Conservation of angular momentum — conditions: Worked examples

Before anything, three plain-word reminders (no symbol used before it is earned):

The scenario matrix

Every problem this topic throws is one of these cells. The examples below are labelled with the cell they hit, and together they fill the whole grid.

#	Cell (case class)	What's tricky about it	Example
A	$I$ shrinks, no mass added (internal reshape)	$ω$ must rise; KE rises (work done)	Ex 1
B	$I$ grows, no mass added	$ω$ must fall; KE falls	Ex 2
C	Mass added by sticking (inelastic)	new $m r^{2}$ appears; KE lost to heat	Ex 3
D	Central force, curved path (planet)	$τ = 0$ because $r ∥ F$ ; $v r$ constant	Ex 4
E	Two spinning bodies couple (opposite spins)	signs of $ω$ matter, can cancel to zero	Ex 5
F	Conserved about ONE axis, not another	must pick the right origin/axis	Ex 6
G	Degenerate / limiting: $r \to 0$ , $I \to \infty$ , $ω \to 0$	check the formula doesn't break	Ex 7
H	Exam twist: energy vs momentum trap	wrong to assume KE conserved	Ex 8

Example 1 — Cell A: internal reshape, $I$ shrinks

Worked example Skater tucks in (revisited with the numbers spelled out)

A skater spins at $ω_{1} = 3 rad/s$ with $I_{1} = 8 kg\cdotm^{2}$ . She pulls her arms in so $I_{2} = 2 kg\cdotm^{2}$ . Find $ω_{2}$ , and compare kinetic energies.

Forecast: guess before reading — does $ω$ go up or down? By roughly how many times?

Condition check. The ice is frictionless; her muscles are internal forces. So no external torque about the vertical spin axis ⟹ $L$ about that axis is conserved. Why this step? We must earn the right to write $I_{1} ω_{1} = I_{2} ω_{2}$ ; that equation is only true when $τ_{e x t} = 0$ .
Write conservation. $I_{1} ω_{1} = I_{2} ω_{2}$ Why this step? Constant $L$ means the number $I ω$ before = the number after.
Solve for $ω_{2}$ . $ω_{2} = \frac{I _{1} ω _{1}}{I _{2}} = \frac{8 \times 3}{2} = 12 rad/s$ Why this step? $I$ dropped to a quarter, so $ω$ must rise fourfold to keep the product fixed.
Kinetic energy (Rotational kinetic energy): $K E = \frac{1}{2} I ω^{2}$ . $K E_{1} = \frac{1}{2} (8) (3)^{2} = 36 J, K E_{2} = \frac{1}{2} (2) (12)^{2} = 144 J$ Why this step? To show $L$ conserved does not mean $K E$ conserved.

Verify: $L_{1} = 8 \times 3 = 24$ , $L_{2} = 2 \times 12 = 24$ ✓ (units $kg\cdotm^{2} / s$ ). $K E$ quadrupled — the muscles did $108$ J of work pulling mass inward. Sanity: arms in ⟹ faster ⟹ matches the chair-and-books feeling.

Example 2 — Cell B: internal reshape, $I$ grows

Worked example Merry-go-round, kids walk outward

A merry-go-round (a disc) with children on it has $I_{1} = 100 kg\cdotm^{2}$ spinning at $ω_{1} = 4 rad/s$ . The children shuffle outward, raising it to $I_{2} = 250 kg\cdotm^{2}$ . Find $ω_{2}$ .

Forecast: faster or slower? This is the mirror image of Ex 1.

Condition check. The children walking are internal to the system; the pivot is smooth. No external torque about the axis ⟹ $L$ conserved. Why this step? Same licence as before — internal motion never changes total $L$ .
Conservation: $I_{1} ω_{1} = I_{2} ω_{2}$ .
Solve: $ω_{2} = \frac{100 \times 4}{250} = 1.6 rad/s$ Why this step? Mass moved out ⟹ $I$ up ⟹ $ω$ must fall to hold $I ω$ fixed.

