Worked examples — Applying E-L equations to various systems

Prerequisites you should already own: 2.1.05-Deriving-the-Euler-Lagrange-equation (where the recipe comes from), Constraints-and-degrees-of-freedom (how to count the coordinates you need), and — for the "conserved quantity" cells — 2.1.07-Cyclic-coordinates-and-conservation-laws.

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The one rule we use on every example

Before we touch a single system, let us pin the tool to the wall so no example has to assume you remember it.

Every example below is just this rule, applied. Read the "Forecast:" line and guess before you scroll.

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The scenario matrix

Every E–L problem lives in one of these case classes. The columns are the kinds of trouble a problem can hide; the examples that follow are tagged with the cell they cover.

Cell	Case class	What makes it tricky	Covered by
A	1 DOF, clean	just turn the crank	Ex 1
B	Sign / direction bookkeeping	which way is "positive"? PE signs	Ex 2
C	Degenerate input (a mass, length, or angle $\to 0$)	formula must survive the limit	Ex 3
D	Limiting / extreme geometry (angle $\to 90°$, $\to 0°$)	check against known physics	Ex 4
E	Non-conservative / driven (external applied force)	plain $L=T-V$ is not enough	Ex 5
F	Coupled 2 DOF (two equations, cross-terms)	${\ddot{q}}_1$ and ${\ddot{q}}_2$ tangle	Ex 6
G	Cyclic coordinate → conserved momentum	spot the missing coordinate	Ex 7
H	Real-world word problem	translate words → $T,V$	Ex 8
I	Exam-style twist (moving support / time in $L$)	$t$ appears explicitly	Ex 9

We will pass through all nine cells with nine examples.

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Example 1 — the clean 1-DOF baseline · Cell A

Step 1 — Count DOF. Why this step? Everything starts by asking "how many independent numbers do I need to locate the system?" The block moves on one straight rail ⇒ 1 DOF, coordinate $x$.

Step 2 — Write $T$. Why? Kinetic energy is $\frac{1}{2}m{v}^2$ and here $v=\dot{x}$: $T=\frac{1}{2}m{\dot{x}}^2.$

Step 3 — Write $V$. Why? A stretched/compressed spring stores $\frac{1}{2}k{x}^2$ — the standard elastic potential: $V=\frac{1}{2}k{x}^2.$

Step 4 — Form $L$ and its two partial derivatives. Why two? The E–L rule needs $\partial L/\partial \dot{x}$ (the "momentum slot") and $\partial L/\partial x$ (the "force slot").

