Moment of inertia I = Σmᵢrᵢ² — concept: Worked examples

This is the worked-example clinic for the parent concept note. There we built the formula $I = \sum_{i} m_{i} r_{i}^{2}$ . Here we drive it through every case class — so that when a problem lands on your desk, you have already seen its shape.

The scenario matrix

Before any arithmetic, here is the full landscape. Every worked example below is tagged with the cell it covers.

Cell	Case class	What's tricky	Example
A	Ordinary discrete sum	just apply the formula cleanly	Ex 1
B	Mass on the axis ( $r = 0$ )	zero contribution — degenerate distance	Ex 2
C	Masses on both sides of axis (negative coordinates)	does the sign of position matter?	Ex 3
D	Axis-dependence — same object, new axis	$I$ is not fixed	Ex 4
E	Limiting behaviour — push mass far out	$I$ grows like $r^{2}$ , unbounded	Ex 5
F	Real-world word problem — figure skater	link $I$ to $ω$ via conservation	Ex 6
G	Exam twist — axis through a corner, 2-D layout	perpendicular distance in a plane	Ex 7
H	Sanity/units + parallel-axis cross-check	catch errors before you submit	Ex 8

Notation reminder (nothing used before it is defined):

$m_{i}$ = the mass of particle number $i$ , in kilograms ( $kg$ ).
$r_{i}$ = the perpendicular distance from particle $i$ to the axis line, in metres ( $m$ ). "Perpendicular" means the shortest straight line from the particle to the axis, meeting it at a right angle.
$I$ = moment of inertia, in $kg m^{2}$ . Read it as "spin-stubbornness."
$ω$ (Greek letter omega) = angular velocity, how fast something turns, in radians per second.

Cell A — the plain vanilla sum

Figure — Moment of inertia I = Σmᵢrᵢ² — concept

Worked example Example 1 (Cell A) — three beads on a wire

Three beads on a straight wire, axis perpendicular to the wire at the left end: $m_{1} = 1 kg$ at $r_{1} = 1 m$ , $m_{2} = 2 kg$ at $r_{2} = 2 m$ , $m_{3} = 3 kg$ at $r_{3} = 3 m$ . Find $I$ .

Forecast: Will the far $3 kg$ bead dominate? Guess whether $I$ is closer to $10$ or to $30 kg m^{2}$ before reading.

Step 1 — write the sum term by term. $I = m_{1} r_{1}^{2} + m_{2} r_{2}^{2} + m_{3} r_{3}^{2} .$ Why this step? For discrete point masses there is no integral — the definition literally is a sum. Look at figure s01: each bead throws a coloured spoke of length $r_{i}$ to the axis.

Step 2 — square each reach, then multiply by its mass. $I = 1 (1)^{2} + 2 (2)^{2} + 3 (3)^{2} = 1 + 8 + 27.$ Why this step? The square is the whole point (inherited from $v^{2}$ in kinetic energy). Notice how $27$ already dwarfs $1$ : the far bead wins.

Step 3 — add. $I = 36 kg m^{2} .$

Verify: Units: $kg \times m^{2} = kg m^{2}$ ✓. Your forecast: it's $36$ , well past $30$ — the far mass dominated exactly as the squared reach predicts.

Cell B — a mass sitting ON the axis

Cell C — masses on both sides (does sign matter?)

Worked example Example 3 (Cell C) — dumbbell straddling the axis

Two $2 kg$ masses on a rod. The axis is at the centre. One mass sits at position $x = + 0.4 m$ , the other at $x = - 0.4 m$ (opposite side).

Forecast: The positions have opposite signs. Will they partly cancel (like in centre of mass), or both add?

Step 1 — find the perpendicular distance $r$ , not the position $x$ . $r_{1} = ∣ + 0.4 ∣ = 0.4 m, r_{2} = ∣ - 0.4 ∣ = 0.4 m .$ Why this step? $r$ is a distance — always $\geq 0$ . In figure s02 both masses are $0.4 m$ from the axis line; which side they're on is irrelevant to how big a circle they swing through.

