Angular Momentum of a Rigid Body

Every satellite is a rigid body whose attitude computer lives inside this article. Spacecraft carry reaction wheels and control-moment gyroscopes precisely because $\mathrm{d}\mathbf{L}/\mathrm{d}t=\boldsymbol{\tau}$ works both ways: spin a wheel one way and the bus turns the other, angular momentum traded internally without a gram of propellant. The flight software integrates Euler's equations — equation (9), with wheel torques on the right — hundreds of times per second, and "momentum management" (dumping accumulated $\mathbf{L}$ with magnetic torquers or thrusters before the wheels saturate) is a standing chore on everything from CubeSats to the International Space Station, whose four massive control-moment gyros store on the order of $2\times10^4\ \mathrm{N\,m\,s}$ of angular momentum for exactly this bookkeeping. The frontier application is grappling things that tumble. Defunct satellites and spent upper stages spin about non-principal axes, nutating exactly as equations (1) and (9) dictate, and a servicing or debris-removal spacecraft — the missions now flying and manifested by ESA's ClearSpace and Astroscale — must estimate the target's inertia tensor and angular momentum from video before daring to touch it: match the tumble, capture, then despin, budgeting thruster torque against the captured $\mathbf{L}$. Closer to the ground, the same physics is why your car's wheels are dynamically balanced (making the spin axis principal, so $\mathbf{L}_\perp$ and the $\boldsymbol{\omega}\times\mathbf{L}$ shake vanish), and why every phone's attitude filter fuses gyroscope data through a model whose beating heart is $\mathbf{L}=\mathbf{I}\boldsymbol{\omega}$.

Think of $\mathbf{I}$ as a machine that stretches different directions by different factors. Feed it $\boldsymbol{\omega}$ and it hands back a vector that is longer along the directions the body resists spinning about, shorter along the easy ones — and therefore, in general, pointing somewhere else. Only if you feed it a direction the machine does not rotate — an eigenvector, a principal axis — do you get back something parallel. That single sentence of linear algebra is the entire phenomenology of wobble.

Make it concrete with a thin book. Spin it about a diagonal of its cover. The far corners carry most of the mass and most of the speed; summing $\mathbf{r}\times\mathbf{v}$ over the body produces an $\mathbf{L}$ canted away from the spin axis, toward the principal direction with the larger moment. Now hold the spin steady. The body turns, its mass distribution turns with it, and the canted $\mathbf{L}$ is dragged around $\boldsymbol{\omega}$ on a cone. But a rotating $\mathbf{L}$ is a changing $\mathbf{L}$, and $\mathrm{d}\mathbf{L}/\mathrm{d}t=\boldsymbol{\tau}$ says changing angular momentum must be paid for in torque. Whatever grips the axle pays it — which is why your hand feels a rhythmic shudder when you spin the book on a skewer, and why an unbalanced wheel shakes a car. A spherical top ($I_1=I_2=I_3$) is the special case where the machine acts like a scalar: then $\mathbf{L}=I\boldsymbol{\omega}$ for every axis, exactly as a point particle's $\mathbf{p}=m\mathbf{v}$, and nothing ever wobbles. The entire richness of rigid-body motion — precession, tumbling, the Dzhanibekov flip — comes from the breaking of that isotropy. One more preview, because it organizes everything downstream. The same tensor that converts spin to angular momentum also prices the energy, $T=\tfrac12\,\boldsymbol{\omega}\cdot\mathbf{I}\,\boldsymbol{\omega}$. Angular momentum and kinetic energy are two contractions of one object, which is why a single geometric picture — the inertia ellipsoid — manages to encode both at once, and why torque-free motion can be drawn rather than integrated.

Everything starts from the defining relation. It deserves to be read slowly, because its shape — matrix times vector, not number times vector — is the physics:

where: $\mathbf{L}$ — angular momentum about the chosen fixed point or the center of mass; $\mathbf{I}$ — the inertia tensor (a symmetric $3\times3$ matrix) about that same point; $\boldsymbol{\omega}$ — the angular velocity vector, instantaneous rotation axis times rate. The one non-negotiable bookkeeping rule: $\mathbf{L}$ and $\mathbf{I}$ must refer to the same origin, and the law in this clean form holds about a genuinely fixed point or about the center of mass (moving or not) — those are the two origins for which the internal motion terms cancel. In components, with the tensor written out explicitly:

