Parallel axis theorem
Short Description of Topic
claimed by Fabian Nunez Fall 2025
The Main Idea
The Parallel Axis Theorem is used to interpret the moment of inertia (I) for any axis parallel to the axis through the center line used to calculate the moment of inertia by using the object's mass and distance between the axes.
A Mathematical Model
The formula for the Parallel Axis Theorem is
[math]\displaystyle{ {I}_{parallel} = {I}_{cm} + Md^2 }[/math]
where:
[math]\displaystyle{ {I}_{cm} }[/math] is the moment of inertia about the center of mass. (Usually original / easier to solve)
[math]\displaystyle{ M }[/math] is the total mass of the object.
[math]\displaystyle{ d }[/math] is the distance between the Parallel Axis and the Center of Mass axis.
Proof
If we start by setting our axis along a normal Cartesian plane, centered around the center of mass, we can define the moment of inertia about a line parallel to the center of mass as an integral with respect the object's mass.
[math]\displaystyle{ {I}_{cm} = \int r^2 dm }[/math]
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, [math]\displaystyle{ r= \surd (x^2+y^2) }[/math]
We can plug this in to redefine our equation as [math]\displaystyle{ {I}_{cm} = \int \surd (x^2+y^2) dm }[/math]
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis
[math]\displaystyle{ {I}_{new axis} = \int \surd ((x-D)^2+y^2) dm }[/math]
This expands into
[math]\displaystyle{ {I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm }[/math]
We can solve this integral by separating the Distance terms, where the term [math]\displaystyle{ \int -2Dx dm }[/math] evaluates to 0 (the x coordinate of the center of mass), and the term [math]\displaystyle{ \int D^2 dm }[/math] is simplifies to [math]\displaystyle{ D^2 \int dm }[/math]. Our final solution is
[math]\displaystyle{ {I}_{new axis} = {I}_{cm} + MD^2 }[/math]
Examples
Remember the formula for the Parallel Axis Theorem is
[math]\displaystyle{ {I}_{parallel} = {I}_{cm} + Md^2 }[/math]
Simple
Problem
A thin rod of mass m and length L has a moment of interia around the center axis as [math]\displaystyle{ {I}_{center} = (1/12)md^2 }[/math] Find the moment of inertia about a line passing through one end of the rod parallel to the center axis.
Solution Distance d = L/2
[math]\displaystyle{ {I}_{parallel} = (1/3)ML^2 }[/math]
Middling
Problem A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as [math]\displaystyle{ {I}_{cm} = (1/2)mR^2 }[/math]. Find the moment of interia of a line parallel to the axis on the edge of the disk
Solution 1st, find distance from center to edge (radius). Radius = R = D/2 = 3
2nd, setup formula
[math]\displaystyle{ {I}_{parallel} = {I}_{cm} + Md^2 }[/math] => [math]\displaystyle{ {I}_{parallel} = (1/2)M(R^2) + M(D/2)^2 }[/math] => [math]\displaystyle{ {I}_{parallel} = (3/2)MR^2 }[/math]
3rd, plug in variables.
[math]\displaystyle{ {I}_{parallel} = 135 kg * m^2 }[/math]
Difficult
Problem I have a thin-rod-shaped stick of length L = 1 m and mass 5 kg. The moment of inertia of this stick about the center of mass is given by [math]\displaystyle{ {I}_{center} = (1/12)mL^2 }[/math] [the axis is perpendicular to the length of the stick]. I apply massless adhesive to one end of a stick, then I touch the center of the top face of a cube with side length =.5 m and mass = 6 kg. What is total moment of inertia of a line parallel to the axis perpendicular to the stick's length of the combined object about the new center of mass? The moment of inertia of the cube is [math]\displaystyle{ I = (1/6)ml^2 }[/math]
Solution 1st, identify variables. [math]\displaystyle{ {I}_{stick} = (1/12){m}_{stick}L^2 }[/math]
[math]\displaystyle{ {m}_{stick} = 5 kg }[/math]
[math]\displaystyle{ {I}_{block} = (1/6){m}_{block}(S)^2 }[/math]
[math]\displaystyle{ {m}_{block} = 6 kg }[/math]
Side length = S = .5 m
2nd, solve unknowns Distance of new center of mass = [math]\displaystyle{ {\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. }[/math] = [math]\displaystyle{ 5*(0.5)+6*(-.25)/11 = 1/11 }[/math] (Remember the block is below the stick)
3rd, apply parallel axis theorem [math]\displaystyle{ {I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2 }[/math]
[math]\displaystyle{ {I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2 }[/math]
4th, plug in values
Total moment of inertia = [math]\displaystyle{ (1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264} }[/math]
Connectedness
- How is this topic connected to something that you are interested in?
- How is it connected to your major?
- Is there an interesting industrial application?
History
The parallel axis theorem was original stated in Christian Huygens' work "Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae". This document was published in 1673, but was adapted and pasteurized by Jacob Steiner into the formula we use today. It was developed by Huygens while he was researching how to track motion of a pendulum in clock-making.
See also
Are there related topics or categories in this wiki resource for the curious reader to explore? How does this topic fit into that context?
Further reading
Books, Articles or other print media on this topic:
Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae (1673)
External links
Internet resources on this topic
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia
References
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA.
GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/
Moebs, W., Ling, S. J., & Sanny, J. (n.d.). 10.5 calculating moments of inertia - university physics volume 1. OpenStax. https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia