Visual walkthrough — Applying E-L equations to various systems

We are studying a single ball hanging from a stiff, weightless stick that can swing. Nothing more. Let us name the picture before we do any mathematics.

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Step 1 — Draw the system and name every letter

WHAT. We have a ball (call its mass $m$) at the end of a rigid rod. The rod has a fixed length; we call that length $\ell$ (the little script "L", read "ell"). The top of the rod is pinned so it cannot move; the ball swings around that pin. The angle the rod makes with the straight-down direction we call $\theta$ (the Greek letter "theta").

WHY. Before writing any equation we must agree what each symbol points at in the picture, otherwise later formulas are just noise. $m$ is "how heavy", $\ell$ is "how long the rod is", $\theta$ is "how far it has swung, measured as an angle from straight down".

PICTURE. In the figure, the black dot is the pin, the line is the rod of length $\ell$, the coloured ball is the mass $m$, and the little wedge angle at the top is $\theta$. The dashed line is "straight down" — the place where $\theta =0$.

Why radians and not degrees? Because when we talk about how fast the angle changes, we will multiply it by the radius $\ell$ to get a real speed — and that "angle × radius = arc length" trick is only true when the angle is in radians. Degrees would need an ugly conversion factor every single time.

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Step 2 — How fast is the ball actually moving?

WHAT. The ball is stuck on a circle of radius $\ell$ (the rod cannot stretch). We want its speed — how many metres per second it covers. We introduce $\dot{\theta }$ (theta with a dot on top), which means "how fast $\theta$ is changing per second", the angular speed.

WHY. To use energy we will need the ball's linear speed $v$. The ball moves along the circle, so its speed is the rate at which arc-length is swept.

PICTURE. Look at the red arc. If the ball moves through a tiny extra angle $d\theta$, it travels along the arc a tiny distance $\ell d\theta$ (radius times angle, radians again). Divide by the tiny time $dt$: the speed is $\ell \dot{\theta }$.

The dot notation ($\dot{\theta }$) is just shorthand for $\frac{d\theta }{dt}$. Why a derivative here? Because "speed" is by definition rate of change of position, and the only thing changing is $\theta$. The derivative is the exact tool that answers "how fast is this quantity moving right now?"

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Step 3 — Write the moving energy $T$

WHAT. Kinetic energy — the "energy of moving" — is $T=\frac{1}{2}m{v}^2$ for any mass. We plug in the speed we just found.

WHY. The whole Lagrangian method (recipe in 2.1.05-Deriving-the-Euler-Lagrange-equation) runs on energies, not forces. Kinetic energy is one of the two ingredients. We need it written using only our coordinate $\theta$ and its speed $\dot{\theta }$, nothing else.

PICTURE. Substituting $v=\ell \dot{\theta }$ into $\frac{1}{2}m{v}^2$: the arrow shows $v$ getting replaced by the arc-speed and then squared.

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Step 4 — Write the height energy $V$

WHAT. Potential energy of gravity is $V=mgh$, where $g$ is gravity's strength ($\approx 9.8$) and $h$ is height above some chosen "zero height". We choose the lowest point of the swing as $h=0$.

WHY. Gravity is a conservative force — its effect is fully captured by a height-energy $V$, so we never need to draw the gravity arrow into the equations. We just need $h$ as a function of $\theta$.

PICTURE. The rod points straight down at length $\ell$. When it swings to angle $\theta$, the ball rises. The vertical drop from the pin is $\ell \cos \theta$ (adjacent side of the right triangle formed by the rod). At the bottom that drop was $\ell$. So the height gained is $\ell -\ell \cos \theta =\ell (1-\cos \theta )$. Look at the green bracket in the figure marking this height.

Edge check — all angles. At $\theta =\pi$ (ball straight up) $\cos \pi =-1$, so $V=mg\ell (2)$ — the maximum height $2\ell$, exactly right. For negative $\theta$ (swinging the other way) $\cos (-\theta )=\cos \theta$, so $V$ is the same on both sides — the pendulum is symmetric, as it should be.

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Step 5 — Build the Lagrangian $L$ and take its two derivatives

WHAT. The Lagrangian is $L=T-V$ (moving energy minus height energy). Then the recipe needs two specific derivatives: one with respect to the speed $\dot{\theta }$, one with respect to the angle $\theta$.

