Lagrangian L = T − V: Worked examples — Hinglish deep dive

2.1.4 · D3 · Physics › Analytical Mechanics › Lagrangian L = T − V

Yeh page ek drill hai. Parent note ne tumhe machinery dikhaayi; yahan hum use har corner mein push karte hain — potential ke signs, curved coordinates, coupled coordinates, woh coordinates jo $L$ mein dikhte bhi nahi, degenerate free particles, aur ek exam twist. Shuru karne se pehle, ek reminder ki har symbol ka kya matlab hai, kyunki hum kuch bhi trust pe earn nahi karenge.

Definition Woh vocabulary jo hum har line pe use karenge

$q$ — ek generalized coordinate: koi bhi single number jo yeh pin down karne mein madad karta hai ki system kahan hai (ek length $x$ , ek height $y$ , ek angle $θ$ ). Dekho Generalized Coordinates and Constraints.
$\overset{q}{˙}$ — generalized velocity: woh number per second kitni tezi se change hota hai. Dot ka matlab hai "uske neeche wali cheez ka time-derivative".
$\overset{q}{¨}$ — us coordinate ka acceleration (do dots = do time-derivatives).
$T$ — kinetic energy, motion ki energy, hamesha $\frac{1}{2} m v^{2}$ hoti hai jahan $v$ real speed hai.
$V$ — potential energy, position ki energy (gravity mein height, spring mein stretch).
$L = T - V$ — Lagrangian. Total energy nahi hai yeh.
Euler–Lagrange equation (engine, har coordinate ke liye ek): $\frac{d}{d t} \frac{\partial L}{\partial q ˙} - \frac{\partial L}{\partial q} = 0$ Ise padho: "( $L$ ki velocity mein slope) ke change ki rate, ( $L$ ki position mein slope) ke barabar hoti hai." Poori derivation Euler–Lagrange Equation mein hai.

Scenario matrix

$L = T - V$ ka har problem inhi cells mein se kisi ek mein aata hai. Neeche ke examples is tarah choose kiye gaye hain ki milke woh sab cover kar lein — koi bhi "tum khud figure out kar loge" wala nahi choraa gaya.

Cell	Kya cheez isko alag banati hai	Covered by
A. Straight-line, positive-slope $V$	$V$ $q$ ke saath badhta hai (chadhai) → restoring force neeche/peeche point karti hai	Ex 1
B. Straight-line, negative-slope $V$	$V$ $q$ ke saath ghatta hai (downhill push) → force aage point karti hai	Ex 2
C. Curved coordinate ( $T$ mein $q$ hai)	$T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ — velocity term khud hi position pe depend karta hai	Ex 3
D. Two coupled coordinates	Do Euler–Lagrange equations jo ek doosre se baat karte hain	Ex 4
E. Cyclic (ignorable) coordinate	$q$ $L$ mein missing hai → ek conserved momentum milta hai	Ex 5
F. Degenerate / zero potential	$V = 0$ , free particle — limiting case	Ex 6
G. Sign-of-force stress test	Same spring, equilibrium ke dono taraf (positive AND negative $x$ )	Ex 7
H. Real-world word problem + exam twist	Rotating hoop pe bead: ek constraint jo $T$ mein add hoti hai	Ex 8

Worked examples

Cell A — potential jo coordinate ke saath badhta hai

Worked example Ex 1 · Block frictionless ramp par chadh raha hai, incline

$α$ Mass $m$ ka ek block ek ramp par slide karta hai jo horizontal se $α$ angle par tilted hai. Maano $s$ = distance slope ke upar measure kiya gaya hai. Iska equation of motion nikalo.

Forecast: $\overset{s}{¨}$ ka sign padhne se pehle guess karo. Gravity ise wapas neeche kheenchni chahiye, toh hum expect karte hain $\overset{s}{¨} < 0$ (negative, matlab down-slope, acceleration). Yeh socha raho.

Figure — the incline triangle.

Ek blue slope jiska length $s$ hai (hypotenuse) ek white horizontal base par uthta hai; laal dashed vertical leg height gained hai, $s sin α$ ; green arc $α$ mark karta hai aur green down-arrow gravity $m g$ hai. Picture hi step 1 ka poora point hai — triangle se height padho.

