Exercises — Generalized coordinates — choosing them, degrees of freedom

Jo hamare paas hai: ek particle, $N=1$, three-dimensional, koi constraint nahi isliye $k=0$. Formula apply karo: $n=3N-k=3(1)-0=3$. Coordinates: Cartesian triple $(x,y,z)$ kaam karta hai. Spherical $(r,\theta ,\varphi )$ bhi — koi bhi complete independent triple chalega. Answer: $n=3$.

Kya hai: $N=1$, raw count $3N=3$. Constraints: bead ko line par force karna 3D mein matlab usse do equations ek saath satisfy karni padti hain (line = do planes ka intersection / do coordinate relations). Toh $k=2$. Apply karo: $n=3-2=1$. Coordinate: wire ke saath measure ki gayi arc-length $s$. Answer: $n=1$. Yahi upar ki figure mein toy-train / bead picture hai.

Raw: $N=1$, $3N=3$. Constraint: surface ek equation hai, $x^2+y^2+z^2=R^2$, toh $k=1$. Apply karo: $n=3-1=2$. Natural coordinates: do spherical angles $(\theta ,\varphi )$ — polar aur azimuthal. Kyun yahi? $r$ ke $R$ par locked rehne se, koi bhi $(\theta ,\varphi )$ automatically sphere par land karta hai, toh constraint construction se hi gayab ho jaati hai — exactly pendulum wali trick, ek dimension upar. Answer: $n=2$.

Raw: $N=1$, $3N=3$. Constraint: rod ki length fixed, $x^2+y^2+z^2={\ell }^2$, toh $k=1$. Apply karo: $n=3-1=2$. Coordinates: $(\theta ,\varphi )$ jaisa 2.1 mein hai — yeh literally radius $\ell$ ke sphere par bead hai. Answer: $n=2$. Planar pendulum ($n=1$) se compare karo: "ek plane mein raho" ki restriction hatane se ek aur freedom mil jaati hai.

Way A — brute force. $N=3$, toh $3N=9$. Rods: $|{\mathbf{r}}_1-{\mathbf{r}}_2|=a$, $|{\mathbf{r}}_2-{\mathbf{r}}_3|=b$, $|{\mathbf{r}}_3-{\mathbf{r}}_1|=c$. Kya yeh independent hain? Haan — teen rods jo teen non-collinear points ke beech teen alag-alag distances fix karti hain woh independently shape pin karti hain. Toh $k=3$. $n=9-3=6.$ Way B — rigid-body shortcut. Teen rigidly connected points ek rigid body banate hain. Ek free rigid body ka $n=6$ hota hai (3 translation + 3 rotation). ✔ Answer: $n=6$, dono tareekon se agree. Yahi reason hai ki rigid body mein chahe kitne bhi particles hon hamesha 6 count hota hai — extra particles extra rigidity constraints laate hain jo unhe exactly cancel kar dete hain.

Raw: $N=4$, $3N=12$. Naive constraint count: ek rigid tetrahedron mein $\left(\genfrac{}{}{0ex}{}{4}{2}\right)=6$ edges hote hain, jisse tum $k=6$ likhne aur $n=12-6=6$ paane ke liye tempt ho sakte ho. Yahan number ittefaq se kaam karta hai — lekin general pattern se savdhaan raho. Independence check: ek rigid body ke liye tum kabhi $3N-6$ se zyada freedoms nahi hata sakte, kyunki 6 bachne hi chahiye. Toh independent distance constraints ki maximum sankhya $3N-6=12-6=6$ hai. 4 points ke liye yeh exactly 6 hai, toh saare edges independent hain. 5 points ke liye, $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ possible distances hain lekin sirf $3(5)-6=9$ independent ho sakte hain — 10th already doosron se forced hai. Answer: $n=6$. Lesson: independent rigid constraints ko hamesha $3N-6$ par cap karo.

Raw: $N=2$, $3N=6$. Constraints: wire ko $x$-axis ke along rakh do. Har bead ko $y_i=0$ aur $z_i=0$ satisfy karna hoga. Yeh $2$ constraints per bead $\times$ 2 beads $=k=4$ hai. (Saari independent: yeh chaar alag-alag numbers $y_1,z_1,y_2,z_2$ pin karti hain.) Apply karo: $n=6-4=2$. Smart coordinates: center of mass $X=\frac{x_1+x_2}{2}$ aur separation $s=x_2-x_1$. Kyun? Spring ki potential energy sirf $s$ par depend karti hai (uski stretch ke through), toh $s$ choose karna ek coordinate ko saari interaction carry karaata hai aur $X$ ko free chhod deta hai — equations decouple ho jaati hain. Answer: $n=2$.