\frac{\partial L}{\partial\dot x}=m\dot x,\qquad \frac{\partial L}{\partial x}=-kx.$$ **Step 5 — Plug into E–L.** *Why the $d/dt$ on the first term only?* Because the rule is $\dfrac{d}{dt}\dfrac{\partial L}{\partial\dot x}-\dfrac{\partial L}{\partial x}=0$: $$m\ddot x-(-kx)=0\;\Rightarrow\;\boxed{m\ddot x=-kx}.$$ > [!formula] Result > $$\ddot x=-\frac{k}{m}\,x,\qquad \omega=\sqrt{\frac{k}{m}}.$$ **Verify:** With $m=2\ \text{kg}$, $k=8\ \text{N/m}$: $\omega=\sqrt{8/2}=2\ \text{rad/s}$. Units: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{\text{s}^{-2}}=\text{s}^{-1}$ ✔. This is exactly Newton's $F=-kx$ — the baseline works. --- ## Example 2 — sign bookkeeping done wrong then right · Cell B > [!example] Two beads, one pulley, chosen the "dangerous" way > Redo the Atwood machine ($m_1,m_2$, frictionless pulley, inextensible string) but this time let $y$ measure how far $m_2$ has **risen**. Get the acceleration of $m_2$. > > **Forecast:** will the sign of the answer flip compared to the parent note's version (which used "$m_1$ descends")? Guess yes/no. **Step 1 — DOF & coordinate.** *Why one coordinate?* The string couples the two masses: fix one and the other is determined ⇒ **1 DOF**. Our chosen $y$ = rise of $m_2$. Then $m_1$ **descends** by the same $y$ (string length fixed). **Step 2 — $T$.** *Why both move at $\dot y$?* Inextensibility forces equal speeds: $$T=\tfrac12 m_1\dot y^2+\tfrac12 m_2\dot y^2=\tfrac12(m_1+m_2)\dot y^2.$$ **Step 3 — $V$ (the sign minefield).** *Why sign care matters here:* PE **increases** when a mass goes **up**. $m_2$ rises by $y$ ⇒ its PE is $+m_2 g y$. $m_1$ falls by $y$ ⇒ its PE is $-m_1 g y$: $$V=m_2 g y-m_1 g y=(m_2-m_1)g\,y.$$ **Step 4 — Partial derivatives.** *Why these two:* the momentum slot $\partial L/\partial\dot y$ feeds the outer $d/dt$, and the force slot $\partial L/\partial y$ is subtracted from it, per the rule. $$\frac{\partial L}{\partial\dot y}=(m_1+m_2)\dot y,\qquad \frac{\partial L}{\partial y}=-(m_2-m_1)g=(m_1-m_2)g.$$ **Step 5 — Assemble into E–L.** *Why we differentiate the first slot in time:* the rule says rate-of-change of momentum minus force $=0$; here $\frac{d}{dt}(m_1+m_2)\dot y=(m_1+m_2)\ddot y$: $$(m_1+m_2)\ddot y-(m_1-m_2)g=0\;\Rightarrow\;\boxed{\ddot y=\frac{(m_1-m_2)g}{m_1+m_2}}.$$ **Verify:** If $m_1>m_2$ (heavier mass descends) the answer is **positive** ⇒ $m_2$ accelerates *upward* — exactly what physical sense demands. With $m_1=3,\,m_2=1,\,g=9.8$: $\ddot y=\frac{2}{4}\cdot 9.8=4.9\ \text{m/s}^2$. Same magnitude as the parent's result — the physics is coordinate-independent, only the *interpretation of the sign* changed. ✔ --- ## Example 3 — degenerate input: a mass goes to zero · Cell C > [!example] Pendulum with a shrinking bob > Take the simple pendulum EOM $\ddot\theta=-\dfrac{g}{\ell}\sin\theta$. What happens as $m\to 0$? As $\ell\to 0$? Does the machine break? > > **Forecast:** guess whether $m\to0$ changes the motion. **Step 1 — Recall where $m$ sits.** *Why look back?* Before taking a limit, see *whether the quantity even appears* in the final law: $$L=\tfrac12 m\ell^2\dot\theta^2-mg\ell(1-\cos\theta).$$ Every term has one factor of $m$. **Step 2 — Factor it out of the E–L equation.** *Why this is the key move:* the E–L equation sets a sum of terms equal to zero, and **every** term carries exactly one factor of $m$. An equation of the form $m\cdot(\text{stuff})=0$ with $m\neq0$ is *equivalent* to $(\text{stuff})=0$ — dividing both sides by the same nonzero number never changes the solutions. That is *why* we are allowed to cancel $m$ before doing anything with limits: $$\frac{d}{dt}(m\ell^2\dot\theta)+mg\ell\sin\theta=0 \;\xrightarrow{\ \div\, m\ (m\neq0)\ }\; \ell^2\ddot\theta+g\ell\sin\theta=0.