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]

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