---
title: Angular Momentum
module: Rotation
moduleNumber: 5
lessonNumber: 4
order: 504
summary: >
  A skater pulls in her arms and spins faster, with no torque acting: that is
  angular momentum conservation, and it lets us answer questions that would be
  hopeless force by force. We build $\vec L=\vec r\times\vec p$, show it obeys
  $\vec\tau_{ext}=\d\vec L/\d t$, and use its conservation under zero external
  torque to link before and after in collisions, reconfigurations, and coupled
  rotors without ever resolving the internal forces. The catch is bookkeeping: the
  origin, the system boundary, and the frame must be fixed first, and a change in
  total $\vec L$ always points to an external impulse someone forgot.
topics: [Rotation]
draft: false
sources:
  - book: Tipler & Mosca
    ref: "Ch. 10 — Angular Momentum; §§10-1–10-3"
---

## Particle angular momentum and torque

Angular momentum is defined relative to an origin. For a particle of linear
momentum $\vec p=m\vec v$ at position $\vec r$ measured from that
origin,

$$
\vec L=\vec r\times\vec p.
$$

Its magnitude is $L=rp\sin\phi=mrv_\perp$, where $v_\perp$ is the velocity
component perpendicular to $\vec r$. The direction follows the right-hand
rule. A particle moving radially has zero angular momentum about that origin,
even when its linear momentum is nonzero. The dimensions are
$\mathrm{kg\,m^2\,s^{-1}}$, equivalently $\mathrm{J\,s}$.

Differentiation establishes the torque law. Since $\vec v\times m\vec v=0$,

$$
\frac{\d\vec L}{\d t}
=\frac{\d\vec r}{\d t}\times\vec p
+\vec r\times\frac{\d\vec p}{\d t}
=\vec r\times\vec F
=\vec\tau.
$$

In a many-particle system, internal forces obeying Newton's third law along the
line joining the particles give cancelling torques. Therefore

$$
\frac{\d\vec L_{\rm sys}}{\d t}=\vec\tau_{\rm ext}.
$$

The reference point is significant. A force can have zero torque about one point
and nonzero torque about another. In an inertial frame, total angular momentum
about any fixed origin obeys the displayed law. A centre-of-mass formulation
separates translational and internal angular momenta.

$$
% caption: Angular momentum $\vec L=\vec r\times\vec p$ about $O$ has magnitude $rp\sin\phi$; only the component of $\vec p$ perpendicular to $\vec r$ contributes, and $\vec L$ points out of the plane (dot) by the right-hand rule.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[fill=white,draw=black] (0,0) circle (2pt) node[below left] {$O$};
\draw[fill=white,draw=black] (3.5,1.45) circle (2.4pt);
\draw[black,thick] (0,0)--(3.5,1.45) node[midway,below right] {$r$};
\draw[->,black,very thick] (3.5,1.45)--(4.55,3.4) node[above] {$p$};
\draw[black,dashed] (3.5,1.45)--(4.62,0.98);
\draw[acc] (0.75,1.1) circle (5pt);
\draw[fill=acc] (0.75,1.1) circle (1.3pt);
\node[acc,right] at (1.05,1.1) {$L$};
\end{tikzpicture}
$$

Resolving the momentum as
$\vec p=p_r\hat r+p_t\hat\theta$ leaves only the
tangential component in the cross product: $\vec L=\vec r\times
\bigl(p_t\hat\theta\bigr)$ and $L=rp_t$.

**Central forces and conservation.**

A central force has the form $\vec F=f(r)\hat r$. Its line of
action passes through the origin, so $\vec\tau=\vec r\times\vec F=0$.
Angular momentum about the force centre is conserved. The motion is consequently
confined to a plane perpendicular to the constant vector $\vec L$. In plane
polar coordinates the scalar statement is

$$
L=mr^2\dot\theta=\text{constant}.
$$

During a short interval $\d t$, a radius vector sweeps area
$\d A=\frac12r^2\d\theta$. Hence the areal velocity is

$$
\frac{\d A}{\d t}=\frac12r^2\dot\theta=\frac{L}{2m}.
$$

Equal areas are swept in equal times. This result is Kepler's second law for
gravitational motion, but it applies to every central force. A particle moves
faster near the force centre so that $r^2\dot\theta$ stays constant.

> **Theorem (Angular-momentum conservation).** If the net external torque about a fixed origin is zero, total angular momentum about that origin is constant in both magnitude and direction.

The theorem differs from force balance. A system may have nonzero external force
and still conserve angular momentum about a point if the force acts through that
point. Conversely, zero net force does not ensure zero torque.

$$
% caption: Under a central force the areal velocity $\d A/\d t=L/2m$ is constant, so the radius vector sweeps equal areas $A$ in equal times (Kepler's second law).
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[fill=white,draw=black] (0,0) circle (2pt);
\draw[black,thick] (0,0) circle (2.0);
\fill[acc!14] (0,0)--(20:2.0) arc (20:80:2.0)--cycle;
\fill[acc!14] (0,0)--(170:2.0) arc (170:230:2.0)--cycle;
\draw[black] (0,0)--(20:2.0);
\draw[black] (0,0)--(80:2.0);
\draw[black] (0,0)--(170:2.0);
\draw[black] (0,0)--(230:2.0);
\node[acc] at (1.35,0.95) {$A$};
\node[acc] at (-1.35,-0.95) {$A$};
\node[black,anchor=west] at (2.15,1.05) {equal times};
\end{tikzpicture}
$$

## Rigid-body angular momentum and gyroscopic motion

The total angular momentum of a rigid body is the sum over its particles. For
rotation about a fixed symmetry axis,

$$
\vec L=I\vec\omega,
$$

provided $\vec\omega$ is along a principal symmetry axis. In general,
$\vec L$ need not be parallel to $\vec\omega$; an asymmetric object
has different moments of inertia about different axes. Fixed-axis problems reduce
to the scalar equation $L=I\omega$ and the torque law $\sum\tau=I\alpha$.

