Kinetic energy in generalized coordinates: Worked examples

2.1.3 · D3 · Physics › Analytical Mechanics › Generalized coordinates mein kinetic energy

Kuch bhi shuru karne se pehle, teen symbols ko plain words mein re-anchor karte hain jo hum baar baar use karenge, taaki koi line one se hi na bhatak jaaye.

Neeche ki table mein do jargon words aate hain; unhe ek baar yahan define karte hain taaki matrix saaf padhta rahe.

Aur master formula jo hum har baar mechanically apply karenge:

The scenario matrix

Is topic ke har problem ka ek cell mein aana zaroori hai. Aakhri column wo example batata hai jo use cover karta hai. ("Scleronomic" aur "rheonomic" upar define kiye gaye hain.)

#	Cell (scenario class)	$\partial_{t} r = 0$ ?	Kaun se pieces survive karte hain	Dhyan rakhne wala twist	Covered by
A	Flat Cartesian, koi constraint nahi	haan (scleronomic)	sirf $T_{2}$ , $M$ constant	trivial baseline	Ex 1
B	Curvilinear, scleronomic (polar/spherical)	haan	sirf $T_{2}$ , $M = M (q)$	$M$ position par depend karta hai	Ex 2
C	Off-diagonal $M_{ij} \neq = 0$ (skewed coords)	haan	cross term ke saath $T_{2}$	factor-2 pairing mat chhodna	Ex 3
D	Rheonomic, rotation → sirf $T_{0}$	nahi (rheonomic)	$T_{2} + T_{0}$ , $T_{1} = 0$	yahan $T_{1}$ kyun vanish karta hai	Ex 4
E	Rheonomic, translation → $T_{1}$ aur $T_{0}$	nahi (rheonomic)	$T_{2} + T_{1} + T_{0}$ teeno present	linear term real hai	Ex 5
F	Curved surface (sphere)	haan	$T_{2}$ , do coupled coords	2-sphere ka metric	Ex 6
G	Degenerate / limiting input ( $r \to 0$ , $ω \to 0$ )	kuch bhi	pieces collapse ho jaate hain	matrix ek direction kho deta hai	Ex 7
H	Real-world word problem (elevator + pendulum)	nahi (rheonomic)	teeno pieces	map mein physics padhna	Ex 8
I	Exam twist (Euler's theorem verify karo / $H$ nikalo)	haan	$T_{2}$ property use karta hai	$\sum \overset{q}{˙} \partial_{\overset{q}{˙}} T = 2 T$	Ex 9

Example 1 — Cell A: flat baseline

Worked example Plane mein free particle, Cartesian coordinates

Ek particle, mass $m$ , coordinates $q = (x, y)$ , toh $r = (x, y)$ , map mein koi time nahi. Forecast: padhne se pehle guess karo $T$ kya hai — tumhe yeh grade-school physics se already pata hai.

Step 1. Direction vectors $\partial r / \partial q_{i}$ compute karo. $\frac{\partial r}{\partial x} = (1, 0), \frac{\partial r}{\partial y} = (0, 1) .$ Ye step kyun? $M_{ij}$ inhe dot karke banta hai, toh pehle inhe chahiye. Har ek keh raha hai "agar sirf $x$ change ho, to point kis direction mein move karta hai" — seedha axis ke saath, jaise picture dikhati hai.

Step 2. $M_{ij}$ banao. $M_{xx} = m (1, 0) \cdot (1, 0) = m, M_{y y} = m, M_{x y} = m (1, 0) \cdot (0, 1) = 0.$ Ye step kyun? Hum direction vectors ko definition mein plug karte hain. Zero keh raha hai axes perpendicular hain — koi cross term nahi.

Step 3. $T$ assemble karo. $T = \frac{1}{2} m \overset{x}{˙}^{2} + \frac{1}{2} m \overset{y}{˙}^{2} .$ Ye step kyun? $r$ mein koi $t$ nahi ⟹ $a_{i} = 0$ , $T_{0} = 0$ . Sirf $T_{2}$ survive karta hai.

