Parallel axis theorem - Revision history

FabianN2 at 23:13, 2 December 2025

2025-12-02T23:13:00Z

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	~~Short Description of Topic~~		Made by Fabian Nunez Fall 2025


	~~claimed~~ by Fabian Nunez Fall 2025
	==The Main Idea==		==The Main Idea==

FabianN2: /* Connectedness */

2025-12-02T23:12:19Z

Connectedness

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	==Connectedness==		==Connectedness==
	#How is this topic connected to something that you are interested in?		#How is this topic connected to something that you are interested in?
			The parallel axis theorem is connected to statics, which is something I am interested in. It is used to calculate the moment of inertia
	#How is it connected to your major?		#How is it connected to your major?
			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).
	#Is there an interesting industrial application?		#Is there an interesting industrial application?
			It is used to calculate the moment of inertia of complex systems, such as hydraulic machinery or robotics.

	==History==		==History==

FabianN2: /* References */

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References

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	GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/		GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/

			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.

	Moebs, W., Ling, S. J., & Sanny, J. (n.d.). 10.5 calculating moments of inertia - university physics volume 1. OpenStax. https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia		Moebs, W., Ling, S. J., & Sanny, J. (n.d.). 10.5 calculating moments of inertia - university physics volume 1. OpenStax. https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia

FabianN2: /* Further reading */

2025-12-02T23:01:03Z

FabianN2: /* History */

2025-12-02T22:59:11Z

History

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	==History==		==History==
			The parallel axis theorem was original stated in Christian Huygens' work "Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae". This document was published in 1673, but was adapted and pasteurized by Jacob Steiner into the formula we use today. It was developed by Huygens while he was researching how to track motion of a pendulum in clock-making.
	~~Put this idea~~ in ~~historical context~~. ~~Give the reader the Who, What, When, Where~~, and ~~Why~~.

	== See also ==		== See also ==

FabianN2: /* References */

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References

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	(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA.		(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA.

			GeeksForGeeks. (2025, July 23). Parallel axis theotem. Retrieved December 2, 2025 from https://www.geeksforgeeks.org/physics/parallel-axis-theorem/

	Moebs, W., Ling, S. J., & Sanny, J. (n.d.). 10.5 calculating moments of inertia - university physics volume 1. OpenStax. https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia		Moebs, W., Ling, S. J., & Sanny, J. (n.d.). 10.5 calculating moments of inertia - university physics volume 1. OpenStax. https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia

FabianN2: /* External links */

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External links

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	Internet resources on this topic		Internet resources on this topic

			https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia

	==References==		==References==

FabianN2: /* References */

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References

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	[[Category: Angular Momentum ~~Priciple~~]]		[[Category: Angular Momentum]]

FabianN2: /* References */

2025-12-02T22:34:56Z

References

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	==References==		==References==

	~~This section contains the the references you used while writing this page~~		(2021). Parallel Axis Theorem Derivation. Retrieved December 2, 2025, from https://www.youtube.com/watch?v=bjF1--cxgrA.

	[[Category:~~Which Category did you place this in?~~]]		Moebs, W., Ling, S. J., & Sanny, J. (n.d.). 10.5 calculating moments of inertia - university physics volume 1. OpenStax. https://openstax.org/books/university-physics-volume-1/pages/10-5-calculating-moments-of-inertia


			[[Category: Angular Momentum Priciple]]

FabianN2: /* Difficult */

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Difficult

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	'''Problem'''		'''Problem'''
	I have a thin-rod-shaped stick of length L = 1 m and mass 5 kg. The moment of inertia of this stick about the center of mass is given by <math>{I}_{center} = (1/12)md^2</math> [the axis is perpendicular to the length of the stick]. I apply massless adhesive to one end of a stick, then I touch the center of the top face of a cube with side length =.5 m and mass = 6 kg. What is total moment of inertia of a line parallel to the axis perpendicular to the stick's length of the combined object about the new center of mass?		I have a thin-rod-shaped stick of length L = 1 m and mass 5 kg. The moment of inertia of this stick about the center of mass is given by <math>{I}_{center} = (1/12)mL^2</math> [the axis is perpendicular to the length of the stick]. I apply massless adhesive to one end of a stick, then I touch the center of the top face of a cube with side length =.5 m and mass = 6 kg. What is total moment of inertia of a line parallel to the axis perpendicular to the stick's length of the combined object about the new center of mass?
	The moment of inertia of the cube is <math>I = (1/6)ml^2</math>		The moment of inertia of the cube is <math>I = (1/6)ml^2</math>

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	2nd, solve unknowns		2nd, solve unknowns
			Distance of new center of mass = <math>{\sum_{i=1}^n m_i {r}_i\over\sum_{i=1}^n m_i }. </math> = <math> 5(0.5)+6(-.25)/11 = 1/11</math>
			(Remember the block is below the stick)

	<math> ~~Position of new center of mass =~~{~~\sum_~~{i=1}~~^n m_i~~ {r}~~_i\over\sum_~~{~~i=1~~}~~^n m_i~~ }. </math>		3rd, apply parallel axis theorem
			<math>{I}_{stick,parallel} = {I}_{stick} + {m}_{stick}(.5-1/11)^2</math>

			<math>{I}_{block,parallel} = {I}_{block} + {m}_{block}(.25+1/11)^2</math>

			4th, plug in values

			Total moment of inertia = <math>(1/12)51^2 + 5(.5-1/11)^2 + (1/6)6(.5)^2 + 6(.25+1/11)^2 = \frac{581}{264}</math>

	~~1st, identify variables~~.
	~~I_cm~~

	==Connectedness==		==Connectedness==

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	===Further reading===		===Further reading===

	Books, Articles or other print media on this topic		Books, Articles or other print media on this topic:

			Horologium Oscillatorium: Sive de Motu Pendulorum ad Horologia Aptato Demonstrationes Geometricae (1673)

	===External links===		===External links===