When no external torque acts, a change in moment of inertia produces an inverse
change in angular speed:

$$
I_i\omega_i=I_f\omega_f.
$$

This conservation equation does not imply conservation of rotational kinetic
energy. Substitution gives $K=L^2/(2I)$. If a rotating body contracts, $I$
decreases and $K$ increases. Internal forces do work during contraction; that
work supplies the kinetic-energy increase. In the reverse process, rotational
kinetic energy can be converted to internal energy or other mechanical energy.

> **Worked example.** A rotating platform with $I_i=4.0\ \mathrm{kg\,m^2}$ spins
> at $2.0\ \mathrm{rad\,s^{-1}}$. A person moves inward so that the combined
> moment of inertia drops to $2.5\ \mathrm{kg\,m^2}$, with negligible external
> torque. The conserved angular momentum is $L=8.0\ \mathrm{kg\,m^2\,s^{-1}}$, so
>
> $$
> \omega_f=\frac{L}{I_f}=3.2\ \mathrm{rad\,s^{-1}}.
> $$
>
> The rotational energy rises from $8.0\ \mathrm{J}$ to $12.8\ \mathrm{J}$. That
> $4.8\ \mathrm{J}$ is work done by the person against the outward centrifugal
> tendency, not a violation of energy conservation.

**Gyroscopic precession.**

A spinning top supported at one point has spin angular momentum approximately
along its axle. Gravity exerts a torque about the support,

$$
\vec\tau=\vec r_{\rm cm}\times M\vec g.
$$

In a rapidly spinning symmetric top, this torque is nearly perpendicular to
$\vec L$. It changes the direction of $\vec L$ while leaving its magnitude
nearly constant. The axis then sweeps around the vertical: steady precession.
A top whose centre of mass is distance $r$ from the support and whose axle
makes angle $\theta$ to the vertical,

$$
\tau=Mgr\sin\theta,
\qquad \left|\frac{\d\vec L}{\d t}\right|
=\Omega L\sin\theta,
\qquad \Omega=\frac{Mgr}{L}=\frac{Mgr}{I\omega}.
$$

The cancellation of $\sin\theta$ applies to the rapid-precession approximation.
It fails when the top has significant nutation or insufficient spin. The direction
must be determined from the vector equation $\d\vec L/\d t=\vec\tau$;
memorized clockwise pictures are unreliable because orientation conventions vary.

$$
% caption: Gravity exerts a horizontal torque $\vec\tau=\vec r_{cm}\times M\vec g$ about the support. Perpendicular to the spin angular momentum $\vec L$, it turns $\vec L$ sideways, sweeping the axle around the vertical in steady precession.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,dashed] (0,0)--(0,3.6) node[above] {vertical};
\draw[fill=white,draw=black] (0,0) circle (2pt) node[below] {support};
\draw[very thick] (0,0)--(1.35,2.2);
\draw[thick] (1.35,2.2) ellipse (0.9 and 0.25);
\draw[fill=white,draw=black] (0.72,1.18) circle (1.8pt);
\draw[->,acc,very thick] (1.35,2.2)--(2.33,3.8) node[above] {$L$};
\draw[->,black,thick] (0.72,1.18)--(0.72,0.28) node[right] {$Mg$};
\draw[->,black,thick] (0,0)--(-0.95,0.55) node[left] {torque};
\draw[->,acc] (1.9,0.35) arc (16:120:1.92);
\node[acc] at (2.55,1.72) {precession};
\end{tikzpicture}
$$

## Angular impulse and collisions

Integration of the torque law gives angular impulse:

$$
\Delta\vec L=\int\vec\tau_{\rm ext}\,\d t.
$$

An impact can have a large torque over a short time. If the net external angular
impulse about a selected point is negligible during the collision, angular
momentum about that point is conserved even when linear momentum is not. A
striking example is a projectile embedding in a freely pivoted rod. About the
pivot, the impulsive support force has zero lever arm, so the preimpact and
postimpact angular momenta about the pivot are equal. Kinetic energy is not
conserved in the inelastic embedding.

An origin at the impulsive contact or pivot can shorten a collision calculation. For a particle
of mass $m$ striking perpendicular to a rod at distance $r$, the initial angular
momentum about the pivot is $mvr$. If the final rigid system has moment of inertia
$I_f$, its immediate angular speed is $\omega=mvr/I_f$. Subsequent motion under
gravity can then be obtained from energy conservation, because the collision and
the later smooth rotation are physically distinct stages.

**Centre-of-mass decomposition.**

In a system of particles, angular momentum about an arbitrary origin separates
into orbital angular momentum of the centre of mass and angular momentum relative
to the centre of mass:

$$
\vec L_O=\vec R_{\rm cm}\times M\vec V_{\rm cm}+\vec L_{\rm cm}.
$$

This relation separates the two contributions when a rigid body translates and
rotates. A wheel
rolling in a straight line has angular momentum about its centre of mass
$I_{\rm cm}\vec\omega$, but about a ground point it also has the orbital
term $\vec R_{\rm cm}\times M\vec V_{\rm cm}$. The reference point must
remain fixed during differentiation; an instantaneous contact point is not
generally a fixed inertial origin for torque calculations.

**Conservation in astronomical and collision problems.**

The Earth-Sun system has mutual gravitational forces, but their internal torques
about the system centre of mass cancel. A planet also conserves angular momentum
about the Sun in the ideal two-body approximation because the solar force is
central. Perturbing bodies create external torque and slowly alter the orbit.

For collisions, angular-momentum conservation is local to a selected point and
short time interval. A pivot is often selected because its impulsive support force
has zero lever arm. Linear momentum may fail to be conserved for a projectile-plus-
rod system because the pivot supplies an external impulse.

