Constraints — holonomic vs non-holonomic, rheonomic vs scleronomic: Exercises

2.1.1 · D4 · Physics › Analytical Mechanics › Constraints — holonomic vs non-holonomic, rheonomic vs scler

Shuru karne se pehle, do pictures hain jo baar baar kaam aayengi. Pehli hai constraint surface: allowed positions ka set. Ek holonomic constraint $f (r, t) = 0$ literally ek surface (ya line) hai space mein jis par tum rehna padhta hai.

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

Doosri hai integrability question picture form mein: ek rule diya hai ki tum kaun si directions mein move kar sakte ho (ek rule $d x, d y, \dots$ par), kya woh rule secretly tumhe kisi surface par pin kar deta hai, ya har point reachable rehta hai?

Level 1 — Recognition

Goal: ek constraint equation padhkar sahi do labels lagana. Abhi koi calculation nahi.

Recall Solution L1·1

Yeh kis tarah ka statement hai? Yeh sirf coordinates $x, y$ mein ek equality hai. Yeh exactly $f (r) = 0$ ki form hai, isliye yeh holonomic hai. Kya time aata hai? Koi $t$ nahi — allowed points ka circle kabhi hilta nahi. Isliye scleronomic. Degrees of freedom. Plane mein $2$ Cartesian coordinates se shuru karo. Ek holonomic equation ek coordinate hata deta hai: $n = 2 - 1 = 1$ . Bacchi hui single coordinate swing angle hai. Dekho Generalized Coordinates. Answer: holonomic, scleronomic, $n = 1$ .

Recall Solution L1·2

Holonomic? Yeh coordinate aur time mein ek equation hai — form $f (r, t) = 0$ . Isliye holonomic. Time? $t$ explicitly aata hai aur allowed line $y = v t$ time ke saath upar khisak jaati hai. Isliye rheonomic. Moving wire bead ko vertically push kar sakta hai aur uske upar kaam kar sakta hai — energy ke baare mein pehli warning, jo baad mein Kinetic Energy and the Hamiltonian mein aayegi. Answer: holonomic, rheonomic.

Recall Solution L1·3

Holonomic? Yeh ek inequality hai, equation nahi. Ek inequality se ek coordinate eliminate nahi ho sakta — andar baitha molecule teeno directions mein free hai. Isliye yeh non-holonomic hai. Time? Shell radius $a$ fixed hai; koi $t$ nahi. Isliye scleronomic. Answer: non-holonomic, scleronomic.

Level 2 — Application

Goal: actually exactness test chalana taaki decide kar sako ki koi velocity relation holonomic hai ya disguise mein.

Recall Solution L2·1

$A, B$ identify karo. Yahan $A = y$ ( $d x$ multiply karta hai) aur $B = x$ ( $d y$ multiply karta hai). Test. $\frac{\partial A}{\partial y} = \frac{\partial y}{\partial y} = 1$ aur $\frac{\partial B}{\partial x} = \frac{\partial x}{\partial x} = 1$ . Dono equal hain → exact. Integrate. $f$ dhundho jahan $\partial f / \partial x = y$ aur $\partial f / \partial y = x$ . Dono $f = x y$ se satisfy hote hain. Toh constraint hai $d (x y) = 0 \Rightarrow x y = c$ . Verdict: holonomic (yeh surface $x y = c$ chhupa raha hai), scleronomic (koi $t$ nahi). Answer: holonomic, $x y = const$ .

Recall Solution L2·2

Identify karo. $A = y$ , $B = - x$ . Test. $\partial A / \partial y = 1$ , $\partial B / \partial x = - 1$ . Equal nahi → exact nahi. Lekin sirf do variables hain. 2 variables mein integrating factor hamesha exist karta hai. Try karo $μ = 1/ x^{2}$ : $\frac{y}{x ^{2}} d x - \frac{1}{x} d y = 0 = - d (\frac{y}{x}) .$ Check: $d (y / x) = \frac{1}{x} d y - \frac{y}{x ^{2}} d x$ , aur uska negative exactly humara expression hai. Toh yeh $y / x = c$ , yaani $y = c x$ — origin se jaati straight lines mein integrate hota hai. Verdict: holonomic (integrating factor ke baad), scleronomic. Answer: holonomic, $y / x = const$ .

