Worked examples — Generalized coordinates — choosing them, degrees of freedom

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The scenario matrix

Before counting anything, let us map the territory. Every degrees-of-freedom problem falls into one of these cells. Each cell is a different way the counting can behave — and each has one worked example below.

Cell	Case class	What is tricky	Example
A	Single particle, one constraint (2D)	the base case, sanity anchor	Ex 1
B	Many particles, several independent constraints	subtracting the right $k$	Ex 2
C	Rigid body (finite $N$ trick)	use $6$, not $3N$	Ex 3
D	Rheonomic (time-dependent) constraint	$t$ is a parameter, not a DOF	Ex 4
E	Degenerate / dependent constraints	writing more equations than count	Ex 5
F	Zero-DOF (fully pinned) limit	$n=0$ means fixed, not broken	Ex 6
G	Non-holonomic (rolling) — formula fails	velocity constraint ≠ position equation	Ex 7
H	Real-world word problem	translating prose into $N$ and $k$	Ex 8
I	Exam-style twist (coordinate pole)	a good coordinate breaks at a special point	Ex 9

The axes of this matrix are: how many particles (one → many → rigid body → ∞), what kind of constraint (none → holonomic scleronomic → rheonomic → non-holonomic), and degenerate limits (dependent constraints, zero DOF, coordinate singularities). Together the nine cells tile the whole space.

Recall The two words you must not confuse

Scleronomic constraint ::: time does not appear: $f(\mathbf{r})=0$. Rheonomic constraint ::: time appears explicitly: $f(\mathbf{r},t)=0$; still holonomic, still removes one DOF.

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Ex 1 — Cell A: single particle, one constraint

1. Count raw numbers. One particle in a plane means $D=2,N=1$, so raw $=DN=2$ Cartesian numbers $(x,y)$. Why this step? Always start from the most wasteful honest description — the parent's rule "start raw".
2. Write the constraint as one equation. The bead stays on the ring: $x^2+y^2=R^2$. This is one holonomic scleronomic constraint, so $k=1$. Why this step? An equation lets us solve for one variable in terms of the other, killing one number.
3. Subtract. $n=DN-k=2-1=1$. Why this step? Each independent holonomic constraint removes exactly one DOF.
4. Choose the smart coordinate. Use the angle $\phi$ with $x=R\cos \phi ,y=R\sin \phi$. Why this step? Any $\phi$ automatically satisfies $x^2+y^2=R^2$: the constraint is baked in and disappears.

Figure 1 — what to look at: the cyan circle is the wire (the constraint $x^2+y^2=R^2$); the amber dot is the bead; the amber arc measured from the positive $x$-axis is the single coordinate $\phi$; the white dashed radius shows that fixing $\phi$ fixes the bead completely. Watch that exactly one thing varies as the bead slides — the size of the amber angle.

Verify: Substitute back: $(R\cos \phi {)}^2+(R\sin \phi {)}^2=R^2({\cos }^2\phi +{\sin }^2\phi )=R^2.$ ✓ The constraint holds for every $\phi$, confirming $\phi$ alone is complete. One coordinate ⇒ $n=1$.

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Ex 2 — Cell B: many particles, several constraints

1. Raw count. $D=2,N=3$ ⇒ raw $=DN=6$. Why this step? Three independent points, two numbers each ($D=2$).
2. List the rigid rods as equations. Write ${\mathbf{r}}_1,{\mathbf{r}}_2,{\mathbf{r}}_3$ for the position vectors of the three masses (the bold-$\mathbf{r}$ symbol defined in the box above). Sides $12,23,31$ fixed: $|{\mathbf{r}}_1-{\mathbf{r}}_2|=a,|{\mathbf{r}}_2-{\mathbf{r}}_3|=b,|{\mathbf{r}}_3-{\mathbf{r}}_1|=c$, where each $|{\mathbf{r}}_i-{\mathbf{r}}_j|$ is the distance between two masses. That is $k=3$. Why this step? Each fixed distance is one holonomic constraint equation.
3. Check independence. Are these three truly independent? For a triangle, yes — fixing three sides fixes the shape but you cannot derive one length from the other two (except at degenerate collapse). So all three count. Why this step? The parent warns: only independent constraints subtract.
4. Subtract. $n=6-3=3$. Why this step? The rules-removed principle.
5. Interpret. Those 3 DOF = 2 for the triangle's centre position $(X,Y)$ + 1 for its orientation angle $\theta$ in the plane. This matches the "rigid body in 2D has 3 DOF" rule (2 translation + 1 rotation). Why this step? Cross-checking against the rigid-body rule confirms the count.

