Applying E-L equations to various systems — Hinglish notes

2.1.6 · Physics › Analytical Mechanics

The recipe (actually kya karte ho)

YEH sach kyun hai (derivation ki sketch): Hamilton's principle kehta hai ki real path action $S = \int L d t$ ko stationary banata hai ( $δ S = 0$ ). Yeh demand karna ki first variation arbitrary wiggles $δ q_{i} (t)$ ke liye zero ho jo endpoints pe vanish karein, aur $\overset{q}{˙}$ -term ko parts mein integrate karna, exactly upar wali equation deta hai. (Poori derivation 2.1.05-Deriving-the-Euler-Lagrange-equation mein hai.)

HOW to apply it — universal 5-step procedure:

DOF count karo → utne hi independent generalized coordinates $q_{i}$ chuno.
$T$ likho $q_{i}, \overset{q}{˙}_{i}$ ke terms mein (har Cartesian velocity ko apne coords se express karo).
$V$ likho $q_{i}$ ke terms mein.
$L = T - V$ banao, $\frac{\partial L}{\partial q _{i}}$ aur $\frac{\partial L}{\partial q ˙ _{i}}$ compute karo.
E–L mein plug karo, time derivative lo, simplify karo → equation of motion.

Figure — Applying E-L equations to various systems

Example 1 — Simple Pendulum (1 DOF)

Worked example Mass

$m$ ek rigid massless rod of length $ℓ$ pe. Step 1 — DOF: bob ek circle pe constrained hai ⇒ 1 DOF. Angle $q = θ$ use karo. Yeh step kyun? Rod planar coordinates mein se 1 ko khatam kar deta hai, toh ek angle kaafi hai.

Step 2 — $T$ : arc ke along speed $ℓ \dot{θ}$ hai, toh $T = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} .$ Kyun? Bob radius $ℓ$ ke circle pe move karta hai; arc-speed $= r \dot{θ}$ .

Step 3 — $V$ : lowest point se height $ℓ (1 - cos θ)$ hai, toh $V = m g ℓ (1 - cos θ) .$ Kyun? Gravity potential $m g h$ hai; lowest point ko zero lo.

Step 4 — $L$ aur derivatives: $L = \frac{1}{2} m ℓ^{2} \dot{θ}^{2} - m g ℓ (1 - cos θ)$

\frac{\partial L}{\partial\theta}=-mg\ell\sin\theta.$$ **Step 5 — E–L:** $$\frac{d}{dt}(m\ell^2\dot\theta) + mg\ell\sin\theta=0 \;\Rightarrow\; \boxed{\ddot\theta=-\frac{g}{\ell}\sin\theta}$$ *Yeh sahi kyun hai:* small angles ⇒ $\sin\theta\approx\theta$ ⇒ SHM with $\omega=\sqrt{g/\ell}$ — exactly textbook result. **Forecast-then-verify ✔.**

Example 2 — Bead on a Rotating Wire / Atwood Machine (2 systems, 1 DOF each)

Worked example Atwood machine: masses

$m_{1}, m_{2}$ ek frictionless pulley ke upar, inextensible string. Step 1: string ek position fix kar deti hai jab doosri pata ho ⇒ 1 DOF. $x$ = $m_{1}$ kitna neeche gaya; toh $m_{2}$ $x$ se upar uthega, toh $\overset{x}{˙}_{2} = - \overset{x}{˙}$ ... lekin dono $∣ \overset{x}{˙} ∣$ speed se chalte hain. Kyun? Inextensibility dono ko couple karti hai — constraint coordinate mein hi built in hai, toh tension kabhi appear hi nahi karta!

