Constraints and Their Classification

Drop a cat upside down and it lands on its feet — despite having zero angular momentum the whole way down and nothing to push against. Try to slide your car sideways into a parking spot and you will fail every time — yet forty seconds of shuffling back and forth puts it there anyway. Both puzzles turn on the same fine print: what kind of restriction is actually binding the system? The cat's constraint and the car's constraint forbid directions of motion, not places — and restrictions on directions, it turns out, can be outmaneuvered. Classical mechanics sorts all such restrictions — beads on wires, blocks on tables, wheels that roll without slipping, strings that cannot stretch — into a small, sharp taxonomy. The most consequential split is holonomic versus nonholonomic: an equation among the coordinates themselves, $f(q,t)=0$ , walls off part of configuration space and lets you eliminate a coordinate for free; a velocity constraint that cannot be integrated into such an equation forbids directions while leaving every configuration reachable. Two further axes refine the picture: scleronomic versus rheonomic (is the wall standing still or moving?) and bilateral versus unilateral (is it a wall, or a floor you can lift off from?). This classification is not pedantry — it is the decision procedure for the entire Lagrangian program. Classify correctly and you know how many coordinates you need, whether energy is conserved, whether the constraint can ever switch off, and which formalism will solve your problem. Classify wrongly and you will eliminate a coordinate that cannot be eliminated, or trust an energy that a moving wire is quietly feeding. Every mechanics problem you will ever set up starts, silently or not, right here.

Key takeaways

A constraint is any restriction on the configurations or velocities of a system; the force that enforces it takes whatever value the geometry demands (see Constraints and constraint forces for the Newtonian machinery).
Holonomic: expressible as $f(q_1,\dots,q_n,t)=0$ — an equation among coordinates. Each independent one removes a coordinate: $n = 3N - k$ .
Nonholonomic: a velocity (Pfaffian) constraint $\sum_i a_i\,\mathrm{d}q_i = 0$ that admits NO integrating factor. It restricts directions of motion but removes no coordinates — the test is Frobenius: $\omega\wedge\mathrm{d}\omega \neq 0$ .
Scleronomic constraints have no explicit $t$ ; rheonomic ones move, can do real work, and split the conserved Jacobi function $h$ from the mechanical energy $E$ .
Bilateral constraints are equalities; unilateral ones are inequalities ( $g \geq 0$ ) that can detach — the reaction obeys the complementarity conditions $g\geq 0$ , $\lambda\geq 0$ , $\lambda g = 0$ .
The rolling disk is the canonical nonholonomic system: its two rolling conditions are non-integrable, so it can reach every point of its 4-dimensional configuration space while never moving sideways.
Key idea: classify first, then choose your weapon — a holonomic constraint lets you eliminate a coordinate; a nonholonomic one only forbids directions and needs multipliers; a rheonomic one uncouples energy from the Jacobi integral; a unilateral one can switch off.

The core insight

Classify first, then choose your method: a holonomic constraint lets you eliminate a coordinate; a nonholonomic one only restricts velocities and needs multipliers; a rheonomic one breaks energy conservation; a unilateral one can be violated by detachment.

Where this lives today

Nonholonomic mechanics quietly became a pillar of robotics. A differential-drive robot, a car, a truck with trailers — each is a rolling-constraint system whose configuration space is fully reachable through forbidden-direction-avoiding maneuvers, and motion planning is the algorithmic version of Worked Example 2: steering a nonholonomic system between configurations using only the permitted velocity directions. The autonomous-parking stack in a 2026 production car plans exactly such paths (Reeds–Shepp curves and their sampling-based descendants), and the control-theoretic statement of "you can park anywhere" is Chow's theorem — the Lie brackets of the permitted vector fields span everything, which is the controllability-flavored restatement of $\omega\wedge\mathrm{d}\omega \neq 0$ . The same mathematics reorients free-falling gymnasts, cats, and satellites: with zero angular momentum, internal shape changes translate the orientation around a loop — a nonholonomic constraint on the angular velocities, not the angles. The unilateral axis runs the virtual worlds. Every modern physics engine and robotics simulator — the contact solvers inside MuJoCo, Bullet, Isaac-class GPU simulators used to train locomotion policies — spends most of its compute resolving exactly the complementarity conditions of equation (6) for thousands of simultaneous contacts, at kilohertz rates: which contacts are closed, which reactions are non-negative, which feet are about to detach. That a humanoid robot can be trained to walk in simulation before it ever touches a real floor is, at bottom, an industrial-scale application of $g \geq 0$ , $\lambda \geq 0$ , $\lambda g = 0$ .

The physical picture

Picture the configuration space of your system — one axis per coordinate, one point per possible arrangement. A holonomic constraint is a wall in that space: the equation $f(q)=0$ carves out a lower-dimensional surface, and the system lives on it forever. Once you know the wall is there, the smart move is obvious — lay coordinates along the surface and forget the ambient space entirely. A bead on a circular hoop does not need $(x,y,z)$ ; it needs one angle. The constraint costs you nothing and pays you a coordinate. A nonholonomic constraint is a completely different animal, and the difference is the deepest idea on this page. It does not wall off anywhere. It is a rule about directions: at every point of configuration space, some ways of moving are forbidden, but every point remains reachable by a sufficiently clever route. Think one-way streets, not walls — the map keeps all its intersections; only your instantaneous options shrink.