Recall Solution L2·3

Kyun 2-variable trick fail hoti hai. Ab teen variables coupled hain, isliye "factor hamesha exist karta hai" ki guarantee gayi; sahi test karna padega. One-form $ω = d z - y d x$ likho. Yeh surfaces ke family mein integrable hai iff $ω \land d ω = 0$ (Frobenius condition). Compute karo. $d ω = - d y \land d x = d x \land d y$ . Phir $ω \land d ω = (d z - y d x) \land (d x \land d y) = d z \land d x \land d y \neq = 0,$ kyunki $d x \land d x = 0$ doosre piece ko khatam kar deta hai. Non-zero → not integrable. Verdict: non-holonomic, scleronomic. Yeh rolling disk ka algebraic skeleton hai. Answer: non-holonomic (koi surface exist nahi karta).

Level 3 — Analysis

Goal: degrees of freedom, energy, aur kyun koi fix kaam karta hai — is par reason karna.

Recall Solution L3·1

Constraints list karo.

Bead 1 circle par: $x_{1}^{2} + y_{1}^{2} - R^{2} = 0$ .
Bead 2 circle par: $x_{2}^{2} + y_{2}^{2} - R^{2} = 0$ .
Fixed rod: $(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} - ℓ^{2} = 0$ . Teeno sirf coordinates mein equalities hain → holonomic, scleronomic. Count karo. Start $3 N = 4$ (do particles, plane). Har independent holonomic constraint se ek coordinate hatao: $n = 4 - 3 = 1$ . Sanity check. Ek baar bead 1 ring par kisi jagah baitha ho (angle $ϕ_{1}$ ), fixed chord bead 2 ko do spots mein se ek par force karta hai — ek continuous coordinate nahi, discrete choice. Toh exactly ek continuous freedom bachti hai. $n = 1$ . ✓ Answer: $3$ holonomic constraints, $n = 1$ .

Recall Solution L3·2

Generalized coordinate chuno. $q$ = wire par bead ki signed distance. Phir $x = q cos (ω t), y = q sin (ω t) .$ Kyunki $t$ map ke andar hai, yeh ek rheonomic transformation hai. Differentiate karo. Product rule use karke, $\overset{x}{˙} = \overset{q}{˙} cos ω t - q ω sin ω t, \overset{y}{˙} = \overset{q}{˙} sin ω t + q ω cos ω t .$ Notice karo har velocity mein ek $\overset{q}{˙}$ piece hai aur ek $q ω$ piece — doosra piece woh $\partial r / \partial t$ term hai jo wire inject karta hai. Kinetic energy. Mass $m$ ke liye, $T = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2})$ . Expand karo ( $sin cos$ cross terms cancel): $T = \frac{1}{2} m \overset{q}{˙}^{2} + \frac{1}{2} m q^{2} ω^{2} .$ Pehla term $\overset{q}{˙}$ mein usual quadratic hai; doosre term mein bilkul bhi $\overset{q}{˙}$ nahi — yeh driven rotation se aaya velocity-independent piece hai. Yahi extra term rheonomic constraint ka fingerprint hai (dekho Kinetic Energy and the Hamiltonian). Energy meaning. Kyunki wire fixed $ω$ par externally driven hai, woh bead par ek sideways force lagaata hai jo kaam karta hai; bead ki akeli mechanical energy conserved nahi rehti. $q^{2} ω^{2}$ term exactly woh energy hai jo wire andar ya bahar pump kar sakta hai. Answer: $T = \frac{1}{2} m \overset{q}{˙}^{2} + \frac{1}{2} m q^{2} ω^{2}$ ; extra non-quadratic term possible non-conservation signal karta hai.

Level 4 — Synthesis

Goal: ek hi problem mein integrability test, geometry, aur classification combine karo.

Recall Solution L4·1

(a) $d θ$ eliminate karo. Do equations se, $\frac{d x}{cos ϕ} = R d θ = \frac{d y}{sin ϕ}$ , isliye $d x sin ϕ - d y cos ϕ = 0.$ Yeh kehta hai coin sirf apni heading ke saath move kar sakta hai, kabhi sideways nahi. (b) Integrability. Teen variables $x, y, ϕ$ mein $ω = sin ϕ d x - cos ϕ d y$ likho. Frobenius test ko $ω \land d ω$ chahiye: $d ω = cos ϕ d ϕ \land d x + sin ϕ d ϕ \land d y .$ $ω = sin ϕ d x - cos ϕ d y$ ke saath wedge karo aur sirf surviving $d x \land d y \land d ϕ$ terms rakho: $ω \land d ω = (sin^{2} ϕ + cos^{2} ϕ) d x \land d y \land d ϕ = d x \land d y \land d ϕ \neq = 0.$ Non-zero → not integrable: koi surface $f (x, y, ϕ) = c$ exist nahi karta. (c) Labels: non-holonomic (non-integrable velocity relation), scleronomic (explicit $t$ nahi). (d) Consequence. Agar har instant par ek direction forbidden hai, phir bhi path choose karke (steer karo, aage roll karo, steer karo, roll karo) tum koi bhi target $(x, y, ϕ)$ tak pahunch sakte ho. Constraint instantaneous velocities ko limit karta hai, reachable configurations ka set nahi — non-holonomy ki defining feature yahi hai. Isliye parallel parking hoti hai. Answer: $d x sin ϕ - d y cos ϕ = 0$ ; non-integrable; non-holonomic, scleronomic.