Distance $s$ chadh ke height gained $s sin α$ hoti hai (figure mein laal vertical leg dekho — yeh right triangle ki opposite side hai jiska hypotenuse $s$ hai). Yeh step kyun? Potential energy sirf height ki parwah karti hai, aur $sin α$ exactly "vertical rise per unit slope-distance" hai.
Energies likho: $T = \frac{1}{2} m \overset{s}{˙}^{2}$ , $V = m g s sin α$ . Yeh step kyun? Parent recipe ke steps 2–3: chosen coordinate mein energies.
$L = \frac{1}{2} m \overset{s}{˙}^{2} - m g s sin α$ banao. Yeh step kyun? Lagrangian define hota hai $T - V$ se; engine ke chew karne se pehle hume ise assemble karna hoga.
Velocity slope: $\frac{\partial L}{\partial s ˙} = m \overset{s}{˙}$ , phir iska time-derivative lo: $\frac{d}{d t} (\frac{\partial L}{\partial s ˙}) = \frac{d}{d t} (m \overset{s}{˙}) = m \overset{s}{¨}$ . Yeh step kyun? Yeh engine ka " $\frac{d}{d t} \frac{\partial L}{\partial s ˙}$ " wala aadha hissa hai — pehle velocity-slope, phir time mein uska change.
Position slope: $\frac{\partial L}{\partial s} = - m g sin α$ . Yeh step kyun? Yeh engine ka doosra aadha hissa hai; sirf $V$ mein $s$ hai, aur iska slope $s$ par push karne wali force ka (negative) hai.
Euler–Lagrange: dono halves ko $\frac{d}{d t} \frac{\partial L}{\partial s ˙} - \frac{\partial L}{\partial s} = 0$ mein daalo, milega $m \overset{s}{¨} - (- m g sin α) = 0 \Rightarrow \overset{s}{¨} = - g sin α .$ Yeh step kyun? Yeh engine khud hai — rate-of-velocity-slope ko position-slope ke barabar set karna exactly equation of motion hai.

Verify: Sign negative hai → forecast se match karta hai (wapas neeche slide karta hai). Limiting checks: $α = 9 0^{\circ}$ par (vertical wall of a ramp), $\overset{s}{¨} = - g$ — pure free fall. ✓ $α = 0$ par (flat table), $\overset{s}{¨} = 0$ — koi push nahi, baith jaata hai. ✓ Units: $g sin α$ $m/s^{2}$ mein hai. ✓

Cell B — potential jo coordinate ke saath girta hai

Worked example Ex 2 · Same ramp, coordinate

neeche slope measure kiya gaya Ab maano $d$ = distance slope ke neeche measure kiya gaya hai (positive matlab aur neeche). Yeh Cell B hai: jaise $d$ badhta hai, height girti hai, toh $V$ ki negative slope hai.

Forecast: block neeche jaate hue speed up karti hai, toh hum expect karte hain $\ddot{d} > 0$ (positive).

Starting point se upar height ab $- d sin α$ hai (neeche jaana = negative height). Yeh step kyun? Sign ke baare mein honest rehna hoga: neeche matlab lower potential.
$T = \frac{1}{2} m \dot{d}^{2}$ , $V = - m g d sin α$ , toh $L = \frac{1}{2} m \dot{d}^{2} + m g d sin α$ . Yeh step kyun? Negative $V$ ke saath $L = T - V$ ek plus ban jaata hai; yeh sign flip hi Ex 1 se poora fark hai.
Velocity half: $\frac{\partial L}{\partial d ˙} = m \dot{d}$ , phir $\frac{d}{d t} (\frac{\partial L}{\partial d ˙}) = \frac{d}{d t} (m \dot{d}) = m \ddot{d}$ . Yeh step kyun? Hamesha wahi engine — velocity-slope, phir iska time-derivative.
Position half: $\frac{\partial L}{\partial d} = + m g sin α$ . Yeh step kyun? $+ V$ contribution position-slope ko positive banata hai — forward push ka source.
EL: $m \ddot{d} - m g sin α = 0 \Rightarrow \ddot{d} = + g sin α .$ Yeh step kyun? Dono halves ko $\frac{d}{d t} \frac{\partial L}{\partial d ˙} - \frac{\partial L}{\partial d} = 0$ mein daalne se equation of motion milti hai.

