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		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47781</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47781"/>
		<updated>2025-12-02T23:13:00Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: &lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Made by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
The parallel axis theorem is connected to statics, which is something I am interested in. It is used to calculate the moment of inertia&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
It is used to in analiss, in stress calculations, in load calculation, and in most parts regarding statics of physics (integral components of structural engineering).&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
It is used to calculate the moment of inertia of complex systems, such as hydraulic machinery or robotics.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The parallel axis theorem was original stated in Christian Huygens&#039; work &amp;quot;Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae&amp;quot;. 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.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic:&lt;br /&gt;
&lt;br /&gt;
Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae (1673)&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/&lt;br /&gt;
&lt;br /&gt;
Huygens, C. (1673). Christ. Hugenii Zulichemii, const. F. Horologium oscillatorium, sive, de Motu pendulorum ad Horologia Aptato Demonstrationes geometricae: Accedunt Ejusdem de Circuli Magnitudine Inventa, et Systema Saturnium. E Theatro Sheldoniano. &lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47777</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47777"/>
		<updated>2025-12-02T23:12:19Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* Connectedness */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
The parallel axis theorem is connected to statics, which is something I am interested in. It is used to calculate the moment of inertia&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
It is used to in analiss, in stress calculations, in load calculation, and in most parts regarding statics of physics (integral components of structural engineering).&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
It is used to calculate the moment of inertia of complex systems, such as hydraulic machinery or robotics.&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The parallel axis theorem was original stated in Christian Huygens&#039; work &amp;quot;Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae&amp;quot;. 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.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic:&lt;br /&gt;
&lt;br /&gt;
Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae (1673)&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/&lt;br /&gt;
&lt;br /&gt;
Huygens, C. (1673). Christ. Hugenii Zulichemii, const. F. Horologium oscillatorium, sive, de Motu pendulorum ad Horologia Aptato Demonstrationes geometricae: Accedunt Ejusdem de Circuli Magnitudine Inventa, et Systema Saturnium. E Theatro Sheldoniano. &lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47772</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47772"/>
		<updated>2025-12-02T23:01:18Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The parallel axis theorem was original stated in Christian Huygens&#039; work &amp;quot;Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae&amp;quot;. 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.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic:&lt;br /&gt;
&lt;br /&gt;
Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae (1673)&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/&lt;br /&gt;
&lt;br /&gt;
Huygens, C. (1673). Christ. Hugenii Zulichemii, const. F. Horologium oscillatorium, sive, de Motu pendulorum ad Horologia Aptato Demonstrationes geometricae: Accedunt Ejusdem de Circuli Magnitudine Inventa, et Systema Saturnium. E Theatro Sheldoniano. &lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47771</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47771"/>
		<updated>2025-12-02T23:01:03Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* Further reading */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The parallel axis theorem was original stated in Christian Huygens&#039; work &amp;quot;Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae&amp;quot;. 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.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic:&lt;br /&gt;
&lt;br /&gt;
Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae (1673)&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/&lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47770</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47770"/>
		<updated>2025-12-02T22:59:11Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* History */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
The parallel axis theorem was original stated in Christian Huygens&#039; work &amp;quot;Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae&amp;quot;. 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.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/&lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47767</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47767"/>
		<updated>2025-12-02T22:50:23Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/&lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47765</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47765"/>
		<updated>2025-12-02T22:35:58Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* External links */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47764</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47764"/>
		<updated>2025-12-02T22:35:42Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47763</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47763"/>
		<updated>2025-12-02T22:34:56Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* References */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA. &lt;br /&gt;
&lt;br /&gt;
Moebs, W., Ling, S. J., &amp;amp; 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 &lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
[[Category: Angular Momentum Priciple]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47762</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47762"/>
		<updated>2025-12-02T22:25:00Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* Difficult */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)mL^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