Step 2 — $T$ : $T = \frac{1}{2} m_{1} \overset{x}{˙}^{2} + \frac{1}{2} m_{2} \overset{x}{˙}^{2} = \frac{1}{2} (m_{1} + m_{2}) \overset{x}{˙}^{2} .$

Step 3 — $V$ : $V = - m_{1} g x - m_{2} g (- x) = - (m_{1} - m_{2}) g x .$ Signs kyun? Agar $m_{1}$ $x$ neeche jaata hai toh PE lose karta hai; $m_{2}$ $x$ upar jaata hai aur PE gain karta hai.

Step 4–5: $\frac{\partial L}{\partial x ˙} = (m_{1} + m_{2}) \overset{x}{˙}$ , $\frac{\partial L}{\partial x} = (m_{1} - m_{2}) g .$ $(m_{1} + m_{2}) \overset{x}{¨} = (m_{1} - m_{2}) g \Rightarrow \overset{x}{¨} = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}}$ Verify: equal masses ⇒ $\overset{x}{¨} = 0$ (balance), ek mass zero ⇒ free fall $g$ . ✔ Tension ki zaroorat nahi.

Example 3 — Block on Frictionless Incline (1 DOF, tilted coordinate)

Example 4 — 2D Central Force (2 DOF, plane polar)

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (answers cover karo!)

$L$ kya hai? → $T - V$ .
Kitne coordinates chahiye? → ek per DOF.
Cyclic coordinate kya deta hai? → ek conserved generalized momentum.
Tension/normal force kyun appear nahi hota? → ideal constraints koi work nahi karte; coordinates unhe absorb kar lete hain.
Pendulum EOM? → $\ddot{θ} = - (g / ℓ) sin θ$ .

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho tum predict karna chahte ho ki ek jhula kaise move karta hai. Mushkil tarika: har dhakka aur kheench track karo, rope ki tug bhi. Aasaan tarika: sirf ek "score" rakho — kitni moving energy hai minus kitni height energy hai. Nature lazy hai aur hamesha woh path chunti hai jo is score ko time ke saath "balanced" rakhta hai. Humne ek magic rule likha (E–L equation) jo, kisi bhi measuring stick mein energy score dene par (angle, distance, kuch bhi), automatically exactly bata deta hai cheez kaise move karegi — aur tumhe rope ke baare mein sochna hi nahi pada.

The Lagrangian

L

equals

T - V

(kinetic minus potential energy)

Number of generalized coordinates required

ek per degree of freedom

Euler–Lagrange equation for coordinate

q_{i}

\frac{d}{d t} \frac{\partial L}{\partial q ˙ _{i}} - \frac{\partial L}{\partial q _{i}} = 0

Why constraint forces (tension, normal) don't appear in E–L

yeh koi work nahi karte aur independent coordinates ki choice se absorb ho jaate hain

Kinetic energy in plane polar coordinates

T = \frac{1}{2} m (\overset{r}{˙}^{2} + r^{2} \dot{θ}^{2})

A cyclic (ignorable) coordinate (absent from

L

) implies

uska conjugate momentum

\partial L / \partial \overset{q}{˙}

conserved hai

EOM of a simple pendulum

\ddot{θ} = - (g / ℓ) sin θ

Atwood machine acceleration

\overset{x}{¨} = (m_{1} - m_{2}) g / (m_{1} + m_{2})

Block on frictionless incline acceleration along slope

g sin α

Conserved quantity for central force motion

angular momentum

L_{z} = m r^{2} \dot{θ}

First step of the E–L recipe

degrees of freedom count karo aur utne hi coordinates chuno

Inside the partial derivatives of

L

, how are

q

and

\overset{q}{˙}

treated

independent variables ki tarah

Connections

2.1.05-Deriving-the-Euler-Lagrange-equation — recipe kahan se aati hai.
2.1.07-Cyclic-coordinates-and-conservation-laws — ignorable coordinates ka faayda uthana.
2.1.08-Generalized-momenta-and-the-Hamiltonian — E–L ke baad ka next step.
Constraints-and-degrees-of-freedom — achhe coordinates chunna.
Noethers-theorem — symmetry ↔ conservation, Example 4 ko generalize karna.