> **Worked example.** A $0.020\ \mathrm{kg}$ projectile moving at
> $300\ \mathrm{m\,s^{-1}}$ embeds at the end of a $1.0\ \mathrm{m}$ uniform rod
> of mass $1.0\ \mathrm{kg}$ pivoted at its other end. Before impact,
> $L_i=mvL=6.0\ \mathrm{kg\,m^2\,s^{-1}}$ about the pivot. After impact,
>
> $$
> I_f=\frac13(1.0)(1.0)^2+(0.020)(1.0)^2=0.353\ \mathrm{kg\,m^2},
> \qquad \omega_f=\frac{L_i}{I_f}=17.0\ \mathrm{rad\,s^{-1}}.
> $$
>
> The kinetic energy drops sharply in the embedding, so angular momentum, not
> energy, is the conserved quantity through the collision. Gravity enters only
> afterward.

## Vector methods and reference frames

The right-hand rule fixes the axial directions of angular velocity, angular
momentum, and torque. Their directions describe an axis of rotation rather than
translation along that direction. Cross-product order matters:
$\vec r\times\vec F$ reverses sign if the factors are exchanged. In
component form,

$$
\vec r\times\vec p=(yp_z-zp_y)\hat\imath+(zp_x-xp_z)\hat\jmath
+(xp_y-yp_x)\hat k.
$$

$$
% caption: In a planar problem the sign of $L_z=xp_y-yp_x$ follows from the coordinates; a dot marks $L_z>0$ (out of the page) and a crossed circle marks $L_z<0$ (into the page).
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (-0.4,0)--(4.3,0) node[right] {$x$};
\draw[->,black] (0,-0.4)--(0,3.15) node[above] {$y$};
\coordinate (P) at (2.65,1.55);
\draw[fill=white,draw=black] (P) circle (1.8pt) node[above left] {particle};
\draw[black,thick,->] (0,0)--(P) node[midway,above left] {$r$};
\draw[black,thick,->] (P)--(3.55,0.78) node[right] {$p$};
\draw[acc] (0.62,2.42) circle (5pt);
\draw[fill=acc] (0.62,2.42) circle (1.3pt);
\node[acc,right] at (0.95,2.42) {$L_z>0$};
\draw[black] (2.25,2.7) circle (5pt);
\draw[black,thick] (2.075,2.525)--(2.425,2.875);
\draw[black,thick] (2.075,2.875)--(2.425,2.525);
\node[black,right] at (2.6,2.7) {$L_z<0$};
\end{tikzpicture}
$$

A rigid object with three unequal principal moments can have $\vec L$ not
parallel to $\vec\omega$. Symmetry axes avoid this complication and justify
$L=I\omega$ as a scalar relation. The general conservation law is vectorial.

**Reference frames and checks.**

Angular momentum is most simply evaluated in an inertial frame. A rotating frame
introduces apparent forces and their torques. The direction of precession follows
from the short increment $\Delta\vec L=\vec\tau\Delta t$, added as a
vector to the existing spin angular momentum. Gravity therefore changes a top's
axis through torque rather than directly specifying the direction of its motion.

Angular momentum $I\omega$ has units $\mathrm{kg\,m^2\,s^{-1}}$; torque times
time has the same units. A torque perpendicular to $\vec L$ changes direction
without changing rotational kinetic energy immediately. In collision calculations,
the same system and reference point must be used immediately before and after the
impact. Otherwise the conservation equation does not compare like quantities.

**Scope of conservation arguments.**

Angular-momentum conservation requires negligible external torque over the stated
interval. External torque may be negligible during a short collision yet important
afterward. A skater, diver, or contracting cloud changes angular speed by changing
moment of inertia through internal work; kinetic energy must be tracked separately.
When external torque vanishes, angular momentum remains constant in both magnitude
and direction.

$$
% caption: With negligible external axial torque, pulling the masses inward from radius $R_i$ to $R_f$ lowers the moment of inertia, so the same angular momentum $I\omega$ gives a larger angular speed.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,dashed] (1.15,-1.2)--(1.15,1.55);
\draw[black,dashed] (4.2,-1.2)--(4.2,1.55);
\draw[fill=white,draw=black] (1.15,0) circle (1.8pt);
\draw[fill=white,draw=black] (4.2,0) circle (1.8pt);
\draw[black,thick] (1.15,0)--(0.18,0.72);
\draw[black,thick] (1.15,0)--(2.12,-0.72);
\draw[fill=acc!12,draw=acc] (0.18,0.72) circle (2.4pt);
\draw[fill=acc!12,draw=acc] (2.12,-0.72) circle (2.4pt);
\draw[black,thick] (4.2,0)--(3.72,0.36);
\draw[black,thick] (4.2,0)--(4.68,-0.36);
\draw[fill=acc!12,draw=acc] (3.72,0.36) circle (2.4pt);
\draw[fill=acc!12,draw=acc] (4.68,-0.36) circle (2.4pt);
\draw[<->,black] (1.15,-1.0)--(2.12,-0.72) node[midway,below] {$R_i$};
\draw[<->,black] (4.2,-0.78)--(4.68,-0.36) node[midway,below] {$R_f$};
\draw[->,black,thick] (0.55,-0.2) arc (205:75:0.62);
\draw[->,black,thick] (3.75,-0.2) arc (205:75:0.4);
\node[black] at (0.3,-0.55) {slow spin};
\node[black] at (3.5,-0.6) {fast spin};
\node[black] at (1.15,1.85) {arms out};
\node[black] at (4.2,1.85) {arms in};
\end{tikzpicture}
$$

For several bodies, the selected system boundary determines which torques are
external. Gravity between included bodies is internal; the same force is external
if only one body is selected. This bookkeeping is as important as selecting the
origin for the cross product.

