Angular momentum L = Iω (fixed axis), L = r × p (general): Visual walkthrough

Step 1 — What "position" and "momentum" even look like

WHAT. Before any spinning, put down two arrows. Pick a dot in space and call it the origin $O$ — this is your choice of "the center I measure everything from." A moving ball sits somewhere; draw the arrow from $O$ straight to the ball. That arrow is the position $r$ . Now draw a second arrow on the ball, pointing where it's going, as long as it is fast — that's the momentum $p = m v$ ( $m$ = mass in kg, $v$ = velocity in m/s).

WHY. Angular momentum is a story about these two arrows and the angle between them. Nothing else. So we must see them clearly first.

PICTURE. The magenta arrow is $r$ (from $O$ to the ball). The orange arrow is $p$ (the direction of travel). The angle $θ$ (violet) between them is the whole game.

Figure — Angular momentum L = Iω (fixed axis), L = r × p (general)

Step 2 — Which part of the motion actually "goes around"?

WHAT. Split the orange momentum arrow into two arrows that add up to it: one pointing along $r$ (straight away from or toward $O$ ), and one pointing perpendicular to $r$ (sideways, curling around $O$ ). Using basic right-triangle trig on that split: $p_{∥} = p cos θ, p_{⊥} = p sin θ .$

$p_{∥} = p cos θ$ — the piece along $r$ . When $θ = 0$ (moving straight out), $cos θ = 1$ : all motion is radial.
$p_{⊥} = p sin θ$ — the piece across $r$ . This is the only piece that swings the ball around $O$ .

WHY here we need $sin$ and not something else. We asked a specific question: "how much of the motion circles the origin?" The sideways leg of the right triangle answers exactly that, and in a right triangle the leg opposite the angle is (hypotenuse) $\times sin θ$ . So $sin θ$ is chosen because it selects the opposite (perpendicular) leg — the going-around part.

PICTURE. Watch the orange arrow break into a faded radial piece (wasted, points away from $O$ ) and a bright perpendicular piece (the real swirl).

Step 3 — The lever arm: the same swirl, seen a second way

WHAT. There are two equally correct ways to isolate the swirl. Instead of trimming $p$ , keep all of $p$ but ask: how far off does the line of motion miss $O$ ? Extend the momentum into a full straight line and drop a perpendicular from $O$ onto it. That shortest distance is the lever arm $r_{⊥}$ : $r_{⊥} = r sin θ .$

$r$ — full length of the position arrow.
$sin θ$ — again picks the perpendicular leg, now of the $r$ -triangle.
$r_{⊥}$ — how far the path passes from $O$ .

The magic: $r p sin θ$ can be read as $p \cdot (r sin θ) = p r_{⊥}$ or $r \cdot (p sin θ) = r p_{⊥}$ . Same number, two viewpoints.

WHY. For a straight-flying particle, $r_{⊥}$ is constant even while $r$ swings — this is what makes free particles carry a fixed angular momentum (proved in Step 8).

PICTURE. The violet dashed segment is $r_{⊥}$ , the closest the orange path gets to $O$ .

Step 4 — The cross product: turning the swirl into a vector

WHAT. The number $r p sin θ$ is the size of the angular momentum. But spinning has a sense — clockwise or counter-clockwise — so $L$ must be a full arrow. The tool that takes two arrows and produces a third arrow, perpendicular to both, with length $r p sin θ$ , is the cross product: $L = r \times p, ∣ L ∣ = r p sin θ .$

The $\times$ automatically contains the $sin θ$ — it kills the radial part for free (when $r ∥ p$ , $sin 0 = 0$ , so $L = 0$ ). That is why nature's formula is a cross product and not a plain multiply.
Direction by the right-hand rule: fingers point along $r$ , curl toward $p$ , thumb gives $L$ . Counter-clockwise swirl (in the page) $\Rightarrow$ $L$ points out of the page.

WHY a new tool at all? A scalar can't tell counter-clockwise from clockwise. We need an object that flips sign when the spin reverses — a vector along the axis does exactly that.

PICTURE. The green arrow $L$ pops straight out of the page; the curved orange arrow shows the sense of circulation your right hand follows.

Step 5 — Now spin a whole rigid body: every particle circles the axis

WHAT. Take a real object (a disc) spinning about a fixed axis — call it the $z$ -axis, poking out of the disc's center. "Rigid" means the shape never deforms, so every particle shares the same turning rate, the angular speed $ω$ (rad/s). Pick one particle $i$ at perpendicular distance $r_{i}$ from the axis. It rides a circle of radius $r_{i}$ , and its speed is $v_{i} = ω r_{i} .$

$ω$ — radians swept per second, identical for all particles (rigidity).
$r_{i}$ — how far particle $i$ sits from the axis.
$v_{i} = ω r_{i}$ — bigger circle $\Rightarrow$ faster particle, for the same $ω$ .

WHY $v = ω r$ . In one full turn a particle covers circumference $2 π r_{i}$ while sweeping $2 π$ radians; dividing distance by time and using $ω =$ radians/time gives $v_{i} = ω r_{i}$ directly.

PICTURE. Two sample particles — inner (magenta) and outer (orange). Their velocity arrows are tangent to their circles; the outer one is visibly longer.

