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Quick question.

  1. Apr 27, 2005 #1
    Moment of inertia for a rod along the x axis centered at the origin is :

    [tex] \int_{-L/2}^{L/2} r^2 \frac{M}{L} dr [/tex]

    M/L denotes mass density, this evaluates to [itex] \frac{1}{12}mR^2 [/itex] as we'd expect, but my questoin is, if the rod was not rotating about an axis through its center, but an axis at L/5 from its edge, would the integral directly translate to:

    [tex] \int_{-L/5}^{4L/5} r^2 \frac{M}{L} dr [/tex] ?

    In which case it would evaluate to

    [tex] \frac{M}{3L}\left(\frac{4L}{5}\right)^3 - \left(\frac{-L}{5}\right)^3 [/tex] expanded?
  2. jcsd
  3. Apr 27, 2005 #2
    Yep, that's it. Parallel axis theorem gives the same thing--(13/75)ML^2
  4. Apr 27, 2005 #3
    I thought parallel axis was only for the edge?
  5. Apr 28, 2005 #4
    far as i know, it works whenever you are measuring with respect to the CM. So if u know the inertia about an axis thru the CM, then the parallel axis theorem gives the inertia about an axis that is parallel to the axis through the CM. If they are perp. distance A apart, then

    I = I_cm + mA^2

    Is this the version you use?
  6. Apr 28, 2005 #5
    Yeah actually your definition is more correct, I knew it as the same except add "at the edge" to "about an axis parallel to the axis thru the CM" with
  7. Apr 28, 2005 #6
    what does the edge have to do with anything?
  8. Apr 28, 2005 #7
    No idea, thats what I remember it as for some reason.

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