$$
% caption: The dashed boundary sets which torques are external. The contact torque the two rotors exchange is internal to the selected pair; a bearing torque crossing the boundary is external and changes the pair's total angular momentum.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black,dashed] (0.25,-1.2) rectangle (4.6,1.4);
\node[black,anchor=west] at (0.4,1.15) {selected system};
\draw[black,thick,fill=black!4] (1.55,0) ellipse (0.6 and 0.82);
\draw[black,thick,fill=black!4] (3.35,0) ellipse (0.6 and 0.82);
\draw[<->,acc,thick] (2.18,0.35)--(2.72,0.35) node[midway,above] {internal};
\draw[->,black,thick] (5.15,0)--(4.05,0);
\node[black,anchor=west] at (4.75,0.34) {bearing};
\node[black] at (1.55,-1.02) {rotor A};
\node[black] at (3.35,-1.02) {rotor B};
\end{tikzpicture}
$$

## Symmetry and torque-free motion

Rotational symmetry implies conservation through Noether's theorem: invariance of
the laws under rotation about an axis corresponds to angular-momentum conservation
about that axis. In introductory mechanics the result follows directly from the
torque law, but the symmetry statement identifies why central forces conserve
angular momentum and why an externally applied torque breaks that conservation.

A torque-free rigid body with unequal principal moments can exhibit changing
orientation even though its total angular momentum is fixed. The simple formula
$\vec L=I\vec\omega$ applies directly only along a principal axis.
Fixed-axis machinery and symmetric tops avoid this complication; general rigid-body
rotation requires tensor moments of inertia and is outside the scalar model used
in these calculations.

The scalar treatment remains reliable when geometry constrains rotation about one
symmetry axis. Outside that scope, vector diagrams and component equations must
replace a single signed angular-momentum equation.

**Angular impulse in short interactions.**

Angular impulse is the accumulated external torque over an interval. It changes
angular momentum in the same way that linear impulse changes linear momentum. Use
the vector integral of torque rather than a peak torque or a force multiplied by an
arbitrary time. A short impact can have very large torque
but modest angular impulse when its duration is extremely small. Conversely, a
small friction torque applied for many revolutions can produce a large cumulative
change in angular momentum.

The choice of origin determines which external impulses have torque. A pivot force
can be large enough to invalidate linear-momentum conservation for a rod and
projectile, yet it has zero torque about the pivot. Angular momentum about that
point can therefore be conserved through a collision if gravity and other external
torques have negligible impulse during the impact. The same collision analysed
about a different point may require an explicit support-torque term.

The system boundary matters as much as the origin. Friction between two included
rotors is internal and transfers angular momentum between them; friction from a
brake pad outside the selected rotor is external and changes that rotor's angular
momentum. A statement of conservation without both a system boundary and origin
is incomplete. It can produce a numerically plausible result that corresponds to
no defined physical system.

Angular impulse directly represents mechanical clutches, impulsive strikes, and
brief actuator pulses. In each case, kinetic energy may be lost or supplied through
deformation, heating, or a power source. Angular-momentum balance does not
assert energy conservation. After an impact, a separate energy calculation can
describe subsequent smooth motion once the new angular speed has been determined.

Experimental angular impulse can be obtained from torque-sensor data by numerical
area integration. The torque baseline must be removed, sensor bandwidth must exceed
the pulse bandwidth, and the time origin must cover the entire interaction. Missing
the leading or trailing part of a narrow pulse biases the inferred momentum change.

**Origin choice in an impact.**

In a projectile embedding in a pivoted rod, selecting the pivot as origin removes
the support impulse from the angular-impulse balance. The incoming particle has an
angular-momentum magnitude equal to its transverse linear momentum times the
perpendicular impact distance. After embedding, the combined object rotates with
one angular speed about the pivot. The collision stage ends before gravitational
motion is analyzed.

$$
% caption: A projectile carries angular momentum $mvr$ about the pivot before impact; the pivot impulse has zero lever arm, so angular momentum about the pivot is conserved through the inelastic embedding.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[very thick] (0,0)--(0,3);
\node at (-0.35,1.5) {rod};
\draw[->,black,very thick] (-1.9,2.6)--(-0.4,2.6) node[midway,above] {$mv$};
\draw[fill=acc!15,draw=acc] (-0.16,2.6) circle (2.8pt);
\draw[fill=white,draw=black] (0,0) circle (2.4pt) node[below left] {pivot};
\end{tikzpicture}
$$

The angular-momentum decomposition about a fixed origin separates orbital motion
of the centre of mass from rotation relative to the centre. A translating and
spinning body therefore has distinct angular-momentum descriptions, provided the
reference origin is kept explicit.

$$
% caption: About $O$ the angular momentum splits into an orbital part $\vec R_{cm}\times M\vec V_{cm}$ from the centre-of-mass motion and a spin part $\vec L_{cm}$ about the centre; both use the same declared origin.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[fill=white,draw=black] (0,0) circle (2pt) node[below left] {$O$};
\draw[acc,thick] (0,0)--(3.5,1.2);
\node[acc] at (1.7,0.35) {$R_{cm}$};
\draw[thick,fill=black!5] (3.5,1.2) circle (0.55);
\draw[fill=white,draw=black] (3.5,1.2) circle (1.8pt);
\draw[->,black,very thick] (3.5,1.2)--(5.1,1.2) node[above] {$MV_{cm}$};
\draw[->,acc,thick] ([shift={(3.5,1.2)}]120:0.35) arc (120:-140:0.35);
\node[acc] at (3.5,2.05) {spin};
\end{tikzpicture}
$$

The figures establish two complementary practices: select an origin that removes
an unknown impulsive torque when possible, and retain the orbital term when the
selected origin is not the centre of mass. Both practices prevent accidental use
of a scalar spin formula in a translating system.

