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Derivation of moment inertia formula

  • Thread starter Quartz
  • Start date
5
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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.
 

Answers and Replies

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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)

So
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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