Applying E-L equations to various systems

The recipe (what you actually do)

WHY this is true (sketch of the derivation): Hamilton's principle says the real path makes the action $S=\int L\,dt$ stationary ( $\delta S = 0$ ). Demanding that the first variation vanish for arbitrary wiggles $\delta q_i(t)$ that vanish at the endpoints, and integrating the $\dot q$ -term by parts, gives exactly the equation above. (Full derivation lives in 2.1.05-Deriving-the-Euler-Lagrange-equation.)

HOW to apply it — the universal 5-step procedure:

Count DOF → pick that many independent generalized coordinates $q_i$ .
Write $T$ in terms of $q_i,\dot q_i$ (express every Cartesian velocity through your coords).
Write $V$ in terms of $q_i$ .
Form $L=T-V$ , compute $\dfrac{\partial L}{\partial q_i}$ and $\dfrac{\partial L}{\partial \dot q_i}$ .
Plug into E–L, take the time derivative, simplify → equation of motion.

Figure — Applying E-L equations to various systems

Example 1 — Simple Pendulum (1 DOF)

Worked example Mass

$m$ on rigid massless rod of length $\ell$ . Step 1 — DOF: the bob is constrained to a circle ⇒ 1 DOF. Use angle $q=\theta$ . Why this step? The rod kills 1 of the 2 planar coordinates, so one angle is enough.

Step 2 — $T$ : speed along the arc is $\ell\dot\theta$ , so $T=\tfrac12 m \ell^2\dot\theta^2.$ Why? The bob moves on a circle of radius $\ell$ ; arc-speed $=r\dot\theta$ .

Step 3 — $V$ : height above lowest point is $\ell(1-\cos\theta)$ , so $V = mg\ell(1-\cos\theta).$ Why? Gravity potential is $mgh$ ; pick lowest point as zero.

Step 4 — $L$ and derivatives: $L=\tfrac12 m\ell^2\dot\theta^2 - mg\ell(1-\cos\theta)$

\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}$$ *Why this is right:* small angles ⇒ $\sin\theta\approx\theta$ ⇒ SHM with $\omega=\sqrt{g/\ell}$ — exactly the textbook result. **Forecast-then-verify ✔.**

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

Worked example Atwood machine: masses

$m_1,m_2$ over a frictionless pulley, inextensible string. Step 1: string fixes one position once the other is known ⇒ 1 DOF. Let $x$ = how far $m_1$ has descended; then $m_2$ rises by $x$ , so $\dot x_2=-\dot x$ ... but both move at speed $|\dot x|$ . Why? Inextensibility couples the two — the constraint is built into the coordinate, so tension never appears!

Step 2 — $T$ : $T=\tfrac12 m_1\dot x^2 + \tfrac12 m_2\dot x^2 = \tfrac12(m_1+m_2)\dot x^2.$

Step 3 — $V$ : $V=-m_1 g x - m_2 g(-x) = -(m_1-m_2)gx.$ Why the signs? If $m_1$ goes down by $x$ it loses PE; $m_2$ goes up by $x$ and gains PE.

Step 4–5: $\dfrac{\partial L}{\partial\dot x}=(m_1+m_2)\dot x$ , $\dfrac{\partial L}{\partial x}=(m_1-m_2)g.$ $(m_1+m_2)\ddot x = (m_1-m_2)g \;\Rightarrow\; \boxed{\ddot x=\frac{(m_1-m_2)g}{m_1+m_2}}$ Verify: equal masses ⇒ $\ddot x=0$ (balance), one mass zero ⇒ free fall $g$ . ✔ No tension needed.

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

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

Worked example Particle in potential

$V(r)$ , plane motion. Coordinates $(r,\theta)$ . Step 2 — $T$ in polars: velocity components are $\dot r$ (radial) and $r\dot\theta$ (tangential): $T=\tfrac12 m(\dot r^2 + r^2\dot\theta^2).$ Why $r^2\dot\theta^2$ ? Tangential speed is $r\dot\theta$ ; square it.

$L = \tfrac12 m(\dot r^2 + r^2\dot\theta^2) - V(r).$

$\theta$ -equation: $L$ has no explicit $\theta$ ⇒ $\partial L/\partial\theta=0$ : $\frac{d}{dt}(m r^2\dot\theta)=0 \;\Rightarrow\; \boxed{L_z=mr^2\dot\theta=\text{const}}$ Why this matters: an ignorable (cyclic) coordinate gives a conserved momentum — here angular momentum, for free. (See 2.1.07-Cyclic-coordinates-and-conservation-laws.)