where: $I_{ij}$ — the $ij$ entry of the inertia matrix: diagonal entries are the familiar moments of inertia, off-diagonal entries are the products of inertia; $m_a$, $x_{a,i}$ — mass and coordinates of particle $a$ relative to the reference point (for a continuum, replace $\sum_a m_a$ by $\int\rho\,\mathrm{d}V$); $\delta_{ij}$ — the Kronecker delta. It is the off-diagonal entries that mix components: a spin fed in along $x$ comes back with $y$ and $z$ parts whenever the products of inertia are nonzero. Where does the tensor come from? From nothing but rigidity. Each particle of a body rotating about the origin moves with $\mathbf{v}_a=\boldsymbol{\omega}\times\mathbf{r}_a$, and summing the particle angular momenta

where: $\mathbf{r}_a$ — position of particle $a$ from the reference point; $\mathbf{v}_a$ — its velocity, fixed entirely by the shared rotation. Expanding the triple product splits each particle's contribution into a piece along $\boldsymbol{\omega}$ and a piece along $\mathbf{r}_a$ — and that second, tilt-producing piece is the whole story. The expansion is built line by line in the moment of inertia tensor article, which owns the construction; here we take the machine as given and run it. Run it first in the frame where it is simplest. The tensor is real and symmetric, so there is always a rotated frame — the principal axes — in which it is diagonal, and the map decouples:

where: $I_1,I_2,I_3$ — the principal moments of inertia (eigenvalues of $\mathbf{I}$); $\omega_i$, $L_i$ — components along the principal axes. Read it as three independent scalar laws, one per axis — but with three different proportionality constants. That is precisely why a generic $\boldsymbol{\omega}$, with components on two or more axes, gets steered: each component is stretched by its own factor, and the reassembled vector points elsewhere. A concrete number: for a thin rectangular plate with sides in ratio $2{:}1$, the in-plane moments differ by a factor of four, and a spin fed in along the $45°$ bisector comes back tilted a full $31°$ away (the derivation below computes it). The misalignment can be quantified in one line. Dotting $\mathbf{L}$ into $\boldsymbol{\omega}$ and dividing by the magnitudes:

where: $\theta$ — the angle between $\boldsymbol{\omega}$ and $\mathbf{L}$. The numerator is $2T$, twice the kinetic energy, so $\theta$ is always acute — $\mathbf{L}$ never leans more than $90°$ from the spin. By the Cauchy–Schwarz inequality $\theta=0$ exactly when the $I_i$ multiplying the nonzero components are all equal: a principal-axis spin, or a spherical top. Energy is the second contraction of the same machine:

where: $T$ — rotational kinetic energy about the reference point. The first form is the rotational cousin of $T=\tfrac12\,\mathbf{v}\cdot\mathbf{p}$, and it carries a physical verdict: only the component of $\mathbf{L}$ along $\boldsymbol{\omega}$ stores energy. The transverse part — the part that sweeps the cone — does no work on the spin; it is pure wobble, shaking bearings without contributing a joule. Dynamics enters through Newton, and Newton speaks only in the inertial frame:

where: $\boldsymbol{\tau}^{\text{ext}}$ — total external torque about the same fixed point or the center of mass. This is exact and completely general — but awkward to use directly, because the lab-frame components of $\mathbf{I}$ change from instant to instant as the body turns. The tensor is a constant of the body, not of the lab. That tension — the law lives in space, the tensor lives in the body — is resolved by Euler's equations below. Finally, the honest scalar. Project (1) onto the instantaneous axis $\hat n=\boldsymbol{\omega}/\omega$:

where: $I_{\hat n}$ — the scalar moment of inertia about the axis $\hat n$, the quadratic form of the tensor evaluated on that direction. This component always obeys the freshman law. What the freshman law misses is everything perpendicular: $\mathbf{L}_\perp=\mathbf{L}-L_{\hat n}\hat n$, the piece that demands torque. Fixed-axis problems survive on (8) because the axle physically absorbs $\mathbf{L}_\perp$; free bodies enjoy no such service.

Which equation answers which question

Equation (7) is Newton's honest statement, but it prices the tensor in the wrong currency: in the lab, $I_{ij}(t)$ is a moving target. The fix is to do the accounting in the body frame, where the tensor is frozen — at the cost of a frame-transport term that turns out to be the physics of gyroscopes.

