Worked examples — Constraints using Lagrange multipliers

2.1.10 · D3 · Physics › Analytical Mechanics › Lagrange Multipliers se Constraints

Kuch bhi karne se pehle, aao un symbols ko fix kar lein jo hum har jagah reuse karenge, taaki notation kabhi surprise na kare:

Gradient $\partial f/\partial q_k$ simply yeh hai ki "agar main sirf $q_k$ ko thoda nudge karun to $f$ kitni tezi se badlega". Constraint ko ek wall ki tarah socho; gradient woh arrow hai jo wall se seedha bahar nikal raha hai, aur wall sirf seedha bahar push kar sakti hai — isliye constraint force hamesha $\lambda$ times us arrow ke barabar honi chahiye.

Scenario matrix

Is method ke saare possible problems inhi cells mein se ek mein aate hain. Har row ek alag case class hai jise hume kam se kam ek baar dikhana hai.

Ab hum har example chalate hain, aur har ek ko us cell(s) se label kiya gaya hai jo woh fill karta hai.

Example 1 — Vertical hoop par bead (Cells A, D)

Forecast: aage padhne se pehle andaaza lagao — $N$ hoop ke bottom par bada hoga ya top par? Likhlo "bigger at ___".

Neeche diya figure (s01) hoop ko violet circle ke roop mein sketch karta hai, bead ko $\theta =55^{\circ }$ par magenta dot ke roop mein, dashed navy radius $r=R$ centre se bead tak, orange arrow jo radial constraint force $N=\lambda \partial f/\partial r$ hai, aur ek chhota magenta $mg$ arrow seedha neeche. Orange arrow par nazar rakho — yeh woh ek hi direction hai jisme ek frictionless hoop push kar sakta hai.

1. Polar coordinates mein Lagrangian likho. $L=T-V=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)-mgR\sin \theta (\text{using }r=R\text{ in }V).$ Yeh step kyun? Plane mein kinetic energy hai $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)$ — $r^2{\dot{\theta }}^2$ term ghumne ki sideways speed hai. Hum polar choose karte hain taaki constraint simple $r=R$ ho, jo deta hai $\partial f/\partial r=1$.

2. Multiplier ke saath $r$-equation likho. $\frac{d}{dt}(m\dot{r})-mr{\dot{\theta }}^2+mg\sin \theta =\lambda \cdot \frac{\partial f}{\partial r}=\lambda .$ Yeh step kyun? Right-hand side $r$ ke along constraint force hai. Kyunki $\partial f/\partial r=1$, jo bhi $\lambda$ ho, woh radial force ke barabar hai. Figure s01 mein orange arrow $r$ ke along point karta hai, seedha hoop se bahar — bilkul wahi direction jahan ek normal force point kar sakti hai.

3. Constraint lagao $r=R\Rightarrow \dot{r}=0,\ddot{r}=0$. $-mR{\dot{\theta }}^2+mg\sin \theta =\lambda .$ Yeh step kyun? Hoop par $r$ kabhi nahi badlata, to uske time-derivatives khatam ho jaate hain. Radial direction ab pure force balance hai.

4. Normal force padho. $N=\lambda =mg\sin \theta -mR{\dot{\theta }}^2.$ Yeh step kyun? $\lambda \cdot 1$ hi generalized radial force hai, aur kyunki $r$ ek ordinary length hai, woh generalized force newtons mein ek asli force hai.

Verify: Hoop ke bottom par $\theta =-90^{\circ }$, to $\sin \theta =-1$: $N=-mg-mR{\dot{\theta }}^2$ (sign hamara convention reflect karta hai ki $+r$ outward point karta hai; magnitude $mg+mR{\dot{\theta }}^2$, yahan sabse bada). Top par $\theta =+90^{\circ }$, $N=mg-mR{\dot{\theta }}^2$, jo chhota hai. $\lambda$ ka position ke saath sign change Cell D in action hai. Units: $mg\sin \theta$ hai $\text{kg}\cdot {\text{m/s}}^2=\text{N}$, aur $mR{\dot{\theta }}^2$ hai $\text{kg}\cdot \text{m}\cdot {\text{s}}^{-2}=\text{N}$. ✓

Example 2 — Cylinder par wound string se latkaa mass (Cells B, C)

Forecast: kya tension $mg=20\text{N}$ se kam hogi, barabar hogi, ya zyada? Andaaza lagao.