The balance is fixed by one declared origin and one system boundary:

$$
\Delta\vec L_O=\int_{t_i}^{t_f}\vec\tau_{{\rm ext},O}\,\d t.
$$

| interaction | selected system and origin | angular-momentum relation | separate physical account |
| --- | --- | --- | --- |
| projectile embedding in a pivoted rod | projectile plus rod, pivot $O$ | pivot impulse has zero moment about $O$ | collision energy becomes deformation and heat |
| clutch engagement | both coaxial rotors, common axis | $I_1\omega_{1i}+I_2\omega_{2i}=(I_1+I_2)\omega_f$ when external axial torque is negligible | speed equalization dissipates rotational energy |
| translating, spinning body | whole body, laboratory origin $O$ | $\vec L_O=\vec r_{\rm cm}\times M\vec v_{\rm cm}+\vec L_{\rm cm}$ | orbital and spin terms use the same origin |
| rapidly precessing top | top, support point $O$ | $\dot{\vec L}_O=\vec\tau_{{\rm grav},O}$ | rapid-spin approximation requires small nutation |

The listed equations apply to different system boundaries and time intervals. The
origin must remain explicit when data or forces are compared across stages.

Angular momentum and torque are inseparable from the selected origin. A force on a
particle has torque determined by the perpendicular distance from that origin to
the force line. Moving the origin changes that distance and can change both torque
and angular momentum, even though the physical force and particle motion have not
changed. No contradiction is involved: the two calculations describe rotation about
different reference points. A force through one origin has zero torque about that
point, but it can rotate the same system about another point.

An inertial frame and a fixed origin provide the cleanest torque law. When the
origin moves with the centre of mass, relative angular momentum isolates spin, but
translational and rotational terms must be separated correctly. A moving origin
chosen only because it simplifies a diagram can introduce terms that are absent
from the fixed-origin equation. The choice should be made from the system geometry
and the external forces, not from a preference for shorter algebra.

Angular impulse is the rotational analogue of linear impulse. It describes the
accumulated effect of torque over a finite interval. A door pushed gently for a long
time and a hammer striking a hinge briefly can produce comparable angular-momentum
changes if their torque-time areas are comparable. Peak torque alone does not
determine the result; duration and sign history are equally important. A torque
that reverses direction can have a large peak value and a small net angular impulse.

Impulsive problems require a timescale comparison. During a collision lasting a
few milliseconds, gravity usually gives negligible angular impulse compared with
contact forces. Over the later swing of a pendulum or rod, gravity can dominate the
angular-momentum change. Splitting the problem into collision and post-collision
stages prevents accidental use of a short-time conservation law over a long-time
motion. The same separation is used in ballistics, machinery impacts, and clutch
engagement.

The support force at a pivot illustrates why linear and angular momentum have
different conservation conditions. A pivot can exert an external impulse, so the
linear momentum of a selected object changes. About the pivot, that same impulse
has no moment arm and contributes no angular impulse. Angular momentum about the
pivot can be conserved through the impact. About a point away from the pivot, the
support force has a nonzero moment and must be included. The origin is therefore a
physical part of the model, not an arbitrary annotation.

Experimental angular impulse is obtained by integrating calibrated torque data over
the actual contact interval. Sensor bandwidth must capture rapid changes, and the
baseline torque should be removed before integration. Comparison with measured
angular-speed change requires a known inertia about the same axis. Differences
between these two measurements can arise from compliance, an unmeasured support
torque, an incorrect axis definition, or random noise.

Origin choice can remove the torque of an unknown external force. A hinge, support,
or contact point often exerts a force whose magnitude is difficult to measure during
an interaction. Taking angular momentum about that point removes
its torque only if the line of action passes through the selected origin. The
selection does not make the force disappear from the physical system. It removes
one moment term from one equation while its linear impulse and force-balance role
remain present elsewhere in the analysis.

The chosen origin must remain fixed in an inertial frame during the interval for
the standard angular-momentum equation to apply without modification. A point
attached to an accelerating body is not automatically an acceptable fixed origin.
Centre-of-mass formulations require the rotational angular momentum relative to
the centre plus the separate orbital contribution when comparison is made with a
laboratory origin.
Writing only a spin term for a translating body discards real angular momentum.

| approximation or choice | required physical condition | direct quantitative check |
| --- | --- | --- |
| omit pivot impulse about $O$ | force line passes through the fixed pivot | zero perpendicular lever arm |
| omit gravity during impact | collision interval is short enough | $\lvert\tau_g\rvert\Delta t\ll\lvert\Delta L_O\rvert$ |
| conserve axial angular momentum in a clutch | external axial torque is negligible over engagement | compare coupling impulse with bearing and motor impulses |
| use a centre-of-mass origin | orbital and spin contributions are kept separate | transform back to one laboratory origin before comparison |

These checks identify the term removed from a balance and the scale comparison that
supports its removal.

Collisions transfer angular momentum between bodies through internal torques when
the bodies are included in one system. A clutch connecting two coaxial rotors has
friction forces that are internal to the pair. One rotor slows, the other speeds
up, and total angular momentum is unchanged if external torque is negligible.
Rotational kinetic energy decreases because microscopic deformation and sliding
convert a portion to internal energy. The same distinction applies to a projectile
that embeds in a rotor, a person stepping onto a turntable, or a spacecraft using a
reaction wheel.

The collision interval is not always short enough to neglect all external torque.
A long clutch engagement can experience bearing drag or motor torque. A vehicle
wheel interacting with the road experiences external contact torque, so angular
momentum of the wheel alone is not conserved. A conservation statement must state
which torques are negligible over which time interval. That restriction is often
more informative than the final algebraic equality.

Angular-momentum transfer also clarifies recoil. If a rotor accelerates internally
within a freely suspended platform, the platform counter-rotates so that total
angular momentum remains constant. The energy source is internal electrical or
chemical energy; conservation of angular momentum does not imply conservation of
rotational kinetic energy of each component. A small platform inertia can produce a
large counter-rotation even when the platform mass is large, because inertia depends
on how mass is distributed relative to the axis.

Measurements of transfer should compare the same axis before and after interaction.
Encoder data from two rotors can be converted to angular momentum only after each
rotor's inertia about the common axis is known. A torque sensor at the coupling
can independently integrate angular impulse. Agreement of speed-based and
torque-based transfer estimates tests the assumption that support and drag torques
are negligible. Persistent disagreement points to unmeasured external torque or a
compliant coupling storing angular momentum temporarily in elastic motion.

In precession problems, torque changes angular momentum direction continuously
rather than transferring it between separate bodies. A support and gravity together
provide the external torque about the support point. The rapid-spin approximation
is valid when the direction changes slowly relative to spin period. When it fails,
nutation and changes of spin-axis angle must be solved from the full vector motion
rather than from a single scalar precession rate.

Conservation applies to the combined rotor pair, not to either
rotor individually. After engagement, the final common angular speed follows from
the sum of initial angular momenta divided by total inertia. The lost rotational
kinetic energy is the energy dissipated during speed equalization.

Angular momentum conservation can apply component by component. If the external
torque has zero component about a vertical axis, vertical angular momentum remains
constant even when horizontal torque components are present. A turntable supported
by a bearing provides a familiar example. The support can exert horizontal forces
and torques, but an axisymmetric bearing may exert negligible torque around its own
vertical axis. The applicable conservation statement is then one scalar component,
not automatically the full three-dimensional vector.

The distinction matters whenever a system has a preferred axis. A satellite in a
gravity-gradient field, a spinning laboratory platform, and atmospheric flow on a
rotating planet can exchange some angular-momentum components with external
supports or fields while retaining another component. The physical symmetry must be
identified before a conservation law is written. A scalar equation is justified by
one symmetry; a vector equation requires all three external torque components to
vanish.

A rigid body spinning about a principal symmetry axis has angular momentum
parallel to angular velocity. An asymmetric body does not generally have this
alignment. Its angular momentum is determined by the entire mass distribution, and
the body can reorient around a fixed angular-momentum vector without any external
torque. Fixed-axis machines avoid this complication because bearings constrain the
rotation axis. A freely tumbling object does not, and its orientation cannot be
deduced from one scalar moment of inertia.

Gyroscopic precession is a geometric response to torque. A rapidly spinning top
has large angular momentum along its axle. Gravity acts at the centre of mass and
creates torque about the support. For rapid spin, the torque is nearly perpendicular
to angular momentum. During a short interval it produces a small vector change
sideways from the original angular-momentum direction. The accumulated sideways
changes sweep the spin axis around the vertical; the full tilt motion is set by the
coupled rotational dynamics.

The common rapid-precession formula assumes steady tilt, negligible nutation, and
spin angular momentum much larger than the angular momentum associated with the
slow sweep. If spin decreases, the predicted sweep rate increases, but eventually
the assumptions fail. The top can wobble, its tilt can change, and the motion must
be treated with full vector dynamics. A memorized clockwise or counterclockwise
diagram is unreliable because the result depends on the chosen axis orientation,
support geometry, and viewing direction.

Precession measurements should record spin speed, tilt angle, support location,
and sweep rate. Comparing the observed vector change with the measured torque
tests the rapid-spin approximation. A disagreement can arise from bearing friction,
an asymmetric rotor, a moving support, or a nutation amplitude too large for steady
precession. Such discrepancies may reflect either a model departure or an
experimental imperfection.

| geometry | protected quantity | scalar relation that remains valid | data needed for the test |
| --- | --- | --- | --- |
| fixed vertical turntable axis | vertical component $L_z$ | $\Delta L_z=\int\tau_z\,\d t$ | spin rate, inertia, axial external torque |
| freely tumbling asymmetric body | full vector $\vec L$ in an inertial frame | tensor relation $\vec L=\mathbf I\vec\omega$ | principal moments and body orientation |
| steady precession | changing direction of $\vec L$ | $\dot{\vec L}=\vec\tau$ | tilt, spin rate, support geometry, sweep rate |
| translating rotor | total $\vec L_O$ about one laboratory origin | orbital plus spin decomposition | centre-of-mass state and spin inertia |

The selected measurement must match the protected quantity. A spin-only encoder
cannot test a total angular-momentum balance for a body that also translates.

Conservation laws are tested by comparing the same angular-momentum component
before and after an interaction. A common experimental error is to compare spin
angular momentum before an event with total angular momentum afterward. If a rotor
also translates, the orbital contribution about the selected origin must be
included. If the origin changes between measurements, the two values are not
directly comparable. A clear diagram of system boundary, origin, and axis removes
most ambiguity before numerical data is processed.

Internal angular-momentum transfer can be observed even when total angular momentum
is constant. A person on a low-friction turntable pulling masses inward increases
spin speed because moment of inertia decreases. The person's muscles do work, so
rotational kinetic energy increases. Angular momentum conservation determines the
speed ratio; energy accounting identifies the source of the kinetic-energy change.
Treating both quantities as conserved would contradict the measured change in spin
speed and obscure the mechanical work done by the person.

Angular momentum also distinguishes a force applied through an axis from a force
applied with a lever arm. A satellite thruster firing through the centre of mass
changes translational momentum but produces negligible spin change. The same thrust
mounted off-axis produces torque and changes angular momentum. Spacecraft attitude
control uses paired thrusters or reaction wheels to produce torque with little net
translation. The force arrangement is designed from the required angular impulse
and the intended line of action.

In a precessing system, direction change of angular momentum is often easier to
measure than the tiny change in its magnitude. High-speed imaging can track the
axis direction while an encoder measures spin rate. The torque estimate requires
mass distribution and support geometry. A precession rate that differs from the
rapid-spin estimate can indicate that the top is not symmetric, the support point
is moving, or the motion includes nutation. These alternatives are physically
distinct and should not be collapsed into a generic measurement error.

The angular-momentum vector is conserved in an inertial frame for an isolated
system, but component values in a rotating coordinate system can appear to change
because the coordinate axes themselves turn. This is a coordinate description
effect rather than a torque. Reporting the frame is essential when comparing
gyroscopes, satellite attitude data, or laboratory turntable measurements.