Verify: Yeh literally $\frac{1}{2} m ∣ v ∣^{2}$ hai jahan $v = (\overset{x}{˙}, \overset{y}{˙})$ . Units: $kg \cdot (m/s)^{2} = J$ . ✓

Example 2 — Cell B: polar, position-dependent metric

Figure — Kinetic energy in generalized coordinates

Step 1. Direction vectors (figure dekho: point par do coloured arrows). $\frac{\partial r}{\partial r} = (cos θ, sin θ), \frac{\partial r}{\partial θ} = (- r sin θ, r cos θ) .$ Ye step kyun? Blue arrow ( $\partial_{r}$ ) bahar ki taraf point karta hai — $r$ change karna tumhe ray ke saath slide karta hai. Orange arrow ( $\partial_{θ}$ ) sideways point karta hai aur centre se door $r$ times lamba hota hai, kyunki ek radian angle zyada bada arc sweep karta hai jab $r$ bada ho.

Step 2. $M_{ij}$ ke liye inhe dot karo. $M_{r r} = m (cos^{2} θ + sin^{2} θ) = m,$ $M_{θ θ} = m r^{2} (sin^{2} θ + cos^{2} θ) = m r^{2},$ $M_{r θ} = m [cos θ (- r sin θ) + sin θ (r cos θ)] = 0.$ Ye step kyun? $M_{r θ} = 0$ isliye kyunki blue aur orange arrows perpendicular hain — geometry hai, luck nahi.

Step 3. $T = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} \dot{θ}^{2} .$ $r^{2}$ kyun? Lamba orange arrow ka matlab hai ki wohi $\dot{θ}$ door par zyada real speed produce karta hai — metric yahi record karta hai.

Verify: Radius $r$ par, tangential speed $r \dot{θ}$ hai, radial speed $\overset{r}{˙}$ ; perpendicular hain, toh $v^{2} = \overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}$ , deta hai $T = \frac{1}{2} m v^{2}$ . ✓ $r^{2} \dot{θ}^{2}$ ke units: $m^{2} \cdot s^{- 2}$ ✓.

Example 3 — Cell C: genuinely off-diagonal metric

Worked example Shared coordinate par do masses (skewed system)

Ek line par do particles: $x_{1} = q_{1}, x_{2} = q_{1} + q_{2}$ , masses $m_{1} = m_{2} = m$ . Toh $q_{1}$ mein change dono particles ko move karta hai. Forecast: kya $M_{12}$ zero hoga? Socho: kya $q_{1}$ move karna $q_{2}$ move karne ke saath koi direction share karta hai?

Step 1. Direction vectors, state ko $(x_{1}, x_{2})$ maanke: $\frac{\partial r}{\partial q _{1}} = (1, 1), \frac{\partial r}{\partial q _{2}} = (0, 1) .$ Ye step kyun? $q_{1}$ change karna particle 1 ko 1 se aur particle 2 ko 1 se push karta hai (dono feel karte hain); $q_{2}$ change karna sirf particle 2 ko push karta hai.

Step 2. Metric (har dot us particle ke mass se weighted — yahan dono $m$ ): $M_{11} = m (1)^{2} + m (1)^{2} = 2 m, M_{22} = m (0)^{2} + m (1)^{2} = m, M_{12} = m (1) (0) + m (1) (1) = m .$ Ye step kyun? Shared direction (particle 2) ki wajah se $M_{12} = m \neq = 0$ : coordinates coupled hain.

Step 3. Assemble karo, yaad rakhte hue ki $M_{12}$ dono $(1, 2)$ aur $(2, 1)$ ke liye aata hai: $T = \frac{1}{2} (M_{11} \overset{q}{˙}_{1}^{2} + M_{22} \overset{q}{˙}_{2}^{2}) + M_{12} \overset{q}{˙}_{1} \overset{q}{˙}_{2} = \frac{1}{2} (2 m) \overset{q}{˙}_{1}^{2} + \frac{1}{2} m \overset{q}{˙}_{2}^{2} + m \overset{q}{˙}_{1} \overset{q}{˙}_{2} .$ $T = m \overset{q}{˙}_{1}^{2} + \frac{1}{2} m \overset{q}{˙}_{2}^{2} + m \overset{q}{˙}_{1} \overset{q}{˙}_{2} .$ Single (double nahi) cross term kyun? $\frac{1}{2} \sum_{i, j}$ alag se $(1, 2)$ aur $(2, 1)$ count karta hai; saath mein dete hain $\frac{1}{2} (M_{12} + M_{21}) \overset{q}{˙}_{1} \overset{q}{˙}_{2} = M_{12} \overset{q}{˙}_{1} \overset{q}{˙}_{2}$ . Aur 2 se mat gunao.

