Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Proof of parallel axis theorem.

  1. Nov 10, 2013 #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.
     
  2. jcsd
  3. Nov 10, 2013 #2

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    Be specific about which proof. Provide details.
     
  4. Nov 10, 2013 #3
  5. Nov 10, 2013 #4

    SteamKing

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper

    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.
     
  6. Nov 10, 2013 #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]
     

    Attached Files:

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook