Angular Momentum

Particle angular momentum and torque

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

L = r \times p .

Its magnitude is $L = r p sin ϕ = m r v_{⊥}$ , where $v_{⊥}$ is the velocity component perpendicular to $r$ . 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 $kg m^{2} s^{- 1}$ , equivalently $J s$ .

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

\frac{d L}{d t} = \frac{d r}{d t} \times p + r \times \frac{d p}{d t} = r \times F = τ .

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

\frac{d L _{sys}}{d t} = τ_{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.

Angular momentum

L = r \times p

about

O

has magnitude

r p sin ϕ

; only the component of

p

perpendicular to

r

contributes, and

L

points out of the plane (dot) by the right-hand rule.

Resolving the momentum as $p = p_{r} \overset{r}{^} + p_{t} \hat{θ}$ leaves only the tangential component in the cross product: $L = r \times (p_{t} \hat{θ})$ and $L = r p_{t}$ .

Central forces and conservation.

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

L = m r^{2} \dot{θ} = constant .

During a short interval $d t$ , a radius vector sweeps area $d A = \frac{1}{2} r^{2} d θ$ . Hence the areal velocity is

\frac{d A}{d t} = \frac{1}{2} r^{2} \dot{θ} = \frac{L}{2 m} .

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 $r^{2} \dot{θ}$ stays constant.

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.

Under a central force the areal velocity

d A / d t = L /2 m

is constant, so the radius vector sweeps equal areas

A

in equal times (Kepler's second law).

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,

L = I ω,

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

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

I_{i} ω_{i} = I_{f} ω_{f} .

This conservation equation does not imply conservation of rotational kinetic energy. Substitution gives $K = L^{2} / (2 I)$ . 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.

Gyroscopic precession.

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

τ = r_{cm} \times M g .

In a rapidly spinning symmetric top, this torque is nearly perpendicular to $L$ . It changes the direction of $L$ 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 $θ$ to the vertical,

τ = M g r sin θ, \frac{d L}{d t} = Ω L sin θ, Ω = \frac{M g r}{L} = \frac{M g r}{I ω} .

The cancellation of $sin θ$ 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 $d L / d t = τ$ ; memorized clockwise pictures are unreliable because orientation conventions vary.

Gravity exerts a horizontal torque

τ = r_{c m} \times M g

about the support. Perpendicular to the spin angular momentum

L

, it turns

L

sideways, sweeping the axle around the vertical in steady precession.

Angular impulse and collisions

Integration of the torque law gives angular impulse:

Δ L = \int τ_{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 $m v r$ . If the final rigid system has moment of inertia $I_{f}$ , its immediate angular speed is $ω = m v r / I_{f}$ . 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:

L_{O} = R_{cm} \times M V_{cm} + L_{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 $I_{cm} ω$ , but about a ground point it also has the orbital term $R_{cm} \times M V_{cm}$ . 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.

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: $r \times F$ reverses sign if the factors are exchanged. In component form,

r \times p = (y p_{z} - z p_{y}) \overset{}{^} + (z p_{x} - x p_{z}) \overset{}{^} + (x p_{y} - y p_{x}) \hat{k} .

In a planar problem the sign of

L_{z} = x p_{y} - y p_{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).

A rigid object with three unequal principal moments can have $L$ not parallel to $ω$ . Symmetry axes avoid this complication and justify $L = I ω$ 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 $Δ L = τ Δ t$ , 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 $I ω$ has units $kg m^{2} s^{- 1}$ ; torque times time has the same units. A torque perpendicular to $L$ 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.

With negligible external axial torque, pulling the masses inward from radius

R_{i}

R_{f}

lowers the moment of inertia, so the same angular momentum

I ω

gives a larger angular speed.

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.

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.

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 $L = I ω$ 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.

A projectile carries angular momentum

m v r

about the pivot before impact; the pivot impulse has zero lever arm, so angular momentum about the pivot is conserved through the inelastic embedding.

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.

About

O

the angular momentum splits into an orbital part

R_{c m} \times M V_{c m}

from the centre-of-mass motion and a spin part

L_{c m}

about the centre; both use the same declared origin.

