Fact-checked by Grok 2 weeks ago

Perpendicular axis theorem

The perpendicular axis theorem, also known as the plane figure theorem, is a fundamental relation in classical mechanics that applies to planar (two-dimensional) objects, stating that the moment of inertia I_z about an axis perpendicular to the plane of the object is equal to the sum of the moments of inertia I_x and I_y about two mutually perpendicular axes lying in that plane and intersecting at the same point.^[1]^[2] Mathematically, this is expressed as I_z = I_x + I_y, where the moments are calculated relative to the common origin, and the theorem holds for any lamina or thin rigid body with mass distribution confined to a single plane, assuming uniform density or arbitrary distribution as long as the axes are properly chosen.^[3]^[4] This theorem is particularly useful in rotational dynamics for simplifying calculations of moments of inertia, especially when the perpendicular-axis moment is known or easier to compute, allowing derivation of in-plane moments by subtraction or symmetry.^[2] For example, it is commonly applied to symmetric shapes like thin disks, rings, or rectangular plates; for a uniform thin disk of mass M and radius R, the moment about the central perpendicular axis is \frac{1}{2}MR^2, so each in-plane axis through the center yields \frac{1}{4}MR^2 due to symmetry.^[3]^[4] The theorem derives from the basic definition of moment of inertia as \int r^2 \, dm, where the perpendicular distance r^2 = x^2 + y^2 in the plane splits naturally into components along the x- and y-axes, making it a direct consequence of vector geometry in rigid body mechanics.^[2] It complements the parallel axis theorem for shifting axes and is essential in engineering applications such as analyzing the rotation of flywheels, blades, or other planar components in machinery.^[1]

Background Concepts

Definition of Moment of Inertia

The moment of inertia, often denoted as I, quantifies a rigid body's resistance to angular acceleration about a specific axis of rotation. For a continuous rigid body, it is defined mathematically as the integral I = \int r^2 \, dm, where r is the perpendicular distance from the axis to the infinitesimal mass element dm, and the integration is taken over the entire mass distribution of the body.^[5] This formulation arises from considering the body as composed of point masses, where the contribution of each is m_i r_i^2, summed or integrated accordingly.^[6] Physically, the moment of inertia serves as the rotational analog to mass in linear motion, determining the torque \tau required to produce a given angular acceleration \alpha via Newton's second law for rotation: \tau = I \alpha.^[7] It depends not only on the total mass but also on how that mass is distributed relative to the axis; masses farther from the axis contribute more significantly to I, increasing the body's reluctance to change its rotational speed. The SI unit of moment of inertia is kilogram square meter (kg·m²).^[8] For rotation about a fixed axis, the moment of inertia is a scalar quantity. However, in general three-dimensional rigid body dynamics, the full description requires the moment of inertia tensor, a 3×3 symmetric matrix that relates the angular velocity vector \vec{\omega} to the angular momentum vector \vec{L} through \vec{L} = \mathbf{I} \vec{\omega}.^[9] This tensor accounts for the coupling between rotations about different axes, with its diagonal elements representing moments about the principal axes and off-diagonal elements (products of inertia) capturing cross-terms. The concept emerged in the 18th century through the work of Leonhard Euler, who formalized it in his 1765 treatise Theoria Motus Corporum Solidorum as "momentum inertiae," building on earlier ideas from János András Segner on gyroscopic motion and rigid body principles.^[10] Further refinements in the 19th century, including vector formulations by Louis Poinsot and William Rankine, solidified its role in classical mechanics.^[10] For planar bodies, where the mass distribution is confined to a single plane, the moment of inertia simplifies, often focusing on the axis perpendicular to that plane.^[8]