Verify: Positive, forecast ke anusaar. Aur notice karo yeh Ex 1 jaisi exact same physics hai — coordinate ki direction flip karne se $V$ ke slope ka sign flip hua, jisne acceleration ka sign flip kiya. Duniya nahi badlti; sirf hamaari bookkeeping badlti hai. Yahi reason hai ki Lagrangian coordinate-agnostic hota hai. Units: $g sin α$ $m/s^{2}$ mein hai, ek acceleration. ✓

Cell C — coordinate kinetic term ke andar rehta hai

Worked example Ex 3 · Pendulum,

$ℓ$ -dependence ke liye re-examined Mass $m$ ka bob ek rod par jo $ℓ$ length ka hai, downward vertical se angle $θ$ par. Yeh Cell C hai kyunki bob ki speed $ℓ \dot{θ}$ hai — length velocity ko multiply karti hai, toh $T$ mein $ℓ^{2}$ aata hai.

Forecast: chhote swings spring jaisi lagni chahiye ( $\ddot{θ} \propto - θ$ ); bade swings mein restoring pull slow honi chahiye (kyunki $sin θ$ bend ho jaata hai).

Figure — pendulum geometry.

White pivot se yellow bob tak ek blue rod; green arc angle $θ$ hai dotted vertical se; green tangent arrow bob ki speed $ℓ \dot{θ}$ hai (rod ke perpendicular); red dashed segment lowest point se upar uthi height hai, $ℓ (1 - cos θ)$ . Steps 1 aur 2 literally is picture se padhe jaate hain.

Bob ki speed $= ℓ \dot{θ}$ (arc-length rate: radius times angular rate — green tangent arrow dekho). Yeh step kyun? $T$ ke liye real speed chahiye, sirf $\dot{θ}$ nahi.
Lowest point se upar height $= ℓ (1 - cos θ)$ (figure mein laal gap). Yeh step kyun? $V$ ko actual height chahiye; $ℓ (1 - cos θ)$ yeh hai ki bob kitna utha hai.
$T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}$ , $V = m g ℓ (1 - cos θ)$ , $L = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} - m g ℓ (1 - cos θ)$ . Yeh step kyun? $L = T - V$ assemble karo; note karo $θ$ ab $T$ aur $V$ dono mein aata hai — Cell C ka feature.
Velocity half: $\frac{\partial L}{\partial θ ˙} = m ℓ^{2} \dot{θ}$ , phir $\frac{d}{d t} (\frac{\partial L}{\partial θ ˙}) = m ℓ^{2} \ddot{θ}$ . Yeh step kyun? Velocity-slope phir iska time-derivative — engine ka pehla half, ab $ℓ^{2}$ carry karte hue.
Position half: $\frac{\partial L}{\partial θ} = - m g ℓ sin θ$ ( $- m g ℓ (1 - cos θ)$ ka derivative; $sin$ is liye aata hai kyunki $\frac{d}{d θ} (- cos θ) = sin θ$ ; yahan $T$ mein koi $θ$ nahi toh sirf $V$ contribute karta hai). Yeh step kyun? Engine ka doosra half — restoring "torque per unit angle".
EL: $m ℓ^{2} \ddot{θ} + m g ℓ sin θ = 0 \Rightarrow \ddot{θ} = - \frac{g}{ℓ} sin θ .$ Yeh step kyun? Dono halves ko $\frac{d}{d t} \frac{\partial L}{\partial θ ˙} - \frac{\partial L}{\partial θ} = 0$ mein combine karo equation of motion paane ke liye.

Verify: Small-angle limit $sin θ \approx θ$ se $\ddot{θ} = - \frac{g}{ℓ} θ$ milta hai, simple harmonic with $ω = g / ℓ$ . ✓ $θ = 0$ par (seedha neeche hanging): $\ddot{θ} = 0$ , ek equilibrium. ✓ $θ = π$ par (seedha upar balanced): $sin π = 0$ toh $\ddot{θ} = 0$ bhi — ek unstable equilibrium, physically correct. ✓ Units: $g / ℓ$ $s^{- 2}$ mein hai, dimensionless $sin θ$ se multiply ho ke → $rad/s^{2}$ . ✓

Cell D — do coordinates jo ek doosre se baat karte hain

Worked example Ex 4 · Frictionless line par spring se jude do masses

Masses $m$ each rail par positions $x_{1}$ aur $x_{2}$ par hain, spring constant $k$ aur natural length $ℓ_{0}$ wali spring se jude hain. Do coordinates → do Euler–Lagrange equations.