Verify: $L_{1} = 400 = L_{2} = 250 \times 1.6 = 400$ ✓. $K E_{1} = \frac{1}{2} (100) (16) = 800$ J, $K E_{2} = \frac{1}{2} (250) (2.56) = 320$ J — KE dropped; the children's legs absorbed the difference. Slower-when-spread matches intuition.

Example 3 — Cell C: mass sticks on (inelastic)

Worked example Putty drops on a turntable

Turntable $I_{t} = 0.5 kg\cdotm^{2}$ spins at $ω_{0} = 10 rad/s$ . Putty of mass $m = 0.2 kg$ falls straight down and sticks at radius $r = 0.5 m$ . Find final $ω_{f}$ and the KE lost.

Forecast: does adding mass speed it up or slow it down?

Condition check. Gravity points down, parallel to the spin axis, so it makes zero torque about that vertical axis ( $F ∥$ axis ⟹ no sideways twist). The sticking force is internal. ⟹ $L$ about the axis conserved. Why this step? Even though the collision is violent, $τ_{e x t} = 0$ about this axis, so $L$ survives the impact.
Add the putty's moment of inertia: a point mass contributes $m r^{2}$ . $I_{new} = I_{t} + m r^{2} = 0.5 + 0.2 (0.5)^{2} = 0.55 kg\cdotm^{2}$
Conservation (putty had no spin before, so it adds no $L$ ): $I_{t} ω_{0} = (I_{t} + m r^{2}) ω_{f}$
Solve: $ω_{f} = \frac{0.5 \times 10}{0.55} = \frac{5}{0.55} \approx 9.09 rad/s$ Why this step? Bigger $I$ ⟹ smaller $ω$ , but only slightly since the putty is light.

Verify: $L$ before $= 5$ , after $= 0.55 \times 9.09 = 5.0$ ✓. $K E_{i} = \frac{1}{2} (0.5) (100) = 25$ J, $K E_{f} = \frac{1}{2} (0.55) (9.09)^{2} \approx 22.7$ J. About $2.3$ J vanished to heat/sound — expected, it's an inelastic stick.

Example 4 — Cell D: central force, curved path

Worked example Planet at perihelion vs aphelion (with numbers)

A comet is $r_{1} = 1$ AU from the Sun moving at $v_{1} = 60 km/s$ at perihelion (closest point). At aphelion it is $r_{2} = 5$ AU away. Find its speed $v_{2}$ there. At the apsides $v ⊥ r$ .

Forecast: far away — faster or slower?

Condition check. Gravity always points straight at the Sun: $F$ is along $r$ . Then $τ = r \times F = 0$ (a vector crossed with a parallel vector is zero — Central forces). ⟹ $L$ conserved. Why this step? This is the whole reason Kepler's 2nd law works.
At the apsides $v ⊥ r$ , so $L = m v r$ (no messy angle). Why this step? Look at the figure: only where velocity is perpendicular to the radius does $L$ collapse to the simple product $m v r$ .
Conservation (mass $m$ cancels): $m v_{1} r_{1} = m v_{2} r_{2} \Rightarrow v_{2} = v_{1} \frac{r _{1}}{r _{2}} = 60 \times \frac{1}{5} = 12 km/s$ Why this step? Five times farther ⟹ one-fifth the speed. Close = fast, far = slow.

Verify: $v_{1} r_{1} = 60 \times 1 = 60$ ; $v_{2} r_{2} = 12 \times 5 = 60$ ✓ (both $km\cdotAU/s$ ). Matches equal-area law: the swept area rate $\frac{1}{2} r v_{⊥}$ is the same at both ends.