Moments for Planar Bodies

In the context of rigid body dynamics, planar bodies, also known as laminae, are idealized as two-dimensional objects with mass distributed uniformly or arbitrarily within the xy-plane and negligible thickness along the z-direction. This assumption simplifies the analysis by confining all mass elements to z = 0, allowing the moment of inertia to be computed via double integrals over the area of the lamina.^[11] The density function ρ(x, y) describes the mass per unit area, which may be constant for uniform laminae or vary for non-uniform distributions, enabling the total mass m to be expressed as m = ∬_D ρ(x, y) dA, where D is the region occupied by the body.^[12] The coordinate system is typically established with the origin at an arbitrary point or, preferably, the center of mass for rotational dynamics calculations, and the z-axis oriented perpendicular to the xy-plane. For axes lying in the plane of the lamina, the moment of inertia quantifies resistance to rotation about those directions: about the x-axis, it is I_x = ∬_D y² ρ(x, y) dA, where y represents the perpendicular distance from the axis to each mass element dm = ρ(x, y) dA; similarly, about the y-axis, I_y = ∬_D x² ρ(x, y) dA, with x as the distance.^[13] In contrast, the moment of inertia about the z-axis, perpendicular to the plane, accounts for distances in both in-plane directions and is given by I_z = ∬_D (x² + y²) ρ(x, y) dA.^[11] To illustrate, consider a single point mass m positioned at coordinates (x, y) within the plane, which serves as the fundamental building block for extended bodies. The moment of inertia of this point mass about the z-axis passing through the origin is simply I_z = m (x² + y²), where √(x² + y²) is the radial distance from the axis; about the x-axis, it would be I_x = m y², and about the y-axis, I_y = m x².^[11] This example highlights how the perpendicular axis configuration leverages the planar geometry, with distances measured solely in the xy-plane.^[14]

Statement of the Theorem

Mathematical Formulation

The perpendicular axis theorem states that for a lamina or planar mass distribution lying in the xy-plane, the moment of inertia I_z about an axis perpendicular to the plane (the z-axis) passing through a given point is equal to the sum of the moments of inertia I_x and I_y about two mutually perpendicular axes (the x- and y-axes) in the plane that intersect at the same point.^[1]^[15] In terms of the inertia tensor for such a two-dimensional mass distribution, the theorem corresponds to the relation among the diagonal elements: I_{zz} = I_{xx} + I_{yy}, where the off-diagonal elements involving the z-direction vanish due to the planarity.^[16]^[17] The axes must be mutually perpendicular and concurrent at the same point, which is often taken as the center of mass for convenience, though the theorem holds for any common intersection point in the plane.^[1]^[4] This relation applies exclusively to planar (two-dimensional) mass distributions; for three-dimensional bodies with thickness or extent along the perpendicular direction, the equality becomes an inequality I_{zz} \leq I_{xx} + I_{yy}, with equality only in the limiting case of zero thickness.^[16]^[18]

Geometric Interpretation

The perpendicular axis theorem provides an intuitive geometric understanding of how the moment of inertia about an axis perpendicular to a plane relates to those about in-plane axes. For a planar mass distribution in the xy-plane, consider a small mass element dm located at coordinates (x, y). The distance from this element to the z-axis (perpendicular to the plane) is the radial distance r = √(x² + y²), so the squared distance r² = x² + y² represents the perpendicular distance contributing to rotational resistance about the z-axis. This decomposition shows that the contributions to the moment of inertia from deviations along the x and y directions add directly, as the total "spread" of mass relative to the z-axis is the quadrature sum of the individual spreads along the perpendicular in-plane axes.^[15] This additive nature illustrates why the theorem holds for planar bodies: each mass element's contribution to the perpendicular moment I_z is the sum of its separate contributions to I_x and I_y, reflecting independent "resistances" to rotation from the x- and y-directions. For instance, a mass element far from the x-axis but close to the y-axis will primarily contribute to I_x (due to its y-deviation) and thus to I_z, but minimally to I_y; conversely, elements offset primarily in x will dominate I_y and I_z. This independent summation underscores the theorem's utility in visualizing how in-plane mass distributions build the overall rotational inertia about the out-of-plane axis.^[1] The geometric rationale draws a direct analogy to the Pythagorean theorem, where the hypotenuse squared (r²) equals the sum of the squares of the legs (x² + y²) in a right triangle formed by the axes and the position vector to the mass element. Just as perpendicular distances combine in quadrature to yield the total distance, the moments of inertia about perpendicular in-plane axes combine additively to give the perpendicular moment, providing a visual bridge between Euclidean geometry and rotational dynamics.^[15] To visualize this, imagine a coordinate system with the xy-plane containing the lamina and the z-axis piercing the origin perpendicularly. For example, mass elements clustered along the x-axis (where y=0) contribute significantly to I_y due to their distances from the y-axis (x-coordinates), while contributions to I_x are zero since y=0, so I_z = I_y. Off-axis elements, such as those in the first quadrant, contribute to both I_x and I_y via their respective distances, with the z-axis seeing the full Pythagorean combination. A conceptual diagram would show arrows from the origin along x, y, and z, with position vectors to sample masses highlighting how x- and y-offsets vectorially sum to r, emphasizing differential contributions to each moment.^[19]