Forecast: spring sirf stretch $x_{2} - x_{1}$ ki parwah karti hai, toh dono equations mirror images honi chahiye jo masses ko ek doosre ki taraf ya door push karti hain.

$T = \frac{1}{2} m \overset{x}{˙}_{1}^{2} + \frac{1}{2} m \overset{x}{˙}_{2}^{2}$ (har mass apni kinetic energy carry karta hai). Yeh step kyun? Kinetic energy independent particles mein additive hoti hai.
Stretch $= (x_{2} - x_{1}) - ℓ_{0}$ , toh $V = \frac{1}{2} k [(x_{2} - x_{1}) - ℓ_{0}]^{2}$ . Yeh step kyun? Ek spring $\frac{1}{2} k (stretch)^{2}$ store karti hai, aur stretch relative separation minus natural length hota hai.
$L = T - V$ . Yeh step kyun? Lagrangian assemble karo; kyunki $V$ mein $x_{1}$ aur $x_{2}$ dono milte hain, do coordinates couple ho jaayenge.
$x_{1}$ ke liye: velocity half $\frac{\partial L}{\partial x ˙ _{1}} = m \overset{x}{˙}_{1}$ , phir $\frac{d}{d t} (\frac{\partial L}{\partial x ˙ _{1}}) = m \overset{x}{¨}_{1}$ ; position half $\frac{\partial L}{\partial x _{1}} = + k [(x_{2} - x_{1}) - ℓ_{0}]$ (chain rule; $- x_{1}$ ka inner derivative $+$ deta hai). Yeh step kyun? Yaad rakho $q$ aur $\overset{q}{˙}$ independent hain — jab hum $x_{1}$ ke respect mein differentiate karte hain toh $\overset{x}{˙}_{1}$ frozen rehta hai, toh sirf $V$ position half mein contribute karta hai.
$x_{1}$ ke liye EL: $m \overset{x}{¨}_{1} = k [(x_{2} - x_{1}) - ℓ_{0}] .$ Yeh step kyun? Engine $\frac{d}{d t} \frac{\partial L}{\partial x ˙ _{1}} - \frac{\partial L}{\partial x _{1}} = 0$ mass-1 ki equation deta hai.
$x_{2}$ ke liye: identical steps velocity half $m \overset{x}{¨}_{2}$ aur position half $\frac{\partial L}{\partial x _{2}} = - k [(x_{2} - x_{1}) - ℓ_{0}]$ dete hain, toh EL: $m \overset{x}{¨}_{2} = - k [(x_{2} - x_{1}) - ℓ_{0}] .$ Yeh step kyun? Har coordinate ko apna khud ka Euler–Lagrange equation milta hai; yahan sign flip hota hai kyunki $x_{2}$ stretch mein $+$ ke saath enter hota hai.

Verify: Dono equations add karo: $m \overset{x}{¨}_{1} + m \overset{x}{¨}_{2} = 0$ , toh center of mass accelerate nahi karta — momentum conserved, isolated pair ke liye correct. ✓ Subtract karo: $r = x_{2} - x_{1}$ ke saath, $m \overset{r}{¨} = - 2 k (r - ℓ_{0})$ , natural length ke baare mein frequency $ω = 2 k / m$ wala clean oscillator. ✓ Mirror symmetry confirmed. Units: $k [(x_{2} - x_{1}) - ℓ_{0}]$ $(N/m) (m) = N$ hai, aur $m \overset{x}{¨}$ $kg \cdot m/s^{2} = N$ hai — dono sides forces hain. ✓

Cell E — ek coordinate jo $L$ se gayab ho jaata hai

Worked example Ex 5 · Free particle plane par, polar coordinates mein

Ek puck horizontal plane par freely slide karta hai (koi potential nahi). Polar coordinates $(r, ϕ)$ mein, $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2})$ , $V = 0$ . Notice karo: $ϕ$ $L$ mein appear hi nahi karta, sirf $\dot{ϕ}$ karta hai. Aisa coordinate cyclic ya ignorable kehlaata hai.

