Goal: evaluate the moment of inertia integral I = ∫ r2 dm
For a uniform rod of length L rotating about an axis passing perpendicularly through one end of the rod, align the rod along the x-axis. Split the rod into pieces of size dx.
The mass of each piece is: dm = λ dx, where λ = M/L is the mass per unit length of the rod.
The integral becomes: I = ∫0L x2 λ dx = λ x3/3 |0L = λ L3/3
Substituting M = λL
|ML2||for a uniform rod rotating about one end.|