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Derivation of moment inertia formula

  1. Apr 10, 2008 #1
    How do I derive the formula 1/12 Ml^2?
    Derive the formula for moment of inertia of a uniform thin rod of length l about an axis through its center perpendicular to the rod.
  2. jcsd
  3. Apr 10, 2008 #2
    There are a few ways to do it. Moment of inertia is calculated by
    [tex]\int R^2.dm[/tex]

    So place x=0 at the centre, the x-axis running along the rod. So you're integrating from -l/2 to l/2.
    We must find dm in terms of our integration variable x. In dx we have an element of mass dm.
    mass = (density)(volume)=(density)(cross-sectional area)(length)

    dm = p.A.dx
    where p is the density and A the cross-sectional area. Our integral is now:
    [tex]\int_{-l/2}^{l/2} pAx^2.dx[/tex]
    If you work it out you find it equals:
    [tex]\frac{1}{12} pAl^3[/tex]

    but if we remember that mass = pAl, then we get 1/12 Ml^2.
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