Forecast: agar $ϕ$ missing hai, toh iska Euler–Lagrange equation hume free mein ek conserved quantity dena chahiye — angular momentum.

$L = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{ϕ}^{2})$ . Yeh step kyun? $V = 0$ ke saath $L = T - V$ assemble karo; note karo $ϕ$ khud absent hai — Cell E ka feature.
$ϕ$ ke liye (velocity half): generalized momentum hai $p_{ϕ} = \frac{\partial L}{\partial ϕ ˙} = m r^{2} \dot{ϕ}$ . (Dekho Generalized Momentum and Conservation.) Yeh step kyun? Yeh $\frac{\partial L}{\partial ϕ ˙}$ hai, woh quantity jiska time-derivative engine track karta hai.
$ϕ$ ke liye (position half): $\frac{\partial L}{\partial ϕ} = 0$ kyunki $ϕ$ literally kabhi appear hi nahi karta. Yeh step kyun? Missing coordinate ka position-slope zero hota hai — yahi ise cyclic banata hai.
$ϕ$ ke liye EL: $\frac{d}{d t} (\frac{\partial L}{\partial ϕ ˙}) - \frac{\partial L}{\partial ϕ} = \frac{d}{d t} (m r^{2} \dot{ϕ}) - 0 = 0$ , isliye $m r^{2} \dot{ϕ} = constant$ . Yeh step kyun? Jab position-slope zero ho, toh engine kehta hai velocity-slope (momentum) conserved hai.
$r$ ke liye (dono halves): $\frac{\partial L}{\partial r ˙} = m \overset{r}{˙} \Rightarrow \frac{d}{d t} (m \overset{r}{˙}) = m \overset{r}{¨}$ ; aur $\frac{\partial L}{\partial r} = m r \dot{ϕ}^{2}$ . EL: $m \overset{r}{¨} - m r \dot{ϕ}^{2} = 0 \Rightarrow m \overset{r}{¨} = m r \dot{ϕ}^{2} .$ Yeh step kyun? Doosra coordinate abhi bhi apna full Euler–Lagrange equation leta hai — $r$ -equation centrifugal term $m r \dot{ϕ}^{2}$ carry karta hai.

Verify: $m r^{2} \dot{ϕ}$ exactly ek particle ka angular momentum $L_{z}$ hai — conservation ek missing variable se drop out ho jaata hai, koi torque analysis ki zaroorat nahi. ✓ Radial equation $m \overset{r}{¨} = m r \dot{ϕ}^{2}$ jaana-pehchaana outward centrifugal effect hai. ✓ Units: $p_{ϕ} = m r^{2} \dot{ϕ}$ ke units $kg \cdot m^{2} \cdot s^{- 1}$ hain, angular momentum ka correct dimension. ✓

Cell F — degenerate case: zero potential (free particle)

Worked example Ex 6 · 1D mein free particle — sabse simple limit

Koi forces nahi, koi potential nahi: $V = 0$ , $T = \frac{1}{2} m \overset{x}{˙}^{2}$ . Yeh matrix ka degenerate end hai, sanity floor.

Forecast: kuch bhi push nahi karta, toh hum expect karte hain $\overset{x}{¨} = 0$ — constant velocity, straight-line motion (Newton's first law).

$L = \frac{1}{2} m \overset{x}{˙}^{2}$ . Yeh step kyun? $V = 0$ ke saath $L = T - V$ assemble karo; kuch subtract karna nahi.
Velocity half: $\frac{\partial L}{\partial x ˙} = m \overset{x}{˙}$ , phir $\frac{d}{d t} (m \overset{x}{˙}) = m \overset{x}{¨}$ . Yeh step kyun? Velocity-slope phir iska time-derivative — engine ka pehla half.
Position half: $\frac{\partial L}{\partial x} = 0$ ( $L$ mein koi $x$ nahi). Yeh step kyun? Koi potential nahi matlab koi position-slope nahi, toh $x$ bhi cyclic hai.
EL: $m \overset{x}{¨} - 0 = 0 \Rightarrow \overset{x}{¨} = 0.$ Yeh step kyun? Zero position-slope wala engine zero acceleration force karta hai.

