Worked examples — Derivation of Euler-Lagrange equations from D'Alembert's principle

2.1.5 · D3 · Physics › Analytical Mechanics › Derivation of Euler-Lagrange equations from D'Alembert's pri

Numbers ko haath lagane se pehle, ek reminder ki symbols ka matlab kya hai, taaki kuch bhi bina samjhe use na ho:

Recall Teen quantities jo kisi bhi problem mein read kar paana zaroori hai

• $T$ ::: kinetic energy, ${\sum }_k\frac{1}{2}{m}_k{\dot{\vec{r}}}_k^2$ — "system mein kitni motion energy hai." Generalized velocities ${\dot{q}}_i$ par depend karta hai (aur kabhi-kabhi $q_i,t$ par bhi).
• $V$ ::: potential energy — stored energy jo sirf position par depend karti hai, yaani sirf $q_i$ par.
• $Q_i$ ::: coordinate $q_i$ par generalized force — har applied (non-constraint) force, us direction mein project ki gayi jisme $q_i$ badalne par system move karta hai. Jab forces kisi potential se aate hain, $Q_i=-\partial V/\partial q_i$.

Yahan $q_i$ koi bhi generalized coordinate hai (dekho Constraints and generalized coordinates): ek number jo configuration pin karta hai aur automatically constraints satisfy karta hai (ek angle, ek distance, ek arc-length). Dot ka matlab hai $d/dt$ — time mein change ki rate.

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The scenario matrix

Cell	Kya stress-test karta hai	Example
A — trivial / sanity	$V=0$, straight line, Newton recover karo	Ex 1
B — restoring force ka sign	$\partial V/\partial q$ equilibrium ke across sign badalta hai	Ex 2 (spring)
C — $\partial T/\partial q\ne 0$	kinetic energy khud position par depend karti hai	Ex 3 (pendulum, full)
D — coupled coordinates ($n>1$)	do brackets, cross terms	Ex 4 (2D polar orbit)
E — velocity-dependent / time-dependent $T$	driven / rotating constraint, "fictitious" terms	Ex 5 (bead on rotating wire, general $\omega$)
F — limiting / degenerate input	$\ell \to \infty$, $m\to 0$, small-angle limit	Ex 6 (small-angle pendulum)
G — real-world word problem	prose → coordinates → EL translate karo	Ex 7 (block sliding on a wedge, both free)
H — exam twist	non-obvious coordinate, ek ignorable (cyclic) coordinate	Ex 8 (Atwood-on-a-cone / conserved momentum)

Neeche diye har worked example par uska cell(s) tag kiya gaya hai.

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Ex 1 — Cell A · sanity check

Forecast: answer padhne se pehle guess karo — koi force nahi ho to particle kya karta hai? Likh lo.

1. $T,V$ likho. $T=\frac{1}{2}m{\dot{x}}^2$, $V=0$, isliye $L=\frac{1}{2}m{\dot{x}}^2$. Yeh step kyun? Har EL problem energies naam karne se shuru hoti hai; jab tak woh exist na ho kuch bhi compute nahi ho sakta.
2. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Yeh step kyun? EL ko $\partial L/\partial \dot{q}$ chahiye — "generalized momentum" — apne pehle ingredient ke roop mein.
3. Time-derivative. $\frac{d}{dt}(m\dot{x})=m\ddot{x}$.
4. Position slot. $\frac{\partial L}{\partial x}=0$ (kuch bhi $x$ par depend nahi karta). Yeh step kyun? $L$ mein $x$ hai hi nahi, isliye $x$ cyclic hai — conservation ka ek preview (Ex 8).
5. EL assemble karo: $m\ddot{x}-0=0\Rightarrow \ddot{x}=0$.

Verify: $\ddot{x}=0$ ka matlab constant velocity — Newton's first law. Units: $[m\ddot{x}]=\text{kg}\cdot {\text{m/s}}^2=\text{N}$, ek force, $0$ ko balance karta hai. ✓

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Ex 2 — Cell B · restoring force ka sign

Forecast: $x>0$ hone par force kis direction mein hai? $x<0$ par? Sign pattern guess karo.

1. Energies. $T=\frac{1}{2}m{\dot{x}}^2$, $V=\frac{1}{2}k{x}^2$. Isliye $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2$. Yeh step kyun? Spring potential "position energy" ka archetype hai — dekho Generalized forces and potentials.
2. Momentum slot & derivative. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}=m\ddot{x}$.
3. Position slot. $\frac{\partial L}{\partial x}=-kx$. Yeh step kyun? Yeh $-\partial V/\partial x=Q_x$ hai, generalized force. Iska sign hi poori baat hai: $x>0$ par yeh negative hai (wapas $0$ ki taraf kheenchta hai); $x<0$ par yeh positive hai (wapas $0$ ki taraf dhakelta hai). Dono signs mein restoring.
4. EL: $m\ddot{x}-(-kx)=0\Rightarrow m\ddot{x}+kx=0$.
5. Frequency padhna. $\ddot{x}=-\frac{k}{m}x$, isliye $\omega =\sqrt{k/m}=\sqrt{4/1}=2\text{rad/s}$.

Verify: guessed solution $x=\cos (\omega t)$ plug karo: $\ddot{x}=-{\omega }^2\cos \omega t=-4x$, aur waqai $-k/m=-4$. ✓ $\omega$ ki units: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{1/{\text{s}}^2}={\text{s}}^{-1}$. ✓

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Ex 3 — Cell C · jab $\partial T/\partial q\ne 0$

Pendulum ka danger yeh hai ki mapping $x=\ell \sin \theta$ se velocity $\theta$ par depend karti hai, isliye $T$ khud $\theta$ carry karta hai. Figure dekho: bob ek circle trace karta hai, aur velocity vector hamesha tangent hota hai.

Forecast: gravity ka torque $\propto \theta$ hoga ya $\propto \sin \theta$? Guess karo.

1. Position → velocity. $x=\ell \sin \theta ,y=-\ell \cos \theta$, isliye $\dot{x}=\ell \cos \theta \dot{\theta },\dot{y}=\ell \sin \theta \dot{\theta }$, aur ${\dot{x}}^2+{\dot{y}}^2={\ell }^2{\dot{\theta }}^2$. Yeh step kyun? Humein $T$ ko chosen coordinate mein express karna hoga; ${\cos }^2+{\sin }^2=1$ ka collapse hi reason hai ki $\theta$ natural choice hai.
2. Energies. $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$, $V=mgy=-mg\ell \cos \theta$. $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta$.
3. Momentum slot. $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$.
4. Position slot. $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$. Yeh step kyun? Yeh "$\partial T/\partial q\ne 0$"-style term hai (yahan yeh $V$ mein hai), restoring torque — aur yeh $\sin \theta$ hai, naki $\theta$.
5. EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$.

Verify: rod mein unknown tension kabhi appear hi nahi hui — yahi D'Alembert ka payoff hai. Dimensionally $g/\ell$ ki units ${\text{s}}^{-2}$ hain, $\ddot{\theta }$ se match karta hai. ✓

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Ex 4 — Cell D · coupled coordinates ($n=2$)

Forecast: do coordinates → do EL equations. Inme se kis mein "${\dot{\varphi }}^2$" term kahin se aa jayega?

1. Polar mein velocity. ${\dot{\vec{r}}}^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$ (radial part + tangential part). Yeh step kyun? Dekho Kinetic energy in generalized coordinates — $r^2{\dot{\varphi }}^2$ wala term hi do coordinates ko couple karta hai.
2. Energies. $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)$, $V=0$.
3. $r$-equation. $\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r}$; $\frac{\partial L}{\partial r}=mr{\dot{\varphi }}^2$. EL: $m\ddot{r}-mr{\dot{\varphi }}^2=0\Rightarrow \ddot{r}=r{\dot{\varphi }}^2$. Yeh step kyun? $r{\dot{\varphi }}^2$ $\partial T/\partial r$ se aaya — centripetal term, geometry se born, kisi real force se nahi.
4. $\varphi$-equation. $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$; $\frac{\partial L}{\partial \varphi }=0$. EL: $\frac{d}{dt}(m{r}^2\dot{\varphi })=0$. Yeh step kyun? $\varphi$ cyclic hai ($L$ mein absent hai), isliye uska momentum conserved hai.
5. Conserved quantity. $m{r}^2\dot{\varphi }=\text{const}$ — angular momentum. (Yeh Noether's theorem miniature mein hai: rotational symmetry → conserved angular momentum.)

Verify: ek free particle straight line mein move karta hai; polar coordinates mein ek straight line ke liye waqai $m{r}^2\dot{\varphi }=$ const hota hai (equal-areas / Kepler's 2nd law) aur $\ddot{r}=r{\dot{\varphi }}^2$. Special case check karo $\dot{\varphi }=0$: tab $\ddot{r}=0$, pure radial straight-line motion. ✓

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Ex 5 — Cell E · $T$ mein time/velocity structure (rotating constraint)

Forecast: koi applied force nahi hone par, kya bead still rahega, ya bahar fly karega? Guess karo.