Verify: Directly, $\overset{x}{˙}_{1} = \overset{q}{˙}_{1}, \overset{x}{˙}_{2} = \overset{q}{˙}_{1} + \overset{q}{˙}_{2}$ , toh $T = \frac{1}{2} m \overset{q}{˙}_{1}^{2} + \frac{1}{2} m (\overset{q}{˙}_{1} + \overset{q}{˙}_{2})^{2} = m \overset{q}{˙}_{1}^{2} + \frac{1}{2} m \overset{q}{˙}_{2}^{2} + m \overset{q}{˙}_{1} \overset{q}{˙}_{2} . ✓$

Example 4 — Cell D: rotating wire ( $T_{0}$ aata hai, $T_{1}$ nahi)

Step 1. Do tarah ke derivatives: $\frac{\partial r}{\partial r} = (cos ω t, sin ω t) (radial), \frac{\partial r}{\partial t} = (- r ω sin ω t, r ω cos ω t) (tangential, spin se) .$ Ye step kyun? $\partial_{r}$ = "agar bead ko bahar slide karoon to position kaise move hoti hai"; $\partial_{t}$ = "wire khud rotate kare aur bead still baitha rahe to position kaise move hoti hai". Figure mein ye perpendicular blue (radial) aur orange (tangential) arrows ke roop mein dikhte hain.

Step 2 — $T_{2}$ piece. $M_{r r} = m (cos^{2} ω t + sin^{2} ω t) = m$ , toh $T_{2} = \frac{1}{2} m \overset{r}{˙}^{2}$ . Ye step kyun? $T_{2}$ ko sirf $q$ -direction vector apne aap se dot chahiye; blue radial arrow ki length $1$ hai, jo plain mass $m$ deta hai — wire ke saath slide karne ki energy.

Step 3 — $T_{1}$ piece. $a_{r} = m \partial_{r} r \cdot \partial_{t} r = m [cos ω t (- r ω sin ω t) + sin ω t (r ω cos ω t)] = 0$ . Ye step kyun? $T_{1}$ "bahar slide karne" aur "spin kiye jaane" ke beech overlap hai. Radial arrow (blue) aur spin arrow (orange) figure mein perpendicular hain, toh unka dot product — aur isliye $a_{r}$ — zero ho jaata hai.

Step 4 — $T_{0}$ piece. $T_{0} = \frac{1}{2} m ∣ \partial_{t} r ∣^{2} = \frac{1}{2} m (r ω)^{2}$ . Ye step kyun? Wire par still baitha bead bhi ( $\overset{r}{˙} = 0$ ) speed $r ω$ par circle mein gheeencha jaata hai; $T_{0}$ exactly wohi "frozen-but-still-moving" energy hai.

Step 5 — assemble karo. $T = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2} .$ $T_{0}$ potential ki tarah kyun kaam karta hai: yeh $r$ par depend karta hai lekin $\overset{r}{˙}$ par nahi; Euler-Lagrange equations mein yeh outward centrifugal push produce karta hai.

Verify: Full speed $v^{2} = \overset{r}{˙}^{2} + (r ω)^{2}$ (radial + tangential, perpendicular), toh $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} ω^{2})$ . ✓

Example 5 — Cell E: sliding frame ( $T_{1}$ aur $T_{0}$ dono nonzero)

Worked example Ek frame par coordinate jo constant

$v$ par move karta hai $x = q + v t, y = w$ ; coordinates $(q, w)$ , mass $m$ , constant speed $v$ . Phir se rheonomic — frame drag kiya ja raha hai. Forecast: ab tumhe teeno pieces expect karne chahiye. Kaun sa term $v$ ko linearly carry karta hai?

Step 1. Direction vectors: $\frac{\partial r}{\partial q} = (1, 0), \frac{\partial r}{\partial w} = (0, 1), \frac{\partial r}{\partial t} = (v, 0) .$ Ye step kyun? Frame $x$ -direction ko speed $v$ par drag karta hai, toh $\partial_{t} r = (v, 0)$ — is baar yeh $\partial_{q} r = (1, 0)$ ke parallel hai, rotating case ke unlike.

Step 2 — $T_{2}$ piece. $M_{q q} = m (1)^{2} = m, M_{w w} = m (1)^{2} = m, M_{q w} = m (1) (0) + m (0) (1) = 0$ , toh $T_{2} = \frac{1}{2} m (\overset{q}{˙}^{2} + \overset{w}{˙}^{2})$ . Ye step kyun? Dono coordinate directions perpendicular axes ke saath unit vectors hain, toh $M$ sirf $m$ times identity hai — drag se unbothered free-particle quadratic core.