Distance of new center of mass = &amp;lt;math&amp;gt;{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt; = &amp;lt;math&amp;gt; 5*(0.5)+6*(-.25)/11 = 1/11&amp;lt;/math&amp;gt;&lt;br /&gt;
(Remember the block is below the stick)&lt;br /&gt;
&lt;br /&gt;
3rd, apply parallel axis theorem&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
4th, plug in values&lt;br /&gt;
&lt;br /&gt;
Total moment of inertia = &amp;lt;math&amp;gt;(1/12)*5*1^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
This section contains the the references you used while writing this page&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47756</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47756"/>
		<updated>2025-12-02T22:11:25Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* Examples */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember the formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of inertia about a line passing through one end of the rod parallel to the center axis. &lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
Distance d = L/2&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = (1/3)ML^2 &amp;lt;/math&amp;gt;&lt;br /&gt;
===Middling===&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
A disk of diameter D= 6 m and mass m = 10 kg has its moment of inertia about its center defines as &amp;lt;math&amp;gt;{I}_{cm} = (1/2)mR^2&amp;lt;/math&amp;gt;. Find the moment of interia of a line parallel to the axis on the edge of the disk&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, find distance from center to edge (radius).&lt;br /&gt;
Radius = R = D/2 = 3&lt;br /&gt;
&lt;br /&gt;
2nd, setup formula&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (1/2)M(R^2) + M(D/2)^2&amp;lt;/math&amp;gt; =&amp;gt; &amp;lt;math&amp;gt;{I}_{parallel} = (3/2)MR^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
3rd, plug in variables.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = 135 kg * m^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Problem&#039;&#039;&#039;&lt;br /&gt;
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 &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; [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&#039;s length of the combined object about the new center of mass?&lt;br /&gt;
The moment of inertia of the cube is &amp;lt;math&amp;gt;I = (1/6)ml^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&#039;&#039;&#039;Solution&#039;&#039;&#039;&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{stick} = (1/12){m}_{stick}L^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{stick} = 5 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{block} = (1/6){m}_{block}(S)^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{m}_{block} = 6 kg&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Side length = S = .5 m&lt;br /&gt;
&lt;br /&gt;
2nd, solve unknowns&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; Position of new center of mass ={\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. &amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
1st, identify variables.&lt;br /&gt;
I_cm&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
This section contains the the references you used while writing this page&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47713</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47713"/>
		<updated>2025-12-02T20:18:55Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* Examples */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Remember&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
A thin rod of mass m and length L has a moment of interia around the center axis as &amp;lt;math&amp;gt;{I}_{center} = (1/12)md^2&amp;lt;/math&amp;gt; Find the moment of interia about a line passing through one end of the rod paralled to the center axis. &lt;br /&gt;
===Middling===&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
This section contains the the references you used while writing this page&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47709</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47709"/>
		<updated>2025-12-02T20:05:42Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* The Main Idea */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
&lt;br /&gt;
where:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
===Proof===&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass.&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int r^2 dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where r is the distance between the center of mass and the point mass. Because of the Pythagorean theorem, &amp;lt;math&amp;gt;r= \surd (x^2+y^2)&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can plug this in to redefine our equation as&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm} = \int \surd (x^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Then, if we know D is the distance between our axes, we can plug this in to calculate our new axis&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x-D)^2+y^2) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This expands into &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = \int \surd ((x^2+D^2+y^2-2Dx) dm&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
We can solve this integral by separating the Distance terms, where the term &amp;lt;math&amp;gt; \int -2Dx dm&amp;lt;/math&amp;gt; evaluates to 0 (the x coordinate of the center of mass), and the term &amp;lt;math&amp;gt; \int D^2 dm &amp;lt;/math&amp;gt; is simplifies to &amp;lt;math&amp;gt;D^2 \int dm&amp;lt;/math&amp;gt;. Our final solution is &lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{new axis} = {I}_{cm} + MD^2&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
===Middling===&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
This section contains the the references you used while writing this page&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47696</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47696"/>
		<updated>2025-12-02T19:13:42Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* The Main Idea */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
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&#039;s mass and distance between the axes.&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
The formula for the Parallel Axis Theorem is&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{parallel} = {I}_{cm} + Md^2&amp;lt;/math&amp;gt; &lt;br /&gt;
where:&lt;br /&gt;
&amp;lt;math&amp;gt;{I}_{cm}&amp;lt;/math&amp;gt; is the moment of inertia about the center of mass. (Usually original / easier to solve)&lt;br /&gt;