$$ So **$m$ has vanished from the EOM** for any $m>0$. **Step 3 — The $m\to 0$ limit.** *Why it's subtle:* $m=0$ exactly means $L\equiv0$ — no dynamics defined. But the *limit* $m\to0^+$ leaves $\ddot\theta=-\frac{g}{\ell}\sin\theta$ unchanged. Physics: a lighter bob swings at the *same* rate (Galileo's insight). **Step 4 — The $\ell\to 0$ limit.** *Why dangerous:* $\ell$ does **not** fully cancel — it appears to *different powers* in the two terms ($\ell^2$ and $\ell^1$). Start from the equation of Step 2, $\ell^2\ddot\theta+g\ell\sin\theta=0$. Both terms share **one** common factor of $\ell$, and $\ell\neq0$, so by the same "divide out a nonzero number" logic we divide both sides by $\ell^2$ (equivalently: pull out $\ell$, cancel it, then isolate $\ddot\theta$): $$\ell^2\ddot\theta+g\ell\sin\theta=0 \;\xrightarrow{\ \div\,\ell^2\ (\ell\neq0)\ }\; \ddot\theta+\frac{g}{\ell}\sin\theta=0 \;\Rightarrow\; \boxed{\ddot\theta=-\frac{g}{\ell}\sin\theta}.$$ Now take the limit: $\ \ell\to0^+\ \Rightarrow\ \ddot\theta\to-\infty$. A vanishingly short pendulum oscillates infinitely fast ($\omega=\sqrt{g/\ell}\to\infty$). The formula survives — it just predicts an extreme. **Verify:** $\omega=\sqrt{g/\ell}$. For $\ell=1\ \text{m},\,g=9.8$: $\omega=\sqrt{9.8}\approx3.130\ \text{rad/s}$. For $\ell=0.01\ \text{m}$: $\omega=\sqrt{980}\approx31.30\ \text{rad/s}$ — ten times faster, matching $\omega\propto\ell^{-1/2}$. ✔ --- ## Example 4 — limiting geometry: incline angle at its extremes · Cell D > [!example] Incline, all the way from flat to vertical > A block slides on a frictionless incline of angle $\alpha$. **Coordinate convention:** let $s$ be the distance travelled **down the slope**, measured positive in the down-slope direction, starting from $s=0$. With this choice the along-slope EOM is $\ddot s=g\sin\alpha$ (parent Ex 3, same down-slope-positive convention). Check the two **limits** $\alpha\to0°$ and $\alpha\to90°$, and the sign of $\ddot s$ everywhere in between. > > **Forecast:** at $\alpha=90°$ what should $\ddot s$ equal? Say it before reading. The figure below shows the setup: a right triangle whose hypotenuse is the slope. The **burnt-orange arrow** labelled $s$ points *down the slope* (our positive direction); the **teal arrow** shows the resulting vertical drop $s\sin\alpha$; the **plum arc** at the base marks the incline angle $\alpha$. ![[deepdives/dd-physics-2.1.06-d3-s01.png]] **Step 1 — Read the geometry off the figure.** *Why a figure here?* The whole result rides on one right triangle: for a down-slope slide of length $s$ (orange arrow), the vertical drop is the **opposite** side of the angle $\alpha$, namely $s\sin\alpha$ (teal arrow). Because $s$ is positive *down-slope*, a positive $\ddot s$ means "accelerating downhill." **Step 2 — The flat limit $\alpha\to 0°$.** *Why check it:* zero slope should give zero acceleration. $\sin 0°=0\Rightarrow\ddot s=0$ — the block just sits. ✔ **Step 3 — The vertical limit $\alpha\to 90°$.** *Why check it:* a vertical "incline" is free fall. $\sin90°=1\Rightarrow\ddot s=g$ — exactly free fall, and since $s$ points down-slope (now straight down), positive means falling. ✔ **Step 4 — In between, the sign.