Angular momentum L = Iω (fixed axis), L = r × p (general): Worked examples

We will lean on ideas from Cross Product, Linear Momentum, Moment of Inertia, Torque, Conservation of Angular Momentum, Rotational Kinetic Energy and Kepler's Second Law, all seeded by the parent topic note.

The scenario matrix

Every problem this topic can throw is one of these case classes. The last column names the worked example that covers it.

#	Case class	What is special / what could trip you	Covered by
C1	Particle, general angle $θ$	must use $r sin θ$ , not $r$	Ex 1
C2	Degenerate: motion straight toward/away from $O$	$θ = 0$ or $18 0^{\circ} \Rightarrow L = 0$	Ex 2
C3	Straight-line motion, $L$ constant in time	$r$ changes but $r_{⊥}$ fixed	Ex 3
C4	Sign / direction (right-hand rule, $+ z$ vs $- z$ )	clockwise = negative	Ex 4
C5	Rigid body, fixed symmetry axis, $L = I ω$	pick correct $I$	Ex 5
C6	Conservation, $I$ shrinks, $ω$ grows	no external torque; energy NOT conserved	Ex 6
C7	Limiting behaviour: $r_{⊥} \to 0$ , and $ω \to 0$	$L \to 0$ smoothly	Ex 7
C8	Real-world word problem (Kepler / planet)	equal-areas from $L$	Ex 8
C9	Exam twist: two particles / net $L$ with cancellation	vectors can cancel	Ex 9

Let us first fix the geometry we keep reusing.

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

$∣ L ∣ = r p sin θ = p r_{⊥} p = m v$

The worked examples

Ex 1 — Cell C1: general angle

Worked example Particle at a slanted angle

A stone of mass $m = 2 kg$ is at position $r = 4 m$ from $O$ , moving at $v = 5 m/s$ . The angle between $r$ and $v$ is $θ = 3 0^{\circ}$ . Find $∣ L ∣$ about $O$ .

Forecast: guess before computing — will using the full $r = 4$ overestimate or match the true $L$ ? (Since $sin 3 0^{\circ} = 0.5$ , the true answer is half of the naive $r p$ .)

Momentum $p = m v = (2) (5) = 10 kg m/s$ . Why this step? $L$ is built from momentum, not raw velocity — mass matters (Linear Momentum).
Lever arm $r_{⊥} = r sin θ = 4 sin 3 0^{\circ} = 4 (0.5) = 2 m$ . Why this step? Only the "going-around" part of the motion counts; $sin θ$ extracts it.
$∣ L ∣ = p r_{⊥} = (10) (2) = 20 kg m^{2} /s$ . Why this step? This is the magnitude form $r p sin θ$ .

Verify: direct $r p sin θ = (4) (10) (0.5) = 20$ . ✓ Units: $kg m^{2} /s$ . ✓ The naive $r p = 40$ is double — confirming our forecast.

Ex 2 — Cell C2: degenerate (radial motion)

Ex 3 — Cell C3: straight line, $L$ stays constant

Worked example Free particle, no force

A puck ( $m = 2 kg$ , $v = 3 m/s$ ) slides in a straight line whose closest approach to $O$ is $r_{⊥} = 0.5 m$ . No forces act. Find $L$ and show it does not change as the puck moves.

Forecast: $r$ keeps changing as the puck slides. Does $L$ change? Guess: no — nothing twists it.

$p = m v = (2) (3) = 6 kg m/s$ . Why this step? Angular momentum is built from momentum $p = m v$ , not raw speed — we need the mass baked in (Linear Momentum).
$∣ L ∣ = p r_{⊥} = (6) (0.5) = 3 kg m^{2} /s$ . Why this step? Use the fixed lever arm, never the shrinking-then-growing $r$ .
Time-derivative: $\frac{d L}{d t} = r \times F$ . With $F = 0$ , this is $0$ . Why this step? Torque is the only thing that changes $L$ ; no force means no torque.

Verify: as the puck moves, both $p$ (constant, no force) and $r_{⊥}$ (geometry of a fixed straight line) stay the same, so $L = 3 kg m^{2} /s$ throughout. ✓ Constant, as forecast.

Ex 4 — Cell C4: sign and direction

Worked example Which way does

$L$ point? A particle in the $x y$ -plane is at $r = (3, 0, 0) m$ with momentum $p = (0, - 4, 0) kg m/s$ (moving in the $- y$ direction). Find $L = r \times p$ , including sign.

Forecast: point of $r$ to the right, motion downward — by right-hand rule the swirl is clockwise. Guess $L$ points into the page ( $- z$ ), i.e. negative.

Cross product in components: $L = (3, 0, 0) \times (0, - 4, 0) = (0 \cdot 0 - 0 \cdot (- 4), 0 \cdot 0 - 3 \cdot 0, 3 \cdot (- 4) - 0 \cdot 0) .$ Why this step? The determinant formula of the Cross Product gives each component.
Simplify: $L = (0, 0, - 12) kg m^{2} /s$ . Why this step? Only the $z$ -component survives for planar motion.

Verify: magnitude $= 12$ , direction $- \overset{z}{^}$ (into page = clockwise = negative). ✓ Matches the forecast. Compare: if $p$ were $+ y$ , we'd get $+ 12 \overset{z}{^}$ (out of page, counter-clockwise, positive). Sign encodes spin direction.

Ex 5 — Cell C5: rigid body, $L = I ω$

Ex 6 — Cell C6: conservation, $I$ shrinks

Worked example Skater pulls arms in

A skater starts with $I_{1} = 4 kg m^{2}$ , $ω_{1} = 2 rad/s$ , then pulls arms to $I_{2} = 1 kg m^{2}$ . Find $ω_{2}$ , and the change in kinetic energy.