**Particle geometry and a declared origin.**

For one particle, the shortest route from geometry to a usable magnitude is the
perpendicular distance from the chosen origin to the particle's line of motion.
If that distance is $b$, then the magnitude is $L=mvb$ at the instant of
interest. The same result follows from $L=rmv\sin\phi$, where $\phi$ is the
angle from the position vector to the velocity. The distance form is usually less
error-prone because $b$ identifies the actual lever arm directly. A particle can
be far from an origin and still have small angular momentum if it is moving nearly
radially; it can be close to an origin and have substantial angular momentum when
its motion is nearly tangential.

The sign of a planar component must be fixed with the coordinate system, not
inferred from the sketch after the calculation. With $x$ to the right and $y$
upward, $L_z=xp_y-yp_x$.

> **Worked example.** A particle at $(0.40\ \mathrm{m},0.30\ \mathrm{m})$ with
> momentum $(3.0\ \mathrm{kg\,m\,s^{-1}},-4.0\ \mathrm{kg\,m\,s^{-1}})$ has
> $L_z=(0.40)(-4.0)-(0.30)(3.0)=-2.5\ \mathrm{kg\,m^2\,s^{-1}}$, pointing into the
> page. Using $rmv$ instead would overstate the magnitude, since it would count the
> radial part of the momentum that carries no angular momentum.

Changing the origin changes both the lever arm and the numerical angular
momentum. If the origin is shifted from $O$ to $O'$ by vector $\vec a$ while
the particle state is unchanged, then $\vec L_{O'}=\vec L_O-\vec a
\times\vec p$. This relation is not a correction term for an error; it states
that angular momentum is an origin-dependent quantity. A force line passing
through $O$ has zero torque about $O$, yet the same force may have nonzero torque
about $O'$. Collision solutions exploit this property by selecting an origin at a
support or contact point only when that point remains suitable throughout the
short interval.

The particle formula also sets the conditions for central-force conservation. A
central force remains on the line joining the particle to the force centre, so
its moment arm about that centre is zero. About a different origin, its torque
need not vanish. Planetary angular momentum is therefore conserved about the
centre of the attracting body under the ideal central-force model, not about an
arbitrary point drawn elsewhere in the orbit diagram. The stated origin is part
of the physical model, alongside the system boundary and the time interval.

$$
% caption: The moment arm $b$ is the perpendicular distance from $O$ to the particle's line of motion, giving the magnitude $L=mvb$ from only the tangential part of the momentum.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\coordinate (O) at (0,0);
\coordinate (P) at (2.2,1.15);
\draw[fill=white,draw=black] (O) circle (1.6pt) node[below left] {$O$};
\draw[black,thick,->] (1.05,2.2)--(3.55,-0.05) node[right] {$p$};
\draw[black,thick,->] (O)--(P) node[midway,below right] {$r$};
\draw[fill=white,draw=black] (P) circle (1.8pt);
\node at (2.55,1.5) {particle};
\draw[acc,dashed,thick] (O)--(1.557,1.729);
\node[acc] at (0.45,1.2) {$b$};
\node[black,anchor=west] at (2.85,-0.42) {line of motion};
\end{tikzpicture}
$$

**Angular impulse and component balances.**

Angular impulse is the time integral of external torque about one fixed origin.
During a short event, the component equation gives
$\Delta L_z=\int \tau_z\,\d t$, the signed area under a torque-time graph. A
torque that changes sign must be integrated algebraically; adding the absolute
areas would report total turning activity rather than the net change in angular
momentum.

> **Worked example.** A triangular torque pulse rising to $12\ \mathrm{N\,m}$ and
> lasting $0.040\ \mathrm{s}$ has angular impulse
> $\tfrac12(0.040)(12)=0.24\ \mathrm{N\,m\,s}$. Applied about the positive $z$
> axis to a rotor starting at $L_z=-0.10\ \mathrm{kg\,m^2\,s^{-1}}$, it leaves
> $L_z=+0.14\ \mathrm{kg\,m^2\,s^{-1}}$. The sign reversal is a real reversal of
> the component direction, not a failure of conservation.

The angular-impulse equation does not require the torque to be constant. It does
require a declared origin and a consistent interval. An impulsive force at a
pivot can be omitted from a balance about that pivot because its moment arm is
zero during the collision. That same impulse cannot be omitted from a balance
about the centre of mass or a remote laboratory origin. Likewise, weight is often
omitted only after comparing its maximum possible angular impulse with the contact
impulse over the same duration. The approximation comes from a scale comparison,
not from the fact that the collision is described as "fast."

Component equations are especially valuable when a support is axisymmetric. A
rotating platform can receive horizontal support forces and still have negligible
external torque about its vertical axis. During an internal rearrangement,
$L_z$ remains constant while $L_x$ and $L_y$ need not have the same protection.
Writing $L_{z,i}=L_{z,f}$ is then stronger and more accurate than writing a full
vector equation without examining the support. The selected component must be
the component along the axis for which the external torque is negligible.

The same bookkeeping separates internal redistribution from external angular
impulse. Two coaxial disks connected by a clutch exert friction torques of equal
magnitude and opposite sign on one another. Each disk changes angular momentum,
but the torque pair is internal to the two-disk system. For the pair,
$I_1\omega_{1i}+I_2\omega_{2i}=(I_1+I_2)\omega_f$ when external axial torque
is negligible. Mechanical energy is not conserved: the difference between the
initial and final rotational kinetic energies becomes internal energy. The
angular-momentum equation determines the common speed; it does not determine the
thermal energy unless the energy balance is also written.

