Constraints — holonomic vs non-holonomic, rheonomic vs scleronomic: Question bank

Before we start, let us make the whole classification visible so every item stands on its own.

Recall The two questions you ask of ANY constraint

Axis 1 — Holonomic? Can I write it as an equation $f (coordinates, t) = 0$ with no velocities left over (or velocities that integrate away)? If yes → holonomic. If not → non-holonomic. Two different things live under "non-holonomic": (a) a velocity rule that refuses to integrate, and (b) an inequality — see the callout below. Axis 2 — Scleronomic vs rheonomic? Does the clock symbol $t$ appear explicitly? No $t$ → scleronomic (frozen). Explicit $t$ → rheonomic (moving/driven). These are separate questions. Every constraint gets one label from each pair.

Formula The 2-variable integrating factor — why one always exists, and how to find it

Claim: every Pfaffian $A (x, y) d x + B (x, y) d y = 0$ in just two variables is integrable, so genuine non-holonomy needs $\geq 3$ . Why: the relation $A d x + B d y = 0$ defines a slope $\frac{d y}{d x} = - \frac{A}{B}$ at every point — a field of directions in the plane. A first-order ODE like this always has a family of solution curves $f (x, y) = c$ (Picard existence), and those curves ARE the integrated constraint. The function $μ$ that makes $μ A d x + μ B d y = df$ exact is the integrating factor, and it must satisfy $\frac{\partial ( μ A )}{\partial y} = \frac{\partial ( μ B )}{\partial x} .$ How to find it (standard cases): if $\frac{1}{B} (\frac{\partial A}{\partial y} - \frac{\partial B}{\partial x})$ depends on $x$ alone, then $μ (x) = exp \int (\dots) d x$ ; symmetrically for $y$ alone. In three or more variables no such $μ$ need exist — that is the mathematical room in which rolling lives. See the picture:

True or false — justify

A pendulum of fixed length swinging in a vertical plane is holonomic.

True — the bob obeys

x^{2} + y^{2} - ℓ^{2} = 0

, a pure coordinate equation, so it is holonomic and scleronomic (no

t

Every constraint that involves velocities is non-holonomic.

False — a velocity (Pfaffian) relation is non-holonomic only if non-integrable. If

\sum a_{i} d q_{i} = df

for some

f

, integrate it back to a coordinate equation and it was holonomic in disguise.

A rheonomic constraint is automatically non-holonomic.

False — the axes are independent. A bead on a wire spun at fixed

ω

obeys

y - x tan (ω t) = 0

, which is an equation → holonomic, yet has explicit

t

→ rheonomic. Holonomic AND rheonomic.

A holonomic constraint always reduces the number of independent coordinates by one.

True for each independent equality constraint:

n = 3 N - k

. The equation lets you solve one coordinate in terms of the others.

A non-holonomic constraint reduces the number of generalized coordinates.

False — it restricts velocity directions, not reachable configurations. A rolling disk still needs all of

(x, y, ϕ)

; none can be eliminated.

If time appears explicitly in a constraint, mechanical energy is generally not conserved.

True — the transformation

r = r (q, t)

gains a

\partial r / \partial t

term, the constraint surface can do work, and energy needn't be conserved (relevant to Kinetic Energy and the Hamiltonian).

In two variables, a Pfaffian velocity constraint

A d x + B d y = 0

can be genuinely non-holonomic.

False — a Pfaffian in two variables always admits an integrating factor

μ

(its direction field always has integral curves

f = c

), so it is always integrable. Genuine non-holonomy needs three or more coupled variables.

A particle resting on top of (not attached to) a smooth sphere is subject to a holonomic constraint while it stays on the surface.

Partly — while on the surface

∣ r ∣ = a

(holonomic equality), but the physical rule is the inequality

∣ r ∣ \geq a

, which the particle may leave. The true constraint is a unilateral (inequality) constraint, which is non-holonomic in the broad sense.

A scleronomic constraint means the system never moves.

False — scleronomic means the constraint surface is frozen (no explicit

t

), not that the particle is static. A bead slides freely along a fixed wire; the wire is frozen, the bead moves.

An inequality constraint and a non-integrable rolling constraint are the same kind of thing.

False — both are grouped as "non-holonomic," but the first is unilateral (allows a whole region, can switch on/off) and the second is a Pfaffian velocity restriction (limits directions on an always-active surface). Different mechanisms, same broad label.

Spot the error

y = x tan (ω t)

contains

t

, so it's non-holonomic." — find the mistake.

Confuses the two axes. Explicit

t

makes it rheonomic, not non-holonomic. It is still an equation in coordinates, hence holonomic. Correct label: holonomic + rheonomic.

"The rolling-disk relation

d x sin ϕ - d y cos ϕ = 0

can be integrated to

f (x, y, ϕ) = 0

, so it's holonomic." — find the mistake.

It fails the exactness test and has no integrating factor (three coupled variables), so it cannot be integrated. It is a genuinely non-holonomic Pfaffian constraint.

"Since the pendulum bob's speed changes, the length constraint depends on velocity and is non-holonomic." — find the mistake.

The constraint

x^{2} + y^{2} - ℓ^{2} = 0

contains no velocities at all. Speed changing is the motion, not the constraint. It stays holonomic.

