Conservation of angular momentum — conditions: Exercises

Before we start, let us pin down the only symbols and operations used on this page so nothing appears un-earned.

Definition The toolkit (all terms defined once, here)

$r$ — the position vector: an arrow from your chosen origin (a dot you pick) to the object. Its length $r$ is a distance in metres.
$v$ — velocity, how fast and which way the object moves.
$m$ — mass in kilograms.
$I$ — moment of inertia: how hard it is to spin something. For a point mass at distance $r$ , $I = m r^{2}$ . For a spread-out body it is bigger the further mass sits from the axis. See Moment of inertia.
$ω$ (omega) — angular velocity: how many radians the body turns per second.
$L$ — angular momentum, the "amount of spin". On this page we always use $L = I ω$ , which holds specifically for a rigid body spinning about a fixed axis that is one of its symmetry (principal) axes — every problem below is of that type. For a single point mass whose velocity is perpendicular to $r$ , this reduces to $L = m v r$ . (The fully general definition $L = \int r \times v d m$ is not needed here, but note $L = I ω$ is a special case of it, not a universal law.)
$τ$ (tau) — torque $= r \times F$ , the twisting effort. It changes $L$ over time exactly the way force changes ordinary momentum. See Torque.
$K E$ — kinetic energy, the energy stored in motion, measured in joules (J). For a spinning body it is $K E = \frac{1}{2} I ω^{2}$ ; for a moving point mass it is $K E = \frac{1}{2} m v^{2}$ . See Rotational kinetic energy.
$W$ — work, energy delivered by a force as it pushes something along the direction it moves. The work–energy theorem says the total work done on a body equals its change in kinetic energy: $W = Δ K E = K E_{after} - K E_{before}$ . This is a separate account from angular momentum — a force can do work (change $K E$ ) while exerting zero torque (leaving $L$ untouched).
The rule that runs the whole page: $\frac{d L}{d t} = τ_{e x t, n e t}$ , so $τ_{e x t, n e t} = 0 \Rightarrow L = constant$ .

Definition Sign convention — how vectors become the

$\pm$ numbers we compute with Every problem on this page has a single fixed axis (the spin axis). A vector along that axis has only one free choice left: which way it points. We fix a positive sense with the right-hand rule: curl the right hand's fingers the way the body turns; the thumb points along $+ L$ . On this page we take counter-clockwise (as seen from above) $= +$ .

A body spinning counter-clockwise has $L = + I ω$ (a positive number); clockwise gives $L = - I ω$ (negative).
A torque that would speed up counter-clockwise spin is $+$ ; one that opposes it is $-$ .
Adding angular momenta then means adding signed numbers. Two objects spinning the same way add; spinning opposite ways they partly cancel — e.g. $L_{total} = I_{A} ω_{A} - I_{B} ω_{B}$ if B turns the other way. Every scalar equation below (like $I_{1} ω_{1} = I_{2} ω_{2}$ ) is really the axis-component of the vector law, with each term carrying its own sign. Where all bodies turn the same way (Examples in L2, L4A) the signs are all $+$ and drop out; where directions could differ (L1·A(d), the "opposite spin" mistake below) you must keep them.

Level 1 — Recognition

L1·A — Which scenario conserves $L$ ?

Recall Solution

Apply one test each time: is there a net external torque about this axis?

(a) Conserved. The muscular pull is internal; ice is frictionless so no external twist. $L$ fixed. ✅
(b) NOT conserved. The child's push is an external force applied off-axis → external torque → $L$ grows. ❌
(c) Conserved. Gravity is a central force pointing straight at the Sun, so $F ∥ r$ ; by the cross-product zero-rule $τ = r \times F = 0$ . $L$ fixed — this is Kepler's 2nd law. ✅
(d) NOT conserved. Friction at the contact acts off-axis and opposes spin → external torque of the opposite sign to $L$ → $L$ decays toward zero. ❌

L1·B — Read the master equation

Recall Solution

If the time-derivative of $L$ is zero, then by $\frac{d L}{d t} = τ_{e x t, n e t}$ the net external torque is zero, and $L$ is constant in both magnitude and direction (it is a fixed arrow in space).

