Constraints — holonomic vs non-holonomic, rheonomic vs scleronomic: Worked examples

Before anything: two words we will use on every line, plus one letter.

Definition The two yes/no questions (recap so nothing is assumed)

Every constraint is classified along two independent questions — we will call them Axis 1 and Axis 2 and use those names on every worked example.

Axis 1 — "where or which way?"
- Holonomic ::: the rule can be squeezed into an equation among positions and time only, written $f (r, t) = 0$ . Here $f$ is just a function — a machine that eats the coordinates $r$ and the time $t$ and spits out a single number; the rule is that this number must be $0$ . It says where you may be.
- Non-holonomic ::: it cannot be squeezed into such an $f (r, t) = 0$ — it is an inequality, or a velocity relation you cannot integrate. It says which directions you may move, not where you may end up.
Axis 2 — "does time show up?"
- Scleronomic ::: the letter $t$ (time) does not appear on its own. The allowed region is frozen.
- Rheonomic ::: $t$ appears explicitly. The allowed region moves as the clock ticks.
The letter == $N$ == on this page always means the number of particles in the system. A single particle in ordinary space needs 3 numbers to locate, so $N$ particles need $3 N$ Cartesian coordinates before any constraint is applied.

Every constraint gets one label from Axis 1 and one label from Axis 2 — the two axes are independent questions.

The scenario matrix

Each row is a kind of constraint you can be handed. The last column names the example that nails it. Both the "Holo?" (Axis 1) and the "Time?" (Axis 2) columns are always filled with a definite label. Two entries need a word of warning:

"test decides" (row D): a raw velocity/differential relation has no label yet — you must run the integrability test (Ex 4, Ex 5) before you can write Holonomic or Non-holonomic.
"depends on the input" (row G) and "mixed" (row I): these are not new labels — they mean this row bundles several sub-cases, and the specific example spells out a definite Holonomic/Non-holonomic and Scleronomic/Rheonomic label for each sub-case. No constraint ever stays "depends" once you look at it closely.

Cell	Fingerprint you see on paper	Holo? (Axis 1)	Time? (Axis 2)	Worked in
A	Fixed distance / rigid link, $=$ sign, no $t$	Holonomic	Scleronomic	Ex 1
B	Equation containing $t$ (moving wall, spun wire)	Holonomic	Rheonomic	Ex 2
C	An inequality ( $\geq, \leq$ )	Non-holonomic	Scleronomic	Ex 3
D	Velocity/differential relation — must run the integrability test	test decides	Scleronomic	Ex 4, Ex 5
E	Velocity relation that passes the test (holonomic in disguise)	Holonomic	Scleronomic	Ex 5
F	Velocity relation that fails the test (genuine rolling)	Non-holonomic	Scleronomic	Ex 4
G	Degenerate / limiting input (zero length, $ω \to 0$ , sphere)	depends on the input	depends on the input	Ex 6
H	Real-world word problem (gas in a moving piston)	Non-holonomic	Rheonomic	Ex 7
I	Exam twist — count degrees of freedom correctly	mixed (see Ex 8)	mixed (see Ex 8)	Ex 8

We now hit each cell. Watch the fingerprint every time.

Example 1 — Cell A · rigid dumbbell (holonomic, scleronomic)

Forecast: guess the label before reading on. Fixed length, no clock — what pair of words?

Write what "rigid" means as an equation. The distance between the masses can never change: $∣ r_{1} - r_{2} ∣^{2} - ℓ^{2} = 0.$ Why this step? "Rigid rod" is a word; mechanics needs a number relation. Squaring the distance avoids a square root (cleaner, same information), and setting it to a constant is exactly the form $f (r) = 0$ .
Test Axis 1 (holonomic?). It is already a clean equation in coordinates only. ✔ Holonomic. Why this step? Axis 1 asks "can I write it as $f (r, t) = 0$ ?" — and we already have exactly that form.
Test Axis 2 (time?). No letter $t$ appears. ✔ Scleronomic. Why this step? Axis 2 asks "does $t$ show up on its own?" — here it does not, so the allowed region is frozen.
Count the lost freedom. One independent equality constraint removes exactly one coordinate. Here $N = 2$ particles, so before constraints there are $3 N = 6$ coordinates; the dumbbell has $6 - 1 = 5$ . Why this step? This matches the parent's rule $n = 3 N - k$ with $k = 1$ .

