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Mass Moment of inertia

  1. Dec 7, 2005 #1

    Fermat

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    Homework Helper

    Can someone explain something to me ?

    I refer you to this page.

    If we consider the rectangular block to be a cube, Then I would expect that we would get the mass moments of inertia all equal, Ixx = Iyy = Izz.
    But, in the page I refered you to, if we let a = b = L, then we get Ixx = Iyy, but Izz is different.

    Can anyone explain why Izz is different from Ixx and Iyy ?
     
  2. jcsd
  3. Dec 7, 2005 #2

    berkeman

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    Staff: Mentor

    Good question. Looks wrong to me too.
     
  4. Dec 7, 2005 #3

    Fermat

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    Thanks. Now I don't feel so bad.

    Someone had asked me for help in calculatring MOI for a rectangular block.

    I did some working and got the Izz format in that page. And posted my working on the forum I was at.

    Then I decided to check it out on the web.

    Imagine my consternation when I found that web page !!

    I hope some more people tell me it is wrong <hopes mightily>
     
  5. Dec 7, 2005 #4

    Fermat

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    Gottit!

    The x-axis and the y-axis don't pass through the COM of the block, but the z-axis does.

    Ixcxc is of the same form as Izz, but Ixx needs the parallel axes theorem giving,

    Ixx = Ixcxc + M(L/2)²

    which gives,

    Ixx = (1/12)Ma² + (1/3)ML²

    So, my working was right after all :smile:

    Oh, yes. And that web page is also correct.
     
  6. Dec 7, 2005 #5
    Are you sure you looked at the right fomulas?

    You had to observe x_C, y_c,z_c, NOT X,y,Z.

    x_c,y_c,z_c ARE equal.

    (You are right in your observation abotu x and y not being in the center.)

    Be aware that the COM is locationally dependent!
     
    Last edited: Dec 7, 2005
  7. Nov 23, 2011 #6
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