$r$ -equation: $\dfrac{\partial L}{\partial\dot r}=m\dot r,\ \dfrac{\partial L}{\partial r}=mr\dot\theta^2 - V'(r)$ : $\boxed{m\ddot r = mr\dot\theta^2 - \frac{dV}{dr}}$ The $mr\dot\theta^2$ is the centrifugal term — it appears automatically.

Common mistakes (steel-manned)

Active recall

Recall Quick self-test (cover the answers!)

What is $L$ ? → $T-V$ .
How many coordinates do you need? → one per DOF.
What does a cyclic coordinate give you? → a conserved generalized momentum.
Why don't tension/normal force appear? → ideal constraints do no work; coordinates absorb them.
Pendulum EOM? → $\ddot\theta=-(g/\ell)\sin\theta$ .

Recall Feynman: explain to a 12-year-old

Imagine you want to predict how a swing moves. The hard way: track every push and pull, including the rope's tug. The easy way: just keep a "score" — how much moving energy it has minus how much height energy it has. Nature is lazy and always picks the path that keeps this score "balanced" over time. We wrote down a magic rule (the E–L equation) that, given the energy score in whatever measuring stick you like (angle, distance, whatever), automatically spits out exactly how the thing moves — and you never had to think about the rope at all.

The Lagrangian

L

equals

T - V

(kinetic minus potential energy)

Number of generalized coordinates required

one per degree of freedom

Euler–Lagrange equation for coordinate

q_i

\frac{d}{dt}\frac{\partial L}{\partial\dot q_i}-\frac{\partial L}{\partial q_i}=0

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

they do no work and are absorbed by the choice of independent coordinates

Kinetic energy in plane polar coordinates

T=\tfrac12 m(\dot r^2 + r^2\dot\theta^2)

A cyclic (ignorable) coordinate (absent from

L

) implies

its conjugate momentum

\partial L/\partial\dot q

is conserved

EOM of a simple pendulum

\ddot\theta = -(g/\ell)\sin\theta

Atwood machine acceleration

\ddot x = (m_1-m_2)g/(m_1+m_2)

Block on frictionless incline acceleration along slope

g\sin\alpha

Conserved quantity for central force motion

angular momentum

L_z=mr^2\dot\theta

First step of the E–L recipe

count degrees of freedom and pick that many coordinates

Inside the partial derivatives of

L

, how are

q

and

\dot q

treated

as independent variables

Connections

2.1.05-Deriving-the-Euler-Lagrange-equation — where the recipe comes from.
2.1.07-Cyclic-coordinates-and-conservation-laws — exploiting ignorable coordinates.
2.1.08-Generalized-momenta-and-the-Hamiltonian — next step after E–L.
Constraints-and-degrees-of-freedom — choosing good coordinates.
Noethers-theorem — symmetry ↔ conservation, generalizing Example 4.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Newton ke method mein har jagah forces, directions aur constraint forces (tension, normal reaction) ka hisaab rakhna padta hai — kaafi headache. Euler–Lagrange (E–L) method bolta hai: bhai, sirf energy sambhal. Tum bas kinetic energy $T$ aur potential energy $V$ likho, $L=T-V$ banao, aur ek fix formula mein daal do. Bas, equation of motion nikal aati hai. Sabse pyaari baat — tension aur normal force kabhi aate hi nahi, kyunki acche coordinates choose karne se woh apne aap "absorb" ho jaate hain (kyunki woh koi work nahi karte).

Recipe yaad rakho — C-T-V-L-E: pehle Count degrees of freedom (utne hi coordinates lo), phir $T$ likho, phir $V$ , phir $L=T-V$ , phir E–L equation $\frac{d}{dt}\frac{\partial L}{\partial \dot q}-\frac{\partial L}{\partial q}=0$ apply karo. Pendulum mein $\theta$ lo, Atwood mein ek $x$ , incline pe slope ke along $s$ , aur central force mein polar $(r,\theta)$ . Polar mein dhyan rakho — $T=\frac12 m(\dot r^2 + r^2\dot\theta^2)$ , sirf $\dot r^2$ likhna bada common mistake hai.

Ek aur jhakaas cheez: agar kisi coordinate (jaise $\theta$ central force mein) $L$ mein explicitly aata hi nahi, toh uska momentum conserve ho jaata hai — angular momentum free mein mil gaya! Isko cyclic coordinate kehte hain. Yeh method itna systematic hai ki ek baar coordinates set kar liye, baaki sab mechanical calculation hai — isliye exam mein speed aur accuracy dono milti hai. Practice karo 3-4 systems, pattern haath mein aa jaayega.

The recipe (what you actually do)

Example 1 — Simple Pendulum (1 DOF)

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

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

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Analytical Mechanics

Connections