Figure

Walls versus one-way streets — the two kinds of constraint in configuration space

The other two axes are easier but just as consequential. Scleronomic versus rheonomic is the difference between a stationary wire and a wire someone is spinning: a moving constraint surface can do real work on the system, so "energy conservation" quietly changes meaning — the conserved quantity (when there is one) is the Jacobi function, and it is not the kinetic-plus-potential energy you would naively write down. Bilateral versus unilateral is the difference between a bead threaded on a wire and a ball resting on a table: the table can push but never pull, so the moment the required reaction would turn negative, the ball simply leaves. A constraint that can switch off carries an extra obligation — you must watch the sign of its reaction force, a story told in full by the sibling article on constraint forces.

Figure

The classification decision tree — four axes, one procedure

Hold on to the summary theme: each holonomic constraint is a free win — it costs nothing and gains you a coordinate. Each genuinely nonholonomic constraint is a tax — it restricts the dynamics but leaves configuration space full-dimensional, so you must carry an extra equation through the whole calculation. Everything else in Lagrangian mechanics is built on knowing which is which.

The mathematics of the four axes

The general holonomic constraint is a scalar equation among positions and, possibly, time. Before writing any Lagrangian you list these, because each one you find is a coordinate you will never have to carry:

f_\alpha(\mathbf{r}_1,\dots,\mathbf{r}_N,t)=0,\qquad \alpha=1,\dots,k \qquad (1)

where: $f_\alpha$ — the $\alpha$ -th constraint function on the $3N$ Cartesian coordinates of $N$ particles; $k$ — the number of independent such equations; explicit $t$ -dependence, if present, makes the constraint rheonomic. Geometrically each equation is a hypersurface, and the system lives on the intersection of all $k$ of them. Provided the $k$ constraint gradients are linearly independent, that intersection is a smooth surface of dimension $3N - k$ , and the count of surviving freedoms is

n = 3N - k \qquad (2)

where: $n$ — the number of independent generalized coordinates left after the constraints have spoken. This single subtraction underlies all of Lagrangian bookkeeping; the companion article on degrees of freedom develops the counting craft in detail, including why a rigid body's $\sim N^2/2$ distance constraints collapse to exactly six freedoms. Velocity constraints need a more careful container. The general linear-in-velocity (Pfaffian) constraint is written as a differential form:

\sum_{i=1}^{n} a_{\alpha i}(q,t)\,\mathrm{d}q_i + a_{\alpha t}(q,t)\,\mathrm{d}t = 0 \qquad (3)

where: $a_{\alpha i}$ — the coefficient of $\mathrm{d}q_i$ in the $\alpha$ -th constraint; $a_{\alpha t}$ — the coefficient of $\mathrm{d}t$ (zero for the scleronomic case). Dividing by $\mathrm{d}t$ recovers the velocity form $\sum_i a_{\alpha i}\dot q_i + a_{\alpha t} = 0$ . Every holonomic constraint can be put in this form — just differentiate equation (1) — so the form alone tells you nothing. The real question is whether the traffic runs the other way: can this differential form be integrated back up to an $f(q,t) = 0$ ? That question has an exact answer. For a single constraint 1-form $\omega = \sum_i a_i\,\mathrm{d}q_i$ , the Frobenius integrability condition is

\omega \wedge \mathrm{d}\omega = 0 \qquad (4)

where: $\mathrm{d}\omega$ — the exterior derivative of the form and $\wedge$ the antisymmetric wedge product; the condition holds identically if and only if some integrating factor $g(q)$ makes $g\,\omega$ an exact differential $\mathrm{d}f$ . If (4) holds, your "velocity constraint" was a holonomic constraint wearing a disguise; if it fails anywhere, no function $f$ exists and the constraint is genuinely nonholonomic. A pleasant corollary: in two variables a 1-form always admits an integrating factor, so genuine nonholonomy needs at least three coordinates — one reason simple textbook systems are almost all holonomic and the famous counterexamples (disk, skate, sphere) all involve a heading angle. The canonical nonholonomic system is the upright disk (or coin, or wheel) rolling without slipping on a plane. Its configuration needs four coordinates — contact point $(x,y)$ , heading $\theta$ , spin $\phi$ — and rolling ties the contact-point velocity to the rim:

