Visual walkthrough — Applying E-L equations to various systems

2.1.6 · D2 · Physics › Analytical Mechanics › Applying E-L equations to various systems

Hum ek simple ball study kar rahe hain jo ek stiff, weightless stick se latki hai aur swing kar sakti hai. Bas itna hi. Mathematics shuru karne se pehle picture ko naam dete hain.

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Step 1 — System draw karo aur har letter ko naam do

KYA HAI. Humare paas ek ball hai (uski mass $m$ kehte hain) jo ek rigid rod ke end par hai. Rod ki ek fixed length hai; us length ko hum $\ell$ kehte hain (chhota script "L", padho "ell"). Rod ka upar wala end pin kiya hua hai taaki woh hil na sake; ball us pin ke around swing karti hai. Rod jo angle straight-down direction se banati hai use hum $\theta$ kehte hain (Greek letter "theta").

KYUN. Koi bhi equation likhne se pehle hume agree karna hoga ki har symbol picture mein kya point kar raha hai, warna baad ke formulas sirf shor hain. $m$ hai "kitna bhaari", $\ell$ hai "rod kitni lambi hai", $\theta$ hai "kitna swing kiya hai, straight down se angle measure karke".

PICTURE. Figure mein, kala dot pin hai, line $\ell$ length ki rod hai, coloured ball mass $m$ hai, aur upar ka chhota wedge angle $\theta$ hai. Dashed line "straight down" hai — woh jagah jahan $\theta =0$ hota hai.

Radians kyun aur degrees kyun nahi? Kyunki jab hum baat karenge ki angle kitni tezi se change ho raha hai, toh hum ise radius $\ell$ se multiply karenge taaki real speed mile — aur yeh "angle × radius = arc length" wali trick tabhi sahi hai jab angle radians mein ho. Degrees ke liye har baar ek ugly conversion factor chahiye hoga.

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Step 2 — Ball actually kitni tezi se move kar rahi hai?

KYA HAI. Ball $\ell$ radius ke circle par stuck hai (rod stretch nahi ho sakti). Hume uski speed chahiye — kitne metres per second cover karta hai. Hum $\dot{\theta }$ introduce karte hain (theta ke upar ek dot), jiska matlab hai "$\theta$ per second kitni tezi se change ho raha hai", yaani angular speed.

KYUN. Energy use karne ke liye hume ball ki linear speed $v$ chahiye. Ball circle ke saath move karti hai, toh uski speed woh rate hai jis par arc-length sweep hoti hai.

PICTURE. Laal arc dekho. Agar ball ek tiny extra angle $d\theta$ se move kare, toh woh arc ke saath tiny distance $\ell d\theta$ travel karti hai (radius times angle, phir se radians). Tiny time $dt$ se divide karo: speed hai $\ell \dot{\theta }$.

Dot notation ($\dot{\theta }$) sirf $\frac{d\theta }{dt}$ ka shorthand hai. Yahan derivative kyun? Kyunki "speed" by definition position ki rate of change hai, aur sirf $\theta$ change ho raha hai. Derivative woh exact tool hai jo answer deta hai "yeh quantity abhi kitni tezi se move kar rahi hai?"

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Step 3 — Moving energy $T$ likho

KYA HAI. Kinetic energy — "chalane ki energy" — kisi bhi mass ke liye $T=\frac{1}{2}m{v}^2$ hai. Hum usme woh speed plug karte hain jo hum abhi mile hain.

KYUN. Poora Lagrangian method (recipe in 2.1.05-Deriving-the-Euler-Lagrange-equation) energies par chal‍ta hai, forces par nahi. Kinetic energy do ingredients mein se ek hai. Hume ise sirf apne coordinate $\theta$ aur uski speed $\dot{\theta }$ use karke likhna hai, kuch nahi.

PICTURE. $v=\ell \dot{\theta }$ ko $\frac{1}{2}m{v}^2$ mein substitute karna: arrow dikhata hai $v$ ki jagah arc-speed aa rahi hai aur phir square ho rahi hai.

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Step 4 — Height energy $V$ likho

KYA HAI. Gravity ki potential energy $V=mgh$ hai, jahan $g$ gravity ki strength hai ($\approx 9.8$) aur $h$ kisi chosen "zero height" se upar ki height hai. Hum swing ke lowest point ko $h=0$ choose karte hain.

KYUN. Gravity ek conservative force hai — uska effect poora ek height-energy $V$ mein capture ho jaata hai, isliye hume kabhi gravity ka arrow equations mein draw nahi karna. Hume sirf $h$ $\theta$ ke function ke roop mein chahiye.