Proof and Derivation

Derivation from First Principles

The perpendicular axis theorem relates the moments of inertia of a planar lamina about three mutually perpendicular axes, with two axes lying in the plane of the lamina and the third perpendicular to it. To derive this from first principles, begin with the general definition of the moment of inertia for a continuous mass distribution about an arbitrary axis, which is the integral of the squared perpendicular distance from the axis to each mass element dm. For a lamina confined to the xy-plane (where the z-coordinate is zero for all mass elements), consider the moments about the coordinate axes originating at some point in the plane. The moment of inertia I_z about the z-axis, perpendicular to the plane, is the integral over the entire mass distribution:

I_z = \int (x^2 + y^2) \, dm,

where x and y are the coordinates of the mass element dm relative to the origin, and the integral extends over the area of the lamina.^[2]^[16] Now, the moment of inertia I_x about the x-axis (lying in the plane) is defined as the integral of the squared distance to this axis. For points in the xy-plane, the perpendicular distance to the x-axis is |y| (with z = 0), so

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

since z = 0. Similarly, the moment of inertia I_y about the y-axis is

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

These expressions follow directly from the general three-dimensional definition of moment of inertia, specialized to the planar case where the thickness in the z-direction is negligible.^[2]^[16] Adding I_x and I_y yields

I_x + I_y = \int y^2 \, dm + \int x^2 \, dm = \int (x^2 + y^2) \, dm = I_z.

This equality holds without additional assumptions beyond the planarity of the lamina and the use of Cartesian coordinates. For a lamina with constant areal mass density \sigma, each mass element is dm = \sigma \, dA, where dA = dx \, dy is the differential area element in the xy-plane, so the integrals become double integrals over the region's boundaries:

I_z = \sigma \iint_R (x^2 + y^2) \, dx \, dy, \quad I_x = \sigma \iint_R y^2 \, dx \, dy, \quad I_y = \sigma \iint_R x^2 \, dx \, dy,

with R denoting the area of the lamina, confirming the relation I_z = I_x + I_y through direct integration. This step-by-step setup demonstrates the theorem's origin in the additive nature of squared distances in the plane.^[16]

Using Coordinate Axes

The perpendicular axis theorem can be derived using the inertia tensor in a Cartesian coordinate system for a planar body confined to the xy-plane, where the z-coordinate is zero for all mass elements. The inertia tensor \mathbf{I} for such a body, with respect to axes passing through a common point (typically the center of mass), has diagonal components given by

I_{xx} = \int (y^2 + z^2) \, dm = \int y^2 \, dm, \quad I_{yy} = \int (x^2 + z^2) \, dm = \int x^2 \, dm, \quad I_{zz} = \int (x^2 + y^2) \, dm,

since z = 0. The off-diagonal products of inertia, such as I_{xy} = \int xy \, dm, may be nonzero unless the axes are chosen as principal axes, where these terms vanish by diagonalization of the tensor.^[20]^[21] The theorem emerges directly from these components: I_{zz} = \int x^2 \, dm + \int y^2 \, dm = I_{yy} + I_{xx}. This relation holds for any pair of orthogonal axes x and y in the plane, as the sum I_{xx} + I_{yy} equals the invariant trace of the in-plane sub-tensor, which is independent of rotation about the z-axis and always matches I_{zz}. In principal axes, where off-diagonal terms are zero, I_{xx} and I_{yy} represent the principal moments, but the theorem's validity does not require this condition.^[22]^[16] For a general three-dimensional body, the relation does not hold as an equality because nonzero z-coordinates contribute to I_{xx} and I_{yy} via the z^2 terms, making I_{xx} + I_{yy} > I_{zz}. The full trace of the inertia tensor, I_{xx} + I_{yy} + I_{zz} = 2 \int (x^2 + y^2 + z^2) \, dm, remains invariant under orthogonal transformations of the axes, but the planar simplification relies on the absence of z-contributions.^[16]^[21]