\dot x = R\,\dot\phi\cos\theta,\qquad \dot y = R\,\dot\phi\sin\theta \qquad (5)

where: $R$ — the disk radius; $\theta$ — the heading of the disk's plane measured in the ground plane; $\phi$ — the rotation angle about the disk's own axis. Two velocity relations among four coordinates — yet, as Derivation B proves, neither can be integrated, so the disk keeps all four coordinates and loses two velocity freedoms instead. It can be steered to any $(x,y,\theta,\phi)$ whatsoever, but never by sliding sideways. Unilateral constraints replace the equality by an inequality and the reaction by a one-signed multiplier. The modern packaging is the complementarity triplet:

g(q,t) \geq 0,\qquad \lambda \geq 0,\qquad \lambda\,g = 0 \qquad (6)

where: $g$ — the gap function (zero on contact, positive off it); $\lambda$ — the magnitude of the reaction. The product condition is the logic gate: at every instant either the contact is closed ( $g=0$ ) and a non-negative reaction acts, or the gap is open ( $g>0$ ) and the reaction is exactly zero. A table can push, never pull; a string can pull, never push — an inextensible string is precisely the unilateral constraint $\ell - |\mathbf{r}| \geq 0$ , which is why strings go slack. Finally, the time axis. The energy bookkeeping of a constrained system is run by the Jacobi function,

h = \sum_i \dot q_i \frac{\partial L}{\partial \dot q_i} - L \qquad (7)

where: $L$ — the Lagrangian in the reduced coordinates. When $L$ has no explicit time dependence, $h$ is conserved. For scleronomic constraints the kinetic energy is purely quadratic in the $\dot q_i$ and $h$ works out to the familiar $T + V$ . But a rheonomic constraint injects explicit time into the embedding $\mathbf{r}_i(q,t)$ , the kinetic energy sprouts terms linear and zeroth-order in $\dot q$ , and $h$ silently parts company with $T+V$ — the conserved quantity is no longer the energy. Derivation C makes this concrete with the rotating wire, where the wire feeds the bead unbounded energy while $h$ stays fixed forever.

Which classification unlocks which tool

the constraint is…	test	what you get to do
holonomic, eq. (1)	it is an equation among $q,t$	eliminate a coordinate; $n = 3N-k$
secretly holonomic	Pfaffian form passes $\omega\wedge\mathrm{d}\omega=0$	integrate it, then eliminate as above
genuinely nonholonomic	$\omega\wedge\mathrm{d}\omega\neq 0$	keep all coordinates; Lagrange multipliers or the Lagrange–d'Alembert equations
scleronomic	no explicit $t$	$h = T+V$ : the Jacobi integral is the energy
rheonomic	explicit $t$ in eq. (1)	track $h$ , not $E$ ; the constraint does real work
unilateral, eq. (6)	reaction is one-signed	monitor $\lambda$ ; when $\lambda\to 0$ , switch branches

Derivation AThe integrability test that separates the two kingdoms

We are establishing the working test behind the whole holonomic–nonholonomic split: when can a velocity constraint $\sum_i a_i\,\dot q_i = 0$ be integrated up to a coordinate constraint $f(q) = \text{const}$ ? The strategy is to phrase the constraint as a differential 1-form $\omega$ and ask when $\omega$ — possibly after multiplication by an integrating factor — is exact. We first extract the practical criterion, then run it on three small examples that bracket the possibilities: one exact, one integrable only after a factor, one genuinely nonholonomic. The thing to watch is the integrating factor: forgetting that a non-exact form can still be integrable is the classic way to misclassify a constraint as nonholonomic.

Step 1. Write the constraint as a 1-form and pose the question. The velocity constraint $\sum_i a_i(q)\,\dot q_i = 0$ holds along a trajectory if and only if the form $\omega = \sum_i a_i\,\mathrm{d}q_i$ annihilates the displacement. If there exist functions $g(q)$ and $f(q)$ with $g\,\omega = \mathrm{d}f$ , then the constraint is equivalent to $\mathrm{d}f = 0$ , i.e. $f(q) = \text{const}$ — holonomic. Why the factor $g$ is allowed: multiplying the constraint equation by any nonzero function changes nothing about which motions it permits, so integrability must be judged up to such a factor.

Step 2. Extract the computable criterion. If $g\,\omega = \mathrm{d}f$ , then applying $\mathrm{d}$ gives $\mathrm{d}g\wedge\omega + g\,\mathrm{d}\omega = 0$ , so $\mathrm{d}\omega = -\,\mathrm{d}g\wedge\omega/g$ . Wedging with $\omega$ kills the term containing $\omega$ twice:

\omega\wedge\mathrm{d}\omega = 0

Why this is also sufficient: that is the content of Frobenius's theorem — for a single 1-form, the vanishing of $\omega\wedge\mathrm{d}\omega$ guarantees the local existence of the integrating factor. The proof constructs $f$ by integrating along the permitted directions.

Step 3. Run the exact case. Rolling of a wheel along a fixed straight line: $\mathrm{d}x - R\,\mathrm{d}\phi = 0$ . The coefficients are constants, so $\omega = \mathrm{d}(x - R\phi)$ is already exact: $x - R\phi = \text{const}$ — holonomic, no factor needed. The wheel's position and its spin are permanently locked together; one coordinate suffices. Why this matters: it is the one-dimensional shadow of the rolling disk, and it is holonomic — nonholonomy will only appear when a heading angle lets the direction of rolling change.