Step 2 — square kills the sign anyway. $I = 2 (0.4)^{2} + 2 (- 0.4)^{2} = 2 (0.16) + 2 (0.16) .$ Why this step? Even if you carelessly used $x = - 0.4$ , the square makes $(- 0.4)^{2} = (+ 0.4)^{2} = 0.16$ . So $I$ cannot be reduced by masses being on opposite sides. They add.

Step 3 — evaluate. $I = 0.32 + 0.32 = 0.64 kg m^{2} .$

Verify: Contrast with centre of mass, which uses $\sum m_{i} x_{i}$ (linear in $x$ ): there the $+ 0.4$ and $- 0.4$ do cancel, putting the centre of mass at $x = 0$ . Moment of inertia uses $x^{2}$ , so nothing cancels — both sides pile on. Units ✓.

Recall Why can inertia never be negative?

Every term is $m_{i} r_{i}^{2}$ : mass is positive, and any real number squared is $\geq 0$ . So the sum is always $\geq 0$ . A negative moment of inertia would mean ::: impossible — it would mean an object helps itself spin, i.e. resistance below zero. Physically forbidden.

Cell D — same object, axis moved

Cell E — limiting behaviour

Worked example Example 5 (Cell E) — push the mass to infinity

A single $1 kg$ mass at distance $r$ . Track $I$ as $r$ goes $0.5 \to 1 \to 2 \to 4 m$ .

Forecast: Each time $r$ doubles, does $I$ double, triple, or quadruple?

Step 1 — tabulate $I = 1 \cdot r^{2}$ . $r = 0.5 : I = 0.25; r = 1 : I = 1; r = 2 : I = 4; r = 4 : I = 16.$ Why this step? With one particle, $I = m r^{2}$ is a clean parabola in $r$ . Figure s03 plots it: the curve steepens the farther out you go.

Step 2 — read the growth rate. Doubling $r$ multiplies $I$ by $2^{2} = 4$ . So $I$ grows without bound as $r \to \infty$ . Why this step? This is the limiting case: there is no ceiling. Real machines exploit it — a flywheel puts mass at large $r$ to store lots of rotational stubbornness.

Step 3 — the other limit. As $r \to 0$ , $I \to 0$ . A point on the axis has zero moment of inertia (matches Cell B).

Verify: From $r = 1$ ( $I = 1$ ) to $r = 2$ ( $I = 4$ ): factor of $4$ , not $2$ ✓. Growth is quadratic — plot s03 shows the curve, not a straight line.

Cell F — real-world word problem

Worked example Example 6 (Cell F) — figure skater pulls in

A skater spins with arms out at $I_{1} = 6 kg m^{2}$ and $ω_{1} = 2 rad/s$ . She pulls her arms in, dropping to $I_{2} = 2 kg m^{2}$ . No external torque. Find her new spin rate $ω_{2}$ .

Forecast: Faster or slower? By what factor?

Step 1 — invoke conservation of angular momentum. With no torque, angular momentum $L = I ω$ is constant: $I_{1} ω_{1} = I_{2} ω_{2} .$ Why this step? Pulling arms in involves only internal forces; they cannot change $L$ . This is the rotational analogue of "no external push, momentum stays put."

Step 2 — solve for $ω_{2}$ . $ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{6 \times 2}{2} .$ Why this step? We rearrange to isolate the unknown; smaller $I_{2}$ in the denominator forces $ω_{2}$ up.

Step 3 — evaluate. $ω_{2} = 6 rad/s .$

Verify: $L$ before $= 6 \times 2 = 12$ ; after $= 2 \times 6 = 12$ ✓ (units $kg m^{2} / s$ both sides). She spins 3× faster because she cut $I$ to one-third by shrinking every $r$ . Physically sensible — this is why skaters accelerate by tucking in.

Cell G — exam twist: axis through a corner of a 2-D layout

Worked example Example 7 (Cell G) — four masses on a square

Four $1 kg$ masses at the corners of a square of side $a = 2 m$ . Axis is perpendicular to the plane and passes through one corner (call it the origin). Find $I$ .