1. Lagrangian. $L=T-V=\frac{1}{2}m{\dot{x}}^2+\frac{1}{2}I{\dot{\varphi }}^2+mgx.$ Yeh step kyun? Do kinetic pieces (girta hua mass, ghoomta hua cylinder) $T$ banate hain; gravitational potential hai $V=-mgx$ (dropping distance $x$ se height kam hoti hai), to $-V=+mgx$.

2. Do multiplier equations. $\partial f/\partial x=+1$ aur $\partial f/\partial \varphi =-a$ ke saath: $m\ddot{x}-mg=\lambda (\text{i}),I\ddot{\varphi }=-a\lambda (\text{ii}).$ Yeh step kyun? Yeh hai Cell B: ek $\lambda$ dono equations mein aata hai. Woh ek shared unknown hi do motions ko ek doosre se baandhta hai.

3. Constraint se accelerations link karo. $x=a\varphi \Rightarrow \ddot{\varphi }=\ddot{x}/a$. (ii) mein daalo: $\frac{I}{a}\cdot \frac{\ddot{x}}{a}=-a\lambda \Rightarrow \lambda =-\frac{I\ddot{x}}{a^2}.$ Yeh step kyun? Ab hum mystery $\lambda$ ko purely $\ddot{x}$ ke through express karte hain, (i) mein daalne ke liye ready.

4. Solve karo. (i) mein $\lambda$ daalo: $m\ddot{x}-mg=-\frac{I\ddot{x}}{a^2}\Rightarrow \ddot{x}=\frac{mg}{m+I/a^2}.$ Numbers: $I/a^2=0.02/0.01=2$, to $\ddot{x}=\frac{2\cdot 10}{2+2}=5{\text{m/s}}^2.$

5. Tension. $T=|\lambda |=\frac{I\ddot{x}}{a^2}=2\cdot 5=10\text{N}.$ Yeh step kyun? $\lambda$ negative aaya (Cell C). Hamare sign rule se, $+x$ neeche point karta hai, to $\lambda <0$ ek force hai jo upar point karta hai — ek tension. Iska magnitude hai $T$.

Verify: $\ddot{x}=5<g=10$, sahi hai — cylinder ki inertia mass ko rok rahi hai. Aur $T=10\text{N}<mg=20\text{N}$: tension weight se kam hai, bilkul yahi mass ko neeche accelerate karne deta hai. Newton cross-check: $mg-T=m\ddot{x}\Rightarrow 20-10=2\cdot 5=10$ ✓. Kya tumne "less than $mg$" forecast kiya tha?

Example 3 — Solid sphere incline par rolling (Cells C, F)

Forecast: ek frictionless block ($\ddot{x}=g\sin \alpha =5{\text{m/s}}^2$) ke comparison mein, kya sphere tezi se accelerate karega ya dheere?

Neeche diya figure (s02) violet incline ko $30^{\circ }$ par dikhata hai, magenta sphere uske upar rakhhi hai, orange arrow friction ke liye slope ke upar point karta hai (yeh hai $|\lambda |$), aur navy arrow gravity component $Mg\sin \alpha$ ke liye slope ke neeche point karta hai. Yeh do opposing arrows hi poori kahani hain ki sphere sliding block se dheere kyun hai.

1. Lagrangian. $L=T-V=\frac{1}{2}M{\dot{x}}^2+\frac{1}{2}I{\dot{\theta }}^2+Mgx\sin \alpha .$ Yeh step kyun? Ex 2 jaisi hi structure, lekin "gravity along the coordinate" sirf slope ke neeche component hai, jo deta hai $-V=+Mgx\sin \alpha$.

2. Multiplier equations. $\partial f/\partial x=1$, $\partial f/\partial \theta =-R$: $M\ddot{x}-Mg\sin \alpha =\lambda ,I\ddot{\theta }=-R\lambda .$

3. Constraint $\ddot{x}=R\ddot{\theta }$, $\lambda$ exactly pehle jaisi tarah eliminate karo: $\lambda =-\frac{I\ddot{x}}{R^2}=-\frac{(2/5)M{R}^2\ddot{x}}{R^2}=-\frac{2}{5}M\ddot{x}.$ Yeh step kyun? $\lambda <0$; kyunki $+x$ down-slope point karta hai, force up-slope point karta hai — yeh friction hai (Cell C/F), exactly s02 mein orange arrow ki tarah.