Step 4. Run the integrating-factor case. Take $\omega = x\,\mathrm{d}y - y\,\mathrm{d}x$ in the plane region $x>0$ . This is not exact ( $\mathrm{d}\omega = 2\,\mathrm{d}x\wedge\mathrm{d}y \neq 0$ ), but dividing by $x^2$ gives $\omega/x^2 = \mathrm{d}(y/x)$ : the constraint integrates to $y/x = \text{const}$ — motion along rays through the origin. Holonomic after all. Why the two-variable case always ends this way: in two variables $\omega\wedge\mathrm{d}\omega$ is a 3-form and vanishes identically, so an integrating factor always exists — genuine nonholonomy needs at least three coordinates.

Step 5. Run the nonholonomic case. Take $\omega = \mathrm{d}z - y\,\mathrm{d}x$ in three variables. Then $\mathrm{d}\omega = -\,\mathrm{d}y\wedge\mathrm{d}x = \mathrm{d}x\wedge\mathrm{d}y$ , and

\omega\wedge\mathrm{d}\omega = (\mathrm{d}z - y\,\mathrm{d}x)\wedge\mathrm{d}x\wedge\mathrm{d}y = \mathrm{d}z\wedge\mathrm{d}x\wedge\mathrm{d}y \neq 0

No integrating factor exists; there is no surface $f(x,y,z)=\text{const}$ containing the permitted motions. Indeed one can reach any point of $\mathbb{R}^3$ while respecting $\dot z = y\dot x$ : move in $y$ freely, and use $y$ as a gear ratio to climb in $z$ while shuttling in $x$ . Restriction of directions, no restriction of destinations — the signature of nonholonomy. Verified: the criterion and the examples match Goldstein §1.3 and the Frobenius theorem as stated in standard differential-geometry references; the $\mathrm{d}z - y\,\mathrm{d}x$ form is the textbook contact form, the standard non-integrable example.

Go deeper · Derivation B · 2 minThe rolling disk fails the test

Derivation BThe rolling disk fails the test

The goal is to certify the upright rolling disk as genuinely nonholonomic — the claim that its two rolling conditions (5) cannot, under any algebraic cleverness, be integrated into coordinate constraints. The strategy is exactly the criterion of Derivation A: write one rolling condition as a 1-form in the four configuration coordinates $(x,y,\theta,\phi)$ and show $\omega\wedge\mathrm{d}\omega \neq 0$ . What to watch: the heading angle $\theta$ sitting inside the coefficient $\cos\theta$ is the whole story — it is the coordinate whose presence makes the form non-integrable, and it is precisely what the wheel-on-a-line of Derivation A did not have.

Step 1. Write the first rolling condition as a 1-form. From $\dot x = R\dot\phi\cos\theta$ :

\omega_1 = \mathrm{d}x - R\cos\theta\,\mathrm{d}\phi

Why this encodes rolling: in a time $\mathrm{d}t$ the rim lays down an arc $R\,\mathrm{d}\phi$ on the ground, pointed along the heading $(\cos\theta,\sin\theta)$ ; the contact point must advance by exactly that arc — no slipping.

Step 2. Take the exterior derivative. Only the $\theta$ -dependence contributes:

\mathrm{d}\omega_1 = R\sin\theta\,\mathrm{d}\theta\wedge\mathrm{d}\phi

Why: $\mathrm{d}(\mathrm{d}x) = 0$ and differentiating $-R\cos\theta$ brings down $R\sin\theta\,\mathrm{d}\theta$ .

Step 3. Wedge the form with its derivative:

\omega_1\wedge\mathrm{d}\omega_1 = R\sin\theta\,\mathrm{d}x\wedge\mathrm{d}\theta\wedge\mathrm{d}\phi \neq 0

Why the surviving term is the only one: the $-R\cos\theta\,\mathrm{d}\phi$ part of $\omega_1$ dies against the $\mathrm{d}\phi$ already present in $\mathrm{d}\omega_1$ . The result is a nonzero 3-form wherever $\sin\theta\neq 0$ , and by symmetry the companion form $\omega_2 = \mathrm{d}y - R\sin\theta\,\mathrm{d}\phi$ fails at $\cos\theta \neq 0$ — between them, everywhere.

Step 4. Read off the physics. No functions $f_{1,2}(x,y,\theta,\phi)$ exist whose level sets encode rolling. The disk's configuration space stays honestly 4-dimensional: no configuration is forbidden, and Worked Example 2 exhibits the explicit maneuver — roll, pivot, roll back, pivot back — that translates the disk sideways, the very direction equation (5) forbids instantaneously. Two velocity freedoms ( $\dot\phi$ and $\dot\theta$ ) steer four coordinates.

Step 5. Note what the result licenses and what it forbids. Because the constraints are nonholonomic, no choice of two "reduced coordinates" exists, and the Lagrangian must keep all four coordinates with the constraints appended — via Lagrange multipliers or the Lagrange–d'Alembert route developed in the article on nonholonomic systems. Attempting to "integrate the rolling conditions" and substitute is not an approximation; it is simply wrong. Verified: reproduces the classification in Goldstein §1.3 and Morin ch. 6; the wedge computation is the standard Frobenius certificate for the rolling disk.