Example 5 — Cell E: two spins couple, signs matter

Worked example Two discs clutch together, opposite spins

Disc A: $I_{A} = 3 kg\cdotm^{2}$ , spinning at $+ 8 rad/s$ (call anticlockwise positive). Disc B: $I_{B} = 1 kg\cdotm^{2}$ , spinning at $- 8 rad/s$ (clockwise). They are pressed onto the same axle and grip together. Find the final common $ω_{f}$ .

Forecast: with these numbers, will they end up spinning fast, slow, or dead still?

Condition check. The clutch/friction between the discs is an internal force pair; the axle is frictionless externally. ⟹ $L$ about the axle conserved. Signs are physics here, so keep them. Why this step? Angular momentum is a vector — opposite spins carry opposite signs and can subtract.
Total $L$ before (add signed contributions): $L = I_{A} ω_{A} + I_{B} ω_{B} = 3 (+ 8) + 1 (- 8) = 24 - 8 = 16$
After gripping they share $I_{A} + I_{B}$ at one $ω_{f}$ : $(I_{A} + I_{B}) ω_{f} = 16 \Rightarrow ω_{f} = \frac{16}{4} = + 4 rad/s$ Why this step? Same conservation, but now the "before" is a signed sum, not a single body.

Verify: $L_{after} = 4 \times 4 = 16 = L_{before}$ ✓. Final sign is $+$ : A's spin won because it carried more $L$ . (If instead $I_{B}$ had spun at $- 24$ , total $L$ would be $0$ ⟹ $ω_{f} = 0$ , they'd stop dead — a valid degenerate outcome.)

Example 6 — Cell F: conserved about one axis, not another

Worked example Cylinder rolling down a ramp — choose the axis

A cylinder rolls down an incline under gravity. Is its angular momentum conserved?

Forecast: yes, no, or "depends on the axis"?

About the cylinder's own centre: gravity acts through the centre (zero torque about it), but the ground's friction acts at the rim — friction has a lever arm, so $τ \neq = 0$ . ⟹ $L$ about the centre is not conserved (it spins up). Why this step? $τ = r F_{⊥}$ : friction at radius $R$ makes a real twist about the centre.
About a fixed point on the incline line of contact: now gravity's downhill component does have a lever arm, so again $τ \neq = 0$ .
Conclusion — no single axis here gives $τ = 0$ , so $L$ is not conserved for this system about any obvious axis. Why this step? This is the honest lesson of Cell F: conservation is a per-axis statement. The parent's warning " $L$ need not be conserved about every axis" is exactly this — and here it fails for all of them because gravity+friction leave a net twist.

Verify (numbers): gravity component along incline for angle $3 0^{\circ}$ , $m = 2$ kg, $g = 10$ : $F_{∥} = m g sin 3 0^{\circ} = 2 \times 10 \times 0.5 = 10 N$ — a nonzero force with a nonzero lever arm about the contact line, so $τ \neq = 0$ and $L$ genuinely changes. ✓ (Contrast Ex 3: there the only external force was parallel to the axis, giving $τ = 0$ .)

Example 7 — Cell G: degenerate and limiting inputs

Worked example What the formula does at the edges

Take the putty problem (Ex 3) and test three extreme choices of $r$ and $ω_{0}$ .

Forecast: which limits break the formula, and which just give a boring but correct answer?

$r \to 0$ (putty lands on the exact axis). $ω_{f} = \frac{I _{t} ω _{0}}{I _{t} + m ( 0 ) ^{2}} = \frac{I _{t} ω _{0}}{I _{t}} = ω_{0} = 10 rad/s$ Why this step? Mass on the axis has zero lever arm ⟹ adds no $I$ ⟹ no slowdown. Formula behaves perfectly.
$r \to \infty$ (putty very far out). $I_{t} + m r^{2} \to \infty$ , so $ω_{f} \to 0$ . Why this step? Infinitely spread mass is infinitely hard to spin ⟹ the spin dies. Sensible limit.
$ω_{0} = 0$ (turntable at rest). $ω_{f} = 0$ for any $r$ . Why this step? No spin in, no spin out — dropping putty on a still table can't start it turning (gravity makes no axis-torque).

