Rotational Inertia by Integration: Rods

Rotational inertia measures how spread out an object's mass is from its spin axis, and it's built from the sum m r² for every bit of mass. A rod isn't a bunch of separate point masses, it's a continuous smear of mass, so the sum becomes an integral: I = integral of r² dm. The hard part is never the integral itself, it's turning dm into something you can integrate over position.

For a uniform rod, mass is spread evenly, so define linear density lambda = M/L, constant along the length. A tiny slice of rod with width dx has mass dm = lambda dx. Placing the axis at one end and letting x run from 0 to L, I = integral from 0 to L of x² (lambda dx) = lambda (L³/3) = (1/3) M L^2. Swap the axis to the center, and symmetry lets you integrate from -L/2 to L/2, giving (1/12) M L² - four times smaller, because every bit of mass sits closer to the middle than it does to the end. When density isn't constant, lambda becomes a function of x, like lambda(x) = kx, and dm = lambda(x) dx still works, you just integrate x² lambda(x) dx instead of pulling lambda out front.

This is the first time Newton's second law for rotation needs calculus instead of a formula sheet. The same dm = lambda dx move reappears for rings, disks, and spheres, just with different shapes for dm and different limits. Once you have I for a rod about its end or its center, the parallel axis theorem lets you shift to any other axis without re-integrating, which is where this topic heads next.