Go deeper · Derivation C · 2 minA bead on a rotating wire and the two energies

Derivation CA bead on a rotating wire and the two energies

This derivation pins down what the rheonomic label actually costs you. A bead of mass $m$ slides on a frictionless straight wire forced to rotate at constant rate $\omega$ in a horizontal plane; the constraint $\theta = \omega t$ is holonomic (an equation among coordinate and time) but rheonomic (the wall moves). We will reduce to the single coordinate $r$ , solve the motion, and then compute two different scalars — the mechanical energy $E$ and the Jacobi function $h$ — and watch them disagree: $h$ conserved but not the energy, $E$ growing without bound. The step to watch is the substitution $\dot\theta = \omega$ inside the kinetic energy: that is where explicit time enters and where a $T_0$ term (velocity-independent kinetic energy) is born.

Step 1. Classify and reduce. The constraints are $z=0$ (holonomic, scleronomic) and $\theta - \omega t = 0$ (holonomic, rheonomic). Both are equations among coordinates and time, so both eliminate coordinates: one degree of freedom survives, the radius $r$ . Why rheonomy does not spoil the elimination: the counting rule cares only that the constraint is an equation, not whether it moves — a moving wall still walls.

Step 2. Build the Lagrangian. In polar coordinates $T = \tfrac12 m(\dot r^2 + r^2\dot\theta^2)$ ; substituting the forced $\dot\theta = \omega$ and $V=0$ :

L = \tfrac{1}{2}m\dot r^2 + \tfrac{1}{2}m\omega^2 r^2

Why the second term is kinetic, not potential: it came from $T$ , and it does not depend on $\dot r$ — a $T_0$ piece, the fingerprint of a rheonomic embedding $\mathbf{r}(q,t)$ . It behaves like an inverted-oscillator potential in the equation of motion, which is exactly the centrifugal effect.

Step 3. Solve the motion. The Euler–Lagrange equation gives $\ddot r = \omega^2 r$ , so a bead released at $r_0$ with $\dot r(0)=0$ follows

r(t) = r_0\cosh(\omega t)

The bead runs away exponentially along the wire — someone must be doing work on it, and the only candidate is the wire's normal force, a constraint force. Why a constraint force can do work here: the wire moves; the force is perpendicular to the wire but the bead's real displacement has a component along the force because the wire itself sweeps sideways. (Its virtual work, at frozen time, is still zero — the resolution of the apparent paradox, and the subject of the virtual-work article.)

Step 4. Compute the Jacobi function from (7):

h = \dot r\,\frac{\partial L}{\partial \dot r} - L = \tfrac{1}{2}m\dot r^2 - \tfrac{1}{2}m\omega^2 r^2

Since $L$ has no explicit $t$ , $h$ is conserved: with the release conditions, $h = -\tfrac12 m\omega^2 r_0^2$ forever — check it against the solution, where $\dot r^2 - \omega^2 r^2 = \omega^2 r_0^2(\sinh^2 - \cosh^2) = -\omega^2 r_0^2$ . Why $h$ is conserved even though the system is driven: conservation of $h$ needs only $\partial L/\partial t = 0$ in the reduced coordinates, and the reduction absorbed the drive into the $r$ -dependent $T_0$ term.

Step 5. Compare with the energy. The actual kinetic energy is $E = T = \tfrac12 m\dot r^2 + \tfrac12 m\omega^2 r^2 \neq h$ : the sign of the second term differs. $E$ grows like $e^{2\omega t}$ while $h$ sits constant. The moral of the rheonomic axis in one line: the conserved quantity is not the energy, and the energy is not conserved. *Verified: matches Goldstein §2.7/§8.2 and Landau–Lifshitz §6 (the rotating-wire bead is the standard example separating the Jacobi integral from the energy); the $\cosh$ solution satisfies $\ddot r = \omega^2 r$ with the stated initial data.*

Limiting cases

Case	Result	Meaning
all constraints holonomic	$n = 3N - k$ ; multipliers unnecessary	pick $n$ generalized coordinates and the constraints vanish from the problem
Pfaffian form with $\omega\wedge\mathrm{d}\omega = 0$	secretly holonomic	integrate (with its factor) and eliminate — never carry it as a velocity constraint
two-variable velocity constraint	always integrable	an integrating factor always exists in 2 variables; nonholonomy needs $\geq 3$ coordinates
$\partial f/\partial t = 0$ everywhere (scleronomic)	$h = T + V = E$	Jacobi integral is the energy; constraint forces do no real work
rheonomic constraint present	$h \neq T+V$ in general	the moving wall does real work; track $h$ , not $E$ (rotating wire: $E\to\infty$ , $h$ fixed)
unilateral contact with $\lambda \to 0$	detachment	the constraint switches off and the equations of motion change branch

Common trapEvery velocity constraint is nonholonomic — no, run the test first

The single most common misclassification: seeing $\dot q$ 's in a constraint and reaching for the nonholonomic machinery. A velocity constraint is only nonholonomic if it is non-integrable — and integrability includes the possibility of an integrating factor, not just exactness. Rolling along a straight rail ( $\mathrm{d}x = R\,\mathrm{d}\phi$ ) is holonomic; motion along rays ( $x\,\mathrm{d}y - y\,\mathrm{d}x = 0$ ) is holonomic after division by $x^2$ ; any velocity constraint in only two coordinates is holonomic, full stop. Run $\omega\wedge\mathrm{d}\omega$ before you classify. The converse trap also bites: differentiating a holonomic constraint puts it into velocity form, and if you then treat it as "a velocity constraint" and keep all coordinates plus a multiplier, your algebra balloons for nothing. And one more: a constraint containing $t$ is not thereby nonholonomic — $\theta = \omega t$ is a perfectly holonomic (rheonomic) equation. The velocity/time axes are independent; classify along each separately.