Verify: at $r = 0$ : $ω_{f} = 5/0.5 = 10$ ✓. As $r \to \infty$ , denominator diverges, $ω_{f} \to 0$ ✓. At $ω_{0} = 0$ , numerator $= 0$ ✓. No division-by-zero ever occurs because $I_{t} > 0$ always.

Example 8 — Cell H: the exam trap (energy vs momentum)

Worked example "Both are conserved, right?" — the classic wrong turn

A student solves Ex 3 by assuming kinetic energy is conserved instead of angular momentum, writing $\frac{1}{2} I_{t} ω_{0}^{2} = \frac{1}{2} (I_{t} + m r^{2}) ω_{f}^{2}$ . What $ω_{f}$ does that give, and why is it wrong?

Forecast: will the bogus answer be bigger or smaller than the true $9.09 rad/s$ ?

The bogus KE equation: $ω_{f}^{bad} = ω_{0} \frac{I _{t}}{I _{t} + m r ^{2}} = 10 \frac{0.5}{0.55} \approx 9.53 rad/s$
Why it's wrong. Sticking putty is inelastic — heat is made, so KE is not conserved. The valid law is $τ_{e x t} = 0 \Rightarrow L$ conserved, giving $9.09 rad/s$ . Why this step? [!mistake]-style trap: conservation laws have different conditions. $L$ needs zero external torque (true here); KE needs no energy loss (false here).
Both can't hold at once — if they did, no energy would be lost, contradicting the sticking.

Verify: bogus $9.53 >$ true $9.09$ ✓ (KE-conservation over-predicts speed because it forbids the energy loss that really happens). The gap $= 25 - 22.7 = 2.3$ J is exactly the heat found in Ex 3.

Active Recall

Skater pulls arms in,

I : 8 \to 2

ω_{1} = 3

— final

ω

ω_{2} = 8 \times 3/2 = 12 rad/s

; KE rises

36 \to 144

J (muscles do work).

Merry-go-round

I : 100 \to 250

ω_{1} = 4

— final

ω

ω_{2} = 400/250 = 1.6 rad/s

Putty

m = 0.2

r = 0.5

sticks on

I_{t} = 0.5

ω_{0} = 10

— final

ω

ω_{f} = 5/0.55 \approx 9.09 rad/s

Comet

v_{1} = 60

r_{1} = 1

AU — speed at

r_{2} = 5

AU?

v_{2} = 60/5 = 12 km/s

(from

v_{1} r_{1} = v_{2} r_{2}

Discs

I_{A} = 3

+ 8

I_{B} = 1

- 8

clutch — final

ω

ω_{f} = (24 - 8) /4 = + 4 rad/s

; signs matter.

Why isn't

L

conserved for a cylinder rolling down a ramp?

Friction at the rim (and gravity's downhill part) give a net external torque about every natural axis.

Putty lands at

r = 0

— why no slowdown?

On the axis it adds zero

m r^{2}

, so

I

is unchanged and

ω_{f} = ω_{0}

Why does assuming KE conservation in the putty problem give a wrong (too-high)

ω

Sticking is inelastic — KE is lost as heat; only

L

is conserved. The KE assumption forbids that loss and over-predicts speed.

Connections

Parent: Conservation of angular momentum — conditions — the rule these examples apply.
Moment of inertia — the $I$ that changes in Ex 1, 2, 3, 5, 7.
Torque — the twist whose absence licenses every conservation step.
Newton's third law — why the internal muscle/clutch/stick forces never change total $L$ .
Kepler's laws — Ex 4 is the equal-area law in disguise.
Central forces — the $r ∥ F$ zero-torque case of Ex 4.
Rotational kinetic energy — why $L$ conserved $\neq =$ KE conserved (Ex 1, 3, 8).
Conservation of linear momentum — the straight-line analogue of the same bookkeeping.