Forecast: Three masses are away from the axis and one is on it. Guess whether $I$ is nearer $8$ or $16 kg m^{2}$ .

Step 1 — place coordinates and read perpendicular distances. Corners at $(0, 0), (2, 0), (0, 2), (2, 2)$ . The axis is the line through $(0, 0)$ perpendicular to the page, so for a particle at $(x, y)$ the perpendicular distance to that axis is $r = x^{2} + y^{2}$ . Why this step? The axis pierces the plane; the shortest line to it lies in the plane, so its length is the ordinary 2-D distance $x^{2} + y^{2}$ . See figure s04: dotted spokes from each corner to the origin.

Step 2 — compute $r^{2}$ for each corner (skip the square root — we need $r^{2}$ ). $(0, 0) : r^{2} = 0; (2, 0) : r^{2} = 4; (0, 2) : r^{2} = 4; (2, 2) : r^{2} = 8.$ Why this step? The formula needs $r^{2}$ , and $r^{2} = x^{2} + y^{2}$ directly — no root necessary. The corner on the axis gives $0$ (Cell B again).

Step 3 — sum $m r^{2}$ . $I = 1 (0) + 1 (4) + 1 (4) + 1 (8) = 16 kg m^{2} .$

Verify: Units ✓. Forecast: it's $16$ , at the top end, because the far diagonal corner ( $r^{2} = 8$ ) contributes half the total on its own. The corner sitting on the axis contributed nothing.

Cell H — sanity check with the parallel-axis theorem

Worked example Example 8 (Cell H) — cross-check Example 4

We claim the dumbbell (two $2 kg$ at $\pm 0.4 m$ , total $M = 4 kg$ ) has $I_{cm} = 0.64 kg m^{2}$ about its centre and $I = 1.28 kg m^{2}$ about the left mass. Confirm with an independent method.

Forecast: Should the two agree exactly, or is one a rounding of the other?

Step 1 — state the parallel-axis theorem. $I = I_{cm} + M d^{2},$ where $d$ is the distance from the centre of mass to the new axis. Why this step? It's an independent relation (proved elsewhere in Parallel axis theorem), so if our direct sums obey it, both are trustworthy.

Step 2 — plug in $d = 0.4 m$ . $I = 0.64 + 4 (0.4)^{2} = 0.64 + 4 (0.16) = 0.64 + 0.64.$ Why this step? The left mass is $0.4 m$ from the centre, so shifting the axis there moves it by $d = 0.4$ .

Step 3 — evaluate. $I = 1.28 kg m^{2} .$

Verify: Matches Example 4's direct computation of $1.28 kg m^{2}$ exactly ✓. Two roads, one answer — the results are self-consistent. Units $kg m^{2}$ throughout ✓.

Recap of the matrix

Recall Which example covered which trap?

Cell A plain sum ::: Example 1 ( $36 kg m^{2}$ ) Cell B mass on axis, $r = 0$ ::: Example 2 ( $14 kg m^{2}$ ) Cell C both sides / sign trap ::: Example 3 ( $0.64 kg m^{2}$ ) Cell D axis moved ::: Example 4 ( $1.28 kg m^{2}$ ) Cell E limiting $r \to \infty$ ::: Example 5 (quadratic growth) Cell F real skater ::: Example 6 ( $ω_{2} = 6 rad/s$ ) Cell G exam twist, 2-D corner ::: Example 7 ( $16 kg m^{2}$ ) Cell H parallel-axis cross-check ::: Example 8 ( $1.28 kg m^{2}$ )

Connections

Parent concept note — where the formula is derived
Kinetic energy ½mv² — the $v^{2}$ that becomes the $r^{2}$
Rotational kinetic energy ½Iω² — where these $I$ values get used
Angular velocity ω — the shared $ω$ in every example
Centre of mass (first moment Σmr) — the linear-in- $r$ contrast (Ex 3)
Parallel axis theorem — the cross-check tool (Ex 4, Ex 8)
Angular momentum L = Iω — the conservation law behind the skater (Ex 6)
Torque τ = Iα — Newton's second law for rotation