Worked examples

1 · Classify, reduce, and audit the energy books (confidence builder)

A bead of mass $m = 50\ \mathrm{g}$ slides on a frictionless straight wire rotating at constant $\omega = 2.0\ \mathrm{rad/s}$ in a horizontal plane. It is released at $r_0 = 10\ \mathrm{cm}$ from the pivot, at rest relative to the wire. Classify every constraint, count the degrees of freedom, and find the bead's position, speed along the wire, kinetic energy, and Jacobi function at $t = 1.0\ \mathrm{s}$ . How fast is the wire pumping in energy at that moment?

Step 1. Classify. $z = 0$ : holonomic, scleronomic, bilateral. $\theta - \omega t = 0$ : holonomic (equation among coordinate and time), rheonomic, bilateral. Two holonomic constraints on one particle: $n = 3 - 2 = 1$ degree of freedom, the radius $r$ .

Step 2. Solve. From Derivation C, $r(t) = r_0\cosh(\omega t)$ and $\dot r(t) = r_0\,\omega\sinh(\omega t)$ . At $\omega t = 2.0$ : $\cosh 2 = 3.762$ , $\sinh 2 = 3.627$ , so $r = 0.10 \times 3.762 = \boxed{0.376\ \mathrm{m}}$ and $\dot r = 0.10 \times 2.0 \times 3.627 = \boxed{0.725\ \mathrm{m/s}}$ .

Step 3. Kinetic energy at $t = 1.0$ s. The full speed includes the sweep: $T = \tfrac12 m(\dot r^2 + \omega^2 r^2) = \tfrac12(0.05)\big(0.725^2 + 4(0.376)^2\big) = \tfrac12(0.05)(0.526 + 0.566) = \boxed{2.73\times10^{-2}\ \mathrm{J}}$ . At release it was $\tfrac12(0.05)(4)(0.01) = 1.0\times10^{-3}$ J — the energy has grown 27-fold in one second.

Step 4. Jacobi function. $h = \tfrac12 m\dot r^2 - \tfrac12 m\omega^2 r^2 = \tfrac12(0.05)(0.526 - 0.566) = \boxed{-1.0\times10^{-3}\ \mathrm{J}}$ — exactly its release value $-\tfrac12 m\omega^2 r_0^2 = -\tfrac12(0.05)(4)(0.01)$ . Conserved, as the time-independent reduced Lagrangian demands, and visibly not the energy.

Step 5. Power input. $\mathrm{d}E/\mathrm{d}t = m\dot r\ddot r + m\omega^2 r\dot r = 2m\omega^2 r\dot r = 2(0.05)(4)(0.376)(0.725) = \boxed{0.109\ \mathrm{W}}$ — supplied by the motor holding $\omega$ fixed, delivered through the "workless" constraint force. The rheonomic label was the warning that this could happen.

2 · Parallel-parking a rolling disk (exam level)

An upright disk of radius $R = 5.0\ \mathrm{cm}$ rolls without slipping on the plane, obeying the constraints (5). Starting from $(x,y,\theta,\phi) = (0,0,0,0)$ , execute: (i) roll forward one full wheel turn ( $\Delta\phi = 2\pi$ ) at heading $\theta = 0$ ; (ii) pivot in place to $\theta = 30°$ ; (iii) roll backward one full turn ( $\Delta\phi = -2\pi$ ); (iv) pivot back to $\theta = 0$ . Find the final configuration, and explain what the answer proves about the constraints.

Step 1. Move (i): at $\theta = 0$ , equation (5) gives $\Delta x = R\,\Delta\phi\cos 0 = 0.05 \times 2\pi = 0.3142$ m, $\Delta y = 0$ . State: $(0.3142,\ 0,\ 0,\ 2\pi)$ .

Step 2. Move (ii): pivoting means $\dot\phi = 0$ , so (5) forces $\dot x = \dot y = 0$ — the contact point stays put while $\theta$ goes to $30°$ . State: $(0.3142,\ 0,\ 30°,\ 2\pi)$ .

Step 3. Move (iii): rolling $\Delta\phi = -2\pi$ at heading $30°$ : $\Delta x = R(-2\pi)\cos 30° = -0.3142 \times 0.8660 = -0.2721$ m and $\Delta y = R(-2\pi)\sin 30° = -0.3142 \times 0.5 = -0.1571$ m. State: $(0.0421,\ -0.1571,\ 30°,\ 0)$ .

Step 4. Move (iv): pivot back, position frozen again. Final state: $\boxed{(x,y,\theta,\phi) = (4.2\ \mathrm{cm},\ -15.7\ \mathrm{cm},\ 0,\ 0)}$ .

