Proof of parallel axis theorem.

  • #1
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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  • #2
SteamKing
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Be specific about which proof. Provide details.
 
  • #4
SteamKing
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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.
 
  • #5
[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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