Applications and Examples

Simple Geometric Shapes

The perpendicular axis theorem provides a practical method to verify moments of inertia for simple planar shapes by relating the polar moment I_z to the in-plane moments I_x and I_y. For these examples, the shapes are assumed to be uniform laminae lying in the xy-plane with the centroid at the origin, and all axes pass through the centroid. The calculations demonstrate the theorem's relation I_z = I_x + I_y, where moments are computed using standard formulas derived from integration over the mass distribution. Consider a rectangular lamina of mass M, length a along the x-axis, and width b along the y-axis. The moment of inertia about the x-axis is I_x = \frac{1}{12} M b^2, as this integrates the squared distances in the y-direction. Similarly, about the y-axis, I_y = \frac{1}{12} M a^2, integrating squared distances in the x-direction. Applying the theorem yields I_z = I_x + I_y = \frac{1}{12} M (a^2 + b^2). To verify independently, the polar moment I_z = \int (x^2 + y^2) \, dm over the rectangle directly computes to the same value, confirming the relation.^[23] For a thin rod of mass M and length L aligned along the x-axis, the negligible thickness implies I_x \approx 0, since distances in the y- and z-directions are zero. The moment about the y-axis is I_y = \frac{1}{12} M L^2, integrating squared x-distances from -\frac{L}{2} to \frac{L}{2}. By symmetry for the thin rod in the plane, I_z = \frac{1}{12} M L^2. The theorem gives I_z = I_x + I_y \approx \frac{1}{12} M L^2, matching the direct computation of I_z = \int x^2 \, dm, which equals I_y due to the rod's linearity. A uniform circular disk of mass M and radius R serves as another illustration. The moments about the diameter axes are I_x = I_y = \frac{1}{4} M R^2, obtained by integrating squared perpendicular distances over the disk area. The theorem predicts I_z = I_x + I_y = \frac{1}{2} M R^2. Independent calculation of the polar moment I_z = \int r^2 \, dm, using polar coordinates, yields \frac{1}{2} M R^2, verifying the equality and highlighting the theorem's utility for symmetric shapes.^[24] For an equilateral triangular lamina of mass M and side length L, the height is H = \frac{\sqrt{3}}{2} L. Due to symmetry, I_x = I_y = \frac{1}{24} M L^2, where I_x is about the axis parallel to the base through the centroid (equivalent to \frac{1}{18} M H^2). The theorem gives I_z = I_x + I_y = \frac{1}{12} M L^2. Direct integration for I_z = \int (x^2 + y^2) \, dm over the triangular area confirms this result, as the contributions from x^2 and y^2 sum symmetrically.^[25]

Practical Uses in Physics and Engineering

The perpendicular axis theorem plays a key role in rotational dynamics by simplifying the computation of moments of inertia for planar objects, thereby facilitating calculations of torque and angular momentum in systems involving flat components such as turbine blades or solar panels. For instance, when analyzing the rotational stability of a thin disk or lamina under applied torques, the theorem allows engineers to derive the perpendicular moment directly from in-plane values, reducing computational complexity in dynamic simulations.^[1] This approach is particularly valuable in scenarios where direct measurement about the perpendicular axis is impractical, enabling efficient prediction of angular acceleration via Newton's second law for rotation.^[26] In engineering applications, the theorem finds an analog in the second moment of area for structural analysis, where it relates the polar moment to in-plane moments, aiding in the design of beams and shafts to resist torsional deflection. For beam deflection under twisting loads, this relation helps quantify shear stress distribution and ensures structural integrity in components like propeller shafts or bridge girders.^[27] Similarly, in flywheel design for energy storage systems, such as those in gyroscopes, the theorem is used to compute the moment of inertia for symmetric planar flywheels, optimizing rotational inertia to maintain stable angular momentum during operation.^[28] In satellite attitude control, it supports modeling the dynamics of planar solar arrays or debris panels, where accurate inertia tensors are essential for predicting tumbling motion and stabilizing orientation in orbit.^[29] For composite bodies, the theorem is applied to individual laminas to determine their in-plane moments before combining with the parallel axis theorem to account for offsets, which is crucial in assembling complex structures like multi-layer panels or rotor assemblies. This method streamlines the inertia calculation for offset components in machinery.^[1] In modern contexts, such as robotics, it aids in computing arm link inertias for cylindrical or planar segments, informing dynamic control algorithms to minimize energy loss during motion.^[30] Software tools like MATLAB leverage the theorem in 2D simulations for verifying moment calculations in robotic or mechanical designs, allowing rapid iteration on prototypes. Despite its utility, the theorem assumes perfect planarity, limiting its direct application to strictly 2D objects; for components with slight thickness, such as thin plates, approximations are used or full 3D inertia tensors are computed to avoid errors in dynamic predictions.^[1] These practical constraints highlight the need for validation against experimental data in real-world implementations. The theorem often builds on moments derived for simple geometric shapes to extend analysis to these engineered systems.

