Question bank — Constraints using Lagrange multipliers

True — embedded (reduced-coordinate) method constraint forces ko zero virtual work karwa ke vanish kar deta hai, jabki multiplier $\lambda \partial f/\partial q_k$ wahi force hai, purpose se raka gaya (figure s02 dekho).

False — ek baar constraint eliminate hone ke baad un coordinates mein koi $f$ nahi rehta, toh $\partial f/\partial q=0$ hai aur koi $\lambda$ nahi hai jiससे force read ki ja sake.

False — $\lambda \partial f/\partial q_k$ ke units generalized force ke hone chahiye, toh $\lambda$ ke units $f$ se set hote hain; agar $f$ ke length units hain, toh $\lambda$ ek force hai.

False — $\lambda (t)$ time ki ek function hai; constraint force motion ke saath change hoti hai (jaise $N=mg\sin \theta -mR{\dot{\theta }}^2$ $\theta$ aur $\dot{\theta }$ ke saath vary karta hai).

False — yeh $\partial f/\partial q$ (yaani $\nabla f$) ki direction mein point karti hai, jo allowed motion (surface $f=0$) ke perpendicular hai; exactly isliye yeh embedding ke under zero virtual work karta hai.

True — tumhe $n$ Euler–Lagrange equations milte hain plus $m$ constraint equations, $n$ coordinates plus $m$ multipliers ke liye: $n+m=n+m$.

Bookkeeping mein True — multiplier method sabhi $n$ coordinates (dependent ones bhi) rakhta hai aur $\lambda$ add karta hai, ek chhote coordinate count ko constraint force tak direct access ke liye trade karta hai. Generalized Coordinates and Degrees of Freedom dekho.

False — genuinely non-holonomic constraint $\sum a_k{\dot{q}}_k=0$ non-integrable hota hai, toh koi position function $f$ exist nahi karta; tumhe velocity form ${\lambda }_a{a}_{ak}$ directly use karni padti hai. Holonomic vs Non-holonomic Constraints dekho.

True — iska sign batata hai constraint force ki direction (jaise ek rope sirf pull kar sakti hai, toh galat sign flag karta hai ki constraint violate hone wali hai / object surface chhod raha hai). Sign-convention figure s04 dekho.

Error — virtual displacements $\delta q_k$ (kalpana mein time-frozen nudges surface ke saath saath) ab independent nahi hain; unhe $\delta f=\sum (\partial f/\partial q_k)\delta q_k=0$ satisfy karni padti hai, toh ek weighted sum ka zero hona har term ko zero hone par majboor nahi karta.

Error — $r=R$ substitute karne ke baad coordinate chala jaata hai aur $r-R\equiv 0$ identically hota hai; vary karne ke liye koi free $r$ nahi hai, toh tum $N$ recover nahi kar sakte. $r$ ko free rakho jab tak differentiate karne ke baad nahi kar lete.

Error — $x$ par generalized force $\lambda \partial f/\partial x=\lambda$ hai, jo $-T$ ke barabar hai kyunki string drop-coordinate $x$ ke opposite pull karti hai; physical tension $T=|\lambda |$ hai (sign diagram s04 dekho).

Error — tum $(n+1)$th equation bhool gaye, constraint $f=0$ khud; woh system ko close karta hai aur $\lambda$ ko pin karta hai.

Subtle — rolling ke liye, constraint $x-R\theta =0$ hai jiska gradient $(x,\theta )$ coordinate space mein exist karta hai, physical space mein nahi; resulting $\lambda$ sahi tarike se tangential friction deliver karta hai kyunki wahi force no-slip enforce karti hai. Normal Force and Tension as Constraint Forces dekho.

Error — virtual work na karna matlab hai force allowed displacements ke perpendicular hai, yeh nahi ki uski magnitude zero hai; $\lambda$ exactly us perpendicular push ki nonzero magnitude hai.

