Constraints using Lagrange multipliers: Exercises — Hinglish deep dive

2.1.10 · D4 · Physics › Analytical Mechanics › Constraints using Lagrange multipliers

Definition Is poore page par

$L$ ka sign convention Hum hamesha $== L = T - V ==$ lete hain (kinetic minus potential). Ek subtlety is page par aadhi sign confusion ki wajah banti hai: agar coordinate $x$ us direction mein badhta hai jis direction mein force point karti hai, toh potential $x$ ke saath ghatती hai, isliye $V$ mein minus aata hai aur $L = T - V$ mein plus dikhta hai. Concretely:

agar $x$ neechay ki taraf badhta hai aur gravity bhi neechay point karti hai, toh $V = - m g x$ , isliye $L = T - (- m g x) = T + m g x$ — yahi wajah hai ki neeche $+ M g x sin α$ ya $+ m_{1} g x - m_{2} g x$ jaise terms plus sign ke saath aate hain;
agar $y$ upar ki taraf badhta hai, toh $V = + m g y$ aur $L = T - m g y$ mein minus dikhta hai. Hamesha poochho: kya coordinate us direction mein badhta hai jis taraf force kheenchti hai? Agar haan, toh $L$ mein plus.

Prerequisites jo tum open rakhna chahoge: Euler-Lagrange Equations, d'Alembert's Principle and Virtual Work, Generalized Coordinates and Degrees of Freedom, Holonomic vs Non-holonomic Constraints, Normal Force and Tension as Constraint Forces, KKT Conditions.

Level 1 — Recognition

Goal: method ke pieces ko bina heavy algebra ke pehchaano.

Exercise L1.1 — Gradient ko naam do

Recall Solution

(a) Sab kuch ek side le jao: $f (x, y) = y - 3 x + 2 = 0$ . (b) $\frac{\partial f}{\partial x} = - 3$ , $\frac{\partial f}{\partial y} = 1$ . (c) Vector $(- 3, 1)$ line ke perpendicular (normal) hai. Yeh isliye matter karta hai kyunki ek rail/wire bead ko track ke side se dhakelta hai — kabhi along nahi. Isliye constraint force $\nabla f$ ke along point karni chahiye, aur exactly yahi $λ \nabla f$ deliver karta hai. (Check karo: line ki direction $(1, 3)$ hai; dot product $(- 3) (1) + (1) (3) = 0$ , confirm karta hai ki perpendicular hai.)

Exercise L1.2 — Equations gino

Recall Solution

Unknowns: $3$ coordinates $q_{1}, q_{2}, q_{3}$ plus $2$ multipliers $λ_{1}, λ_{2}$ = $5$ unknown functions of time. Equations: $3$ Lagrange equations (ek har coordinate ke liye) plus $2$ constraint equations $f_{1} = 0, f_{2} = 0$ = $5$ equations. Balanced: $5 = 5$ . ✓ Constraints extra multipliers ki "rent pay" karte hain.

Level 2 — Application

Goal: ek clean single-constraint system par recipe end-to-end chalao.

Exercise L2.1 — Horizontal rotating-free hoop par bead, normal force

Recall Solution

Horizontal plane ⇒ koi gravity term nahi. $L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ . $r$ -equation: $\frac{d}{d t} (m \overset{r}{˙}) - m r \dot{θ}^{2} = λ \cdot 1$ . $r = R$ impose karo toh $\overset{r}{˙} = \overset{r}{¨} = 0$ : $- m R \dot{θ}^{2} = λ$ . $\dot{θ} = ω$ ke saath: $λ = - m R ω^{2}$ . Interpretation: $λ$ generalized radial force hai. Negative matlab yeh decreasing $r$ ki taraf point karta hai, yaani inward — hoop bead ko center ki taraf kheenchta hai. Iska magnitude $m R ω^{2}$ exactly centripetal force hai. Method familiar $m v^{2} / R$ reproduce karta hai jahan $v = R ω$ . ✓

Exercise L2.2 — Vertical hoop, $N$ derive karo phir pata karo bead kahan chhod deta hai