Example 2 — Cell B · bead on a spinning wire (holonomic, rheonomic)

Forecast: the wire is a moving line. Does time hide inside, or can we scrub it out?

Figure — Constraints — holonomic vs non-holonomic, rheonomic vs scleronomic

The figure shows two wires through the origin: a white dashed line along the $x$ -axis (the wire at $t = 0$ ) and a pink line tilted at $4 5^{\circ}$ (the wire once $ω t = 4 5^{\circ}$ ). The small yellow arc at the origin marks the turned angle $ω t$ , and the blue dot on the pink line is the bead at position $(x, y)$ .

Describe the wire's direction. After time $t$ the wire has turned through angle $ω t$ (angle = rate $\times$ time) — this is the yellow arc in the figure. A line through the origin at angle $ω t$ has slope $tan (ω t)$ — "rise over run" of that line. Why this step? We need the tilt of the wire, and slope is exactly the number that encodes tilt: $tan (angle) = \frac{vertical rise}{horizontal run}$ .
Force the bead onto that line. The blue dot must sit on the wire, so any point $(x, y)$ on it obeys $\frac{y}{x} = tan (ω t)$ , i.e. $y - x tan (ω t) = 0.$ Why this step? This is an equation in coordinates and $t$ — the holonomic form. So the moving-line word-picture is genuinely holonomic.
Classify. It fits $f (x, y, t) = 0$ → holonomic (Axis 1). But $t$ sits inside $tan (ω t)$ and cannot be removed (the wire really does move) → rheonomic (Axis 2). Why this step? Axis 2 is decided purely by "is $t$ ineliminably present?" — here yes.
Evaluate the special instant $t = \frac{π}{4 ω}$ . Then $ω t = π /4$ and $tan (π /4) = 1$ , so the constraint becomes $y - x = 0$ , the $4 5^{\circ}$ line — this is exactly the pink line drawn in the figure. Why this step? Plugging a value turns the abstract rule into a concrete picture, confirming the wire is where we expect at that moment.

Example 3 — Cell C · particle outside a dome (non-holonomic, scleronomic)

Forecast: while touching, is it " $=$ " or " $\geq$ "? Which one refuses to remove a coordinate?

The figure draws the dome as a blue semicircle $∣ r ∣ = a$ ; a yellow dot sits on it (the particle while it is in contact), with a white dotted radius from the centre to that dot; and a pink dashed curve peeling away from the dome shows the free-flight path once the particle has left, out in the region $∣ r ∣ > a$ .

On the dome. Contact means the distance from the centre equals the radius: $∣ r ∣ = a ⟹ ∣ r ∣ - 1 = 0.$ Why this step? Touching = "distance equals radius" — this is the yellow dot sitting exactly on the blue semicircle, the equality picture.
After leaving. Once it flies off, it is outside: $∣ r ∣ > a$ — the pink dashed curve leaving the circle. The single rule covering both phases is the inequality $∣ r ∣ \geq a .$ Why this step? A particle can only be pushed out by the dome, never pulled in — a one-sided (normal) reaction. One-sided pushes always give inequalities.
Classify. An inequality cannot be turned into a coordinate-fixing equation for all time (it switches off when the particle leaves). → Non-holonomic (Axis 1). No $t$ appears → scleronomic (Axis 2). Why this step? Axis 1 (holonomic) demands a genuine equality that holds always; an inequality fails that test.

Example 4 — Cell D & F · rolling disk (non-holonomic, scleronomic)

Forecast: three or more coupled variables in a differential — does the integrability test pass or fail?

The figure shows a wavy white dashed curve — the path traced by the contact point $(x, y)$ — with the disk drawn as a blue thin ellipse (a circle seen almost edge-on) sitting on that path. A pink arrow points along the direction the disk faces (its heading $ϕ$ ), and the yellow caption carries the rolling relation we are about to test.