&amp;lt;math&amp;gt;M&amp;lt;/math&amp;gt; is the total mass of the object.&lt;br /&gt;
&amp;lt;math&amp;gt;d&amp;lt;/math&amp;gt; is the distance between the Parallel Axis and the Center of Mass axis.&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
===Middling===&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
This section contains the the references you used while writing this page&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47586</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47586"/>
		<updated>2025-12-01T16:47:34Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: /* The Main Idea */&lt;/p&gt;
&lt;hr /&gt;
&lt;div&gt;Short Description of Topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
claimed by Fabian Nunez Fall 2025&lt;br /&gt;
==The Main Idea==&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
State, in your own words, the main idea for this topic&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
===A Mathematical Model===&lt;br /&gt;
&lt;br /&gt;
What are the mathematical equations that allow us to model this topic.  For example &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}&amp;lt;/math&amp;gt; where &#039;&#039;&#039;p&#039;&#039;&#039; is the momentum of the system and &#039;&#039;&#039;F&#039;&#039;&#039; is the net force from the surroundings.&lt;br /&gt;
&lt;br /&gt;
===A Computational Model===&lt;br /&gt;
&lt;br /&gt;
How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]&lt;br /&gt;
&lt;br /&gt;
==Examples==&lt;br /&gt;
&lt;br /&gt;
Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
&lt;br /&gt;
===Simple===&lt;br /&gt;
===Middling===&lt;br /&gt;
===Difficult===&lt;br /&gt;
&lt;br /&gt;
==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
&lt;br /&gt;
==History==&lt;br /&gt;
&lt;br /&gt;
Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
&lt;br /&gt;
== See also ==&lt;br /&gt;
&lt;br /&gt;
Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
&lt;br /&gt;
===Further reading===&lt;br /&gt;
&lt;br /&gt;
Books, Articles or other print media on this topic&lt;br /&gt;
&lt;br /&gt;
===External links===&lt;br /&gt;
&lt;br /&gt;
Internet resources on this topic&lt;br /&gt;
&lt;br /&gt;
==References==&lt;br /&gt;
&lt;br /&gt;
This section contains the the references you used while writing this page&lt;br /&gt;
&lt;br /&gt;
[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
	</entry>
	<entry>
		<id>http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47585</id>
		<title>Parallel axis theorem</title>
		<link rel="alternate" type="text/html" href="http://www.physicsbook.gatech.edu/index.php?title=Parallel_axis_theorem&amp;diff=47585"/>
		<updated>2025-12-01T16:45:50Z</updated>

		<summary type="html">&lt;p&gt;FabianN2: Created page with &amp;quot;Short Description of Topic  ==The Main Idea==  Created by: Fabian Nunez Fall 2025  State, in your own words, the main idea for this topic Electric Field of Capacitor  ===A Mathematical Model===  What are the mathematical equations that allow us to model this topic.  For example &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}&amp;lt;/math&amp;gt; where &amp;#039;&amp;#039;&amp;#039;p&amp;#039;&amp;#039;&amp;#039; is the momentum of the system and &amp;#039;&amp;#039;&amp;#039;F&amp;#039;&amp;#039;&amp;#039; is the net force from the surroundings.  ===A Computational Model===  How do we...&amp;quot;&lt;/p&gt;
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&lt;div&gt;Short Description of Topic&lt;br /&gt;
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==The Main Idea==&lt;br /&gt;
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Created by: Fabian Nunez Fall 2025&lt;br /&gt;
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State, in your own words, the main idea for this topic&lt;br /&gt;
Electric Field of Capacitor&lt;br /&gt;
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===A Mathematical Model===&lt;br /&gt;
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What are the mathematical equations that allow us to model this topic.  For example &amp;lt;math&amp;gt;{\frac{d\vec{p}}{dt}}_{system} = \vec{F}_{net}&amp;lt;/math&amp;gt; where &#039;&#039;&#039;p&#039;&#039;&#039; is the momentum of the system and &#039;&#039;&#039;F&#039;&#039;&#039; is the net force from the surroundings.&lt;br /&gt;
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===A Computational Model===&lt;br /&gt;
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How do we visualize or predict using this topic. Consider embedding some vpython code here [https://trinket.io/glowscript/31d0f9ad9e Teach hands-on with GlowScript]&lt;br /&gt;
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==Examples==&lt;br /&gt;
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Be sure to show all steps in your solution and include diagrams whenever possible&lt;br /&gt;
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===Simple===&lt;br /&gt;
===Middling===&lt;br /&gt;
===Difficult===&lt;br /&gt;
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==Connectedness==&lt;br /&gt;
#How is this topic connected to something that you are interested in?&lt;br /&gt;
#How is it connected to your major?&lt;br /&gt;
#Is there an interesting industrial application?&lt;br /&gt;
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==History==&lt;br /&gt;
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Put this idea in historical context. Give the reader the Who, What, When, Where, and Why.&lt;br /&gt;
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== See also ==&lt;br /&gt;
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Are there related topics or categories in this wiki resource for the curious reader to explore?  How does this topic fit into that context?&lt;br /&gt;
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===Further reading===&lt;br /&gt;
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Books, Articles or other print media on this topic&lt;br /&gt;
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===External links===&lt;br /&gt;
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Internet resources on this topic&lt;br /&gt;
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==References==&lt;br /&gt;
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This section contains the the references you used while writing this page&lt;br /&gt;
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[[Category:Which Category did you place this in?]]&lt;/div&gt;</summary>
		<author><name>FabianN2</name></author>
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