Verify: Independent check via the rigid-body shortcut: a rigid body in a 2D plane has $2+1=3$ DOF regardless of how many particles compose it. $6-3=3$ agrees. ✓

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Ex 3 — Cell C: the rigid-body trick

1. Do NOT use $DN$ literally. With $D=3$ and $N\to \infty$ the rigid constraints also become infinite, and $\infty -\infty$ is meaningless. Why this step? The naive formula chokes on rigid bodies — this is the trick the parent flags.
2. Use the rigid-body shortcut. Every finite rigid body needs exactly $6$ numbers: 3 to say where it is (centre of mass $X,Y,Z$) and 3 to say how it is oriented. Why this step? Rigidity pins every particle once you know position + orientation of a reference frame glued to the body.
3. Name the 3 orientation numbers. These are the Euler angles $(\alpha ,\beta ,\gamma )$ — see Rigid body kinematics — Euler angles. Why this step? You must be able to name the coordinates, not just count them.
4. State their ranges and watch for their singularity. The three Euler angles do not roam freely: typically $\alpha \in [0,2\pi )$ (first spin), $\beta \in [0,\pi ]$ (tilt), $\gamma \in [0,2\pi )$ (final spin). At the tilt extremes $\beta =0$ or $\beta =\pi$ the first and third axes line up, so $\alpha$ and $\gamma$ turn the body about the same axis — you cannot tell them apart. This is gimbal lock: two of the three angle-coordinates momentarily merge, exactly like the pole trap of Ex 9. Why this step? Orientation coordinates have edge cases just like position ones; ignoring them makes you think a DOF was lost.
5. The DOF is still $6$ despite gimbal lock. Gimbal lock is a defect of the chart $(\alpha ,\beta ,\gamma )$, not of the body: the cube genuinely still has 3 free rotations there. As with Ex 9 you switch to a different chart (or use quaternions) near the singular tilt. Why this step? Never confuse a broken coordinate label with a lost degree of freedom.
6. Total. $n=3+3=6$. Why this step? Translation DOF plus rotation DOF.

Verify: $n=6$ is independent of particle count — the defining property of a rigid body. Consistent with the parent's "$6$ for a rigid body" box; gimbal lock does not reduce it. ✓

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Ex 4 — Cell D: rheonomic (time-dependent) constraint

1. Raw count. One bead in a plane: $D=2,N=1$ ⇒ raw $=DN=2$. Why this step? Base description.
2. Write BOTH constraints.
   • Angle forced: $\theta (t)=\omega t$. Rather than the risky ratio form, write it as the clean equation $x\sin (\omega t)-y\cos (\omega t)=0$ (this is just "the point lies on the line at angle $\omega t$").
   • Radius forced: $\sqrt{x^2+y^2}=r_0-vt$. Each contains $t$ explicitly — both are rheonomic but still holonomic. So $k=2$. Why this step? Time-dependent equations still count as constraints; $t$ is a parameter we cannot vary.
3. Note why the pole-free angle form is used, not $y/x=\tan (\omega t)$. The ratio form is only valid where $x\ne 0$; it blows up (division by zero) whenever the wire points straight up, i.e. $\omega t=\frac{\pi }{2},\frac{3\pi }{2},\dots$, where $\tan (\omega t)$ is infinite. The form $x\sin (\omega t)-y\cos (\omega t)=0$ has no pole: it stays finite and correct for every angle including the vertical. Why this step? We must confirm the constraint we subtract is valid in every case, including the degenerate $x=0$ orientation — otherwise the count silently fails at those instants.
4. Subtract. $n=2-2=0$. Why this step? Two independent equations remove two numbers.
5. Interpret. $n=0$ means the bead's position is a fully determined function of time — no freedom at all: $x(t)=(r_0-vt)\cos \omega t,y(t)=(r_0-vt)\sin \omega t$. Why this step? Zero DOF does not mean "no motion" — it means motion is prescribed, not free (contrast Ex 6).