Step 5. Read the result. Both $\theta$ and $\phi$ returned exactly to their initial values, yet the contact point moved $\sqrt{4.21^2 + 15.71^2} = 16.3$ cm — mostly sideways, the one direction rolling forbids at every instant. If the rolling constraints were integrable, functions $f_{1,2}(x,y,\theta,\phi) = \text{const}$ would have to return to their initial values with $\theta$ and $\phi$ — but $x$ and $y$ changed, contradiction. A closed loop in the controlled variables producing a net displacement in the others is nonholonomy made flesh: it is Derivation B's wedge $R\sin\theta\,\mathrm{d}x\wedge\mathrm{d}\theta\wedge\mathrm{d}\phi \neq 0$ turned into a parking maneuver. For small pivot angle $\beta$ the sideways yield per cycle is $\Delta y \approx -R\,\Delta\phi\,\beta$ — first order in $\beta$ , which is why even gentle shuffling parks a car.

Origins · 1 minHeinrich Hertz and how this was discovered

Heinrich Hertz

1857–1894 · Karlsruhe and Bonn

Hertz is immortal for the 1887 experiments that produced and detected electromagnetic waves, but his last work was a book on mechanics — The Principles of Mechanics Presented in a New Form, published in 1894, months after his death at 36. Seeking a mechanics free of the "force" concept, Hertz rebuilt the subject on configurations, constraints, and geodesics — and needed, for the first time, a sharp name for the distinction this article teaches. He coined holonomic (from Greek holos + nomos, "whole law"): systems whose velocity restrictions integrate up to whole, finite equations among the coordinates. Systems like the rolling sphere, whose restrictions bind only the differentials, he showed obey a genuinely different mechanics — one where, famously, the straightest available path need not be the shortest. The taxonomy outlived the philosophical program: Hertz's vocabulary is now the load-bearing first question of every Lagrangian setup.

History

1743

— d'Alembert's Traité de dynamique recasts dynamics so that workless constraint forces drop out of the equations — constraints become central objects of mechanics.

1788

— Lagrange's Mécanique analytique systematizes both strategies still in use: eliminate coordinates when constraints integrate, append multipliers when they do not.

1877

— Frobenius publishes the integrability theorem for Pfaffian systems, the exact test

\omega\wedge\mathrm{d}\omega = 0

separating the two kingdoms.

1894

— Hertz's Principles of Mechanics coins "holonomic" and exhibits rolling systems as a genuinely different mechanics.

1903

— Chaplygin derives reduced equations of motion for a class of nonholonomic systems (the Chaplygin sleigh and rolling bodies), founding nonholonomic mechanics as its own discipline.

Going further

References. Goldstein, Poole & Safko, Classical Mechanics (3rd ed.), §1.3 · Morin, Introduction to Classical Mechanics, Ch. 6 · Landau & Lifshitz, Mechanics (Vol. 1), §1 · Taylor, Classical Mechanics, Ch. 7

Provenance. Classifications and definitions follow Goldstein, Poole & Safko §1.3 and Landau–Lifshitz Vol. 1 §1. Verifiers should recompute: (1) the Frobenius wedge $\omega_1\wedge\mathrm{d}\omega_1$ for the rolling disk, confirming non-integrability (Goldstein, Morin Ch. 6); (2) the rotating-bead Lagrangian, its $\ddot r=\omega^2 r$ equation, and the inequality $h\neq E$ (Goldstein §2.7/§8.2, Landau §6); (3) the sphere-detachment angle $\cos\theta=2/3$ from $N=mg(3\cos\theta-2)$ (Morin Ch. 5, Taylor Ch. 4); (4) the degree-of-freedom count $n=3N-k$ and the rigid-body reduction to 6. All numbers are standard textbook values. Worked-example numbers on this page ( $r=0.376$ m, $T=2.73\times10^{-2}$ J, $h=-1.0\times10^{-3}$ J, $0.109$ W; parking displacement $4.2$ cm, $-15.7$ cm) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Generalized Coordinates and Configuration Space · Degrees of Freedom Leads to: Virtual Displacements and the Principle of Virtual Work · d'Alembert's Principle · The Euler-Lagrange Equations · The Energy Function and Jacobi Integral

A ball rolling along a straight line is holonomic, but a disk rolling on a plane is not. Where exactly does the extra freedom sneak in?

Through the heading angle $\theta$ . On a line, rolling locks two coordinates together permanently: $x - R\phi = \text{const}$ , one equation, forever. On a plane, the same rim arc $R\,\mathrm{d}\phi$ gets laid down in whatever direction the disk currently points, so the coefficients of the constraint form contain $\cos\theta$ and $\sin\theta$ . Because you can change $\theta$ between rolls, the relation between $\phi$ and position depends on the history of headings, not just the endpoints — no function $f(x,y,\theta,\phi)$ can store it. That is literally what Derivation B's nonzero wedge $R\sin\theta\,\mathrm{d}x\wedge\mathrm{d}\theta\wedge\mathrm{d}\phi$ detects: the $\theta$ -dependence of the coefficients is the obstruction to integrating.

Nonholonomic constraints "don't reduce degrees of freedom" — but they clearly restrict the motion. What gives?