1. Constraint $\varphi$ ko khatam karta hai. Wire force karta hai $\varphi =\omega t$, isliye $\dot{\varphi }=\omega$ ek coordinate nahi hai — sirf $r$ free hai. Yeh step kyun? Ek moving (time-dependent) constraint freedom reduce karta hai; $\omega$, $T$ mein ek fixed number ki tarah enter karta hai.
2. Energies. $T=\frac{1}{2}m({\dot{r}}^2+r^2{\omega }^2)$, $V=0$.
3. Momentum slot. $\frac{\partial T}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r}$.
4. Position slot. $\frac{\partial T}{\partial r}=mr{\omega }^2$. Yeh step kyun? Yahi woh term hai jo parent note ne flag kiya tha: kinetic energy position par depend karti hai, isliye $\partial T/\partial r\ne 0$ ek real dynamical push produce karta hai.
5. EL: $m\ddot{r}-mr{\omega }^2=0\Rightarrow \ddot{r}={\omega }^{2}r$.

Verify: solution $r=r_0\cosh (\omega t)$ (rest se $r_0$ par release ke liye) deta hai $\ddot{r}={\omega }^2{r}_0\cosh (\omega t)={\omega }^{2}r$. ✓ Bead bahar accelerate karta hai — "centrifugal" effect, koi constraint force dikhaye bina. Degenerate check: $\omega =0$ ⇒ $\ddot{r}=0$, fixed wire par ek free bead. ✓

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Ex 6 — Cell F · limiting / degenerate input

Forecast: bottom ke paas, kya pendulum spring jaisa lagta hai (Ex 2)? Period guess karo.

1. Small-angle expansion. $|\theta |\ll 1$ ke liye, $\sin \theta \approx \theta$ (series ka leading term). Yeh step kyun? Hum EL result ko input space ke ek degenerate corner mein test karte hain, jahan geometry linearize ho jaati hai.
2. Reduced equation. $\ddot{\theta }=-\frac{g}{\ell }\theta$ — Ex 2 ke $\ddot{x}=-\frac{k}{m}x$ se identical form.
3. Frequency & period. $\omega =\sqrt{g/\ell }$, isliye $T_{\text{period}}=2\pi \sqrt{\ell /g}$. Numeric: $\ell =1\text{m},g=9.8{\text{m/s}}^2\Rightarrow T_{\text{period}}=2\pi \sqrt{1/9.8}\approx 2.007\text{s}$.
4. $\ell \to \infty$ limit. $g/\ell \to 0\Rightarrow \ddot{\theta }\to 0$: infinitely long pendulum mein koi restoring pull nahi hota — locally arc flat lagti hai, motion uniform ho jaati hai. Ex 1 (free motion) se consistent. ✓

Verify: period formula 1 m pendulum ke liye $\approx 2.007\text{s}$ deta hai (classic "seconds pendulum" $\ell \approx 0.994$ m hai). Units: $\sqrt{\text{m}/({\text{m/s}}^2)}=\text{s}$. ✓

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Ex 7 — Cell G · real-world word problem (block on a free wedge)

Forecast: jab block down-right slide karta hai, wedge kis taraf recoil karta hai? Momentum kehta hai…?

1. Coordinates → positions. Wedge horizontal position $X$. Block position: $x_m=X+s\cos \alpha ,y_m=-s\sin \alpha$. Yeh step kyun? Do moving objects, do free coordinates $(X,s)$; incline constraint in formulas mein baked hai isliye normal force gayab ho jaati hai.
2. Velocities. ${\dot{x}}_m=\dot{X}+\dot{s}\cos \alpha ,{\dot{y}}_m=-\dot{s}\sin \alpha$.
3. Energies. $T=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m[(\dot{X}+\dot{s}\cos \alpha {)}^2+{\dot{s}}^2{\sin }^2\alpha ]$, aur $V=mg{y}_m=-mgs\sin \alpha$.
4. $X$-equation (cyclic — $X$ $L$ mein absent hai): $\frac{\partial L}{\partial \dot{X}}=M\dot{X}+m(\dot{X}+\dot{s}\cos \alpha )=\text{const}$. Yeh step kyun? Pure system ka horizontal momentum conserved hai (floor frictionless hai). Differentiate karo: $(M+m)\ddot{X}+m\ddot{s}\cos \alpha =0$.
5. $s$-equation. $\frac{\partial L}{\partial \dot{s}}=m(\dot{X}+\dot{s}\cos \alpha )\cos \alpha +m\dot{s}{\sin }^2\alpha$, aur $\frac{\partial L}{\partial s}=mg\sin \alpha$. EL: $m\ddot{X}\cos \alpha +m\ddot{s}=mg\sin \alpha$.
6. Pair solve karo. (4) se: $\ddot{X}=-\frac{m\cos \alpha }{M+m}\ddot{s}$. Substitute karo: $\ddot{s}=\frac{g\sin \alpha }{1-\frac{m{\cos }^2\alpha }{M+m}}=\frac{(M+m)g\sin \alpha }{M+m{\sin }^2\alpha }.$ Aur $\ddot{X}=-\frac{m\cos \alpha g\sin \alpha }{M+m{\sin }^2\alpha }$ (negative: wedge block ke horizontal motion ke opposite recoil karta hai — momentum forecast se match karta hai).