References

[1]
Perpendicular Axis Theorem - HyperPhysics
Perpendicular Axis Theorem. For a planar object, the moment of inertia about an axis perpendicular to the plane is the sum of the moments of inertia of two ...
[2]
Center of Mass; Moment of Inertia - Feynman Lectures - Caltech
If the object is a plane figure, the moment of inertia about an axis perpendicular to the plane is equal to the sum of the moments of inertia about any two ...<|control11|><|separator|>
[3]
[PDF] Classical Mechanics LECTURE 23: MoI THEOREMS AND EXAMPLES
The perpendicular axis theorem links Iz (MoI about an axis perpendicular to the plane) with Ix , Iy (MoI about two perpendicular axes lying within the plane). ▻ ...
[4]
Moment of inertia - Richard Fitzpatrick
The first of these theorems is called the perpendicular axis theorem, and only applies to uniform laminar objects. ... Let us use the perpendicular axis theorem ...
[5]
Moment of Inertia - HyperPhysics
The moment of inertia of a point mass with respect to an axis is defined as the product of the mass times the distance from the axis squared.Moment of Inertia Concepts · Uniform rod · Parallel Axis Theorem · Sphere
[6]
[PDF] Chapter 10: Moments of Inertia
Mass moment of inertia is the mass property of a rigid body that determines the torque needed for a desired angular acceleration () about an axis of rotation ( ...
[7]
[PDF] Rotational Dynamics Moment of Inertia.
Nov 18, 2013 · Moment of inertia (I) is the rotational equivalent of mass, and it depends on mass distribution and the axis of rotation.
[8]
10.4 Moment of Inertia and Rotational Kinetic Energy
We use the definition for moment of inertia for a system of particles and perform the summation to evaluate this quantity. The masses are all the same so we ...
[9]
The Feynman Lectures on Physics Vol. II Ch. 31: Tensors
The moment of inertia, then, is a tensor of the second rank whose terms are a property of the body and relate L to ω by Li=∑jIijωj. For a body of any shape ...
[10]
[PDF] A Historical Discussion of Angular Momentum and its Euler Equation
This paper discusses the history of angular momentum and its Euler equation, aiming to provide a short outline for teaching purposes.
[11]
Mass, center of mass, and moment of inertia
Since the moment of inertial of a little box of size dA at position (x,y) is (x2+y2)ρ(x,y)dA, the moment of inertia of the entire lamina is I=∬D(x2+y2)ρ(x,y)dA.
[12]
[PDF] MTH-201 Calculus with Analytic Geometry III Lecture
(2) Second moments are also called the moment of inertia of a lamina about a line: Ix = y2𝜌 x,y dA. R and Iy = x2𝜌 x,y dA. R. The sum of Ix ...
[13]
[PDF] MATH 233 LECTURE 30: APPLICATIONS OF DOUBLE INTEGRALS ...
I := mr2 is called the moment of inertia. By integrating, we can define moments of inertia of a lamina about the y-axis, x-axis, and origin: Iy = ˜. D x2ρ(x ...
[14]
[PDF] moments of inertia
where r is the perpendicular distance from the axis to an increment of mass dm, k is the radius of gyration, and m is the total mass of the body.<|control11|><|separator|>
[15]
Perpendicular Axis Theorem
Perpendicular Axis Theorem. In general, for any 2D distribution of mass, the moment of inertia about an axis orthogonal to the plane of the mass equals the ...
[16]
[PDF] Perpendicular-Axis Theorem for Volumes