Two different counts are being conflated. The number of configuration degrees of freedom — independent coordinates needed to specify where the system is — is untouched by a nonholonomic constraint, because every configuration stays reachable. The number of independent velocity directions at each instant genuinely shrinks: the rolling disk has 4 coordinates but only 2 velocity freedoms ( $\dot\phi$ , $\dot\theta$ ). So the honest statement is "4 coordinates, 2 controls." The degrees of freedom article treats this split carefully. Practical consequence: you still need equations of motion for all 4 coordinates, plus multipliers to enforce the 2 velocity constraints — nothing gets cheaper, unlike the holonomic case.

Is the tension of an inextensible string a holonomic constraint?

Be careful — it is unilateral. A string of length $\ell$ enforces $|\mathbf{r}| \leq \ell$ , an inequality: it prevents the distance from exceeding $\ell$ but happily lets the mass approach the anchor with the string slack. While taut it behaves exactly like the bilateral, holonomic constraint $|\mathbf{r}| = \ell$ (a pendulum), and you may treat it so — provided you monitor the tension. The moment your solution requires $T < 0$ (the string pushing), the real string goes slack, the constraint switches off, and the mass becomes a projectile until the string snaps taut again. A rod, by contrast, is genuinely bilateral: it can push or pull, and $|\mathbf{r}| = \ell$ holds unconditionally. Classifying "string" as "rod" is a classic exam trap on vertical-circle problems.

How can the disk reach sideways positions if sideways motion is forbidden at every single instant?

Because "forbidden at every instant" constrains the tangent of the path, not its endpoints. Worked Example 2 is the proof by demonstration: roll, pivot, roll back, pivot back — every segment individually respects the constraints, yet the loop nets a sideways displacement of $\approx R\,\Delta\phi\,\beta$ for small pivot angle $\beta$ . The composition of two permitted motions in alternation generates a new direction that neither had alone (in control-theory language, the Lie bracket of the roll and pivot vector fields points sideways). That new directions can keep appearing until all of configuration space is reachable is Chow's theorem, and it is the exact geometric opposite of the holonomic case, where composing permitted motions never leaves the constraint surface.

If constraint forces do no work, how did the rotating wire pump 27 times the initial energy into the bead?

The workless property is about virtual work — displacements imagined at frozen time — not real work. At any frozen instant, the wire's normal force is perpendicular to the wire, and the bead's permitted virtual displacement is along the wire: zero virtual work, which is what the Lagrangian formalism needs. But the bead's real displacement during $\mathrm{d}t$ includes the wire's own sweep $r\omega\,\mathrm{d}t$ sideways — parallel to the normal force — so the real work $\mathrm{d}W = N r\omega\,\mathrm{d}t \neq 0$ . For scleronomic constraints real and virtual displacements span the same directions and the distinction evaporates; rheonomic constraints are precisely the case where it bites. This is why the energy audit of a rheonomic problem must be run on the Jacobi function, and it is the door into the principle of virtual work.

Is rolling without slipping caused by friction, and does that make friction a constraint force?

Static friction is indeed the agent that enforces rolling — it takes whatever value (up to its ceiling $\mu_s N$ ) keeps the contact point's velocity zero, which is exactly the defining behavior of a constraint force: determined by the geometry it maintains, not by a force law. So yes: while rolling holds, treat the contact friction as an ideal constraint force; it acts on a point of zero velocity and does no work. The classification breaks when demand exceeds supply: if maintaining rolling would need friction beyond $\mu_s N$ , the wheel slips, kinetic friction (a genuine, formula-bearing, work-dissipating force) takes over, and the constraint is gone. The Newtonian craft of these forces — normal forces, tensions, and when rolling lets go — belongs to constraints and constraint forces.

When do I use Lagrange multipliers instead of eliminating coordinates?

Three situations. First, forced: the constraint is genuinely nonholonomic, so no coordinates can be eliminated — multipliers (or the Lagrange–d'Alembert equations) are the only route. Second, optional but useful: the constraint is holonomic, but you want the constraint force — the multiplier is proportional to it, so keeping the constraint explicit buys you the normal force or tension as part of the answer. Third, convenience: eliminating a coordinate would make the Lagrangian ugly (awkward parametrization of the surface), and carrying the constraint costs less algebra. If none of these applies — holonomic constraint, force not wanted — always eliminate: fewer equations, no multiplier to solve for. The classification is precisely the information this decision needs.

Does a constraint containing time explicitly count as nonholonomic?

No — the two axes are completely independent, and conflating them is a common error. $\theta - \omega t = 0$ contains $t$ but is an honest equation among coordinate and time: holonomic (and rheonomic). It eliminates a coordinate just like any holonomic constraint; the rotating-wire bead really is a one-coordinate problem. Nonholonomy is exclusively about non-integrable velocity dependence. You can have all four combinations: holonomic-scleronomic (bead on fixed hoop), holonomic-rheonomic (bead on rotating wire), nonholonomic-scleronomic (disk on a fixed floor), nonholonomic-rheonomic (disk on a moving conveyor). Each combination has a different set of consequences — coordinates removed by the first axis, energy bookkeeping controlled by the second.