Lessons · rotation

Moment of inertia: where the table of shapes comes from

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 2 modeling assumptions (author-asserted)

A uniform rod of mass $M = 2$ kg and length $L = 1.2$ m spins about one end. The algebra course just hands you $I = \tfrac13 ML^2$ from a table — but where does the $\tfrac13$ come from? Each slice at radius $r$ adds $r^2\,dm$ , so the moment is the area under $dI/dr = \lambda r^2$ . Drag the cursor out from the pivot and watch $I$ build up — slowly near the axis, fast near the tip.

Moment of inertia

I = \tfrac13 M L^2

(the shaded area)

0.96 kg·m²

If all the mass were at the tip:

I = M L^2

(the hoop/point rectangle)

2.88 kg·m²

Linear mass density

\lambda = M/L

1.667 kg/m

Integrand at the tip

\lambda L^2

2.4 kg·m

Moment of inertia is the integral of r² over the mass

I = \int r^2\,dm

Every slice of mass $dm$ at radius $r$ contributes $r^2\,dm$ to the moment of inertia. The total is the integral over the whole body — and for a body whose mass is spread out, that is the area under a curve, not a single term.

check $\tfrac{dI}{dr} = \lambda r^2$ ; check $I = \int_0^r \lambda r'^2\,dr'$ ; recover $\tfrac13 M L^2$ at $r=L$ ; collapse the hoop case to $M R^2$

✓ The moment's slope is the integrand: $\tfrac{dI}{dr} = \lambda r^2$ — the area's growth rate is the curve's height. [structural]
✓ The accumulated moment is the area: $I(r) = \int_0^r \lambda r'^2\,dr'$ . [structural]
✓ At the rod's end the area is the memorized $I = \tfrac13 M L^2$ . [structural]
✓ If all the mass sat at one radius $R$ (a hoop), the integral collapses to $I = M R^2$ — the rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “Moment of inertia just depends on how much mass there is, like ordinary mass does.”

It depends on where the mass is, not just how much. Each slice contributes $r^2\,dm$ , so mass far from the axis counts far more — the area under $dI/dr = \lambda r^2$ piles up near the tip. That is why the same mass gives $I = \tfrac13 ML^2$ as a rod about its end but $I = ML^2$ if you bunch it all at the tip (a hoop). Move the mass outward and $I$ grows even though $M$ is unchanged.

Modeling assumptions — author-asserted, disclosed not discharged

Uniform thin rod: mass spread evenly at linear density $\lambda = M/L$ , thickness negligible.
Rotation about a fixed axis through one end; $r$ is the distance from that axis.

The dI/dr–r graph, fully annotated

A static rendering (Matplotlib): the shaded area under dI/dr is the accumulated integral I, and the slope of I is dI/dr. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):