Recall Solution

(a) $N$ scratch se derive karo. Plane polar coordinates mein position height $y = r sin θ$ hai, isliye $V = m g y = m g r sin θ$ (yahan $θ$ gravity ke against upward badhta hai, isliye potential mein hamare convention se plus aata hai). Toh $L = T - V = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}) - m g r sin θ .$ Constraint $f = r - R = 0$ hai, isliye $\partial f / \partial r = 1$ , $\partial f / \partial θ = 0$ . Recipe ka $r$ -equation hai

\;\Rightarrow\; \frac{d}{dt}(m\dot r)-\big(mr\dot\theta^2-mg\sin\theta\big)=\lambda\cdot 1.$$ Isliye $m\ddot r-mr\dot\theta^2+mg\sin\theta=\lambda$. Ab — *aur sirf ab* — constraint $r=R\Rightarrow\dot r=\ddot r=0$ impose karo: $$-mR\dot\theta^2+mg\sin\theta=\lambda.$$ Physical normal force $N=\lambda\,\partial f/\partial r=\lambda\cdot 1=\lambda$ hai, isliye $$\boxed{N=mg\sin\theta-mR\dot\theta^2}.$$ **Sign convention check:** bilkul bottom par $\theta=-90^\circ$, $\sin\theta=-1$, milta hai $N=-mg-mR\dot\theta^2<0$; hamare convention mein $\lambda\partial f/\partial r$ negative matlab force *decreasing* $r$ ki taraf point karta hai (inward), yaani hoop bead ko center ki taraf dhakelta hai — correct, yahi ek floor-like surface karta hai. **(b)** Bead *inside* par press karta hai jab tak hoop inward push kar sakta hai, yaani jab tak $N$ (inward magnitude ke roop mein) available hai; jab required inward push reverse ho jaaye toh chhod deta hai. Top par $\theta=+90^\circ$, $\sin\theta=1$, critical case $N=0$ hai: $$N=mg(1)-mR\dot\theta^2_{\text{top}}=0\ \Rightarrow\ R\dot\theta^2_{\text{top}}=g\ \Rightarrow\ v_{\text{top}}^2=(R\dot\theta)^2=gR.$$ Energy conservation bottom ($y=-R$) se top ($y=+R$) tak, height gain $2R$: $$\tfrac12 v_0^2=\tfrac12 v_{\text{top}}^2+g(2R)=\tfrac12 gR+2gR=\tfrac52 gR.$$ $$\boxed{v_0=\sqrt{5gR}}.$$ $N=\lambda$ khud derive karna hi tha jo hamein "leaves the surface" condition $N=0$ directly likhne diya — multiplier method ka poora payoff yahi hai.

Level 3 — Analysis

Goal: multi-equation systems, physical force extract karo, signs check karo.

Exercise L3.1 — Incline par cylinder, friction as multiplier

Neeche di gayi figure hamare sign convention ko fix karti hai: $x$ slope ke neechay measure kiya jaata hai (amber arrow), isliye disk ka center $+ x$ mein move karta hai jab yeh descend karta hai; friction (cyan arrow) contact point par slope ke upar act karti hai, aur gravity $M g sin α$ along $+ x$ resolve hoti hai. Yeh picture dhyan mein rakho — neeche har sign isi se paḍha jaata hai.

Figure — Constraints using Lagrange multipliers

Recall Solution

$x$ -equation ( $\partial f / \partial x = 1$ ): $M \overset{x}{¨} - M g sin α = λ$ . …(i) $θ$ -equation ( $\partial f / \partial θ = - R$ ): $I \ddot{θ} = - R λ$ . …(ii)

Rolling constraint ko do baar differentiate kyun karein? Equations (i) aur (ii) mein do unknown accelerations $\overset{x}{¨}, \ddot{θ}$ plus unknown $λ$ hain — teen unknowns, sirf do equations. Recipe ek teesra equation promise karti hai: constraint khud, $f = x - R θ = 0$ . Lekin $f = 0$ positions relate karta hai, jabki (i)–(ii) accelerations ke baare mein hain. Constraint ko same language mein bolane ke liye, isko time mein do baar differentiate karo: $f = x - R θ = 0 d / d t \overset{x}{˙} - R \dot{θ} = 0 d / d t \overset{x}{¨} - R \ddot{θ} = 0.$ Yeh $\ddot{θ} = \overset{x}{¨} / R$ deta hai, woh missing teesra relation jo $\ddot{θ}$ eliminate karne aur $λ$ solve karne deta hai.