Step 6 — One particle's angular momentum about the axis

WHAT. Feed particle $i$ into $∣ L ∣ = r p sin θ$ with its momentum $p_{i} = m_{i} v_{i} = m_{i} ω r_{i}$ and $θ = 9 0^{\circ}$ : $L_{i} = r_{i} p_{i} sin 9 0^{\circ} = r_{i} (m_{i} ω r_{i}) (1) = m_{i} r_{i}^{2} ω .$

Term by term, watch it build:

$r_{i}$ — the lever arm (here it is the full radius, since motion is perpendicular).
$p_{i} = m_{i} ω r_{i}$ — substitute Step 5's speed.
$sin 9 0^{\circ} = 1$ — the swirl is maximal; nothing wasted.
Result $m_{i} r_{i}^{2} ω$ — notice the $r_{i}$ appears twice (once from the lever arm, once from the speed). That squared distance is the seed of moment of inertia.

WHY the $r^{2}$ matters. Distance from the axis counts twice over — far mass is punished hard. That's why holding books at arm's length makes you so slow to spin.

PICTURE. A single particle's $L_{i}$ arrow along $+ z$ , with the two roles of $r_{i}$ labelled: one as lever arm, one hidden inside the speed.

Step 7 — Sum over all particles: $L = I ω$ appears

WHAT. Every particle's $L_{i}$ points the same way (along $+ z$ ), so their sizes simply add: $L = \sum_{i} L_{i} = \sum_{i} m_{i} r_{i}^{2} ω = (\sum_{i} m_{i} r_{i}^{2}) ω .$ The $ω$ is common to all (rigidity!), so it factors out. The bracket is a pure property of how the mass is arranged — the moment of inertia $I = \sum_{i} m_{i} r_{i}^{2}$ . Hence: $L = I ω .$

$\sum_{i}$ — add up every chunk of the body.
$m_{i} r_{i}^{2}$ — each chunk's contribution; far chunks dominate.
$ω$ pulled out — legal only because every chunk shares it.
$I$ — the collected bracket; measured once, reused forever (for this axis).

WHY this is not a new law. It is literally $r \times p$ summed. $I ω$ is what the general definition becomes under two conditions: rigid (common $ω$ ) and fixed symmetry axis (all $L_{i}$ parallel, so magnitudes add).

PICTURE. Many little $L_{i}$ arrows stacking into one fat total $L$ along the axis.

Step 8 — Edge & degenerate cases (never get surprised)

WHAT. Check the corners of the formula so no scenario catches you off guard.

Straight-line, force-free particle. No spin at all, yet $L = p r_{⊥} \neq = 0$ about an off-line origin. As it flies, $r$ swings but $r_{⊥}$ and $p$ stay fixed $\Rightarrow$ $L$ is constant (no torque, conserved).
Motion straight through $O$ ( $θ = 0$ or $18 0^{\circ}$ ): $sin θ = 0 \Rightarrow L = 0$ . Pure radial motion carries no angular momentum. ✔
Particle on the axis ( $r_{i} = 0$ ): contributes $m_{i} \cdot 0 \cdot ω = 0$ . Mass at the center is "free" — it adds nothing to $L$ or $I$ .
$ω = 0$ (not spinning): $L = I \cdot 0 = 0$ . A resting rigid body has no spin angular momentum, whatever its shape.
Reverse the spin ( $ω \to - ω$ ): $L$ flips to point the other way along the axis — the sign lives in the direction, exactly what the cross product was built to track (Step 4).

WHY. Every one of these is a limit of the same $r p sin θ$ ; seeing them confirms the formula degrades gracefully everywhere.

PICTURE. Four mini-panels: (i) straight-line ball with constant $r_{⊥}$ , (ii) radial motion $\Rightarrow L = 0$ , (iii) on-axis particle $\Rightarrow 0$ , (iv) reversed spin flipping $L$ .

The one-picture summary

WHAT. One frame stitches the whole journey: the single-particle triangle ( $r$ , $p$ , $θ$ , $r_{⊥}$ ) on the left, an arrow of implication, and the summed rigid body on the right giving $L = I ω$ along the axis.

Recall Feynman: tell the walkthrough to a friend

Put a dot down and call it home. Draw an arrow from home to a moving ball (that's $r$ ), and an arrow showing where the ball heads (that's $p$ ). Only the sideways part of the ball's motion swirls around home — the part heading straight at or away from home does nothing. We measure that swirl as (momentum) times (how far the path misses home), which is the same as (distance) times (sideways momentum), and it equals $r p sin θ$ . To remember which way it spins, we make it an arrow poking along the axis — that's the cross product. Now spin a whole disc: every tiny bit rides its own circle, all at the same turning rate $ω$ , each moving perfectly sideways so nothing's wasted. Each bit's swirl is its mass times its distance-squared times $ω$ . Add them all: the $ω$ is shared so it steps outside, and what's left — mass times distance-squared, summed — is the moment of inertia $I$ . So $L = I ω$ is just a billion little $r \times p$ 's added up, and it's only this clean when the object is rigid and spins about a nice symmetric axis.

Connections

Parent topic — the full statement this page draws in pictures
Cross Product — the tool of Step 4 that makes $L$ a vector
Moment of Inertia — the $\sum m_{i} r_{i}^{2}$ born in Step 7
Linear Momentum — $p = m v$ , the arrow we started with
Torque — $τ = d L / d t$ , why zero torque freezes $L$ (Step 8)
Conservation of Angular Momentum — the constant- $L$ of the free particle
Kepler's Second Law — the same $p r_{⊥}$ constancy for planets