Eliminate the roll $d θ$ . Divide the two rolling equations (or cross-multiply): $d x sin ϕ = d y cos ϕ$ , i.e. $d x sin ϕ - d y cos ϕ = 0.$ Why this step? $θ$ is the disk's own spin; the shape of the path in the $(x, y, ϕ)$ world is what decides holonomy, so we hide $θ$ .
Name it as a Pfaffian $A d x + B d y + C d ϕ = 0$ with $A = sin ϕ$ , $B = - cos ϕ$ , $C = 0$ . Why this step? The exactness test (parent, "Integrability test") is stated for coefficients $A, B, C$ ; naming them lets us apply it mechanically.
Run the exactness test. For integrability of a genuine 3-variable Pfaffian we need the vector $a = (A, B, C)$ to satisfy the Frobenius condition $a \cdot (\nabla \times a) = 0, a = (sin ϕ, - cos ϕ, 0),$ where the variables are $(x, y, ϕ)$ . Compute the curl. Its three components are $(\nabla \times a)_{x} = \partial_{y} C - \partial_{ϕ} B = 0 - sin ϕ = - sin ϕ,$ $(\nabla \times a)_{y} = \partial_{ϕ} A - \partial_{x} C = cos ϕ - 0 = cos ϕ,$ $(\nabla \times a)_{ϕ} = \partial_{x} B - \partial_{y} A = 0 - 0 = 0,$ so $\nabla \times a = (- sin ϕ, cos ϕ, 0)$ . Then $a \cdot (\nabla \times a) = sin ϕ (- sin ϕ) + (- cos ϕ) cos ϕ + 0 = - sin^{2} ϕ - cos^{2} ϕ = - 1.$ Why this step? In three variables a single exactness pair is not enough; the Frobenius scalar $a \cdot (\nabla \times a)$ is the honest integrability detector. If it is not identically zero, no integrating factor exists.
Read the verdict. The scalar equals $- 1$ everywhere — a non-zero constant, never zero for any $ϕ$ . So the relation cannot be integrated to any $f (x, y, ϕ) = 0$ . → Non-holonomic (Axis 1). No $t$ → scleronomic (Axis 2). Why this step? A non-zero Frobenius scalar is precisely the parent's "no integrating factor exists (needs ≥3 variables)" statement, made numeric.

Example 5 — Cell D & E · a velocity relation that IS holonomic in disguise

Forecast: it looks like a velocity constraint — but does it integrate? Guess before step 3.

Name the coefficients. $A d x + B d y = 0$ with $A = y$ , $B = x$ . Why this step? Same machinery as Ex 4 — get $A, B$ before testing.
Two variables, so use the plane exactness test $\frac{\partial A}{\partial y} = \frac{\partial B}{\partial x}$ . $\frac{\partial A}{\partial y} = \frac{\partial ( y )}{\partial y} = 1, \frac{\partial B}{\partial x} = \frac{\partial ( x )}{\partial x} = 1.$ They are equal → exact → integrable. Why this step? In two variables an integrating factor always exists, so any 2-variable Pfaffian is holonomic — this example shows the parent's "genuine non-holonomy needs ≥3 variables" from the other side.
Integrate to the coordinate equation. A function whose differential is $y d x + x d y$ is $f = x y$ (since $d (x y) = y d x + x d y$ ). So the constraint is $x y = const .$ → Holonomic (Axis 1). It fixes the particle to a hyperbola. No $t$ appears → scleronomic (Axis 2). Why this step? Recognising $y d x + x d y$ as the product rule $d (x y)$ is the integration — no integrating factor even needed.

Example 6 — Cell G · degenerate & limiting inputs

Forecast: which of these secretly loses its "moving" or "non-holonomic" character in the limit?

