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Momentum of the angular inertia of two solids.

  1. Apr 8, 2009 #1
    The angular inertia of a rectangular prism.

    http://www.turboimagehost.com/p/1595403/untitled.JPG.html
    http://www.turboimagehost.com/p/1595403/untitled.JPG.html
    (how do i put an image)


    I don't know any math, besides algebra.
    I know some physics.


    http://en.wikipedia.org/wiki/List_of_moments_of_inertia - not there

    How do I calculate the angular moments of inertia for a rectangular prism that has it's axis on the side, 1/4 it's length off the side, 3/2 it's length off the side, etc.

    What to do if i don't know the mass of the material, but i know it's density.
    density / volume = mass ?



    if the axis of rotation is in the center, that dimension*mass is/ by 12
    if the axis of rotation is on the edge, that dimension*mass is/ by 3

    Is it a linear relationship?
     
    Last edited: Apr 9, 2009
  2. jcsd
  3. Apr 9, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi cavemen! Welcome to PF! :smile:

    Just use the http://en.wikipedia.org/wiki/Parallel_axis_theorem" [Broken] :wink:
     
    Last edited by a moderator: May 4, 2017
  4. Apr 9, 2009 #3
  5. Apr 9, 2009 #4
    http://www.turboimagehost.com/p/1597687/2649824.GIF.html][img=http://s1d2.turboimagehost.com/t/1597687_2649824.GIF
    [img=http://s1d2.turboimagehost.com/t/1597687_2649824.GIF]
    I don't know how to put up pictures.

    The parallel theorem and perpendicular theorem are to abstract pieces of BS that make a good excuse not to make formulas for specific cases i have to deal with.

    Well, how about this shape?
     
  6. Apr 10, 2009 #5

    tiny-tim

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    Do you mean the url link? that was in my last post.

    Or do you mean how do you apply it? you take the moment of inertia listed for the centre of a box, and add the md2 term.

    Read the link again, and just apply it!
    Find the moment of inertia for the complete box, and then subtract the moment of inertia for the part cut out. :smile:
     
  7. Apr 10, 2009 #6
    my question:
    http://s1d2.turboimagehost.com/t/1597687_2649824.GIF


    Is it:
    (mh^2)*[1/3] + m(w^2+d^2) * [1/12]
    for the box spinning over an axis on it's side?

    Do those momentums easily add and subtract?
    If I have a cube with a cilyndrical hole spinning about it's center, is it just
    I cube- I hole ?
     
  8. Apr 10, 2009 #7

    tiny-tim

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    m(w^2+d^2) * [1/12] is correct, (mh^2)*[1/3] isn't …

    where did you get 1/3 from? :confused:
    (it's moments, not momentums)

    yes, just Icube - Ihole :smile:
     
  9. Apr 10, 2009 #8
    I got it from the Wikipedia list of moments.
    A rectangular plane rotating against an axis on it's side has a coefficient of [1/3] next to the dimension that the axis of rotation crosses.

    If it is not true, then what is true.

    I am not a physics or a math person. THis is the first time in years, that I have to deal with this kind of problem.

    Thanks.
     
  10. Apr 10, 2009 #9

    tiny-tim

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    No, wikipedia says mh2/3 + mw2/12

    (this is because it applies the parallel axis theorem, and gets mh2/4 + m(w2 + h2)/12, which is the same thing)

    You wrote (mh^2)*[1/3] + m(w^2+d^2) * [1/12], which is different.

    This is one reason why you need to understand these formulas, and not just try to copy them.

    Are you doing this for a course, with an exam at the end, or is this just a one-off project?

    If it's for a course, then you must learn how to apply the parallel axis theorem.
     
  11. Apr 10, 2009 #10
    I finally found a hyperphysics.edu explanation on parallel axis theorem.
    I needed this for a machine that I am building.
    Thank you very much for your time and your help.
     
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