Verify: Plug back at any $t$: $\sqrt{x^2+y^2}=|r_0-vt|\sqrt{{\cos }^2\omega t+{\sin }^2\omega t}=|r_0-vt|$ ✓, and the pole-free angle equation $x\sin \omega t-y\cos \omega t=(r_0-vt)(\cos \omega t\sin \omega t-\sin \omega t\cos \omega t)=0$ ✓. Both hold with no free coordinate left, confirming $n=0$.

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Ex 5 — Cell E: degenerate / dependent constraints

1. List what they wrote. $k_{\text{claimed}}=4$: three sides + one area. Why this step? We must audit independence, not just count equations.
2. Test independence. The area of a triangle is fully determined by its three side lengths (Heron's formula: $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}$ with $s=\frac{a+b+c}{2}$). So the area equation is a consequence of the three side equations, not new information. Why this step? The parent's rule: a constraint that follows logically from others does not reduce DOF.
3. Recount. Independent constraints $k=3$. Therefore $n=6-3=3$, matching Ex 2. Why this step? Only independent constraints subtract.

Verify: With $a=3,b=4,c=5$ (a right triangle), Heron gives $s=6$, Area $=\sqrt{6\cdot 3\cdot 2\cdot 1}=\sqrt{36}=6$. The area is pinned once the sides are — so it carries no new restriction. Correct DOF $=3$, not $2$. ✓

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Ex 6 — Cell F: the zero-DOF limit

1. Raw count. Two moving particles in a plane: $D=2,N=2$ ⇒ raw $=DN=4$. (The pivots ${\mathbf{p}}_1,{\mathbf{p}}_2$ are fixed given points, not particles — they contribute no coordinates.) Why this step? Only the free masses carry coordinates; anchored points are constants.

2. List the three independent rod equations — exactly $k=3$, each a genuine written equation.

   • Rod pivot-1 → mass 1: $|{\mathbf{r}}_1-{\mathbf{p}}_1|={\ell }_1$ (1 equation)
   • Rod mass 1 → mass 2: $|{\mathbf{r}}_1-{\mathbf{r}}_2|={\ell }_2$ (1 equation)
   • Rod mass 2 → pivot-2: $|{\mathbf{r}}_2-{\mathbf{p}}_2|={\ell }_3$ (1 equation) So $k=3$, not 4 — there is no fourth equation, and I am not inventing one. Why this step? Honest accounting: every constraint subtracted must be an equation I can actually write down.

3. Subtract. $n=4-3=1$. Why this step? Three independent holonomic constraints remove three numbers from four.

4. Interpret — this is a one-DOF linkage, not zero. Equations (1) and (3) confine each mass to its own circle about its pivot; equation (2) links them. For generic pivot spacing this "four-bar-like" mechanism still has one freedom: turning the input rod at ${\mathbf{p}}_1$ drives the whole chain. So $n=1$. Why this step? We must report the number the honest count gives, even when intuition guessed "locked".

5. Now truly reach zero DOF (the pinned limit). Add a fourth rigid rod, from mass 1 directly to a third fixed pivot ${\mathbf{p}}_3$: $|{\mathbf{r}}_1-{\mathbf{p}}_3|={\ell }_4$ (equation 4). With mass 1 now tied to two fixed pivots (${\mathbf{p}}_1$ and ${\mathbf{p}}_3$), its position is pinned; equation (2) then pins mass 2. Now $k=4$ independent equations and $n=4-4=0.$ Why this step? $n=0$ is only legitimate when four honest independent equations are present. This delivers the genuine locked-truss limit the matrix promised.

6. Read what $n=0$ means. Zero DOF = the whole structure is rigidly fixed in place: there is no number left to vary. Unlike Ex 4 (where $n=0$ meant motion prescribed by time), here $n=0$ means genuinely no motion at all. Why this step? The same number $n=0$ has two flavours — prescribed-by-time versus truly static — and the reader must be able to tell them apart.