PICTURE. Rod straight down point karti hai $\ell$ length par. Jab yeh $\theta$ angle par swing kati hai, ball upar uthti hai. Pin se vertical drop $\ell \cos \theta$ hai (rod se bane right triangle ka adjacent side). Bottom par woh drop $\ell$ tha. Toh gained height hai $\ell -\ell \cos \theta =\ell (1-\cos \theta )$. Figure mein is height ko mark karta green bracket dekho.

Edge check — sab angles. $\theta =\pi$ par (ball straight up) $\cos \pi =-1$, toh $V=mg\ell (2)$ — maximum height $2\ell$, bilkul sahi. Negative $\theta$ ke liye (doosri taraf swing) $\cos (-\theta )=\cos \theta$, toh $V$ dono sides par same hai — pendulum symmetric hai, jaisa hona chahiye.

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Step 5 — Lagrangian $L$ banao aur uske do derivatives lo

KYA HAI. Lagrangian hai $L=T-V$ (moving energy minus height energy). Phir recipe ko do specific derivatives chahiye: ek speed $\dot{\theta }$ ke respect mein, ek angle $\theta$ ke respect mein.

KYUN. Euler–Lagrange machine (proven in 2.1.05-Deriving-the-Euler-Lagrange-equation) ko exactly yeh do partial derivatives chahiye. Partial derivative $\frac{\partial L}{\partial \theta }$ ka matlab hai "differentiate karo treating everything else — including $\dot{\theta }$ — as frozen constants". Hum doosre variable ko freeze karte hain taaki ek baar mein ek influence isolate ho sake.

PICTURE. Figure $L$ ko uske do pieces mein split karta hai aur dikhata hai ki har derivative mein kaun sa piece survive karta hai: ${\dot{\theta }}^2$ wala piece $\partial L/\partial \dot{\theta }$ ko feed karta hai; $\cos \theta$ wala piece $\partial L/\partial \theta$ ko feed karta hai.

$L=\underset{T}{\underbrace{\frac{1}{2}m{\ell }^2{\dot{\theta }}^2}}-\underset{V}{\underbrace{mg\ell (1-\cos \theta )}}$

Ab momentum-derivative ($\theta$ freeze karo, $\dot{\theta }$ vary karo): $\frac{\partial L}{\partial \dot{\theta }}=\frac{1}{2}m{\ell }^2\cdot 2\dot{\theta }=\underset{\text{"angular momentum"}}{\underbrace{m{\ell }^2\dot{\theta }}}$ Yahan $V$ part mein koi $\dot{\theta }$ nahi hai, toh woh zero contribute karta hai — sirf $T$ survive karta hai. ${\dot{\theta }}^2$ par power rule se 2 neeche aata hai, $\frac{1}{2}$ cancel ho jaata hai.

Ab angle-derivative ($\dot{\theta }$ freeze karo, $\theta$ vary karo): $\frac{\partial L}{\partial \theta }=0-mg\ell \cdot \frac{d}{d\theta }(1-\cos \theta )=-mg\ell \sin \theta$ $T$ part mein koi plain $\theta$ nahi hai, toh woh zero deta hai. $-\cos \theta$ differentiate karne par $+\sin \theta$ milta hai; $V=-\dots$ mein leading minus ise flip karta hai, $-mg\ell \sin \theta$ par land karta hai.

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Step 6 — Dono pieces ko Euler–Lagrange rule mein daalo

KYA HAI. Rule hai $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)-\frac{\partial L}{\partial \theta }=0.$ Hum Step 5 ke do derivatives damate hain.

KYUN. Yeh single equation Newton's second law ka disguise hai — lekin kabhi tension ya pin ki reaction force draw kiye bina. Woh constraint forces ne koi kaam nahi kiya, toh jab humne $\theta$ apna ek coordinate choose kiya, woh automatically gayab ho gaye.

PICTURE. Figure dikhata hai ki dono pieces apne slots mein slide ho rahe hain, aur outer $\frac{d}{dt}$ pehle slot par act kar raha hai.

Pehle momentum piece ka time derivative (ab $\dot{\theta }$ genuinely time ke saath change hota hai, toh yeh $\ddot{\theta }$, angular acceleration, ban jaata hai): $\frac{d}{dt}\left(m{\ell }^2\dot{\theta }\right)=m{\ell }^2\ddot{\theta }.$ Phir assemble karo: $m{\ell }^2\ddot{\theta }-(-mg\ell \sin \theta )=0\Rightarrow m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0.$ Common factor $m{\ell }^2$ se divide karo (dono nonzero hain): $\boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta }$ Term by term: $\ddot{\theta }$ hai swing kitni tezi se speed up ya slow down hoti hai; minus sign kehta hai pull hamesha bottom ki taraf wapas hai; $\frac{g}{\ell }$ set karta hai woh restoring pull kitni strong hai; $\sin \theta$ kehta hai pull badhti hai jitna zyada swing kiya ho.