Perpendicular-Axis Theorem for Volumes—C.E. Mungan, Spring 2005. Reference ... provided the origin of the axes is chosen to be at the center of the object.
[17]
[PDF] PDF - 8.012 Physics I: Classical Mechanics
Verify your answers by showing that these moments of inertia satisfy the perpendicular axis theorem, Ix + Iy = Iz, where Iz is the moment of inertia about an ...
[18]
[PDF] Chapter 4 Rigid Body Motion - Rutgers Physics
For an object which has some thickness, with non-zero z components, the perpendicular axis theorem becomes an inequality, Izz ≤ Ixx + Iyy. Principal axes.
[19]
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Variational_Principles_in_Classical_Mechanics_(Cline)/13%3A_Rigid-body_Rotation/13.09%3A_Perpendicular-axis_Theorem_for_Plane_Laminae
[20]
[PDF] variational principles - classical mechanics
Jan 15, 2019 · ... Inertia tensor ... Perpendicular-axis theorem for plane laminae . . . . . . . . . . . . . . . . . . . . . . . . . . . 318. 13.10General ...
[21]
[PDF] 3D Rigid Body Dynamics: The Inertia Tensor - MIT OpenCourseWare
The parallel axis theorem introduced in lecture L22 for the two dimensional moments of inertia can be extended and applied to each of the components of the ...
[22]
[PDF] The Tensor of Inertia
Feb 19, 2006 · An aside – note that for the transformed matrix the perpendicular axis theorem still holds and the trace of the matrix was unchanged under the ...
[23]
3D Centroid and Mass Moment of Intertia Table - Mechanics Map
Center of Mass and Mass Moments of Inertia for Homogeneous Bodies ; Flat Rectangular Plate. Centroid of a Flat Rectangular Plate, Ixx=112mh2 I x x = 1 12 m h 2 I ...Missing: lamina | Show results with:lamina
[24]
Circular Disk Rotating About Its Diameter - Stanford CCRMA
Thus, the uniform disk's moment of inertia in its own plane is twice that about its diameter.
[25]
[PDF] 1 Moments of Inertia of Regular Polygons—C.E. Mungan, Fall 2024 ...
Finally, the parallel-axis theorem gives the moment of inertia about an axis perpendicular to the plane of the triangle but passing through point O in Fig.
[26]
Perpendicular Axis Theorem - (Principles of Physics I) - Fiveable
The perpendicular axis theorem states that for a flat, planar object, the moment of inertia about an axis perpendicular to the plane is equal to the sum of the ...
[27]
Understanding the Area Moment of Inertia | The Efficient Engineer
May 3, 2024 · This is called the perpendicular axis theorem. Perpendicular Axis Theorem. J = I x + I y. The derivation for this theorem is pretty simple ...
[28]
https://phys.libretexts.org/Bookshelves/Classical_Mechanics/Classical_Mechanics_(Dourmashkin)/22%3A_Three_Dimensional_Rotations_and_Gyroscopes/22.02%3A_Gyroscope
[29]
[PDF] Light Curve Inversion for Attitude Reconstruction of Tumbling Space ...
Then, thanks to the perpendicular axis theorem [20]: Iz = Ix + Iy = 1/6. (2.6). Figure 2.4: Flat Square Plate: B and PA frames coincide; the plate has an ...
[30]
[PDF] Web enabled robot design and dynamic control simulation software ...
Using the perpendicular axis theorem the Inertia along the other two axes is given by: 1. 2. 12=I,=-m,5. 4. --- (4). --- (5) where mi is a Mass of the ith link.
[31]
Rotation about CM - Can I use Moment of Inertia? - Physics Forums
Mar 27, 2012 · The simulink finally compiles the MATLAB code into C++ code, so ... Moment of inertia (Perpendicular axis theorem). Oct 15, 2020. Replies ...