Parallel axis theorem: Difference between revisions

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==Examples==
==Examples==


Be sure to show all steps in your solution and include diagrams whenever possible
Remember


===Simple===
===Simple===
A thin rod of mass m and length L has a moment of interia around the center axis as <math>{I}_{center} = (1/12)md^2</math> Find the moment of interia about a line passing through one end of the rod paralled to the center axis.
===Middling===
===Middling===
===Difficult===
===Difficult===

Revision as of 15:18, 2 December 2025

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

Simple

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 interia about a line passing through one end of the rod paralled to the center axis.

Middling

Difficult

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