Level 2 — Application

L2·A — Skater, general form

Recall Solution

Condition: frictionless, internal pull → $L$ conserved about the spin axis (fixed vertical axis, so $L = I ω$ applies). She spins one way throughout, so every term is positive and signs drop out. $I_{1} ω_{1} = I_{2} ω_{2} \Rightarrow ω_{2} = \frac{8 \times 1.5}{3} = 4 rad/s .$ Kinetic energy (from the toolkit) $K E = \frac{1}{2} I ω^{2}$ . Since $I ω$ is fixed, write $K E = \frac{1}{2} (I ω) ω = \frac{1}{2} L ω$ , so $K E \propto ω$ : $\frac{K E _{2}}{K E _{1}} = \frac{ω _{2}}{ω _{1}} = \frac{4}{1.5} = \frac{8}{3} \approx 2.67.$ Energy rose — her muscles did work. See Rotational kinetic energy.

L2·B — Putty drop

Recall Solution

Condition: gravity is parallel to the axis (no torque about it); impact is internal → $L$ about the axis conserved. Disc and putty end up turning the same way → all terms positive. Point-mass addition: $m r^{2} = 0.4 \times 0. 6^{2} = 0.144 kg\cdotm^{2}$ . $I_{d} ω_{0} = (I_{d} + m r^{2}) ω_{f} \Rightarrow ω_{f} = \frac{1.2 \times 6}{1.2 + 0.144} = \frac{7.2}{1.344} \approx 5.36 rad/s .$

Level 3 — Analysis

L3·A — Elliptical orbit speeds

Recall Solution

Gravity is central → $F ∥ r$ → $τ = r \times F = 0$ → $L = m v r$ conserved (at apsides $v ⊥ r$ , so full $v r$ counts). The orbit circulates one way, so both sides keep the same sign. $v_{1} r_{1} = v_{2} r_{2} \Rightarrow v_{2} = \frac{54 \times 0.6}{5.4} = 6 km/s .$ Nine times farther → one-ninth the speed. Slow and far, fast and close: Kepler's 2nd law in numbers.

L3·B — Off-axis catch (component thinking)

The figure below is the whole argument in one picture. Trace it with your finger as you read.

Figure — Conservation of angular momentum — conditions

Recall Solution

Torque is $τ = r \times F$ , and its size is $r F sin θ$ where $θ$ is the angle between $r$ and $F$ .

Vertical axis (red weight arrow in the figure): the weight $F = m g$ points straight down, parallel to the vertical axis. Any turn about that axis has its $r$ lying flat, and the vertical component of $r \times F$ is zero because the force has no sideways push around the axis. Look at how the red arrow runs parallel to the gray axis arrow — parallel means $sin θ = 0$ → zero vertical torque → vertical $L$ conserved (condition 3: component-wise conservation).
Horizontal rim axis: now the same red weight arrow sits a horizontal distance out from that axis (the black $r$ in the figure) and points perpendicular to it. Here $θ = 9 0^{\circ}$ , $sin θ = 1$ , so $r \times F \neq = 0$ → there is a torque → $L$ about the horizontal axis is not protected. Conservation is axis-by-axis, never automatically "about everything".

Level 4 — Synthesis

L4·A — Two discs coupled

Recall Solution

Condition: the friction that couples them is internal to the two-disc system; no external axial torque → $L$ conserved. Both spin the same way (B starts at rest), so all terms are positive. $I_{A} ω_{A} + I_{B} ω_{B} = (I_{A} + I_{B}) ω_{f} \Rightarrow ω_{f} = \frac{2 \times 9 + 0}{6} = 3 rad/s .$ $K E_{i} = \frac{1}{2} (2) (9)^{2} = 81 J$ . $K E_{f} = \frac{1}{2} (6) (3)^{2} = 27 J$ . Fraction lost $= \frac{81 - 27}{81} = \frac{54}{81} = \frac{2}{3} \approx 0.667$ . Two-thirds gone to friction heat — yet $L$ is exactly conserved.

L4·B — Radial pull on a whirling puck

The figure below shows the two circles (before/after) and, crucially, the three arrows whose directions decide everything: $r$ , the velocity $v$ , and the string tension $T$ . Compare the direction of $T$ with $r$ .