Error — reduced equations pehle hi constraint use kar chuki hain; $N$ sirf tab appear karta hai jab redundant coordinate aur uska $\lambda$ term retain kiya jaaye.

EL-bracket vector left-hand sides $(\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_k}-\frac{\partial L}{\partial q_k})$ ka collection hai, ek vector mein stacked, ek entry per coordinate. Allowed nudge $\delta q$ is bracket vector aur $\nabla f$ dono ke perpendicular hai (figure s03); upar wale linear-algebra fact se (do arrows jo same line ke perpendicular hain parallel hote hain), bracket $=\lambda \nabla f$ kisi scalar $\lambda$ ke liye.

Ek zero add karna (kyunki $\delta f=0$) physically kuch nahi badalta lekin $\lambda$ choose karne deta hai dependent coordinate ke bracket ko kill karne ke liye, jiske baad remaining $\delta q$ independent hote hain. Yeh wahi trick hai jaise d'Alembert's Principle and Virtual Work.

$f$ ka gradient surface $f=0$ ke perpendicular point karta hai, aur ek constraint force exactly wahi hai jo ek body ko us surface se off/onto push karti hai — uske perpendicular — toh force $\nabla f$ ke proportional honi chahiye.

Hoop chhodna $N=0$ ki event hai; embedded method ne $N$ ko bilkul discard kar diya hai, toh uske paas zero set karne ke liye koi expression nahi hai.

$n$ EL equations (har retained coordinate ke liye ek) plus constraint equation $f=0$, $n$ coordinates plus extra unknown $\lambda$ se match karta hai.

Dono ek constraint ko $\lambda \times (\text{constraint})$ add karke aur stationarize karke enforce karte hain; multiplier constraint ke liye sensitivity measure karta hai. KKT Conditions mein inequality-constraint generalization dekho — aur neeche edge cases mein concrete mechanical analogue.

Humein $\partial f/\partial q_k$ exist karna chahiye constraint-force direction form karne ke liye; ek non-smooth constraint ka well-defined gradient nahi hota aur method kink par break ho jaata hai.

Yeh zero se guzarta hai aur negative ho jaata agar hoop sirf andar push kar sakta; sign flip signal karta hai ki bead physically separate ho rahi hai, toh model sirf $\lambda =0$ tak valid hai (yeh exactly ek mechanical KKT "inactive constraint" hai: one-sided contact drop out ho jaata hai jab $\lambda$ sign change karta).

Constraint ab active nahi hai — string slack ho jaati hai — aur tumhe constraint drop karni padti hai (set $T=0$) instead of ek nonphysical negative tension par trust karne ke; yeh KKT constraint ke switch off hone ka mechanics version hai (figure s04 dekho).

$T=\frac{mg(I/a^2)}{m+I/a^2}\to 0$; ek massless cylinder react karne ke liye koi rotational inertia offer nahi karta, toh koi tension build nahi hota aur mass $g$ par girta hai.

$\ddot{x}=\frac{mg}{m+I/a^2}\to 0$ aur $T\to mg$; cylinder nahi ghoomega, toh mass essentially static equilibrium mein hang karta hai aur tension gravity balance karti hai.

Surface khud time mein sweep karti hai, toh constraint force $\lambda \nabla f$ ab ek moving wall par act karta hai aur real work kar sakta hai; energy conserved nahi hoti aur $\lambda$ ek moving push track karta hai (figure s05 dekho — wall bead ko slide hote hue displace karti hai).

Yeh utna hi hai jo bead ko line par rakhta hai (jaise uska weight component support karna); agar koi transverse applied force exist nahi karta, toh $\lambda =0$ — ek degenerate lekin bilkul valid answer.

Do constraint surfaces tangentially touch karte hain toh unke normals coincide karte hain (figure s06); combination ${\lambda }_1\nabla f_1+{\lambda }_2\nabla f_2$ infinitely many ways mein split ho sakta hai, toh ${\lambda }_1$ aur ${\lambda }_2$ individually undetermined hain — ek badly posed (redundantly constrained) problem.