$\ddot{θ} = \overset{x}{¨} / R$ ko (ii) mein daalo: $\frac{I}{R} \cdot \frac{x ¨}{R} = - R λ \Rightarrow λ = - \frac{I x ¨}{R ^{2}}$ . (i) mein substitute karo: $M \overset{x}{¨} - M g sin α = - \frac{I x ¨}{R ^{2}}$ . $I = \frac{1}{2} M R^{2}$ ke saath, toh $I / R^{2} = \frac{1}{2} M$ : $M \overset{x}{¨} + \frac{1}{2} M \overset{x}{¨} = M g sin α \Rightarrow \frac{3}{2} \overset{x}{¨} = g sin α \Rightarrow \overset{x}{¨} = \frac{2}{3} g sin α .$ (b) $λ = - \frac{I x ¨}{R ^{2}} = - \frac{1}{2} M \cdot \frac{2}{3} g sin α = - \frac{1}{3} M g sin α .$ Friction magnitude $∣ λ ∣ = \frac{1}{3} M g sin α$ . $x$ par physical friction $λ \partial f / \partial x = λ \cdot 1 = λ < 0$ hai, yaani $- x$ direction mein = slope ke upar (descent oppose karte hue) — exactly figure mein cyan arrow.

Exercise L3.2 — Rigid rod par do beads (shared constraint)

Recall Solution

$L = \frac{1}{2} m \overset{x}{˙}_{1}^{2} + \frac{1}{2} m \overset{x}{˙}_{2}^{2} + F x_{1}$ . $\partial f / \partial x_{1} = - 1, \partial f / \partial x_{2} = + 1$ . $x_{1}$ : $m \overset{x}{¨}_{1} - F = λ (- 1) = - λ$ . …(i) $x_{2}$ : $m \overset{x}{¨}_{2} - 0 = λ (+ 1) = λ$ . …(ii) Constraint do baar differentiate kiya: $\overset{x}{¨}_{2} = \overset{x}{¨}_{1} \equiv a$ . (i)+(ii) add karo: $ma - F + ma = - λ + λ = 0 \Rightarrow 2 ma = F \Rightarrow a = \frac{F}{2 m} .$ (ii) se: $λ = ma = \frac{F}{2}$ . Interpretation: $λ$ rod mein force hai. Bead 2 rod ke through $+ λ = F /2$ force se forward drag hota hai, aur bead 1 rod ko apne par $- λ = - F /2$ ke saath back kheenchte mahsoos karta hai. Isliye rod tension $F /2$ mein hai — rod applied force ka aadha second bead tak transmit karta hai, exactly wahi jo Newton do equal masses ke liye predict karta hai. $λ = F /2$ .

Level 4 — Synthesis

Goal: constraints combine karo, degenerate/limiting cases, setups change karo.

Exercise L4.1 — Heavy pulley wali Atwood machine, saare limits

Figure geometry dikhata hai: $m_{1}$ (amber block) $x$ se neeche jaata hai jabki $m_{2}$ $x$ se upar jaata hai kyunki string inextensible hai, aur pulley of moment $I$ $ϕ$ (white curved arrow) se ghoomta hai bina slip ke, constraint $f = x - a ϕ = 0$ deta hai. Isko track karne ke liye use karo ki kaun sa mass kis taraf jaata hai.

Recall Solution

Note karo ki $x$ already dono masses carry karta hai (string inextensible hai), isliye $x$ ko ek coordinate aur $ϕ$ ke saath treat karo. $\partial f / \partial x = 1, \partial f / \partial ϕ = - a$ . $x$ : $(m_{1} + m_{2}) \overset{x}{¨} - (m_{1} - m_{2}) g = λ$ . …(i) $ϕ$ : $I \ddot{ϕ} = - aλ$ . …(ii) Constraint do baar differentiate kiya (same reasoning as L3.1: $\ddot{ϕ}$ eliminate karne ke liye hamen acceleration-level relation chahiye): $\overset{x}{¨} = a \ddot{ϕ} \Rightarrow \ddot{ϕ} = \overset{x}{¨} / a$ , toh (ii) ⇒ $λ = - \frac{I x ¨}{a ^{2}}$ . (i) mein daalo: $(m_{1} + m_{2}) \overset{x}{¨} - (m_{1} - m_{2}) g = - \frac{I x ¨}{a ^{2}}$ . $\overset{x}{¨} = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2} + I / a ^{2}}, λ = - \frac{I / a ^{2} ( m _{1} - m _{2} ) g}{m _{1} + m _{2} + I / a ^{2}} .$ (b) $I \to 0$ : $\overset{x}{¨} \to \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ — classic Atwood result. ✓ Aur $λ \to 0$ : ek massless pulley ko koi net torque nahi chahiye, isliye dono sides ke beech tension difference zero ho jaata hai. ✓ (c) $I \to \infty$ : $\overset{x}{¨} \to 0$ . Ek infinitely heavy pulley spin nahi ho sakta, isliye kuch bhi move nahi karta. System freeze ho jaata hai — degenerate limit sensibly behave karta hai. ✓

