## Derivation of moment inertia formula

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.

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 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.