Exercises — Generalized coordinates — choosing them, degrees of freedom

WHAT we have: one particle, $N=1$, three-dimensional, no constraint so $k=0$. Apply the formula: $n=3N-k=3(1)-0=3$. Coordinates: the Cartesian triple $(x,y,z)$ works. So does spherical $(r,\theta ,\varphi )$ — any complete independent triple. Answer: $n=3$.

WHAT: $N=1$, raw count $3N=3$. The constraints: forcing the bead onto a line in 3D means it must satisfy two equations at once (a line = intersection of two planes / two coordinate relations). So $k=2$. Apply: $n=3-2=1$. Coordinate: the arc-length $s$ measured along the wire. Answer: $n=1$. This is exactly the toy-train / bead picture in the figure above.

Raw: $N=1$, $3N=3$. Constraint: the surface is one equation, $x^2+y^2+z^2=R^2$, so $k=1$. Apply: $n=3-1=2$. Natural coordinates: the two spherical angles $(\theta ,\varphi )$ — polar and azimuthal. WHY these? With $r$ locked at $R$, any $(\theta ,\varphi )$ automatically lands on the sphere, so the constraint vanishes by construction — exactly the pendulum trick, one dimension up. Answer: $n=2$.

Raw: $N=1$, $3N=3$. Constraint: rod length fixed, $x^2+y^2+z^2={\ell }^2$, so $k=1$. Apply: $n=3-1=2$. Coordinates: $(\theta ,\varphi )$ as in 2.1 — this is literally a bead on a sphere of radius $\ell$. Answer: $n=2$. Contrast with the planar pendulum ($n=1$): removing the "stay in a plane" restriction hands you one more freedom.

Way A — brute force. $N=3$, so $3N=9$. Rods: $|{\mathbf{r}}_1-{\mathbf{r}}_2|=a$, $|{\mathbf{r}}_2-{\mathbf{r}}_3|=b$, $|{\mathbf{r}}_3-{\mathbf{r}}_1|=c$. Are these independent? Yes — three rods fixing three distinct distances among three non-collinear points independently pin the shape. So $k=3$. $n=9-3=6.$ Way B — rigid-body shortcut. Three rigidly connected points form a rigid body. A free rigid body has $n=6$ (3 translation + 3 rotation). ✔ Answer: $n=6$, both ways agree. This is why a rigid body counts as 6 no matter how many particles — extra particles bring extra rigidity constraints that exactly cancel them.

Raw: $N=4$, $3N=12$. Naive constraint count: a rigid tetrahedron has $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ edges, tempting you to write $k=6$ and get $n=12-6=6$. Here the number happens to work out — but watch out for the general pattern. The independence check: for a rigid body you can never remove more than $3N-6$ freedoms, because 6 must survive. So the maximum number of independent distance constraints is $3N-6=12-6=6$. For 4 points that's exactly 6, so all edges are independent. For 5 points, $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ possible distances but only $3(5)-6=9$ can be independent — the 10th is forced by the others. Answer: $n=6$. Lesson: always cap independent rigid constraints at $3N-6$.

Raw: $N=2$, $3N=6$. Constraints: put the wire along the $x$-axis. Each bead must satisfy $y_i=0$ and $z_i=0$. That's $2$ constraints per bead $\times$ 2 beads $=k=4$. (All independent: they pin four distinct numbers $y_1,z_1,y_2,z_2$.) Apply: $n=6-4=2$. Smart coordinates: center of mass $X=\frac{x_1+x_2}{2}$ and separation $s=x_2-x_1$. WHY? The spring's potential energy depends only on $s$ (through its stretch), so choosing $s$ makes one coordinate carry all the interaction and leaves $X$ free — the equations decouple. Answer: $n=2$.

Raw (planar): $N=1$, $2N=2$. Constraint: the wire's angle is forced to be $\omega t$, so the bead obeys $y/x=\tan (\omega t)$. This is a rheonomic (time-dependent) but still holonomic constraint — it is an equation $f(x,y,t)=0$. So $k=1$. Apply: $n=2-1=1$. The single coordinate is $r$, the distance along the wire. Position: $x=r\cos (\omega t),y=r\sin (\omega t).$ Note $\mathbf{r}$ depends explicitly on $t$ — that's the fingerprint of a driven constraint. Answer: $n=1$; time is a parameter, not a coordinate, so it does not add freedom.

(a) Configuration coordinates: $x$ (where the contact point is) and $\phi$ (how much it has spun). So the configuration space is 2-dimensional — you genuinely need both numbers to describe a state, e.g. after a slip and recovery. (b) Rolling constraint: with radius $R$, "no slip" means the contact-point speed matches the rim speed: $\dot{x}=R\dot{\phi }.$ This links velocities. But for a straight-line roll it can be integrated: $x=R\phi +C$, which is a genuine position equation $f(x,\phi )=0$. So this particular constraint IS holonomic (integrable). It removes one DOF: $n=2-1=1$. (c) Instantaneous freedom: because the constraint integrates, once you know $\phi$ you know $x$ — truly 1 degree of freedom. Answer: configuration is 2D but the integrable rolling constraint reduces it to $n=1$.

(a) Rolling constraints (no sideways slip, and rim speed = ground speed): $\dot{x}=R\dot{\phi }\cos \theta ,\dot{y}=R\dot{\phi }\sin \theta .$ (b) Holonomic? No. These couple $\dot{x},\dot{y},\dot{\phi }$ through the variable heading $\theta$, and they cannot be integrated into equations $f(x,y,\theta ,\phi )=0$. They are non-holonomic (velocity) constraints. The configuration space stays 4-dimensional: you can maneuver the coin (roll forward, turn, roll back, turn back) and end up at the same spot with a different spin $\phi$ — so all four numbers are genuinely reachable independently. (c) Why $n=3N-k$ fails: the formula assumes each constraint kills one configuration coordinate. Non-holonomic constraints restrict only velocities at each instant (here, 2 of them: you can't slip sideways, can't slip forward-back). They cut the instantaneous directions of motion from 4 to $4-2=2$, without shrinking the reachable configuration space, which stays 4D. So "position DOF" $=4$ but "velocity DOF" $=2$ — two different counts, and the naive subtraction $4-2=2$ describes neither the config space correctly. Answer: config space is 4-dimensional; 2 non-holonomic constraints; $n=3N-k$ does not apply.

Free particle in 3D → DOF? ::: 3 Bead on a fixed straight wire in 3D → DOF? ::: 1 (a curve removes 2) Mass on a fixed sphere surface → DOF? ::: 2 (a surface removes 1) Spherical pendulum → DOF? ::: 2 Rigid triangle (3 rods) free in 3D → DOF? ::: 6 Two beads + spring on one 3D wire → DOF? ::: 2 Rheonomic (rotating) wire bead → DOF? ::: 1 ($t$ is a parameter) Coin rolling straight (integrable) → DOF? ::: 1 (holonomic) Coin steering on a floor → config space dimension? ::: 4 (constraints are non-holonomic)