4. Solve karo. $M\ddot{x}-Mg\sin \alpha =-\frac{2}{5}M\ddot{x}\Rightarrow \ddot{x}(1+\frac{2}{5})=g\sin \alpha$, to $\ddot{x}=\frac{5}{7}g\sin \alpha =\frac{5}{7}\cdot 10\cdot \frac{1}{2}=\frac{25}{7}\approx 3.571{\text{m/s}}^2.$

5. Friction. $|\lambda |=\frac{2}{5}M\ddot{x}=\frac{2}{5}\cdot 3\cdot \frac{25}{7}=\frac{30}{7}\approx 4.286\text{N}.$ "$\frac{2}{7}Mg\sin \alpha$" formula cross-check: $\frac{2}{7}\cdot 3\cdot 10\cdot \frac{1}{2}=\frac{30}{7}$ ✓.

Verify: $\ddot{x}=3.571<5$: sphere frictionless block se dheera hai, kyunki energy usse spin bhi karti hai — forecast "slower" se match karta hai. Friction $\frac{30}{7}\text{N}$ up-slope point karta hai, jo exactly woh torque supply karta hai jo isse roll karta hai. Units: $Mg\sin \alpha$ hai N. ✓

Example 4 — Fly-off: bead hoop chhod deta hai (Cells D, E)

Forecast: top se chhota angle ya bada? Andaaza lagao "leaves at ___ degrees below the top".

1. Degenerate condition. Constraint force ek push hai, kabhi pull nahi. Bead us instant fly off karta hai jab $N=0$: $mg\sin \theta =mR{\dot{\theta }}^2\Rightarrow g\sin \theta =R{\dot{\theta }}^2.$ Yeh step kyun? Cell E woh limiting case hai jab $\lambda \to 0$. Jab multiplier zero ho jaata hai, constraint kuch nahi kar raha — surface ne jaane diya. Yeh exactly woh sawaal hai jo embedded method kabhi answer nahi kar sakta. (Aur yeh Cell D bhi hai: is angle se thoda upar $\lambda$ positive tha, thoda neeche negative ho jaata — sign flip.)

2. Energy se ${\dot{\theta }}^2$ nikalo. Top se girne par, bead height $R(1-\sin \theta )$ neeche aata hai — figure s01 mein sketch kiya gaya wahi $y=R\sin \theta$ relation use karte hue: $\frac{1}{2}m{R}^2{\dot{\theta }}^2=mgR(1-\sin \theta )\Rightarrow R{\dot{\theta }}^2=2g(1-\sin \theta ).$ Yeh step kyun? Hume ${\dot{\theta }}^2$ sirf $\theta$ ke terms mein chahiye solve karne ke liye. Energy conservation isse cleanly deta hai.

3. Combine karo. $g\sin \theta =2g(1-\sin \theta )\Rightarrow \sin \theta =\frac{2}{3}.$ $\theta =\arcsin \frac{2}{3}\approx 41.8^{\circ }.$ Yeh step kyun? $R{\dot{\theta }}^2$ ke do expressions barabar set karo; $g$ aur $R$ cancel ho jaate hain — fly-off angle universal hai, size aur gravity se independent.

Verify: $\sin \theta =2/3$ deta hai $R{\dot{\theta }}^2=2g(1-2/3)=\frac{2}{3}g$ aur $g\sin \theta =\frac{2}{3}g$ — dono match karte hain, to wahan exactly $N=0$ ✓. $\theta =41.8^{\circ }$ se neeche bead pehle hi ja chuka hai. Classic result $\cos (\text{angle from top})=2/3$. ✓

Example 5 — Real-world: rope tension ke saath do masses ek pulley par (Cells C, F)

Forecast: system kis direction mein accelerate karta hai, aur kya $T$ $m_{2}g$ aur $m_{1}g$ ke beech hai?

1. Linking constraint ke saath $L$ likho. Dono coordinates neeche measure karte hue: $L=T-V=\frac{1}{2}{m}_1{\dot{x}}_1^2+\frac{1}{2}{m}_2{\dot{x}}_2^2+m_{1}g{x}_1+m_{2}g{x}_2.$ Yeh step kyun? Dono coordinates rakhna $\lambda$ ko appear karne deta hai — yeh woh tension hai jo hum chahte hain (Cell F, real-world payoff).