Verify: The three-rod chain gives $n=4-3=1\ge 0$ (a working linkage). The four-rod version gives $n=4-4=0$ (rigidly pinned). If any count returned $n<0$ you have over-counted constraints (revisit independence, Ex 5). Both results are $\ge 0$. ✓

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Ex 7 — Cell G: where the formula FAILS (non-holonomic)

1. Describe the configuration. To locate the coin you need: contact point $(x,y)$, heading angle $\varphi$ (which way it points), and spin angle $\psi$ (how far it has rotated). That is 4 configuration numbers. Why this step? Before we can ask whether a constraint reduces anything, we must first fix the complete list of numbers that place the object — configuration means every number needed to pose it, ignoring for now how it got there.
2. Write the rolling condition. Rolling-without-slipping ties velocities: $\dot{x}=R\dot{\psi }\cos \varphi ,\dot{y}=R\dot{\psi }\sin \varphi$. These involve $\dot{x},\dot{y},\dot{\psi }$ — rates, not positions. (The dot means "rate of change with time".) Why this step? This is the crux: the constraint is on velocities.
3. Try to integrate it into a position equation. You cannot: there is no function $f(x,y,\varphi ,\psi )=0$ whose derivative gives these. The constraint is non-holonomic (see Constraints — holonomic vs non-holonomic). Why this step? $n=DN-k$ subtracts position equations. A velocity constraint that will not integrate is not a position equation.
4. Conclude the split. The configuration space stays 4-dimensional (the coin can be brought to any $(x,y,\varphi ,\psi )$ by clever manoeuvres), yet only 2 velocity directions are free at each instant. So "DOF" splits: 4 accessible configurations, 2 instantaneous freedoms. Why this step? This is exactly the parent's warning that $n=3N-k$ "may not apply directly."

Figure 2 — what to look at: the cyan dashed curve is a path the contact point can trace on the table; the amber dot is the current contact $(x,y)$; the amber arrow is the heading $\varphi$ (the direction the coin faces); the small white circle above shows the spin angle $\psi$. Read the caption text on the figure — it states the key fact: 4 configuration numbers exist, but only 2 of their velocities are independently free, so the rolling law cannot be subtracted like a position equation.

Verify: Test integrability. If $\dot{x}=R\dot{\psi }\cos \varphi$ came from $f=0$, then $x-R\psi \cos \varphi$ would have to stay constant — but $\varphi$ itself changes as you steer, so $R\psi \cos \varphi$ is not a clean total differential. No integrating factor exists ⇒ genuinely non-holonomic ⇒ formula does not subtract it. Configuration DOF $=4$. ✓

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Ex 8 — Cell H: real-world word problem

1. Translate prose to particles + constraints. Treat the arm as its three link endpoints beyond the shoulder. Easier: recognise each joint is one free rotation angle. Why this step? Robot arms are cleanest counted joint-by-joint.
2. Count the free angles directly. Shoulder angle ${\theta }_1$, elbow angle ${\theta }_2$, wrist angle ${\theta }_3$ — three independent angles, each freely choosable. Why this step? Every free joint that changes the configuration is one generalized coordinate.
3. Cross-check with $n=DN-k$. Three moving endpoints in a plane: $D=2,N=3$ ⇒ raw $=DN=6$. Constraints: each link has fixed length (3 equations), so $k=3$: $n=6-3=3$. Why this step? Two methods agreeing is your safety net.
4. Answer. $n=3$: you need three motors, one per joint. Why this step? DOF = number of independent controls.

Verify: Both methods give $3$. A serial planar arm with $J$ revolute joints has $J$ DOF; here $J=3$ ⇒ $3$. ✓ Units: angles are dimensionless (radians) — legal generalized coordinates, confirming the parent's "coordinates need not be lengths."