Verify — do limits.

• $M\to \infty$ (immovable wedge): $\ddot{s}\to g\sin \alpha$, textbook block-on-fixed-incline result, aur $\ddot{X}\to 0$. ✓
• $\alpha \to 90^{\circ }$: $\ddot{s}\to g$ (free fall) aur $\ddot{X}\to 0$. ✓ Numeric check $M=3,m=1,\alpha =30^{\circ },g=9.8$ ke saath: $\ddot{s}=\frac{4\cdot 9.8\cdot 0.5}{3+1\cdot 0.25}=\frac{19.6}{3.25}\approx 6.031{\text{m/s}}^2$; $\ddot{X}=-\frac{1\cdot (\sqrt{3}/2)\cdot 9.8\cdot 0.5}{3.25}\approx -1.305{\text{m/s}}^2$. (VERIFY dekho.)

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Ex 8 — Cell H · exam twist (cyclic coordinate → conservation)

Forecast: $r,\varphi$ mein se kaun sa cyclic hai? Uski conservation se kya milta hai?

1. Energies. $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)$, $V=V(r)$. $L=T-V$.
2. $\varphi$ cyclic hai. $\frac{\partial L}{\partial \varphi }=0$ kyunki na $T$ ki $\varphi$-independence, na $V$, $\varphi$ ka zikr karta hai. Yeh step kyun? Ek missing coordinate ek conserved momentum signal karta hai — yeh variational shadow hai Noether ka.
3. Conserved momentum. $p_{\varphi }=\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }={\ell }_0$ (constant angular momentum).
4. $r$-equation. $\frac{d}{dt}(m\dot{r})-(mr{\dot{\varphi }}^2-V^{\prime }(r))=0\Rightarrow m\ddot{r}=mr{\dot{\varphi }}^2-V^{\prime }(r)$.
5. $\dot{\varphi }$ eliminate karo. $\dot{\varphi }={\ell }_0/(m{r}^2)$ use karo: $m\ddot{r}=\frac{{\ell }_0^2}{m{r}^3}-V^{\prime }(r).$ Yeh step kyun? Ek conserved quantity ek 2-D problem ko effective potential $V_{\text{eff}}=V(r)+\frac{{\ell }_0^2}{2m{r}^2}$ wale 1-D mein badal deta hai.

Verify — Ex 4 recover karo. $V=0$ set karo: $m\ddot{r}={\ell }_0^2/(m{r}^3)$. ${\ell }_0=m{r}^2\dot{\varphi }$ ke saath, right side hai $m{r}^2{\dot{\varphi }}^2/r=mr{\dot{\varphi }}^2$, yaani $\ddot{r}=r{\dot{\varphi }}^2$ — exactly Ex 4. ✓ Numeric spot-check ($m=1,r=2,\dot{\varphi }=3$): ${\ell }_0=1\cdot 4\cdot 3=12$, aur ${\ell }_0^2/(m{r}^3)=144/8=18=r{\dot{\varphi }}^2=2\cdot 9$. ✓

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Recall

Recall Kaun se cell ko kaun sa move chahiye?

Cyclic coordinate ($L$ mein absent) ::: uska generalized momentum $\partial L/\partial \dot{q}$ conserved hota hai (Ex 4, 8). $\partial T/\partial q\ne 0$ ::: kinetic energy position par depend karti hai → "fictitious" (centrifugal/centripetal) terms aate hain (Ex 3, 4, 5). Restoring force sign check ::: $-\partial V/\partial q$ ko $q$ ke dono signs ke liye equilibrium ki taraf point karna chahiye (Ex 2). Do free bodies ::: do coordinates → do EL equations, simultaneously solve karo (Ex 7). Immovable / infinite limit ::: kisi mass ya length ko $\infty$ bhejo aur textbook special case confirm karo (Ex 6, 7).