2. Multiplier equations. $\partial f/\partial x_1=\partial f/\partial x_2=1$: $m_1{\ddot{x}}_1-m_{1}g=\lambda ,m_2{\ddot{x}}_2-m_{2}g=\lambda .$

3. Constraint ${\ddot{x}}_1+{\ddot{x}}_2=0\Rightarrow {\ddot{x}}_2=-{\ddot{x}}_1$. Do equations subtract karo: $m_1{\ddot{x}}_1-m_2{\ddot{x}}_2-(m_1-m_2)g=0\Rightarrow (m_1+m_2){\ddot{x}}_1=(m_1-m_2)g.$ ${\ddot{x}}_1=\frac{m_1-m_2}{m_1+m_2}g=\frac{2}{4}\cdot 10=5{\text{m/s}}^2.$ Yeh step kyun? Subtract karna shared $\lambda$ cancel kar deta hai (ek common tension dono sides par equally act karta hai), aur ${\ddot{x}}_2=-{\ddot{x}}_1$ substitute karne se ek equation mein ek unknown bachta hai — isliye subtraction, addition nahi, sahi move hai.

4. Tension. Pehli equation se, $T=-\lambda =m_{1}g-m_1{\ddot{x}}_1=3\cdot 10-3\cdot 5=15\text{N}.$ Yeh step kyun? $\lambda$ rope ka generalized force hai; $+x_1$ neeche point karta hai, to $\lambda <0$ (Cell C) ek force hai jo upar point karta hai — tension. Iska magnitude hai $15\text{N}$.

Verify: $T=15\text{N}$ $m_{2}g=10$ aur $m_{1}g=30$ ke beech hai ✓ (ek real rope aisa hona chahiye). $m_2$ check karo: $T-m_{2}g=m_2{\ddot{x}}_2\Rightarrow 15-10=1\cdot 5=5$ ✓. Forecast: $m_1$ neeche jaata hai (heavier), $T$ beech mein.

Example 6 — Ek saath do constraints (Cell G)

Forecast: kya ${\lambda }_1$ (vertical) weight $mg$ ke barabar hoga? Kya ${\lambda }_2$ nonzero hoga even though drive purely $x$ ke along hai?

1. Many constraints ke liye general form. $m$ constraints ke saath RHS unpar sum karta hai: $\frac{d}{dt}\frac{\partial L}{\partial {\dot{q}}_k}-\frac{\partial L}{\partial q_k}={\sum }_{a=1}^2{\lambda }_a\frac{\partial f_a}{\partial q_k}.$ Yeh step kyun? Cell G: har constraint ko apna honest helper ${\lambda }_a$ milta hai; $q_k$ par total constraint force unke gradients ka sum hai — yeh multipliers ka wahi stacking hai jo tum KKT Conditions mein dekhte ho.

2. $z$-equation. $L=T-V=\frac{1}{2}m({\dot{x}}^2+{\dot{y}}^2+{\dot{z}}^2)-mgz+Fx$. Gradients: $\partial f_1/\partial z=1$, $\partial f_2/\partial z=0$: $m\ddot{z}+mg={\lambda }_1.$ Yeh step kyun? Sirf $f_1$ hi $z$ mention karta hai, to sirf ${\lambda }_1$ yahan bachta hai. Shelf par $z=2$ constant hai, to $\ddot{z}=0$, jo deta hai ${\lambda }_1=mg=10\text{N}.$

3. $x$- aur $y$-equations. $\partial f_2/\partial x=-1$, $\partial f_2/\partial y=+1$; $f_1$ kuch contribute nahi karta: $m\ddot{x}-F=-{\lambda }_2,m\ddot{y}=+{\lambda }_2.$ Yeh step kyun? Drive $F$ $x$-equation mein enter karta hai ($\partial (Fx)/\partial x=F$ se); diagonal wall ka gradient $x$ mein $-1$ aur $y$ mein $+1$ ke roop mein split hota hai.