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:

Δ L_{O} = t_{i} \int t_{f} τ_{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	$I_{1} ω_{1 i} + I_{2} ω_{2 i} = (I_{1} + I_{2}) ω_{f}$ when external axial torque is negligible	speed equalization dissipates rotational energy
translating, spinning body	whole body, laboratory origin $O$	$L_{O} = r_{cm} \times M v_{cm} + L_{cm}$	orbital and spin terms use the same origin
rapidly precessing top	top, support point $O$	$\dot{L}_{O} = τ_{grav, O}$	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	$∣ τ_{g} ∣ Δ t ≪ ∣ Δ L_{O} ∣$
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 $L_{z}$	$Δ L_{z} = \int τ_{z} d t$	spin rate, inertia, axial external torque
freely tumbling asymmetric body	full vector $L$ in an inertial frame	tensor relation $L = I ω$	principal moments and body orientation
steady precession	changing direction of $L$	$\dot{L} = τ$	tilt, spin rate, support geometry, sweep rate
translating rotor	total $L_{O}$ 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 $L = m v b$ at the instant of interest. The same result follows from $L = r m v sin ϕ$ , where $ϕ$ 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, $L_{z} = x p_{y} - y p_{x}$ .

Changing the origin changes both the lever arm and the numerical angular momentum. If the origin is shifted from $O$ to $O^{'}$ by vector $a$ while the particle state is unchanged, then $L_{O^{'}} = L_{O} - a \times p$ . 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.

The moment arm

b

is the perpendicular distance from

O

to the particle's line of motion, giving the magnitude

L = m v b

from only the tangential part of the momentum.

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 $Δ L_{z} = \int τ_{z} d t$ , 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.

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, $I_{1} ω_{1 i} + I_{2} ω_{2 i} = (I_{1} + I_{2}) ω_{f}$ 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.

The signed area under a torque-time graph is the angular impulse

Δ L_{z} = \int τ_{z} d t

; equal positive and negative areas leave zero net change in the angular-momentum component.

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 $L ≃ I_{s} ω_{s}$ . At a fixed tilt $θ$ , gravity gives a torque magnitude $M g r sin θ$ about the support. If the axle sweeps around the vertical at angular speed $Ω_{p}$ , the angular-momentum vector traces a horizontal circle. Its rate of change has magnitude $Ω_{p} L sin θ$ . Equating this geometric rate with the torque gives $Ω_{p} = M g r / L ≃ M g r / (I_{s} ω_{s})$ . The sine factors cancel only because the tilt is assumed constant and the large spin contribution dominates the angular momentum.

The vector construction fixes the direction. At any instant, draw $L$ along the axle and draw $τ = r \times M g$ about the support. The small change $d L = τ d t$ points along the torque, perpendicular to the old angular momentum in the rapid-spin limit. The new axle direction is parallel to $L + d L$ . Repeating this construction gives the sense of precession for the specified viewing direction. Reversing the spin reverses $L$ 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 $0.80 kg m^{2}$ carries a coaxial reaction wheel of moment of inertia $0.20 kg m^{2}$ . Initially the platform turns at $+ 3.0 rad s^{- 1}$ and the wheel at $- 12 rad s^{- 1}$ , so their angular-momentum components are $+ 2.4$ and $- 2.4 kg m^{2} s^{- 1}$ 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 $+ 0.15 N m$ on the combined platform-and-wheel system. Its angular impulse is $(0.15) (4.0) = + 0.60 N m s$ , and since $1 N m s = 1 kg m^{2} s^{- 1}$ the total angular momentum must become $+ 0.60 kg m^{2} s^{- 1}$ . 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 $+ 8.0 rad s^{- 1}$ , the wheel carries $+ 1.6 kg m^{2} s^{- 1}$ , so the platform must carry $- 1.0 kg m^{2} s^{- 1}$ for the components to sum to $+ 0.60$ . Dividing by the platform inertia gives a final platform speed of $- 1.25 rad s^{- 1}$ . 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.