Derivation of moment inertia formula

[Quartz]

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.

 

[qspeechc]

There are a few ways to do it. Moment of inertia is calculated by
\int R^2.dm

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:
\int_{-l/2}^{l/2} pAx^2.dx
If you work it out you find it equals:
\frac{1}{12} pAl^3

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