Notice karo $m$ cancel ho gaya — ek bhaari aur ek halka bob identically swing karte hain. Gravity sab masses ko equally accelerate karti hai, aur Lagrangian dikhata hai ki yeh apne aap nikl jaata hai.

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Step 7 — Har regime mein sanity-check karo

KYA HAI. Jo formula tum check nahi kar sakte, uspar trust nahi kar sakte. Hum boxed result ko un limits mein test karte hain jahan hum already answer jaante hain.

KYUN. Edge cases cover karna hi fark hai "maine derive kiya" aur "mujhe believe hai" ke beech.

PICTURE. Teen mini-panels: tiny swing (spring jaisi lagti hai), hanging still (koi motion nahi), aur balanced upright (unstable).

• Chhoti swings ($\theta \approx 0$). Tiny angles ke liye $\sin \theta \approx \theta$, toh $\ddot{\theta }\approx -\frac{g}{\ell }\theta$. Yeh spring-jaisi wobble ki equation hai — simple harmonic motion — $\omega =\sqrt{g/\ell }$ rhythm ke saath. Yahi textbook pendulum period $T_{\text{period}}=2\pi \sqrt{\ell /g}$ hai. ✔
• Bottom par rest ($\theta =0$, $\dot{\theta }=0$). Tab $\sin 0=0$, toh $\ddot{\theta }=0$: woh wahan ruka rahega. Bottom par hanging pendulum equilibrium mein hai. ✔
• Straight up balanced ($\theta =\pi$). $\sin \pi =0$ toh phir $\ddot{\theta }=0$ — ek equilibrium — lekin ek bahut chhoti nudge se $\sin \theta$ wapas nahin, door push karega. Yeh unstable equilibrium hai, exactly jo real experience kehta hai. ✔
• Lamba rod (bada $\ell$). $\frac{g}{\ell }$ chhota ho jaata hai, toh restoring push kamzor hoti hai aur swing slow hoti hai — lambe pendulums slow tick karte hain. ✔

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Ek picture mein summary

Upar sab compress karke: picture $\theta ,\ell ,m$ define karti hai; arc speed $\ell \dot{\theta }$ deta hai; usse $T$ banta hai; geometry height $\ell (1-\cos \theta )$ deti hai; usse $V$ banta hai; $L=T-V$; do derivatives; E–L rule; boxed law — sab ek canvas par.

Recall Feynman retelling — poora walkthrough plain words mein

Ek ball imagine karo ek stiff stick par, swing kar rahi hai. Pehle hum names par agree karte hain: kitna bhaari ($m$), stick kitni lambi ($\ell$), kitna swing kiya ($\theta$). Kyunki ball ek circle par ride karti hai, uski speed sirf "angle kitni tezi se ghoomta hai" times "stick kitni lambi hai" hai — woh hai $\ell \dot{\theta }$. Speed se hume uski moving energy milti hai, $T$. Phir hum notice karte hain ki swing karne se ball upar uthti hai; lift hai $\ell (1-\cos \theta )$, aur gravity lift ko height energy mein badal deti hai, $V$. Nature ek running score rakhti hai moving-minus-height energy ka, jise $L$ kehte hain. Ek magic rule hai jo is score ko motion ka law banata hai: dekho score swing ki speed par kaisi react karta hai, dekho woh time mein kaise badalti hai, aur subtract karo score angle par kaisi react karta hai. Woh bookkeeping karne par — aur mass politely cancel ho jaata hai — ek clean sentence baqayi hai: swing bottom ki taraf wapas accelerate karti hai, aur jitna zyada jhukti hai utna strong pull hota hai, $\sin \theta$ follow karte hue. Check karo: tiny swings spring jaisi behave karti hain, bottom par resting wahan ruka rahega, upar balance karna ek knife-edge hai. Aur kabhi bhi humne stick ka tug mention nahi kiya — achhe coordinates ne woh force gayab kar di.

Recall Quick self-test

Bob ki speed $\theta$ ke terms mein ::: $v=\ell \dot{\theta }$ Kinetic energy $T$ ::: $\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ Bob ki lowest point se upar height ::: $\ell (1-\cos \theta )$ Potential energy $V$ ::: $mg\ell (1-\cos \theta )$ Final equation of motion ::: $\ddot{\theta }=-\frac{g}{\ell }\sin \theta$ Mass $m$ kyun gayab ho jaata hai? ::: woh divide out ho jaata hai — gravity sab masses ko equally accelerate karti hai

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Related: 2.1.05-Deriving-the-Euler-Lagrange-equation · 2.1.07-Cyclic-coordinates-and-conservation-laws · 2.1.08-Generalized-momenta-and-the-Hamiltonian · Constraints-and-degrees-of-freedom · Noethers-theorem