plane axes through centroid	Perpendicular axis I_z
Thin ring, radius R	I_x = I_y = MR²/2	MR²
Circular disc, radius R	I_x = I_y = MR²/4	MR²/2
Annular disc, R₁ and R₂	I_x = I_y = M(R₁² + R₂²)/4	M(R₁² + R₂²)/2
Rectangular plate, b × d	I_x = Md²/12, I_y = Mb²/12	M(b² + d²)/12
Square plate, side a	I_x = I_y = Ma²/12	Ma²/6
Elliptical plate, semi-axes a and b	I_x = Mb²/4, I_y = Ma²/4	M(a² + b²)/4
Thin rod, length L (lying along x)	I_x = 0, I_y = ML²/12	ML²/12

Every row satisfies I_z = I_x + I_y. Use that as your check before writing an answer: if the three numbers in a row do not add up, one of them is wrong.

Perpendicular axis theorem vs parallel axis theorem

The parallel axis theorem states that the moment of inertia about any axis equals the moment of inertia about a parallel axis through the centre of mass, plus Md², where d is the perpendicular distance between the two axes:

I = I_cm + Md²

Point of difference	Perpendicular axis theorem	Parallel axis theorem
Formula	I_z = I_x + I_y	I = I_cm + Md²
Applies to	Plane laminae only	Any body, 2D or 3D
Axes involved	Three mutually perpendicular axes meeting at one point	Two parallel axes, one through the centre of mass
Reference axis	Any common point, centroid not required	Must start from the centroidal axis
Shifts the axis?	No, it changes direction at the same point	Yes, it moves the axis sideways by d
Typical use	Disc about a diameter, polar moment of a section	Rod about one end, disc about a tangent

They answer different questions. The perpendicular axis theorem turns one direction into another at a fixed point. The parallel axis theorem moves a fixed direction to a new point. Most exam problems need both, one after the other.

Worked example using both theorems

Problem. A uniform circular disc has mass M = 2 kg and radius R = 0.1 m. Find its moment of inertia about a tangent lying in the plane of the disc.

Step 1, start from the known value. About the central perpendicular axis, I_z = MR²/2 = 0.5 × 2 × (0.1)² = 0.01 kg·m².

Step 2, perpendicular axis theorem. A diameter is an in-plane centroidal axis, and I_x = I_y by symmetry, so I_diameter = I_z/2 = 0.005 kg·m².

Step 3, parallel axis theorem. The tangent in the plane is parallel to a diameter and a distance d = R = 0.1 m away from it, so I_tangent = 0.005 + 2 × (0.1)² = 0.005 + 0.02 = 0.025 kg·m².

In symbols, MR²/4 + MR² = 5MR²/4, and 1.25 × 2 × 0.01 = 0.025 kg·m². The two agree.

If instead the tangent is perpendicular to the plane of the disc, skip step 2 and go straight from I_z: I = MR²/2 + MR² = 3MR²/2 = 0.03 kg·m².

Second worked example: plate about a corner

Problem. A rectangular plate of mass M = 5 kg has sides b = 0.3 m and d = 0.2 m. Find the moment of inertia about an axis through one corner, perpendicular to the plate.

Centroidal in-plane values: I_x = Md²/12 = 5 × 0.04/12 = 0.01667 kg·m², I_y = Mb²/12 = 5 × 0.09/12 = 0.0375 kg·m².

Perpendicular axis theorem at the centroid: I_z,cm = 0.01667 + 0.0375 = 0.05417 kg·m².

Distance from centroid to a corner: d² = (0.15)² + (0.10)² = 0.0325 m².

Parallel axis theorem: I_z,corner = 0.05417 + 5 × 0.0325 = 0.2167 kg·m².

Check by applying the perpendicular axis theorem at the corner instead. About the two edges meeting at that corner, I_x = Md²/3 = 0.06667 kg·m² and I_y = Mb²/3 = 0.15 kg·m². Their sum is 0.2167 kg·m², the same answer. This works because the theorem is valid at any point of the lamina, not just the centroid.

Radius of gyration

The radius of gyration k is the distance from the axis at which the whole mass could be concentrated without changing the moment of inertia:

I = Mk², so k = √(I/M)

Divide the theorem through by M and it transfers directly to radii of gyration for a lamina:

k_z² = k_x² + k_y²

For the 2 kg, 0.1 m disc above, k_z = √(0.01/2) = 0.0707 m, which is R/√2, and about a diameter k = √(0.005/2) = 0.05 m = R/2. Check: 0.05² + 0.05² = 0.005 = 0.0707².

Area moment of inertia and the polar moment

The same theorem holds for the area (second) moment of inertia of a plane cross-section, because a section is a lamina by definition. In that form it gives the polar moment of inertia J used in torsion:

J = I_x + I_y

Solid circular shaft, diameter D: I_x = I_y = πD⁴/64, so J = πD⁴/32.
Hollow shaft, outer D and inner d: J = π(D⁴ − d⁴)/32.
Rectangular section b × d: I_x = bd³/12 and I_y = db³/12, so J = bd(b² + d²)/12.

Units differ from the mass version: area moments are in m⁴ or mm⁴, mass moments in kg·m². Keep the two apart in the same problem, and keep both apart from the strain relations you use in the stress part of the question.

Common mistakes

Using it on a solid body. The single biggest error. A sphere, cone, cylinder or thick block is out of scope.
Axes not concurrent. Taking I_x about the centroid and I_y about an edge and adding them gives a meaningless number. Shift both to one point with the parallel axis theorem first.
In-plane axes not at 90°. Two arbitrary in-plane axes do not add to I_z.
Making z an in-plane axis. The z axis is the one sticking out of the sheet; I_z is always the largest of the three for a lamina.
Adding instead of subtracting for a composite body. For a plate with a hole, compute the full plate and subtract the removed disc, each transferred to the common axis.

References

FAQs

What is the perpendicular axis theorem?

For a plane lamina, the moment of inertia about an axis perpendicular to the lamina equals the sum of the moments of inertia about two perpendicular axes lying in the plane of the lamina and passing through the same point: Iz = Ix + Iy. The z axis is normal to the sheet, and x and y lie in it.

Why does the perpendicular axis theorem apply only to plane laminae?

Because the proof needs z = 0 for every mass element. For a general body, Ix + Iy = Iz + 2∫z² dm, and that extra term vanishes only when all the mass lies in a single plane. For a solid sphere, Ix + Iy = 4MR²/5 while Iz = 2MR²/5, so the theorem clearly fails.

What is the difference between the parallel and perpendicular axis theorems?

The perpendicular axis theorem relates three mutually perpendicular axes meeting at one point and works only for flat laminae. The parallel axis theorem, I = Icm + Md², shifts an axis sideways from the centre of mass by a distance d and works for any body. Many problems use both in sequence.

What is the moment of inertia of a disc about its diameter?

MR²/4. The disc has Iz = MR²/2 about the central perpendicular axis, and by symmetry Ix = Iy, so each diameter carries half of Iz. For a thin ring the same argument gives MR²/2 about a diameter.

Can the perpendicular axis theorem be applied at a point other than the centre of mass?

Yes. It holds at any point of the lamina as long as all three axes pass through that point, the x and y axes lie in the plane and are perpendicular to each other. For a rectangular plate at a corner, Md²/3 + Mb²/3 = M(b² + d²)/3, which matches the parallel axis result.

Perpendicular Axis Theorem: Statement, Proof and Applications