The assembled result deserves its number, because it anchors everything downstream — free precession, the tennis-racket instability, the heavy top:

where: all components live along the body's principal axes, and $\tau_i$ are the torque components in that same rotating basis. Two readings. First, the coupling is quadratic in spin: doubling the rotation rate quadruples the gyroscopic terms, which is why misalignment problems escalate viciously with speed. Second, the equations are nonlinear even for zero torque — rigid-body dynamics is genuinely richer than the linear oscillators of earlier chapters, and its full solution set (stable spins about extreme axes, tumbling about the intermediate one, the rolling constructions of Poinsot) is taken up in the articles this one leads to.

Before the general machinery, one family of shapes where every number is computable by hand: thin plates. For a lamina in the $x$–$y$ plane the perpendicular-axis theorem fixes $I_z=I_x+I_y$, and the in-plane moments follow from the mass layout. The figure shows the case worked in the derivation below.

How $\mathbf{L}$ relates to $\boldsymbol{\omega}$ is a question about the symmetry of the body and the choice of axis:

1 · The tilted plate, with numbers (confidence builder)

A uniform rectangular plate, $M=1.2\ \mathrm{kg}$, sides $a=0.30\ \mathrm{m}$ (along $x$) and $b=0.15\ \mathrm{m}$ (along $y$), spins at $\omega=20\ \mathrm{rad/s}$ about the in-plane $45°$ bisector through its center. Find $\mathbf{L}$, its tilt from the axis, and the rotational kinetic energy.

Step 1. Principal moments. $I_x=\tfrac{1}{12}Mb^2=\tfrac{1}{12}(1.2)(0.15)^2=2.25\times10^{-3}\ \mathrm{kg\,m^2}$ and $I_y=\tfrac{1}{12}Ma^2=\tfrac{1}{12}(1.2)(0.30)^2=9.00\times10^{-3}\ \mathrm{kg\,m^2}$ — a factor 4 apart, as $a=2b$ demands.

Step 2. Components of the spin: $\omega_x=\omega_y=20/\sqrt2=14.14\ \mathrm{rad/s}$. Apply equation (4): $L_x=I_x\omega_x=0.0318\ \mathrm{kg\,m^2/s}$, $L_y=I_y\omega_y=0.1273\ \mathrm{kg\,m^2/s}$. So $\mathbf{L}$ has magnitude $L=\sqrt{L_x^2+L_y^2}=\boxed{0.131\ \mathrm{kg\,m^2/s}}$ and points at $\arctan(0.1273/0.0318)=76.0°$ from the $x$-axis.

Step 3. The tilt. $\theta=76.0°-45°=31.0°$ — or directly from the derivation's formula, $\tan\theta=(I_y-I_x)/(I_y+I_x)=(9.00-2.25)/(9.00+2.25)=0.600$, giving $\boxed{\theta=31.0°}$. The naive scalar route, $L=I_{\hat n}\,\omega$ with $I_{\hat n}=(I_x+I_y)/2=5.625\times10^{-3}\ \mathrm{kg\,m^2}$, predicts $0.1125\ \mathrm{kg\,m^2/s}$ — that is only the component along the axis; it misses the transverse $L\sin\theta=0.0675\ \mathrm{kg\,m^2/s}$ entirely.

Step 4. Energy, two ways as an audit. From $T=\tfrac12\boldsymbol{\omega}\cdot\mathbf{L}=\tfrac12(14.14\times0.0318+14.14\times0.1273)=\boxed{1.13\ \mathrm{J}}$. Check against $T=\tfrac12(I_x\omega_x^2+I_y\omega_y^2)=\tfrac12(2.25\times10^{-3}+9.00\times10^{-3})(200)=1.13\ \mathrm{J}$ ✓. Note it is also $\tfrac12 I_{\hat n}\omega^2$ — the energy sees only $L_\parallel$, exactly as equation (6) promised.

2 · The misaligned rotor — why driveshafts are balanced (exam level)

A rigid dumbbell — two point masses $m=0.50\ \mathrm{kg}$ at the ends of a light rod of half-length $\ell=0.20\ \mathrm{m}$ — is welded at its midpoint to a straight shaft, with the rod at angle $\alpha=30°$ to the shaft. The shaft spins at $3000\ \mathrm{rpm}$. Find $\mathbf{L}$, the torque the bearings must supply, and the load on bearings $d=0.50\ \mathrm{m}$ apart.