(a) $ℓ \to 0$ . The constraint $∣ r_{1} - r_{2} ∣^{2} - ℓ^{2} = 0$ becomes $∣ r_{1} - r_{2} ∣^{2} = 0$ , i.e. $r_{1} = r_{2}$ . Still an equation in coordinates, still no $t$ → holonomic (Axis 1), scleronomic (Axis 2). But now it fuses the two particles into one point: they share all 3 coordinates, removing 3 constraints, not 1. Why this step? A degenerate length collapses two masses into one — a useful check that the count $n = 3 N - k$ still works ( $6 - 3 = 3$ freedoms, the freedoms of a single particle), and that the labels do not change under the limit.
(b) $ω \to 0$ . The wire stops turning. $tan (ω t) \to tan (0) = 0$ , so $y - x tan (ω t) = 0$ becomes $y = 0$ , the fixed $x$ -axis. Time has vanished → the constraint is now scleronomic (Axis 2), and still holonomic (Axis 1). Why this step? Axis 2 can flip in a limit: kill the driving rate $ω$ and the moving surface freezes, turning a rheonomic constraint scleronomic — showing "depends on the input" was a genuine warning.
(c) $∣ r ∣ = a$ (equality this time). Unlike Ex 3's inequality, this is a genuine equality holding for all time → holonomic (Axis 1), and no $t$ appears → scleronomic (Axis 2). A particle glued to a sphere therefore has $3 - 1 = 2$ degrees of freedom (two angles $θ, φ$ ). Why this step? This isolates the single symbol — equality vs inequality — that decides Axis 1, holding all the geometry fixed: swap ≥ for = and the very same dome flips from non-holonomic (Ex 3) to holonomic.

Example 7 — Cell H · real-world word problem (gas in a moving piston)

Forecast: two features fight here — a wall (inequality) and a moving wall (time). Predict both labels.

The molecule stays between the walls. $0 \leq x \leq L_{0} + v t$ . Why this step? Walls only push inward — one-sided pushes → inequalities, as in Ex 3.
Classify the inequality. An inequality cannot fix a coordinate → non-holonomic (Axis 1). Why this step? Same reasoning as the dome: ≤ is not an equality.
Does time appear? The right wall is $x = L_{0} + v t$ — the boundary moves with $t$ → rheonomic (Axis 2). Why this step? The moving piston literally does work on the gas (this is how a gas is compressed/expanded), so energy of the enclosed molecules is not conserved — the rheonomic signature.
Verdict: each molecule's confinement is non-holonomic (Axis 1) AND rheonomic (Axis 2). Why this step? This is the one combined-label cell; both axes are non-trivial simultaneously.

Example 8 — Cell I · exam twist (count the degrees of freedom)

Forecast: how many does each type of constraint subtract, and from what?

Subtract holonomic constraints from the coordinate count. With $N = 3$ we start from $3 N = 9$ . Two rigid rods → $k = 2$ holonomic equalities → coordinates drop by 2: $n = 3 N - k = 9 - 2 = 7.$ Why this step? Only holonomic (Axis 1) equalities remove generalized coordinates (parent rule $n = 3 N - k$ ).
Do NOT subtract the rolling constraint from $n$ . Non-holonomic constraints remove accessible velocity directions, not coordinates. The number of generalized coordinates stays 7. Why this step? This is the parent's third common mistake — a non-holonomic constraint does not reduce the configuration count.
Account for the velocity restriction separately. The single rolling relation removes 1 instantaneous velocity direction, so the system has 7 coordinates but only $7 - 1 = 6$ independent velocity directions at each instant. Why this step? Rolling ties $\overset{x}{˙}, \overset{y}{˙}, \dot{ϕ}$ together at every moment (Ex 4) without shrinking the reachable set — handled with a Lagrange multiplier in the Lagrangian formalism rather than by eliminating a coordinate. This is why the "mixed" row I mixes an Axis-1-holonomic count with an Axis-1-non-holonomic effect.

Recall One-line fingerprints (test yourself)

Equality, no $t$ ::: holonomic + scleronomic (Ex 1) Equality with $t$ inside ::: holonomic + rheonomic (Ex 2) Inequality ::: non-holonomic (Ex 3, 7) Differential relation ::: run the integrability test — could be either (Ex 4, 5) Frobenius scalar $a \cdot (\nabla \times a) \neq = 0$ ::: genuinely non-holonomic (Ex 4) Non-holonomic constraints subtract from ::: velocity directions, not coordinate count (Ex 8)