Raw (planar): $N=1$, $2N=2$. Constraint: wire ka angle forced hai ki $\omega t$ ho, toh bead $y/x=\tan (\omega t)$ satisfy karta hai. Yeh ek rheonomic (time-dependent) lekin phir bhi holonomic constraint hai — yeh ek equation $f(x,y,t)=0$ hai. Toh $k=1$. Apply karo: $n=2-1=1$. Single coordinate $r$ hai, wire ke along distance. Position: $x=r\cos (\omega t),y=r\sin (\omega t).$ Note karo ki $\mathbf{r}$ explicitly $t$ par depend karta hai — yahi driven constraint ki pehchaan hai. Answer: $n=1$; time ek parameter hai, coordinate nahi, isliye yeh freedom add nahi karta.

(a) Configuration coordinates: $x$ (contact point kahan hai) aur $\phi$ (kitna ghuma hai). Toh configuration space 2-dimensional hai — tumhe genuinely dono numbers chahiye kisi state describe karne ke liye, jaise slip aur recovery ke baad. (b) Rolling constraint: radius $R$ ke saath, "no slip" matlab contact-point speed rim speed se match kare: $\dot{x}=R\dot{\phi }.$ Yeh velocities link karta hai. Lekin seedhi line par roll ke liye yeh integrate kiya ja sakta hai: $x=R\phi +C$, jo ek genuine position equation $f(x,\phi )=0$ hai. Toh yeh particular constraint IS holonomic (integrable) hai. Yeh ek DOF remove karta hai: $n=2-1=1$. (c) Instantaneous freedom: kyunki constraint integrate hoti hai, ek baar jab $\phi$ pata ho toh $x$ bhi pata hai — sach mein 1 degree of freedom. Answer: configuration 2D hai lekin integrable rolling constraint isse $n=1$ tak reduce kar deti hai.

(a) Rolling constraints (koi sideways slip nahi, aur rim speed = ground speed): $\dot{x}=R\dot{\phi }\cos \theta ,\dot{y}=R\dot{\phi }\sin \theta .$ (b) Holonomic? Nahi. Yeh $\dot{x},\dot{y},\dot{\phi }$ ko variable heading $\theta$ ke through couple karte hain, aur inhe equations $f(x,y,\theta ,\phi )=0$ mein integrate nahi kiya ja sakta. Yeh non-holonomic (velocity) constraints hain. Configuration space 4-dimensional rehta hai: tum coin ko maneuver kar sakte ho (aage roll karo, turn karo, peeche roll karo, wapas turn karo) aur usi jagah par alag spin $\phi$ ke saath pahunch sakte ho — toh saare chaar numbers genuinely independently reachable hain. (c) $n=3N-k$ kyun fail karta hai: formula assume karta hai ki har constraint ek configuration coordinate khatam karta hai. Non-holonomic constraints sirf velocities ko har instant par restrict karte hain (yahan 2: tum sideways slip nahi kar sakte, aage-peeche slip nahi kar sakte). Yeh motion ki instantaneous directions ko 4 se $4-2=2$ tak cut karte hain, bina reachable configuration space ko chhote kiye, jo 4D rehta hai. Toh "position DOF" $=4$ lekin "velocity DOF" $=2$ — do alag-alag counts, aur naive subtraction $4-2=2$ kisi bhi configuration space ko sahi describe nahi karta. Answer: config space 4-dimensional hai; 2 non-holonomic constraints hain; $n=3N-k$ apply nahi hota.

Free particle in 3D → DOF? ::: 3 Bead on a fixed straight wire in 3D → DOF? ::: 1 (curve 2 remove karti hai) Mass on a fixed sphere surface → DOF? ::: 2 (surface 1 remove karti hai) 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 ($t$ ek parameter hai) Coin rolling straight (integrable) → DOF? ::: 1 (holonomic) Coin steering on a floor → config space dimension? ::: 4 (constraints non-holonomic hain)