Exercise L4.2 — Jab constraint force switch off ho jaaye (degenerate case)

Recall Solution

Effective gravity kyun? Hum elevator ke apne (accelerating, non-inertial) frame mein solve karte hain, kyunki us frame mein constraint $y = 0$ simply "block floor par rehta hai" hai $y$ fixed ke saath. Newton's laws as-is ek accelerating frame mein nahi holde; standard fix yeh hai ki har mass par ek fictitious force $- m a_{frame}$ add karo. Frame downward $a$ se accelerate karta hai, isliye fictitious force upward magnitude $ma$ ke saath point karta hai. Real gravity $m g$ downward ke saath combine hokar, har mass $m (g - a)$ ka net downward pull feel karta hai — yeh effective gravity $g_{eff} = g - a$ hai. Yeh ek genuine derivation hai, assertion nahi: fictitious force + real gravity = effective gravity. $y$ up ke saath, effective-gravity potential $V = m g_{eff} y$ hai (block ke rise karne par badhta hai), isliye $L = \frac{1}{2} m \overset{y}{˙}^{2} - m g_{eff} y, f = y = 0, \partial f / \partial y = 1.$ $y$ -equation: $m \overset{y}{¨} + m g_{eff} = λ$ . $y = 0 \Rightarrow \overset{y}{¨} = 0$ impose karo: $λ = m (g - a) = N .$ (b) Contact ke liye $N \geq 0 \Rightarrow g - a \geq 0 \Rightarrow a \leq g$ chahiye. $a = g$ par (free fall), $λ = 0$ : multiplier zero ho jaata hai, matlab floor koi force se push nahi karta — block weightless hai aur float hone ki verge par hai. $a > g$ par block ko floor ko kheenchna pad jaata ( $λ < 0$ ), jo floor nahi kar sakta, isliye constraint break ho jaata hai aur $y = 0$ ab valid nahi raha. Yeh exactly " $N = 0$ ⇒ leaves the surface" logic hai. ✓

Level 5 — Mastery

Goal: gradient geometry, saare cases, aur interpretation ko joḍne wala full research-style problem.

Exercise L5.1 — Parabolic wire par bead, full multiplier aur "flying off" analysis

Figure parabola $y = \frac{1}{2} k x^{2}$ (cyan), uspar ek bead, aur constraint gradient $\nabla f = (- k x, 1)$ (amber arrow) dikhata hai jo wire se off, normal direction mein point karta hai — woh ek hi direction jisme ek frictionless wire push kar sakti hai. Note karo ki gradient ek unit vector nahi hai: iska length $1 + k^{2} x^{2}$ vertex se door jaane par badhta hai, aur yahi wajah hai ki part (c) ko physical force paane ke liye $λ$ ko $∥\nabla f ∥$ se multiply karna padta hai.

Recall Solution

Yahan $y$ upar point karta hai aur gravity neechay pull karti hai, isliye hamare convention se $V = m g y$ $L$ mein minus ke saath enter karta hai: $L = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) - m g y .$ Constraint ka gradient: $\partial f / \partial x = - k x, \partial f / \partial y = 1$ .

