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Moment of inertia

  1. Apr 14, 2005 #1
    I have to show what the moment of inertia of a rectangular plate with mass M and sides A and B is about its centre of mass. I have come up with
    [tex]\frac{1}{12}M(a^2 + b^2)[/tex]

    Now I have to show what the moment of inertia of the same plate is except this time about an axis perpedicular to the plate and passes through one corner. I know it is:
    [tex]\frac{1}{3}M(a^2 + b^2)[/tex] But having some problems proving it
     
  2. jcsd
  3. Apr 14, 2005 #2
  4. Apr 14, 2005 #3
    Yes it is.

    I = I (at centre of mass) + [tex]Md^2[/tex]

    But I can't see how that can get me from the first to second equation
     
  5. Apr 14, 2005 #4
    Easy,

    [tex]I_{cm} = \frac{1}{12}M(a^2 + b^2)[/tex]

    The displacement is half the diagonal, that is,

    [tex]\frac{a^2 + b^2}{4}[/tex]

    So add them up and you get:

    [tex]I = \frac{1}{3}M(a^2 + b^2)[/tex]

    :smile:
     
  6. Apr 14, 2005 #5

    dextercioby

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    Steiner's theorem.That's the name i learnt once with the theorem itself...

    Daniel.
     
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