Recall Solution

Condition (read it off the figure): the red tension arrow $T$ lies exactly along the black position arrow $r$ (both point to the central hole). Parallel arrows → $sin θ = 0$ → $τ = r \times T = 0$ → $L = m v r$ conserved. $v_{1} r_{1} = v_{2} r_{2} \Rightarrow v_{2} = \frac{3 \times 0.8}{0.4} = 6 m/s .$ The pull is radial so it exerts no torque, but as the puck spirals inward it moves toward the hole, i.e. along the tension, so the tension does work. Using the work–energy theorem $W = Δ K E$ from the toolkit: $W = K E_{2} - K E_{1} = \frac{1}{2} (0.5) (6^{2} - 3^{2}) = 0.25 \times 27 = 6.75 J .$ $L$ constant, energy increased by your pulling — the same lesson as the skater. (Notice in the figure that $v$ is drawn perpendicular to $r$ : that perpendicularity is what makes $L = m v r$ with the full $v$ .)

Level 5 — Mastery

L5·A — Choosing the origin to rescue conservation

Recall Solution

Origin choice: put the axis at the hinge. The hinge exerts a big unknown reaction force during impact, but that force acts at the hinge where $r = 0$ , so its torque $r \times F = 0 \times F = 0$ — zero torque about the hinge. Gravity acts over the tiny collision time → negligible impulse-torque. Hence $L$ about the hinge is conserved (the "pick the smart origin" rule from the parent note). Take the bullet's incoming swing sense as $+$ . Bullet's incoming $L = m v r = 0.01 \times 400 \times 0.9 = 3.6 kg\cdotm^{2} / s$ (positive). After embedding, total $I = I_{r o d} + m r^{2} = 1.6 + 0.01 \times 0. 9^{2} = 1.6081 kg\cdotm^{2}$ . $ω = \frac{L}{I} = \frac{3.6}{1.6081} \approx 2.239 rad/s .$ Positive $ω$ → the rod swings the same way the bullet was heading, as expected.

L5·B — Full audit: what is and isn't conserved

Recall Solution

$K E_{i} = \frac{1}{2} m v^{2} = \frac{1}{2} (0.01) (400)^{2} = 800 J$ . $K E_{f} = \frac{1}{2} I ω^{2} = \frac{1}{2} (1.6081) (2.239)^{2} \approx 4.03 J$ .

Linear momentum: not conserved — the hinge delivers an external horizontal force impulse. (See Conservation of linear momentum for why the hinge breaks it.)
Angular momentum about the hinge: conserved — hinge force has zero moment arm there.
Kinetic energy: not conserved — embedding is inelastic, $\approx 796 J$ became heat. This is the whole point of the topic: the three conservation questions have three independent answers, decided by three different conditions.

Active Recall

Recall One-line reflexes

Inelastic spin collision — which quantity survives? ::: Angular momentum ( $L_{i} = L_{f}$ ); kinetic energy is lost. Radial/central force does what to torque and to energy? ::: Zero torque (so $L$ constant) but it can still do work (energy changes via $W = Δ K E$ ). Why pick the axis at a hinge/pivot during impact? ::: The reaction force there has zero moment arm, so zero torque, so $L$ about that axis is conserved. Skater/puck speeds up — is energy conserved? ::: No; work is done by muscles/your pull. $L$ is conserved, energy is not. When is the cross product $r \times F$ zero? ::: When $r$ and $F$ are parallel ( $sin θ = 0$ ), i.e. a central force or a force applied at the axis. How do you add two angular momenta on the same axis? ::: As signed numbers (right-hand rule): same sense adds, opposite sense subtracts.

Connections

Conservation of angular momentum — conditions — the parent rule these problems drill.
Moment of inertia — the $I$ that changes in skater/putty/disc problems.
Torque — zero net external torque is the single gating condition.
Central forces — string-pull and gravity give $τ = 0$ .
Kepler's laws — the orbit exercise is the equal-area law.
Rotational kinetic energy — why $L$ conserved need not mean KE conserved.
Conservation of linear momentum — contrast: the hinge breaks it but not angular momentum.