Proof of parallel axis theorem.

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Discussion Overview

The discussion revolves around the proof of the parallel axis theorem, focusing on the interpretation of distances in the formula and the specifics of the proof itself. Participants explore the mathematical formulation and implications of the theorem in relation to axes displacement.

Discussion Character

  • Technical explanation, Debate/contested, Mathematical reasoning

Main Points Raised

  • One participant questions the assertion that the distance between the two axes in the formula is always perpendicular, citing a proof where it appears as a hypotenuse.
  • Another participant requests clarification on which specific proof is being referenced, indicating a need for more details.
  • A link to the Wikipedia page on the parallel axis theorem is provided, suggesting that it may contain relevant information for the discussion.
  • One participant asserts that the original and transformed axes are parallel and that the theorem accounts for displacements in both x and y directions, but does not clarify the initial question.
  • A mathematical expression for the moment of inertia is presented, showing the relationship between the moment of inertia about a new axis and the center of mass, but without further context or explanation.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus, as there are differing interpretations of the proof and the conditions under which the theorem applies. Multiple competing views remain regarding the nature of the distances involved in the theorem.

Contextual Notes

There are limitations in the discussion regarding the specific assumptions made in the proof and the definitions of the axes involved. The mathematical steps presented may also depend on further clarification that has not been provided.

Who May Find This Useful

This discussion may be useful for students and professionals interested in the mathematical foundations of the parallel axis theorem, as well as those exploring different proofs and interpretations of the theorem in physics and engineering contexts.

AakashPandita
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Everybody says that the distance ,between the two axis, used in the formula, is perpendicular. But in the proof it was a hypotenuse. It was not perpendicular.
 
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Be specific about which proof. Provide details.
 
The original axes and the transformed axes are parallel to one another. The theorem covers a displacement not only in the x direction, but the y direction as well. I don't see what your question is about.
 
[tex]I=\int r^2 dm = \int [(x-a)^2 + (y-b)^2]dm[/tex]

[tex]I= \int (x^2 + y^2) dm - 2a \int x dm - 2b \int y dm + \int (a^2 + b^2) dm[/tex]

[tex]I = Icom + Mh^2[/tex]
 

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