Step 3 — $T_{1}$ piece. $a_{q} = m \partial_{q} r \cdot \partial_{t} r = m (1) (v) = m v$ , aur $a_{w} = m (0) (v) = 0$ , toh $T_{1} = m v \overset{q}{˙}$ . Ye step kyun? $T_{1}$ ab nonzero hai kyunki $\partial_{q} r$ aur $\partial_{t} r$ parallel hain — poora overlap, Ex 4 ke perpendicular rotating case ke bilkul opposite.

Step 4 — $T_{0}$ piece. $T_{0} = \frac{1}{2} m ∣ \partial_{t} r ∣^{2} = \frac{1}{2} m v^{2}$ . Ye step kyun? Frame mein still baitha particle bhi ( $\overset{q}{˙} = \overset{w}{˙} = 0$ ) speed $v$ par carry hota hai; wohi carried energy $T_{0}$ hai.

Step 5 — assemble karo. $T = \frac{1}{2} m (\overset{q}{˙}^{2} + \overset{w}{˙}^{2}) + m v \overset{q}{˙} + \frac{1}{2} m v^{2} = \frac{1}{2} m [(\overset{q}{˙} + v)^{2} + \overset{w}{˙}^{2}] .$ Verify: $\overset{x}{˙} = \overset{q}{˙} + v, \overset{y}{˙} = \overset{w}{˙}$ , toh $T = \frac{1}{2} m [(\overset{q}{˙} + v)^{2} + \overset{w}{˙}^{2}]$ . ✓ $\frac{1}{2} m \overset{x}{˙}^{2}$ ki pure bookkeeping, koi naya physics nahi.

Example 6 — Cell F: sphere par particle (curved surface)

Step 1. Direction vectors: $\frac{\partial r}{\partial θ} = R (cos θ cos ϕ, cos θ sin ϕ, - sin θ),$ $\frac{\partial r}{\partial ϕ} = R (- sin θ sin ϕ, sin θ cos ϕ, 0) .$ Ye step kyun? $\partial_{θ}$ tumhe ek meridian ke saath move karta hai (pole ki taraf/pole se door); $\partial_{ϕ}$ tumhe latitude ke ek circle ke saath move karta hai — aur wo circle pole ke paas choti ho jaati hai, figure mein upar shorter orange arrow se dikhaya gaya hai.

Step 2 — diagonal metric entries. $M_{θ θ} = m R^{2} (cos^{2} θ cos^{2} ϕ + cos^{2} θ sin^{2} ϕ + sin^{2} θ) = m R^{2},$ $M_{ϕϕ} = m R^{2} sin^{2} θ (sin^{2} ϕ + cos^{2} ϕ) = m R^{2} sin^{2} θ .$ $M_{ϕϕ} = m R^{2} sin^{2} θ$ kyun? Latitude circle ka radius $R sin θ$ hai, toh uski speed $(R sin θ) \dot{ϕ}$ hai; squaring karne par $sin^{2} θ$ milta hai.

Step 3 — off-diagonal entry. $M_{θ ϕ} = m R^{2} (- cos θ sin ϕ sin θ cos ϕ + cos θ cos ϕ sin θ sin ϕ + 0) = 0.$ $M_{θ ϕ} = 0$ kyun? Meridian direction ( $\partial_{θ}$ ) aur latitude direction ( $\partial_{ϕ}$ ) sphere par har jagah perpendicular hain — north-south east-west ko right angles par cross karta hai — toh unka dot product term by term cancel ho jaata hai.

Step 4 — assemble karo. $T = \frac{1}{2} m R^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) .$ Verify: Equator par $θ = π /2$ , $sin^{2} θ = 1$ : full-size latitude circle, term $\frac{1}{2} m R^{2} \dot{ϕ}^{2}$ hai. Pole par $θ \to 0$ , $sin^{2} θ \to 0$ : $ϕ$ spin karne mein koi energy nahi kyunki tum axis par khade ho. ✓ (Yeh limiting behaviour exactly Cell G hai jo neeche hai.)

Example 7 — Cell G: degenerate aur limiting inputs

Worked example Jab ek coordinate collapse ho jaata hai tab kya hota hai

Polar result $T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})$ lo (Ex 2) aur edges probe karo. Forecast: metric matrix $M = (m 0 0 m r^{2})$ $r \to 0$ par kya karta hai?