4. Constraint lagao aur solve karo. $y=x\Rightarrow \ddot{y}=\ddot{x}$. $y$-equation ko $x$-equation se subtract karo: $m\ddot{x}-F-m\ddot{y}=-{\lambda }_2-{\lambda }_2\Rightarrow -F=-2{\lambda }_2\Rightarrow {\lambda }_2=\frac{F}{2}=2\text{N}.$ Yeh step kyun? Subtraction ${\lambda }_2$ isolate karta hai kyunki do equations mein woh opposite signs ke saath hai. Unhe add karne par ${\lambda }_2$ cancel ho jaata hai: $m(\ddot{x}+\ddot{y})-F=0\Rightarrow 2\ddot{x}=F\Rightarrow \ddot{x}=2{\text{m/s}}^2$ line ke along.

Verify: ${\lambda }_1=mg=10\text{N}$ (shelf weight hold karti hai, forecast ✓). ${\lambda }_2=2\text{N}\ne 0$ (diagonal wall ek axis-aligned push par bhi react karta hai ✓). $x$-equation check karo: $1\cdot 2-4=-2=-{\lambda }_2$ ✓.

Example 7 — Trap: already-eliminated coordinate par multiplier (Cell H, degenerate)

Forecast: andaaza lagao $\partial f/\partial \theta$ yahan kya hai.

1. Kaunsa coordinate bachta hai? $r=R$ substitute karne ke baad, sirf $\theta$ bachta hai. Constraint $f=r-R$ kisi bhi live coordinate ka zikar nahi karta. Yeh step kyun? Yeh hai Cell H, woh degenerate case jahan constraint fold away ho gaya hai.

2. Gradient compute karo. $\partial f/\partial \theta =0$ ($r-R$ mein koi $\theta$ nahi hai), aur differentiate karne ke liye koi live $r$ nahi hai. $\lambda \frac{\partial f}{\partial \theta }=\lambda \cdot 0=0.$ Yeh step kyun? Multiplier term identically vanish ho jaata hai — $\lambda$ kuch bhi ho aur kuch nahi badlega. Yeh undetermined aur meaningless hai.

3. Sabak. Constraint force paane ke liye tumhe redundant coordinate (yahan $r$) rakhhna hoga taaki $\partial f/\partial r=1$ $\lambda$ ko carry kar sake. Ise eliminate karna exactly woh information destroy kar deta hai jo tum chahte the.

Verify: $r$ retain karne par (Ex 1) humne paaya $N=mg\sin \theta -mR{\dot{\theta }}^2$, ek real force; $r$ eliminate karne par milta hai $\lambda \cdot 0=0$, koi information nahi. Same physics, lekin sirf retained-coordinate route $N$ deta hai. Yeh exactly parent ke mistake list mein Error 1 hai. ✓

Example 8 — Exam twist: ek moving constraint (Cell I, time-dependent $f$)

Forecast: bead outward push aur spin karne ke saath, kya $\lambda$ positive hai (push out) ya negative (pull in)?

1. Lagrangian (no gravity). $L=T-V=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\theta }}^2)-0.$ Yeh step kyun? Sirf kinetic energy; wire constraint ke through jo bhi force chahiye woh supply karta hai.

2. Time-dependent constraint ke saath $r$-equation. $\partial f/\partial r=1$ ($t$-part spatial gradient ko affect nahi karta): $m\ddot{r}-mr{\dot{\theta }}^2=\lambda .$ Yeh step kyun? Cell I: even though $f$ mein $t$ explicitly hai, coordinates mein gradient wahi hai jo $\lambda$ ko multiply karta hai. Time sirf constraint equation $r=r_0+vt$ ke through enter karta hai.

3. Moving constraint lagao. $r=r_0+vt\Rightarrow \dot{r}=v$, $\ddot{r}=0$. $t=0$ par, $r=r_0=1$: $\lambda =m\cdot 0-m{r}_0{\omega }^2=-m{r}_0{\omega }^2=-0.5\cdot 1\cdot 9=-4.5\text{N}.$ Yeh step kyun? Steady outward push $v$ mein $\ddot{r}=0$ hai, to usse koi force nahi chahiye; jo kuch bachta hai woh sirf $-mr{\omega }^2$ term hai.

Verify: $\lambda =-4.5\text{N}$ negative hai → hamare sign rule se, $+r$ outward point karta hai to $\lambda <0$ ka matlab wire bead ko inward pull kar rahi hai (centripetal), kyunki ek rotating bead "chaahta hai" baahir udna aur mechanism iska resist karta hai. Units: $m{r}_0{\omega }^2={\text{kg⋅m⋅s}}^{-2}=\text{N}$ ✓. Forecast check: negative, inward pull.

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