** *Why sign matters:* $\sin\alpha>0$ for every $0°<\alpha<90°$, so $\ddot s>0$ throughout — with our down-slope-positive convention the block always accelerates *down-slope*, never up. There is no quadrant where the formula flips sign because a real incline lives only in $(0°,90°)$. **Verify:** $\alpha=30°$: $\ddot s=g\cdot0.5=4.9\ \text{m/s}^2$. $\alpha=90°$: $\ddot s=9.8\ \text{m/s}^2$. Units: $g\sin\alpha=\text{m/s}^2\times(\text{dimensionless})=\text{m/s}^2$ ✔. --- ## Example 5 — driven / non-conservative system · Cell E > [!example] Pushed cart on a spring > The Example-1 block ($m$, spring $k$) is now pushed by an external, time-dependent horizontal force $F(t)$ (say from your hand). This force is **not** derivable from a potential. Find the EOM. > > **Forecast:** where does $F(t)$ enter — into $T$, into $V$, or somewhere new? **Step 1 — Why plain $L=T-V$ is not enough.** *Why:* $L=T-V$ only captures forces that come from a stored potential. A hand-push has no potential; we need the **generalized-force** form of E–L: $$\frac{d}{dt}\frac{\partial L}{\partial\dot q}-\frac{\partial L}{\partial q}=Q_q,$$ where $Q_q$ is the non-conservative generalized force in the $q$ slot. *This is the standard extension* — the left side is still $L=T-V$ of the conservative parts. **Step 2 — Build $L$ from the conservative parts only.** $T=\tfrac12 m\dot x^2$, $V=\tfrac12 k x^2$: $$L=\tfrac12 m\dot x^2-\tfrac12 k x^2.$$ **Step 3 — Find $Q_x$.** *Why it equals $F$ here:* $Q_x$ is the work done per unit displacement of $x$. Since $F$ acts *along* $x$, $Q_x=F(t)$ directly (no projection needed). **Step 4 — Assemble.** $\dfrac{\partial L}{\partial\dot x}=m\dot x$, $\dfrac{\partial L}{\partial x}=-kx$: $$m\ddot x-(-kx)=F(t)\;\Rightarrow\;\boxed{m\ddot x+kx=F(t)}.$$ **Verify:** This is the driven-oscillator equation from ordinary Newtonian mechanics ($ma = -kx + F$) ✔. Turn the drive off, $F=0$, and it collapses back to Example 1. With $m=2,\,k=8,\,F=4\ \text{N}$ held constant, the equilibrium shift is $x_{\text{eq}}=F/k=4/8=0.5\ \text{m}$ ✔. --- ## Example 6 — coupled two-DOF system · Cell F > [!example] Double-spring chain (two masses, two springs) > Masses $m$ and $m$ on a frictionless rail. Wall–spring($k$)–mass$_1$–spring($k$)–mass$_2$ (free end). Let $x_1,x_2$ be displacements from equilibrium. Find both EOMs. > > **Forecast:** will each mass's equation contain *only its own* coordinate, or the other's too? The figure below shows the chain: a wall on the left, then a spring (teal) to the first mass (orange dot, displacement $x_1$), then a middle spring (plum) to the second mass (orange dot, displacement $x_2$). The key visual is that the **middle spring stretches by the difference $x_2-x_1$**, not by either displacement alone. ![[deepdives/dd-physics-2.1.06-d3-s02.png]] **Step 1 — DOF.** *Why two:* two masses free to slide ⇒ **2 DOF**, coordinates $x_1,x_2$ (the two orange dots in the figure). **Step 2 — $T$.** *Why this step?* We need kinetic energy in the chosen coordinates; the two masses move independently so their kinetic energies simply add: $$T=\tfrac12 m\dot x_1^2+\tfrac12 m\dot x_2^2.$$ **Step 3 — $V$ (where the coupling hides).** *Why $(x_2-x_1)$:* the middle spring stretches by the **difference** of the two displacements, not by either alone (see the plum spring in the figure): $$V=\tfrac12 k x_1^2+\tfrac12 k (x_2-x_1)^2.$$ **Step 4 — Partial derivatives, one per coordinate.** *Why one E–L equation each:* the rule fires **separately** for every degree of freedom, so we need a momentum slot and force slot for both $x_1$ and $x_2$. $$\frac{\partial L}{\partial\dot x_1}=m\dot x_1,\quad \frac{\partial L}{\partial x_1}=-kx_1+k(x_2-x_1),$$ $$\frac{\partial L}{\partial\dot x_2}=m\dot x_2,\quad \frac{\partial L}{\partial x_2}=-k(x_2-x_1).$$ **Step 5 — Two E–L equations.** *Why differentiate each momentum slot in time:* $\frac{d}{dt}(m\dot x_i)=m\ddot x_i$, then subtract each force slot: $$\boxed{m\ddot x_1=-2k x_1+k x_2},\qquad \boxed{m\ddot x_2=k x_1-k x_2}.