Step 1. Geometry and principal frame. The dumbbell's principal axes at its center: $\hat e_3$ along the rod with $I_3=0$ (point masses on the axis), and any transverse direction with $I_1=I_2=2m\ell^2=2(0.50)(0.20)^2=0.040\ \mathrm{kg\,m^2}$. The spin is $\omega=2\pi(3000/60)=314.2\ \mathrm{rad/s}$ along the shaft, which splits as $\omega_3=\omega\cos\alpha$ along the rod and $\omega_\perp=\omega\sin\alpha$ across it.

Step 2. Angular momentum from equation (4): the rod-axis component contributes $I_3\omega_3=0$; everything lives in the transverse direction, $L=I_1\,\omega\sin\alpha=0.040\times314.2\times0.500=\boxed{6.28\ \mathrm{kg\,m^2/s}}$, pointing perpendicular to the rod in the rod–shaft plane. Since the rod sits $30°$ off the shaft, $\mathbf{L}$ sits $90°-30°=60°$ off the shaft: the angular momentum leans twice as far from the axis as the rod does from principal alignment, a dramatic $\mathbf{L}\nparallel\boldsymbol{\omega}$.

Step 3. Torque for steady spin. $\mathbf{L}$ is body-fixed in direction, so it sweeps a $60°$ cone at $314.2\ \mathrm{rad/s}$, and $\boldsymbol{\tau}=\boldsymbol{\omega}\times\mathbf{L}$ has magnitude $\tau=\omega L\sin60°=314.2\times6.28\times0.866=\boxed{1.71\times10^{3}\ \mathrm{N\,m}}$. Equivalently $\tau=(I_1-I_3)\,\omega^2\sin\alpha\cos\alpha=0.040\times(314.2)^2\times0.433=1.71\times10^{3}\ \mathrm{N\,m}$ ✓ — note the $\omega^2$: doubling the speed quadruples the demand.

Step 4. Bearing load. The torque is delivered as a force couple across the bearing span: $F=\tau/d=1709/0.50=\boxed{3.4\ \mathrm{kN}}$ — on a rotor whose total weight is $Mg=(1.0)(9.81)\approx9.8\ \mathrm{N}$. The bearings carry 350 times the rotor's weight, rotating at $50\ \mathrm{Hz}$: that is the shudder of an unbalanced driveshaft, and the case for balancing machines.

Step 5. Audits. $\alpha\to0$: rod along shaft, a principal spin, $\tau\to0$ ✓. $\alpha\to90°$: rod perpendicular to shaft — also principal (spin along a transverse eigenvector), $\tau\to0$ ✓. The demand peaks at $\alpha=45°$, where $\sin\alpha\cos\alpha=\tfrac12$ gives $\tau_{\max}=1.97\times10^3\ \mathrm{N\,m}$: maximal misalignment between what the geometry wants and what the shaft imposes.

References. Morin, Introduction to Classical Mechanics, Ch. 9 · Goldstein, Classical Mechanics, 3rd ed., Ch. 5 · Landau & Lifshitz, Mechanics, Vol. 1, §32-33 · Taylor, Classical Mechanics, Ch. 10 Key results underlying this article include: the BAC-CAB expansion yielding $I_{ij}=\sum_a m_a(r_a^2\delta_{ij}-x_{a,i}x_{a,j})$; the lamina moments $I=\tfrac{1}{12}Ma^2$ and their perpendicular-axis sum; and the Euler-equation derivation via the rotating-frame derivative. Worked-example numbers on this page (plate tilt $31.0°$, $L=0.131\ \mathrm{kg\,m^2/s}$, $T=1.13\ \mathrm{J}$; rotor $L=6.28\ \mathrm{kg\,m^2/s}$, $\tau=1.71\times10^3\ \mathrm{N\,m}$, bearing load $3.4\ \mathrm{kN}$) follow directly from the boxed formulas.

Builds on: Moment of Inertia Tensor · Products of Inertia and the Inertia Matrix · Angular velocity vector Leads to: Principal Axes and Principal Moments of Inertia · Rotational Kinetic Energy · Euler's Equations of Motion · Free Symmetric Top and Body-Cone/Space-Cone Precession