Verify: $x$ yahan bhi cyclic hai, toh momentum $p = m \overset{x}{˙}$ conserved hai — constant velocity, exactly Newton's first law aur woh base case jis par poora framework reduce hona chahiye. ✓ Units: $m \overset{x}{¨}$ $kg \cdot m/s^{2} = N$ hai; ise zero set karna kehta hai "zero net force". ✓ $V \neq = 0$ generalization ke liye dekho Newton's Second Law aur yeh samajhne ke liye ki straight line stationary path kyun hai Principle of Least Action.

Cell G — equilibrium ke across sign stress test

Worked example Ex 7 · Mass spring par,

$x = 0$ ke DONO sides check kiya gaya $L = \frac{1}{2} m \overset{x}{˙}^{2} - \frac{1}{2} k x^{2}$ . Hum pehle se jaante hain EL equation $m \overset{x}{¨} + k x = 0$ hai. Cell G ka point yeh confirm karna hai ki force sign har case mein restoring hai: $x > 0$ , $x < 0$ , aur $x = 0$ .

Forecast: acceleration hamesha $x = 0$ ki taraf wapas point karna chahiye.

EL equation $m \overset{x}{¨} = - k x$ se. Sign behaviour ko concrete aur hand-wave se pare banane ke liye, hum specific numbers daalte hain $k = 4 N/m$ , $m = 1 kg$ — yahan ek numeric example behtar hai kyunki pedagogical claim cases mein signs ke baare mein hai, aur koi cheez sign error ko numbers ke flip hote dekhe bina itni jaldi expose nahi karti. Yeh step kyun? Clean numbers choose karna ek abstract " $- k x$ " ko teen checkable arithmetic facts mein badal deta hai.
Case $x = + 3$ : $\overset{x}{¨} = - 4 (3) = - 12$ — left kheencha (negative), origin ki taraf wapas. ✓ Yeh step kyun? Positive-displacement quadrant test karta hai.
Case $x = - 3$ : $\overset{x}{¨} = - 4 (- 3) = + 12$ — right push kiya (positive), origin ki taraf wapas. ✓ Yeh step kyun? Negative-displacement quadrant test karta hai — sign flip hona chahiye.
Case $x = 0$ : $\overset{x}{¨} = 0$ — equilibrium par koi force nahi. ✓ Yeh step kyun? Degenerate point test karta hai taaki koi scenario na chhoota rahe.

Verify: Teeno cases mein $sign (\overset{x}{¨}) = - sign (x)$ , restoring force ki defining property. Frequency $ω = k / m = 4/1 = 2 rad/s$ . ✓ Units: $\overset{x}{¨}$ $m/s^{2}$ mein aaya ( $N/m$ over $kg$ times $m$ ), aur $ω$ $s^{- 1}$ mein. ✓

Cell H — real-world word problem with an exam twist

Worked example Ex 8 · Fixed rate

$Ω$ par spinning hoop par bead (the twist) Mass $m$ ka ek bead ek frictionless circular wire of radius $R$ par slide karta hai. Hoop ko ek vertical diameter ke baare mein constant angular speed $Ω$ par spin karne ke liye forced kiya gaya hai. Maano $θ$ = hoop ke bottom se bead ka angle. Yeh exam twist hai: spin ek extra kinetic term inject karta hai aur ek effective potential create karta hai.

Forecast: slow spin mein bead bottom par baithna chahiye ( $θ = 0$ ); zyada tez spin mein, centrifugal effects ek naya stable angle side par upar push karna chahiye.

Figure — bead on the spinning hoop.

Ek blue circle (the hoop) jisme white dotted vertical spin-axis hai; top par yellow arrow forced spin $Ω$ hai; red dot angle $θ$ par bead hai (green arc) bottom tak dotted radius se; yellow dashed segment bead ki axis se distance hai, $R sin θ$ , yahi reason hai ki spin iske speed mein $R sin θ Ω$ contribute karta hai.