Recall Solution L4·2

Pehla axis. Yeh ek inequality hai — jab particle interior mein ho toh yeh ek coordinate eliminate nahi kar sakta. Isliye non-holonomic. Doosra axis. Time $a (t) = a_{0} + u t$ ke through explicitly aata hai; boundary move karti hai. Isliye rheonomic. Physical note. Jab particle wall se touch karta hai toh use moving boundary feel hoti hai (L1·2 jaisi), jo kaam kar sakti hai — particle ki akeli energy conserved nahi rehti. Axes combine karke: yeh constraint non-holonomic aur rheonomic hai — proof ki dono axes sach mein independent hain (compare L1·2 jo holonomic-rheonomic tha aur L1·3 jo non-holonomic-scleronomic tha). Answer: non-holonomic, rheonomic.

Level 5 — Mastery

Goal: ek complete classification argument build karo aur ek false counterexample ke khilaf defend karo.

Recall Solution L5·1

(a) Total differential pehchano. Left side exactly $d (x^{2} + y^{2} + z^{2})$ hai. Zero set karne par $d (x^{2} + y^{2} + z^{2}) = 0$ milta hai, isliye $x^{2} + y^{2} + z^{2} = c .$ Agar yeh velocity/differential relation bhi hai, toh bina kisi factor ke cleanly integrate hua — isliye yeh pehle se hi holonomic tha. (b) Geometry. $x^{2} + y^{2} + z^{2} = c$ radius $c$ ka ek sphere hai: particle ek fixed spherical shell par pin hai. (c) DOF aur labels. Start $3 N = 3$ ; ek holonomic equation ek coordinate hatata hai → $n = 2$ (latitude aur longitude). Koi $t$ nahi → scleronomic. Answer: sphere $x^{2} + y^{2} + z^{2} = c$ mein integrable; holonomic, scleronomic, $n = 2$ .

Recall Solution L5·2

(i) Fixed incline surface $f (r) = 0$ , koi $t$ nahi → holonomic, scleronomic. Yeh ek equality surface hai jo kabhi move nahi karti. (ii) $r - (r_{0} - v t) = 0$ coordinate aur time mein ek equation hai → holonomic, rheonomic. Allowed radius $t$ ke saath shrink hota hai, toh time explicit hai. (iii) Blade sideways velocity forbid karta hai: $\geq 3$ variables mein $d x sin ϕ - d y cos ϕ = 0$ jaisa relation, non-integrable → non-holonomic, scleronomic (L4·1 jaisa skeleton; rink fixed hai, toh koi $t$ nahi). (iv) Rolling without slipping → non-integrable velocity relation → non-holonomic; bowl time ke saath tilt hota hai isliye constraint coefficients $t$ par depend karte hain → non-holonomic, rheonomic. Answer: (i) H, S · (ii) H, R · (iii) NH, S · (iv) NH, R — $2 \times 2$ table ke charon boxes fill ho gaye.

Recall Close karne se pehle quick self-check

Classification decision, har ek ek line mein. Kisi bhi constraint ke liye tumhara PEHLA sawaal kya hai? ::: Kya ise equality $f (r, t) = 0$ ki tarah likha ja sakta hai (possibly ek differential relation integrate karne ke baad)? Exactly 2 variables mein ek differential relation hamesha...? ::: Integrable hota hai (integrating factor hamesha exist karta hai) → holonomic. Doosra, independent label decide hota hai...? ::: Kya $t$ explicitly aata hai — haan → rheonomic, nahi → scleronomic. Non-holonomic constraints...? ::: Accessible velocity directions kam karte hain, coordinates ya reachable configurations nahi.

Related builds: Lagrangian Mechanics · D'Alembert's Principle · Lagrange Multipliers · Virtual Displacements · Generalized Coordinates.