$$ *Why "coupled":* $x_2$ sits in mass-1's equation and vice versa — they cannot move independently, exactly what the shared middle spring should do. **Verify:** A genuine motion of a coupled chain is a **normal mode** — a pattern $x_1=A_1\cos\omega t,\ x_2=A_2\cos\omega t$ that oscillates at a single frequency. Substituting $\ddot x_i=-\omega^2 x_i$ turns the two boxed equations into $-m\omega^2 A_1=-2kA_1+kA_2$ and $-m\omega^2 A_2=kA_1-kA_2$. Setting the determinant of this $2\times2$ system to zero gives $m^2\omega^4-3km\omega^2+k^2=0$, i.e. $\omega^2=\frac{k}{2m}(3\pm\sqrt5)$ — two real positive frequencies, as a stable coupled system must have. Symmetry check: swapping the labels is *not* a symmetry because only mass 1 touches the wall, and indeed the two equations are asymmetric ✔. --- ## Example 7 — cyclic coordinate → conserved momentum · Cell G > [!example] Free particle sliding in the horizontal plane (no potential) > A puck slides freely in the horizontal plane, no forces, described in polar coordinates $(r,\theta)$. Find the conserved quantity **without solving the full motion**. > > **Forecast:** name the conserved quantity before deriving it. **Step 1 — $L$ for a free particle in polars.** *Why polars:* they expose the hidden symmetry. From the parent, $T=\tfrac12 m(\dot r^2+r^2\dot\theta^2)$ and $V=0$: $$L=\tfrac12 m(\dot r^2+r^2\dot\theta^2).$$ **Step 2 — Spot the missing coordinate.** *Why this is the whole trick:* $\theta$ **does not appear** in $L$ (only $\dot\theta$ does). A coordinate absent from $L$ is called **cyclic**; see [[2.1.07-Cyclic-coordinates-and-conservation-laws]]. **Step 3 — E–L for $\theta$.** *Why it gives conservation:* the force slot is $\partial L/\partial\theta=0$, so the rule collapses to $\frac{d}{dt}(\partial L/\partial\dot\theta)=0$ — the momentum slot is constant in time: $$\frac{\partial L}{\partial\dot\theta}=mr^2\dot\theta\;\Rightarrow\;\boxed{p_\theta=mr^2\dot\theta=\text{const}}.$$ This is the **angular momentum** — conserved for free by the symmetry, tying to [[Noethers-theorem]] (every continuous symmetry ⇒ a conservation law). **Verify (numeric, straight-line free motion).** A free puck moves in a straight line at constant speed $v$. Let its perpendicular distance from the origin (impact parameter) be $b$. The angular momentum per unit mass in polars is $r^2\dot\theta$, and geometry says $r^2\dot\theta = r\cdot(r\dot\theta) = r\cdot v_\perp$, where $v_\perp$ is the tangential speed. At closest approach $r=b$ and the whole velocity is tangential, so $v_\perp=v$ giving $r^2\dot\theta=bv$. Farther out at $r=2b$, the tangential component drops to $v_\perp = v\cdot\frac{b}{2b}=\frac{v}{2}$, so $r^2\dot\theta=(2b)\cdot\frac{v}{2}=bv$ — **the same value** $bv$. Take $b=1,\,v=1$: both give $1$. Constant confirmed ✔. --- ## Example 8 — real-world word problem · Cell H > [!example] A child's slide at the playground > A child (mass $m=20\ \text{kg}$) starts from rest at the top of a straight, frictionless slide of length $L=3\ \text{m}$ inclined at $\alpha=40°$. Using E–L, how fast are they going at the bottom? Take $g=9.8\ \text{m/s}^2$. > > **Forecast:** estimate the speed (m/s) before computing. **Step 1 — Translate words to coordinate.** *Why:* "straight slide" ⇒ 1 DOF, $s$ = distance travelled **down** the slope (same down-slope-positive convention as Example 4), starting at $s=0$. **Step 2 — $T,V$.** *Why:* build the two energies in the chosen coordinate. $T=\tfrac12 m\dot s^2$; height dropped $=s\sin\alpha$ so $V=-mg s\sin\alpha$. **Step 3 — EOM (reuse Cell D result).