Step 1 — limit $r \to 0$ . $M_{θ θ} = m r^{2} \to 0$ . Ye step kyun? Hum origin ki taraf input slide karte hain yeh test karne ke liye ki coordinate system kahan toot ta hai. Kisi bhi $\dot{θ}$ ka energy cost vanish ho jaata hai, aur $M$ ban jaata hai $(m 0 00)$ — not positive-definite (iska ek zero eigenvalue hai). Geometrically, origin par angle $θ$ undefined hai: saari directions ek ho jaati hain. Coordinate system wahan degenerate hai.

Step 2 — rotating wire mein limit $ω \to 0$ (Ex 4). $T = \frac{1}{2} m \overset{r}{˙}^{2} + \frac{1}{2} m r^{2} ω^{2} \to \frac{1}{2} m \overset{r}{˙}^{2}$ . Ye step kyun? Hum spin band karte hain check karne ke liye ki rheonomic machinery sahi se reduce hoti hai ya nahi. Koi spin nahi matlab koi moving constraint nahi: $T_{0}$ gayab ho jaata hai aur hum ek scleronomic (Cell A-jaisi) system par wapas aa jaate hain. Rheonomic apparatus quietly simple ban jaata hai — ek reassuring consistency check.

Step 3 — Ex 6 mein bounded scaling, phir ek determinant test. Fixed sphere par $M_{ϕϕ} = m R^{2} sin^{2} θ$ , $m R^{2}$ se bounded hai; kuch blow up nahi hota jaise unbounded plane mein hota. Polar metric ke liye, exactly determine karo degeneracy kahan hai — determinant compute karo: $det M = m \cdot m r^{2} = m^{2} r^{2} .$ Ye step kyun? Vanishing determinant ek precise, sign-independent signal hai ki matrix ek direction kho chuka hai (zero eigenvalue). $det M > 0$ for $r > 0$ (invertible, healthy) aur $det M = 0$ exactly $r = 0$ par (coordinate singularity).

Verify: $M$ har jagah positive-definite hai sivaay single point $r = 0$ ke, jo geometric kahani se match karta hai ki sirf origin degenerate hai. ✓

Example 8 — Cell H: real-world word problem

Worked example Elevator mein hanging pendulum jo accelerate kar raha hai

Length $ℓ$ ka pendulum, bob mass $m$ , ek elevator ki ceiling se latka hua jiska floor height $Y (t) = \frac{1}{2} A t^{2}$ hai (constant upward acceleration $A$ ). Vertical se angle $φ$ coordinate hai. Pivot position: $(0, Y (t))$ . Bob position: $x = ℓ sin φ, y = Y (t) - ℓ cos φ .$ Forecast: teeno pieces ya nahi? Elevator apne schedule par chal raha hai (rheonomic) — $T_{1}$ aur $T_{0}$ expect karo.

Step 1. Derivatives: $\frac{\partial r}{\partial φ} = (ℓ cos φ, ℓ sin φ), \frac{\partial r}{\partial t} = (0, \dot{Y}) = (0, A t) .$ Ye step kyun? $\partial_{φ}$ bob ko swing karta hai; $\partial_{t}$ pure pendulum ko elevator ke saath uthata hai (sirf ek $y$ -component).

Step 2 — $T_{2}$ piece. $M_{φφ} = m (ℓ^{2} cos^{2} φ + ℓ^{2} sin^{2} φ) = m ℓ^{2}$ , toh $T_{2} = \frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2}$ . Ye step kyun? $T_{2}$ swinging energy hai; bob radius $ℓ$ ka circle trace karta hai, toh uski speed $ℓ \overset{φ}{˙}$ hai aur $cos^{2} + sin^{2} = 1$ do components ko ek clean $m ℓ^{2}$ mein collapse kar deta hai.

Step 3 — $T_{1}$ piece. $a_{φ} = m \partial_{φ} r \cdot \partial_{t} r = m (ℓ cos φ) (0) + m (ℓ sin φ) (A t) = m A t ℓ sin φ$ , toh $T_{1} = m A t ℓ \overset{φ}{˙} sin φ$ . $T_{1} \neq = 0$ yahan kyun lekin rotating wire ke liye $0$ tha? Lift direction $(0, A t)$ , swing direction ke saath apne $y$ -part $ℓ sin φ$ ke zariye overlap karta hai — nonzero dot product jab bhi bob vertical se hata hua ho.