Bead ki velocity ke do perpendicular pieces hain: wire ke along, speed $R \dot{θ}$ ; aur spin axis ke around, speed $R sin θ \cdot Ω$ (axis se uski distance $R sin θ$ hai). Kyunki woh perpendicular hain, $v^{2} = R^{2} \dot{θ}^{2} + R^{2} sin^{2} θ Ω^{2}$ , toh $T = \frac{1}{2} m (R^{2} \dot{θ}^{2} + R^{2} sin^{2} θ Ω^{2}) .$ Yeh step kyun? Forced spin ek constraint hai jo hamare coordinate mein work karta hai, toh yeh $T$ mein land karta hai, $V$ mein nahi; hum do perpendicular speeds ko Pythagoras se add karte hain.
Bottom se upar height $= R (1 - cos θ)$ , toh $V = m g R (1 - cos θ)$ . Yeh step kyun? Gravity ek genuine potential hai; height exactly pendulum ki tarah padhi jaati hai.
$L = \frac{1}{2} m R^{2} \dot{θ}^{2} + \frac{1}{2} m R^{2} Ω^{2} sin^{2} θ - m g R (1 - cos θ)$ banao. Yeh step kyun? $L = T - V$ assemble karo; beech wala term spin ka gift hai aur ek negative potential ki tarah act karega.
Velocity half: $\frac{\partial L}{\partial θ ˙} = m R^{2} \dot{θ}$ , phir $\frac{d}{d t} (\frac{\partial L}{\partial θ ˙}) = m R^{2} \ddot{θ}$ . Yeh step kyun? $Ω$ -term mein koi $\dot{θ}$ nahi hai, toh velocity-slope plain pendulum jaisa hi hai.
Position half: $\frac{\partial L}{\partial θ} = m R^{2} Ω^{2} sin θ cos θ - m g R sin θ$ . Yeh step kyun? Ab $θ$ dono mein hai — spin-kinetic term aur $V$ mein — toh dono position-slope mein contribute karte hain (chain rule on $sin^{2} θ$ gives $2 sin θ cos θ$ , $\frac{1}{2}$ cancel ho jaata hai).
EL: $m R^{2} \ddot{θ} - (m R^{2} Ω^{2} sin θ cos θ - m g R sin θ) = 0$ , toh $\ddot{θ} = Ω^{2} sin θ cos θ - \frac{g}{R} sin θ = sin θ (Ω^{2} cos θ - \frac{g}{R}) .$ Yeh step kyun? Engine $\frac{d}{d t} \frac{\partial L}{\partial θ ˙} - \frac{\partial L}{\partial θ} = 0$ equation of motion deliver karta hai; equilibria expose karne ke liye $sin θ$ factor karo.

Verify: Equilibria jahan $\ddot{θ} = 0$ : ya toh $sin θ = 0$ (bottom $θ = 0$ ) ya $cos θ = \frac{g}{R Ω ^{2}}$ . Woh doosra solution tabhi exist karta hai jab $\frac{g}{R Ω ^{2}} \leq 1$ , yaani jab $Ω^{2} \geq \frac{g}{R}$ — ek critical spin rate $Ω_{c} = g / R$ , exactly forecast. ✓ $Ω_{c}$ se neeche bracket $Ω^{2} cos θ - g / R < 0$ $θ = 0$ ke paas hota hai, toh bottom stable hai. ✓ Units: $Ω^{2}$ aur $g / R$ dono $s^{- 2}$ mein hain, aur $\ddot{θ}$ $rad/s^{2}$ mein aata hai. ✓

Active recall

Recall Har naya problem kis cell mein hai? (answers cover karo)

Ek block hill se neeche slide karta hai, coordinate down-slope ::: Cell B — negative-slope potential. Ek double pendulum ::: Cell D — coupled coordinates (do EL equations). Ek planet freely orbit karta hai, angle coordinate ::: Cell E — cyclic coordinate, angular momentum conserved. Koi forces nahi wala ek puck ::: Cell F — degenerate, $V = 0$ , constant velocity. Driven pivot par ek pendulum ::: Cell H — constraint jo $T$ mein add karti hai.

Connections

Euler–Lagrange Equation — woh single engine jo upar har example mein use hua.
Generalized Coordinates and Constraints — kyun $s$ , $θ$ , $(r, ϕ)$ choose karne ke liye free the.
Generalized Momentum and Conservation — Cell E ka conserved $p_{ϕ}$ .
Principle of Least Action — kyun free particle (Cell F) straight line leta hai.
Hamiltonian H = T + V — jahan energy $T + V$ wapas aati hai.
Newton's Second Law — woh base case jis par har example reduce hona chahiye.