Moment of inertia I = Σmᵢrᵢ² — concept: Visual walkthrough

Everything below is built from zero. If a symbol appears, it was drawn first.

Step 1 — One dot, going in a circle

WHAT. Picture a single tiny lump of matter — call its mass $m$ (mass just means "how much stuff"). Tie it to a fixed centre point and let it whirl around in a perfect circle.

WHY. Complicated objects are just lots of dots stuck together. If we understand one dot completely, we can add them up later. Always start with the simplest piece.

PICTURE. The green dot below rides a circle. The straight line from the centre to the dot is its radius, which we name $r$ — the distance from the spin-centre out to the dot.

Nothing squared yet. Nothing summed yet. Just a dot, a radius, and a circle.

Step 2 — How fast is the dot actually moving?

WHAT. As the dot goes around, it has a real straight-line speed at every instant — the speed you'd feel if the string snapped and it flew off. Call that speed $v$ .

WHY. We already trust one law completely from straight-line physics: a moving mass carries kinetic energy $K E = \frac{1}{2} m v^{2}$ . To use it, we first need $v$ — the actual speed of the dot.

PICTURE. Two arrows in the figure. The teal arrow $v$ points along the circle (the direction the dot is heading). Notice: it is always at a right angle to the radius $r$ .

Step 3 — Trade "speed" for "spin rate"

WHAT. Instead of the straight-line speed $v$ , describe the motion by how fast the angle sweeps around. That is the angular velocity $ω$ (Greek letter "omega") — turns of angle per second, measured in radians per second ( $rad/s$ ).

WHY do we switch tools here? Because $v$ is awkward: a dot far from the centre and a dot near the centre have different $v$ even in the same rigid spin. But they share the same $ω$ — the whole rigid body turns as one. Angular velocity is the natural language of spinning. See Angular velocity ω.

PICTURE. The figure shows the angle $θ$ sweeping out. In one second the angle grows by $ω$ . The outer edge of that swept slice — the arc length — is what the dot physically travels.

Recall Where does

$v = ω r$ come from, in one line? Arc length ::: In one full turn the dot travels the circumference $2 π r$ ; the angle swept is $2 π$ radians. Speed = arc per time = $r \times$ (angle per time) $= r ω$ .

Step 4 — Substitute the bridge into the energy brick

WHAT. Take the kinetic energy of our one dot, $\frac{1}{2} m v^{2}$ , and replace $v$ with $ω r$ .

WHY. We want the energy written in spin language ( $ω, r$ ) instead of straight-line language ( $v$ ), so that later every dot speaks the same $ω$ .

PICTURE. The figure shows the swap happening: the $v$ in the energy box is crossed out and $ω r$ slotted in. Watch the square in $v^{2}$ land on both factors.

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} m (ω r)^{2} = \frac{1}{2} m ω^{2} r^{2}$

Term by term on the last piece:

$\frac{1}{2}$ — the same constant, untouched.
$m$ — the dot's mass, untouched.
$ω^{2}$ — the spin rate, squared. It got squared because $v^{2}$ squared everything inside $(ω r)$ .
$r^{2}$ — here it is: the distance is now squared, and it was forced to be, purely because $v$ was squared and $v = ω r$ .

Step 5 — Now add up every dot in the body

WHAT. A real object is a crowd of dots, dot number $i$ having mass $m_{i}$ and distance $r_{i}$ . Total rotational energy is the sum of each dot's energy.

WHY. Energy just adds — one dot's kinetic energy plus the next plus the next. And crucially, all dots share the same $ω$ (rigid body), so $ω$ is a common factor we can pull out front.

PICTURE. Several dots on one rigid spinning body, near and far. Same $ω$ everywhere (one shared sweep), but each has its own $r_{i}$ (so its own speed and its own energy).