Forecast: no external torque → $L$ fixed. $I$ drops by $4 \times$ , so $ω$ should rise by $4 \times$ to $8 rad/s$ . Energy? Muscles do work, so KE should increase.

Conservation: $I_{1} ω_{1} = I_{2} ω_{2}$ (Conservation of Angular Momentum). Why this step? Internal muscle forces make no external torque.
$ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{( 4 ) ( 2 )}{1} = 8 rad/s$ . Why this step? Rearranging the conservation equation for the unknown final spin — divide both sides by $I_{2}$ .
$K E_{1} = \frac{1}{2} I_{1} ω_{1}^{2} = \frac{1}{2} (4) (2)^{2} = 8 J$ ; $K E_{2} = \frac{1}{2} I_{2} ω_{2}^{2} = \frac{1}{2} (1) (8)^{2} = 32 J$ . Why this step? Rotational Kinetic Energy $= \frac{1}{2} I ω^{2}$ ; check it separately since it is NOT conserved.

Verify: $L_{1} = (4) (2) = 8$ , $L_{2} = (1) (8) = 8$ — equal. ✓ $Δ K E = 32 - 8 = 24 J$ (supplied by muscles). ✓

Ex 7 — Cell C7: limiting behaviour

Ex 8 — Cell C8: real-world (Kepler's second law)

Worked example A comet sweeps faster near the Sun

A comet at aphelion (farthest point) is $r_{a} = 5 \times 1 0^{11} m$ from the Sun, moving at $v_{a} = 1 \times 1 0^{4} m/s$ perpendicular to $r$ . At perihelion (closest point) $r_{p} = 1 \times 1 0^{11} m$ . Find its perihelion speed $v_{p}$ .

Forecast: gravity points along $r$ (toward Sun), so its torque about the Sun is zero → $L$ conserved. Smaller $r$ must mean larger $v$ . Guess $v_{p}$ is $5 \times$ bigger.

At both apsides $v ⊥ r$ , so $L = m v r$ (with $sin 9 0^{\circ} = 1$ ). Why this step? At closest/farthest points the motion is purely tangential.
Conservation: $m v_{a} r_{a} = m v_{p} r_{p}$ ; mass cancels. Why this step? Central (radial) force ⇒ zero torque ⇒ constant $L$ — this is Kepler's Second Law.
$v_{p} = \frac{v _{a} r _{a}}{r _{p}} = \frac{( 1 0 ^{4} ) ( 5 \times 1 0 ^{11} )}{1 \times 1 0 ^{11}} = 5 \times 1 0^{4} m/s$ . Why this step? Solve the conservation equation for the unknown $v_{p}$ by dividing through by $r_{p}$ , then plug in the numbers.

Verify: $L \propto v_{a} r_{a} = 1 0^{4} \cdot 5 \times 1 0^{11} = 5 \times 1 0^{15}$ and $v_{p} r_{p} = 5 \times 1 0^{4} \cdot 1 0^{11} = 5 \times 1 0^{15}$ — equal. ✓ Speed is $5 \times$ larger, as forecast.

Ex 9 — Cell C9: exam twist, vector cancellation

Worked example Net angular momentum of two particles

Particle A: $r_{A} = (0, 2, 0) m$ , $p_{A} = (3, 0, 0) kg m/s$ . Particle B: $r_{B} = (0, - 2, 0) m$ , $p_{B} = (3, 0, 0) kg m/s$ . Find the total $L = L_{A} + L_{B}$ about $O$ . (Exam trap: students add magnitudes and forget sign.)

Forecast: A is above $O$ moving right → swirls one way; B is below $O$ moving right → swirls the opposite way. Guess they cancel to zero.

$L_{A} = r_{A} \times p_{A} = (0, 2, 0) \times (3, 0, 0)$ . Compute $z$ -component: $r_{x} p_{y} - r_{y} p_{x} = 0 \cdot 0 - 2 \cdot 3 = - 6$ . So $L_{A} = (0, 0, - 6)$ . Why this step? Planar motion → only $z$ survives.
$L_{B} = (0, - 2, 0) \times (3, 0, 0)$ : $z = 0 \cdot 0 - (- 2) \cdot 3 = + 6$ . So $L_{B} = (0, 0, + 6)$ . Why this step? The flipped position sign flips the swirl direction.
$L = L_{A} + L_{B} = (0, 0, - 6 + 6) = (0, 0, 0)$ . Why this step? Angular momentum is a vector — opposite swirls cancel.

Verify: magnitudes are each $∣ - 6 ∣ = ∣ + 6 ∣ = 6$ , but signs are opposite so the vector sum is $0$ . ✓ Adding magnitudes ( $12$ ) would be the trap.

Recall

Recall Which formula for which cell?

Radial motion toward $O$ , what is $L$ ? ::: Zero — $sin θ = 0$ , lever arm is zero Straight-line free particle, does $L$ change in time? ::: No — no force means no torque, $r_{⊥}$ and $p$ fixed Clockwise spin in the $x y$ -plane, sign of $L_{z}$ ? ::: Negative (into page, $- \overset{z}{^}$ ) Skater pulls arms in — is energy conserved? ::: No, only $L$ is; muscles add KE Comet near the Sun moves faster because... ::: gravity is radial → zero torque → $L = m v r$ constant Two equal particles swirling oppositely — net $L$ ? ::: Can be zero; $L$ is a vector, signs matter

Connections

Cross Product — the machinery behind sign and cancellation
Conservation of Angular Momentum — Ex 6, Ex 8
Kepler's Second Law — Ex 8 is exactly this law
Torque — why free particles keep constant $L$