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Ex 9 — Cell I: exam twist — a coordinate that breaks at a pole

1. Count DOF (coordinate-free). One particle in 3D: $D=3,N=1$ ⇒ raw $=DN=3$; one constraint $x^2+y^2+z^2={\ell }^2$ ⇒ $k=1$; so $n=3-1=2$. Why this step? DOF is a property of the system, computed before choosing any coordinates.
2. Choose coordinates. Polar angle $\theta$ (down from vertical) and azimuth $\varphi$ (around): $x=\ell \sin \theta \cos \varphi ,y=\ell \sin \theta \sin \varphi ,z=-\ell \cos \theta$. Why this step? These two angles bake in the length constraint automatically.
3. Find the singularity. At $\theta =0$ (mass hanging straight down) or $\theta =\pi$ (straight up), $\sin \theta =0$, so every value of $\varphi$ gives the same point. The azimuth $\varphi$ becomes meaningless — the coordinate chart degenerates. Why this step? See the figure: at the pole the amber $\varphi$-circle shrinks to a single point, so $\varphi$ stops labelling anything.
4. Resolve the trap. The DOF is still 2 everywhere. The system genuinely has two freedoms; only the labelling $(\theta ,\varphi )$ fails at the poles, not the physics. Near a pole you would switch to a different chart. Why this step? Exam punchline: coordinate singularities are artefacts of the choice, never of the DOF count.

Figure 3 — what to look at: the cyan circle is the sphere of radius $\ell$ the mass is confined to; the white dashed vertical line is the axis; the tilted amber ellipse is the circle of constant latitude that the azimuth $\varphi$ sweeps; the white dot is the mass at angles $(\theta ,\varphi )$; the amber star at the top marks the pole $\theta =0$. Watch the pole: that amber $\varphi$-circle collapses to a single point there, showing visually that $\varphi$ carries no information while the DOF count stays $2$.

Verify: Constraint check: $x^2+y^2+z^2={\ell }^2{\sin }^2\theta ({\cos }^2\varphi +{\sin }^2\varphi )+{\ell }^2{\cos }^2\theta ={\ell }^2({\sin }^2\theta +{\cos }^2\theta )={\ell }^2$ ✓ for all $\theta ,\varphi$ — including at $\theta =0$, where the answer is ${\ell }^2$ regardless of $\varphi$, confirming $\varphi$ is redundant there. DOF $=2$ throughout. ✓

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Recall

Recall Rapid-fire scenario recall

Bead on a fixed hoop (2D) — DOF? ::: $1$ (angle $\phi$). Rigid planar triangle of 3 masses — DOF? ::: $3$ ($2$ translation $+1$ rotation). Free rigid body in 3D — DOF and why not $3N$? ::: $6$; particle count is infinite so use position $3$ + orientation $3$. What is gimbal lock, and does it lower the DOF? ::: At Euler tilt $\beta =0$ or $\pi$ the angles $\alpha ,\gamma$ turn about the same axis and merge; it is a chart defect, DOF stays $6$. Bead forced to $r(t)=r_0-vt$ AND $\theta =\omega t$ — DOF? ::: $0$; both rheonomic constraints remove one each; motion fully prescribed by $t$. Adding "area = const" to the three side-length constraints of a triangle changes $k$ how? ::: Not at all — area follows from sides (Heron), so it is dependent and does not count. Three-rod chain pinned at two fixed points — DOF? ::: $1$ ($n=4-3$); a working linkage, not locked. A fourth independent rod is needed to reach $n=0$. Rolling upright disk — configuration DOF, and does $n=DN-k$ apply? ::: $4$ configuration numbers; formula does not subtract the rolling condition because it is non-holonomic (velocity constraint). Planar 3-joint robot arm — DOF? ::: $3$ (one angle per joint). Spherical pendulum — DOF, and what breaks at the pole? ::: $2$; the azimuth $\varphi$ becomes meaningless at $\theta =0,\pi$, but DOF stays $2$ — only the chart fails.

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Connections

• Parent: Generalized coordinates & DOF
• Constraints — holonomic vs non-holonomic
• Configuration space and phase space
• Lagrangian mechanics — the Lagrangian L = T - V
• Euler–Lagrange equations
• Rigid body kinematics — Euler angles
• Principle of virtual work and d'Alembert's principle