(a) Multiplier equations:

$x$ : $m \overset{x}{¨} = λ (- k x) = - λk x$ . …(i)
$y$ : $m \overset{y}{¨} + m g = λ (1) = λ$ . …(ii)

(b) Constraint $y = \frac{1}{2} k x^{2}$ do baar differentiate kiya (again acceleration level tak pahunchne ke liye, jaise L3.1 mein): $\overset{y}{˙} = k x \overset{x}{˙}, \overset{y}{¨} = k \overset{x}{˙}^{2} + k x \overset{x}{¨} .$ (i) se: $\overset{x}{¨} = - \frac{λk x}{m}$ . (ii) se: $λ = m (\overset{y}{¨} + g) = m (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g)$ . Yeh $λ$ (i) mein substitute karo: $m \overset{x}{¨} = - k x \cdot m (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g)$ $\overset{x}{¨} = - k x (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g)$ $\overset{x}{¨} (1 + k^{2} x^{2}) = - k x (k \overset{x}{˙}^{2} + g)$ $\overset{x}{¨} = - \frac{k x ( g + k x ˙ ^{2} )}{1 + k ^{2} x ^{2}} .$

(c) $λ$ paane ke liye back-substitute karo: $λ = m (k \overset{x}{˙}^{2} + k x \overset{x}{¨} + g) .$ Boxed $\overset{x}{¨}$ use karke: $k x \overset{x}{¨} = - \frac{k ^{2} x ^{2} ( g + k x ˙ ^{2} )}{1 + k ^{2} x ^{2}}$ , isliye

=m\cdot\dfrac{(k\dot x^2+g)(1+k^2x^2)-k^2x^2(g+k\dot x^2)}{1+k^2x^2}.$$ Numerator mein $k^2x^2$ terms cancel ho jaate hain, bachtaa hai $$\boxed{\lambda=\dfrac{m(g+k\dot x^2)}{1+k^2x^2}}.$$ **Normal force** magnitude $N=|\lambda|\cdot\|\nabla f\|=|\lambda|\sqrt{(kx)^2+1}=\dfrac{m(g+k\dot x^2)}{\sqrt{1+k^2x^2}}$ hai, kyunki *physical* force vector $\lambda\nabla f$ hai aur $\|\nabla f\|=\sqrt{k^2x^2+1}$. **Units:** $[k]=\text{m}^{-1}$ (kyunki $\tfrac12 kx^2$ ek length hai). Toh $k\dot x^2$ ke units $\text{m}^{-1}(\text{m/s})^2=\text{m/s}^2$ hain, $g$ se match karte hue. Isliye $m(g+k\dot x^2)$ $\text{kg}\cdot\text{m/s}^2=$ **newtons** hai, aur denominator dimensionless hai. $\lambda$ ek force hai. ✓ **(d)** Vertex $x=0$ par: $\lambda=\dfrac{m(g+k\dot x^2)}{1}=m(g+k\dot x^2)>0$. Zyada generally boxed $\lambda=\dfrac{m(g+k\dot x^2)}{1+k^2x^2}$ ka **numerator strictly positive** hai (dono $g>0$ aur $k\dot x^2\ge0$) aur **denominator strictly positive** hai har $x$ aur har $\dot x$ ke liye. Isliye $\lambda>0$ **hamesha**. Kyunki physical force $\lambda\nabla f$ hai aur $\nabla f$ bowl se bahar point karta hai, wire hamesha bead ko **inward/upward** press karti hai: bead concave-up parabola se **kabhi fly off nahi kar sakta**. Yeh complete flying-off analysis hai — ek formula, saare $x$ ke liye valid, $\dot x$ ke dono signs ke liye, aur degenerate vertex ke liye bhi. Hoop ke top se contrast karo, jahan corresponding $\lambda$ *zero tak pahunch sakta* hai aur bead nikal jaata hai; parabola ek bowl hai, isliye yeh nahi kar sakta.

Recap ladder

Recall Har level ka ek-line summary

L1 — $\nabla f$ constraint ke normal hai; equations = coordinates + multipliers. L2 — redundant coordinate rakho, differentiate karo, tabhi constraint impose karo; $λ = N$ . L3 — physical force $λ \partial f / \partial q$ hai, sign sahit; friction/tension nikal aate hain. L4 — $I \to 0, \infty$ limits check karo; $λ$ mein sign change matlab (one-sided) constraint break ho gayi. L5 — normal force magnitude $∣ λ ∣ ∥\nabla f ∥$ hai, $∣ λ ∣$ nahi; ek formula saare cases cover karta hai.