"A moving-wall gas is holonomic because the wall's position is a known function of time." — find the mistake.

The constraint is an inequality (molecules stay inside the box) → a unilateral, hence non-holonomic, constraint. Knowing the wall's motion only makes it rheonomic; it doesn't turn an inequality into an equation. (This is a non-holonomic AND rheonomic case.)

A d x + B d y = 0

with

\partial A / \partial y \neq = \partial B / \partial x

is non-holonomic." — find the mistake.

Failing exactness as written is not enough — you must also check for an integrating factor

μ

, which in two variables always exists (find it via

μ (x) = exp \int \frac{1}{B} (\partial_{y} A - \partial_{x} B) d x

when that group is

x

-only). So it is still holonomic.

"Because the rolling disk cannot slip sideways, some of its coordinates are eliminated." — find the mistake.

Non-holonomic constraints kill velocity directions, not coordinates. All three coordinates

(x, y, ϕ)

remain independent and reachable; nothing is eliminated.

Why questions

Why does a non-integrable velocity constraint restrict directions but not reachable positions?

Because there is no function

f

it forces to be constant. With no conserved

f

, no region of configuration space is forbidden — you can still reach anywhere, just not by moving in every direction instantly (e.g. parallel-park a rolling coin).

Why must we run the exactness / integrating-factor test rather than eyeballing a velocity constraint?

Because a Pfaffian relation can be holonomic in disguise: if it equals

df

(perhaps after multiplying by

μ

), integrating gives a coordinate equation. Only the test reveals whether it truly refuses to integrate.

Why does genuine non-holonomy require at least three variables?

In two variables the constraint is a direction field in the plane, which always has integral curves

f = c

(an integrating factor always exists). You need

\geq 3

coupled variables for a differential form that no factor can make exact — exactly why a rolling object in a plane (three coordinates) qualifies.

Why does the rheonomic case threaten energy conservation but the scleronomic case doesn't?

A frozen surface (

f (r) = 0

) does no work; its constraint force is perpendicular to allowed motion. A moving surface (

f (r, t) = 0

) can push along the motion via

\partial r / \partial t

, injecting or removing energy.

Why do constraints let us avoid solving for unknown reaction forces?

Each holonomic equation replaces one unknown force with one geometric equation, and the constraint force does no virtual work — which is precisely what D'Alembert's Principle and Virtual Displacements exploit to eliminate them.

Why can non-holonomic constraints not simply reduce the coordinate count the way holonomic ones do?

Because they never yield a solvable equation

f = 0

to substitute one coordinate out. Instead they are carried alongside the equations of motion, typically via Lagrange Multipliers.

Why is the bead-on-spinning-wire example the standard illustration that the two axes are independent?

It is simultaneously holonomic (a clean equation

y - x tan ω t = 0

) and rheonomic (explicit

t

), proving a constraint can carry one label from each pair independently.

Why is a coin rolling on a floor that is being tilted over time the missing fourth box?

The no-slip rolling relation is a non-integrable Pfaffian (non-holonomic) AND the floor's orientation is an explicit function of

t

(rheonomic), so it fills the non-holonomic + rheonomic corner of the grid.

Edge cases

What is the constraint status of a particle sliding on a smooth sphere at the exact instant it leaves the surface?

The equality

∣ r ∣ = a

ceases to hold and switches to

∣ r ∣ > a

; the constraint effectively turns off. This on/off switching is the signature of a unilateral (inequality) constraint → non-holonomic in the broad sense.

Two constraints on a system are not independent (one implies the other). How many coordinates are removed?

Only one — the degrees-of-freedom formula

n = 3 N - k

counts independent constraints. A redundant constraint removes nothing extra.

A wire is rotated at angular speed

ω

, then the motor is switched off so

ω = 0

afterward. What changes about the constraint?

While driven it is rheonomic (

y - x tan ω t = 0

); once frozen the angle is constant and

t

vanishes, so it becomes scleronomic. The holonomic label is unchanged throughout.

Is the constraint

f (r) = 0

with

f

identically zero (vacuously true) a real constraint?

No — a constraint must genuinely restrict configurations. A vacuous equation removes no coordinate and imposes nothing;

k

counts only effective, independent restrictions.

Can a single physical setup have both a holonomic and a non-holonomic constraint at once?

Yes — e.g. a disk rolling on a fixed horizontal table has the holonomic contact condition (

z = R

, staying on the table) plus the non-holonomic no-slip rolling relations. Each is classified separately.

What happens to the scleronomic "energy = Hamiltonian" guarantee if you add even one rheonomic constraint?

It breaks — the explicit

t

adds velocity-linear and constant terms to the kinetic energy, so the Hamiltonian no longer equals total mechanical energy in general (see Kinetic Energy and the Hamiltonian).

A constraint is written with velocities but they all cancel after algebra, leaving

g (r, t) = 0

. Holonomic or not?

Holonomic — if the velocities disappear (or integrate away) leaving a pure coordinate-and-time equation, the constraint restricts configuration, which is the definition of holonomic.

Give a concrete member of each of the four boxes of the grid.

Holonomic+scleronomic: rigid rod. Holonomic+rheonomic: bead on spinning wire. Non-holonomic+scleronomic: rolling disk on a fixed floor. Non-holonomic+rheonomic: coin rolling on a floor tilted as a function of time.