** *Why reuse:* it's the same incline geometry, so the E–L machine gives the same law: $$\ddot s=g\sin\alpha=9.8\sin40°.$$ $\sin40°\approx0.6428$, so $\ddot s\approx6.299\ \text{m/s}^2$ (constant). **Step 4 — Find the exit speed.** *Why kinematics now:* constant acceleration from rest over distance $L$: $v^2=2\,\ddot s\,L$: $$v=\sqrt{2(6.299)(3)}=\sqrt{37.79}\approx6.147\ \text{m/s}.$$ **Verify (energy cross-check):** Energy says $\tfrac12 m v^2=mg L\sin\alpha\Rightarrow v=\sqrt{2gL\sin\alpha}=\sqrt{2(9.8)(3)(0.6428)}=\sqrt{37.80}\approx6.148\ \text{m/s}$ ✔. Notice $m$ cancelled — a heavier child arrives at the *same* speed. Units: $\sqrt{\text{m/s}^2\cdot\text{m}}=\text{m/s}$ ✔. --- ## Example 9 — exam twist: time appears explicitly (moving support) · Cell I > [!example] Pendulum whose pivot is dragged sideways at constant speed > The pivot of a simple pendulum (length $\ell$, bob $m$) is forced to move horizontally as $X(t)=ut$ (constant speed $u$). Coordinate: the swing angle $\theta$. Find the EOM. > > **Forecast:** does a *constant-velocity* moving pivot change the pendulum's equation? Guess. **Step 1 — Position of the bob.** *Why in Cartesian first:* with a moving pivot the velocity has two contributions we must add carefully. With pivot at $(ut,0)$: $$x=ut+\ell\sin\theta,\qquad y=-\ell\cos\theta.$$ **Step 2 — Velocities.** *Why differentiate both terms:* the pivot drift $u$ adds to the swing: $$\dot x=u+\ell\dot\theta\cos\theta,\qquad \dot y=\ell\dot\theta\sin\theta.$$ **Step 3 — $T$.** Square and add. *Why the cross term appears:* expanding $\dot x^2$ mixes $u$ and $\ell\dot\theta\cos\theta$: $$T=\tfrac12 m\big(u^2+2u\ell\dot\theta\cos\theta+\ell^2\dot\theta^2\big).$$ **Step 4 — $V$ and $L$.** $V=mg y=-mg\ell\cos\theta$: $$L=\tfrac12 m u^2+m u\ell\dot\theta\cos\theta+\tfrac12 m\ell^2\dot\theta^2+mg\ell\cos\theta.$$ **Step 5 — E–L for $\theta$.** *Why be careful with the cross term:* $$\frac{\partial L}{\partial\dot\theta}=mu\ell\cos\theta+m\ell^2\dot\theta,\qquad \frac{d}{dt}\!\left(\frac{\partial L}{\partial\dot\theta}\right)=-mu\ell\dot\theta\sin\theta+m\ell^2\ddot\theta.$$ $$\frac{\partial L}{\partial\theta}=-mu\ell\dot\theta\sin\theta-mg\ell\sin\theta.$$ Subtract: the two $-mu\ell\dot\theta\sin\theta$ terms **cancel**: $$m\ell^2\ddot\theta+mg\ell\sin\theta=0\;\Rightarrow\;\boxed{\ddot\theta=-\frac{g}{\ell}\sin\theta}.$$ **Verify:** *Identical* to the stationary pendulum! *Why it must be:* a pivot moving at **constant** velocity is just an inertial frame (Galilean relativity) — no new force. Had $X(t)=\tfrac12 a t^2$ (accelerating pivot), the cross term would leave an extra $-\frac{a}{\ell}\cos\theta$ and the equation would change. So the twist teaches: only the *acceleration* of the support matters. ✔ --- ## Active recall > [!recall]- Which cell is which? (cover the answers) > - A mass or length $\to 0$ and the formula still holds — which cell? ::: Cell C (degenerate input) > - A hand-push with no potential — which cell, which extra symbol? ::: Cell E; generalized force $Q_q$ > - $\theta$ missing from $L$ gives you what? ::: a conserved momentum $p_\theta$ (Cell G) > - Two coordinates appearing in each other's equations — name it. ::: coupled system (Cell F) > - A constant-velocity moving pivot changes the pendulum EOM? ::: No — inertial frame; only acceleration of the support matters (Cell I) > [!mnemonic] The nine cells > **A**–clean, **B**–signs, **C**–zero, **D**–extremes, **E**–external force, **F**–coupled, **G**–cyclic, **H**–word, **I**–time. *"A Bad Cat Dashed, Every Fluffy Gato Hid Inside."*