$K E_{rot} = \sum_{i} \frac{1}{2} m_{i} ω^{2} r_{i}^{2} = \frac{1}{2} ω^{2} depends only on mass & geometry (i \sum m_{i} r_{i}^{2})$

Reading it:

$\sum_{i}$ = "add over all dots", $i = 1, 2, 3, \dots$
$\frac{1}{2} ω^{2}$ — pulled out front because it is the same for every dot.
The bracket $\sum_{i} m_{i} r_{i}^{2}$ — pure property of where the mass sits, no $ω$ in it at all.

Step 6 — Give the bracket a name by demanding the analogy

WHAT. Compare the shape of what we built to the original linear law.

WHY. Linear physics said $K E = \frac{1}{2} m v^{2}$ . Our result is $K E_{rot} = \frac{1}{2} (bracket) ω^{2}$ . If we want these to be twins — with $ω$ playing the role of $v$ — then the bracket must play the role of $m$ . So we name it.

PICTURE. Side-by-side twins: mass $m$ pairs with the bracket, velocity $v$ pairs with $ω$ .

Compare with Centre of mass (first moment Σmr): that sum uses $r$ to the first power. Moment of inertia uses $r$ squared — and now you have seen exactly why.

Step 7 — The degenerate & extreme cases (so nothing surprises you)

Every honest derivation must survive its edge cases. Here they are, drawn.

Case A — a dot on the axis ( $r = 0$ ). Its term is $m_{i} \cdot 0^{2} = 0$ . It contributes nothing to $I$ , no matter how heavy it is. A heavy hub at the centre is "free" to spin.

Case B — a dot far out vs near in. Double the distance ( $r \to 2 r$ ) and its term goes from $m r^{2}$ to $m (2 r)^{2} = 4 m r^{2}$ — a 4× jump, not 2×. Far mass dominates.

Case C — mass all at the rim (hoop) vs spread inward (disk). Same total mass and radius, but the hoop parks every gram at the biggest $r$ , so $I_{hoop} = M R^{2}$ beats $I_{disk} = \frac{1}{2} M R^{2}$ .

PICTURE. Three panels: the zero-contribution centre dot, the near-vs-far pair with a $1 \times$ vs $4 \times$ label, and the hoop-vs-disk comparison.

The one-picture summary

Everything on this page in a single flow: from one dot's straight-line kinetic energy, through the bridge $v = ω r$ , to the summed rotational energy, to the boxed definition of $I$ .

Recall Feynman retelling — the whole walkthrough in plain words

Imagine one tiny bead whirling on a string. It has energy just from moving fast — that's $\frac{1}{2} m v^{2}$ , the oldest law we know. But "how fast" is a clumsy way to talk about spinning, because a bead far out on the string moves faster than one near the centre even though the whole thing turns together. So we describe the turning by one shared number, $ω$ — how quickly the angle sweeps. The speed of any bead is then $v = ω r$ : farther out, longer to travel each turn, faster. Slide that into the energy law and the $v^{2}$ squares both pieces, dropping an $r^{2}$ into our lap — we never chose to square the distance, the physics did. Now glue millions of beads into one solid object; energies just add, and the shared $ω$ pulls out front, leaving behind a pure "mass-and-where-it-sits" number: $\sum m r^{2}$ . Demand it look like the old law $\frac{1}{2} m v^{2}$ , and that leftover is the new "mass" for spinning — the moment of inertia $I$ . Centre beads count for nothing ( $r = 0$ ), far beads count quadruple, and that's why a skater pulling her arms in whips around faster: she shrinks $r$ , shrinks $I$ , and speed must climb.

Connections

Kinetic energy ½mv² — Step 2, the foundation brick
Angular velocity ω — Step 3, supplies $v = ω r$
Rotational kinetic energy ½Iω² — Step 6, the twin law
Centre of mass (first moment Σmr) — the $r^{1}$ cousin (contrast the square)
Parallel axis theorem — what happens to $I$ when the axis shifts
Angular momentum L = Iω — why the skater speeds up
Torque τ = Iα — Newton's 2nd law for rotation

Visual walkthrough — Moment of inertia I = Σmᵢrᵢ² — concept