$$
% caption: The signed area under a torque-time graph is the angular impulse $\Delta L_z=\int\tau_z\,\d t$; equal positive and negative areas leave zero net change in the angular-momentum component.
\begin{tikzpicture}[>=stealth,font=\footnotesize]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->,black] (0,0)--(5.5,0) node[right] {time};
\draw[->,black] (0,-1.55)--(0,2.05) node[above] {torque};
\fill[acc!14] (0.55,0)--(1.65,1.45)--(2.75,0)--cycle;
\fill[black] (2.75,0)--(3.65,-0.9)--(4.55,0)--cycle;
\draw[acc,thick] (0.55,0)--(1.65,1.45)--(2.75,0)--(3.65,-0.9)--(4.55,0);
\node[acc] at (1.65,1.78) {positive};
\node[black] at (3.66,-1.25) {negative};
\node[black,below] at (0.55,0) {$t_1$};
\node[black,below] at (4.55,0) {$t_2$};
\end{tikzpicture}
$$

**Steady precession as a vector calculation.**

Steady precession requires a separation of time scales. Let a symmetric top of
mass $M$ have its centre of mass a distance $r$ from the support and spin rapidly
about its axle. Its spin angular momentum has magnitude $L\simeq I_s\omega_s$.
At a fixed tilt $\theta$, gravity gives a torque magnitude $Mgr\sin\theta$ about
the support. If the axle sweeps around the vertical at angular speed
$\Omega_p$, the angular-momentum vector traces a horizontal circle. Its rate of
change has magnitude $\Omega_p L\sin\theta$. Equating this geometric rate with
the torque gives $\Omega_p=Mgr/L\simeq Mgr/(I_s\omega_s)$. The sine factors
cancel only because the tilt is assumed constant and the large spin contribution
dominates the angular momentum.

> **Worked example.** For $M=0.30\ \mathrm{kg}$, $r=0.12\ \mathrm{m}$,
> $I_s=2.2\times10^{-3}\ \mathrm{kg\,m^2}$, and
> $\omega_s=200\ \mathrm{rad\,s^{-1}}$, the spin angular momentum is
> $L=I_s\omega_s=0.44\ \mathrm{kg\,m^2\,s^{-1}}$. The rapid-precession estimate is
> $\Omega_p=(0.30)(9.81)(0.12)/0.44=0.80\ \mathrm{rad\,s^{-1}}$, a precession
> period near $7.9\ \mathrm{s}$. Doubling the spin speed halves the rate; doubling
> the support-to-centre distance doubles it. These proportionalities follow from
> the torque and angular-momentum magnitudes, not from a direction mnemonic.

The vector construction fixes the direction. At any instant, draw
$\vec L$ along the axle and draw $\vec\tau=\vec r\times M\vec g$
about the support. The small change $\d\vec L=\vec\tau\,\d t$ points
along the torque, perpendicular to the old angular momentum in the rapid-spin
limit. The new axle direction is parallel to $\vec L+\d\vec L$. Repeating
this construction gives the sense of precession for the specified viewing
direction. Reversing the spin reverses $\vec L$ and therefore reverses the
precession sense for the same gravitational torque.

Nutation marks the limit of the steady-precession approximation. A top released
with an unsuitable initial sweep rate changes its tilt while its axis precesses,
so the angular momentum has an additional contribution from the motion of the
axle. Friction at the support reduces spin angular momentum and can change the
tilt. A measured precession rate should therefore be compared with the estimate
only after recording the spin rate, tilt range, and any visible wobble. The full
torque law remains valid; the simplified scalar formula no longer contains all of
the relevant angular-momentum components.

A component balance is clearest when every rotation is assigned a sign before
numbers are substituted. Take the upward vertical as the positive $z$ axis. A
platform rotating counterclockwise when viewed from above has positive angular
velocity and positive $L_z$; a wheel rotating clockwise has negative values. The
choice is arbitrary, but it must remain unchanged for the initial state, every
torque impulse, and the final state. Mixing a top-view convention with a
side-view sketch is a frequent source of an otherwise unexplained minus sign.

> **Worked example.** A horizontal platform of moment of inertia
> $0.80\ \mathrm{kg\,m^2}$ carries a coaxial reaction wheel of moment of inertia
> $0.20\ \mathrm{kg\,m^2}$. Initially the platform turns at
> $+3.0\ \mathrm{rad\,s^{-1}}$ and the wheel at $-12\ \mathrm{rad\,s^{-1}}$, so
> their angular-momentum components are $+2.4$ and $-2.4\ \mathrm{kg\,m^2\,s^{-1}}$
> and the combined component is zero. The opposite signs are not an error: the two
> bodies carry equal and opposite angular momentum about the same axis.
>
> Over the next four seconds a weak external magnetic brake exerts a constant
> $+0.15\ \mathrm{N\,m}$ on the combined platform-and-wheel system. Its angular
> impulse is $(0.15)(4.0)=+0.60\ \mathrm{N\,m\,s}$, and since
> $1\ \mathrm{N\,m\,s}=1\ \mathrm{kg\,m^2\,s^{-1}}$ the total angular momentum must
> become $+0.60\ \mathrm{kg\,m^2\,s^{-1}}$. The motor torque between wheel and
> platform is internal to this two-body system and does not enter that total.
>
> If an encoder then records a final wheel speed of $+8.0\ \mathrm{rad\,s^{-1}}$,
> the wheel carries $+1.6\ \mathrm{kg\,m^2\,s^{-1}}$, so the platform must carry
> $-1.0\ \mathrm{kg\,m^2\,s^{-1}}$ for the components to sum to $+0.60$. Dividing
> by the platform inertia gives a final platform speed of
> $-1.25\ \mathrm{rad\,s^{-1}}$. The reversal is consistent with the positive
> external impulse: the wheel took a larger positive internal transfer while the
> platform supplied the compensating negative component.

The numerical check is performed on angular momentum, not on angular speeds.
Adding the final speeds would mix quantities with different moments of inertia and
would have no conservation meaning. The same calculation also shows why one
component can be conserved while another is not. Here the balance applies only to
the vertical component because the stated external torque and the two rotation
axes are vertical. A horizontal support torque would require its own component
equation and would not invalidate the vertical calculation.