Step 4 — $T_{0}$ piece. $T_{0} = \frac{1}{2} m ∣ \partial_{t} r ∣^{2} = \frac{1}{2} m (A t)^{2}$ . Ye step kyun? Seedha latka aur still baid bob bhi elevator ki wajah se speed $A t$ par utha uthaya jaata hai; wohi carried energy $T_{0}$ hai.

Step 5 — assemble karo. $T = \frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2} + m A t ℓ \overset{φ}{˙} sin φ + \frac{1}{2} m A^{2} t^{2} .$ Verify: Directly $\overset{x}{˙} = ℓ \overset{φ}{˙} cos φ, \overset{y}{˙} = A t + ℓ \overset{φ}{˙} sin φ$ . $T = \frac{1}{2} m (\overset{x}{˙}^{2} + \overset{y}{˙}^{2}) = \frac{1}{2} m [ℓ^{2} \overset{φ}{˙}^{2} cos^{2} φ + (A t + ℓ \overset{φ}{˙} sin φ)^{2}] .$ Expand karne par: $\frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2} + m A t ℓ \overset{φ}{˙} sin φ + \frac{1}{2} m A^{2} t^{2}$ . ✓ ( $cos^{2} + sin^{2} = 1$ use karta hai).

Example 9 — Cell I: exam twist (Euler's theorem, aur $H$ )

Worked example Confirm karo

$\sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i} = 2 T_{2} + T_{1}$ aur energy function nikalo Ex 8 ke elevator pendulum ka $T$ use karo. Dekho Energy function and Hamiltonian aur Generalized momentum and conserved quantities. Forecast: Ek pure quadratic $T_{2}$ ke liye, Euler's theorem exactly $2 T$ deta hai. $T_{1}$ present hone par, correction guess karo.

Step 1. $T$ ko $\overset{φ}{˙}$ ke respect mein differentiate karo: $\frac{\partial T}{\partial φ ˙} = m ℓ^{2} \overset{φ}{˙} + m A t ℓ sin φ .$ Ye step kyun? Yeh generalized momentum $p_{φ}$ hai; $\overset{φ}{˙}$ se multiply karne par Euler's theorem mein sum banta hai.

Step 2. $\overset{φ}{˙} \partial T / \partial \overset{φ}{˙}$ form karo: $\overset{φ}{˙} \frac{\partial T}{\partial φ ˙} = m ℓ^{2} \overset{φ}{˙}^{2} + m A t ℓ \overset{φ}{˙} sin φ = 2 T_{2} + T_{1} .$ Ye step kyun? Degree $k$ ki homogeneous function ke liye Euler's theorem $k \times (us piece)$ deta hai: degree 2 → $T_{2}$ par factor 2, degree 1 → $T_{1}$ par factor 1, degree 0 → $T_{0}$ se kuch nahi.

Step 3. Energy function $h = \sum_{i} \overset{q}{˙}_{i} \partial T / \partial \overset{q}{˙}_{i} - L$ jahan $L = T - V$ hai, ban jaata hai $h = (2 T_{2} + T_{1}) - (T_{2} + T_{1} + T_{0}) + V = T_{2} - T_{0} + V .$ Ye step kyun? Sirf pure-quadratic $T_{2}$ apna poora weight contribute karta hai; $- T_{0}$ dikhata hai ki energy function $T + V$ nahi hoti jab constraints move karein (rheonomic).

Verify: Scleronomic system ke liye ( $T_{1} = T_{0} = 0$ ), $h = T_{2} + V = T + V$ — jaana-pehchana total energy. ✓ Hamare elevator ke liye, $h = \frac{1}{2} m ℓ^{2} \overset{φ}{˙}^{2} - \frac{1}{2} m A^{2} t^{2} + V$ , jo explicitly time-dependent hai, toh energy conserved nahi — physically sahi, elevator motor energy inject kar raha hai.

Recall Har surviving-term pattern kaun sa cell signal karta hai?

Sirf $T_{2}$ ::: scleronomic (map mein koi $t$ nahi) — Cells A, B, C, F, G(ω→0). $T_{2} + T_{0}$ , koi $T_{1}$ nahi ::: moving constraint jiska motion har coordinate direction ke perpendicular hai (rotating wire) — Cell D. $T_{2} + T_{1} + T_{0}$ ::: moving constraint jiska ek coordinate direction ke saath component hai (sliding frame, lifting elevator) — Cells E, H. Energy function $h = T + V$ ::: sirf tab jab scleronomic ( $T_{1} = T_{0} = 0$ ) — Cell I.