WHY. The Euler–Lagrange machine (proven in 2.1.05-Deriving-the-Euler-Lagrange-equation) needs exactly these two partial derivatives. A partial derivative $\frac{\partial L}{\partial \theta }$ means "differentiate treating everything else — including $\dot{\theta }$ — as frozen constants". We freeze the other variable so we isolate one influence at a time.

PICTURE. The figure splits $L$ into its two pieces and shows which piece survives each derivative: the ${\dot{\theta }}^2$ piece feeds $\partial L/\partial \dot{\theta }$; the $\cos \theta$ piece feeds $\partial L/\partial \theta$.

$L=\underset{T}{\underbrace{\frac{1}{2}m{\ell }^2{\dot{\theta }}^2}}-\underset{V}{\underbrace{mg\ell (1-\cos \theta )}}$

Now the momentum-derivative (freeze $\theta$, vary $\dot{\theta }$): $\frac{\partial L}{\partial \dot{\theta }}=\frac{1}{2}m{\ell }^2\cdot 2\dot{\theta }=\underset{\text{"angular momentum"}}{\underbrace{m{\ell }^2\dot{\theta }}}$ Here the $V$ part has no $\dot{\theta }$ in it, so it contributes zero — only $T$ survives. Power rule on ${\dot{\theta }}^2$ brings down the 2, cancelling the $\frac{1}{2}$.

Now the angle-derivative (freeze $\dot{\theta }$, vary $\theta$): $\frac{\partial L}{\partial \theta }=0-mg\ell \cdot \frac{d}{d\theta }(1-\cos \theta )=-mg\ell \sin \theta$ The $T$ part has no plain $\theta$, so it gives zero. Differentiating $-\cos \theta$ gives $+\sin \theta$; the leading minus in $V=-\dots$ flips it, landing on $-mg\ell \sin \theta$.

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Step 6 — Feed both pieces into the Euler–Lagrange rule

WHAT. The rule is $\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta }}\right)-\frac{\partial L}{\partial \theta }=0.$ We drop in the two derivatives from Step 5.

WHY. This single equation is Newton's second law in disguise — but reached without ever drawing tension or the pin's reaction force. Those constraint forces did no work, so they vanished automatically when we chose $\theta$ as our one coordinate.

PICTURE. The figure shows the two pieces sliding into their slots, and the outer $\frac{d}{dt}$ acting on the first slot.

First the time derivative of the momentum piece (now $\dot{\theta }$ genuinely changes with time, so it becomes $\ddot{\theta }$, angular acceleration): $\frac{d}{dt}\left(m{\ell }^2\dot{\theta }\right)=m{\ell }^2\ddot{\theta }.$ Then assemble: $m{\ell }^2\ddot{\theta }-(-mg\ell \sin \theta )=0\Rightarrow m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0.$ Divide through by the common factor $m{\ell }^2$ (both nonzero): $\boxed{\ddot{\theta }=-\frac{g}{\ell }\sin \theta }$ Term by term: $\ddot{\theta }$ is how the swing speeds up or slows down; the minus sign says the pull is always back toward the bottom; $\frac{g}{\ell }$ sets how strong that restoring pull is; $\sin \theta$ says the pull grows with how far you have swung.

Notice $m$ cancelled — a heavy and a light bob swing identically. Gravity accelerates all masses alike, and the Lagrangian shows it falling out on its own.

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Step 7 — Sanity-check every regime

WHAT. A formula you cannot check is a formula you cannot trust. We test the boxed result in the limits where we already know the answer.

WHY. Covering edge cases is the difference between "I derived it" and "I believe it".

PICTURE. Three mini-panels: tiny swing (looks like a spring), hanging still (no motion), and balanced upright (unstable).

• Small swings ($\theta \approx 0$). For tiny angles $\sin \theta \approx \theta$, so $\ddot{\theta }\approx -\frac{g}{\ell }\theta$. This is the equation of a spring-like wobble — simple harmonic motion — with rhythm $\omega =\sqrt{g/\ell }$. That is the textbook pendulum period $T_{\text{period}}=2\pi \sqrt{\ell /g}$. ✔
• At rest at the bottom ($\theta =0$, $\dot{\theta }=0$). Then $\sin 0=0$, so $\ddot{\theta }=0$: it stays put. A hanging pendulum at the bottom is in equilibrium. ✔
• Balanced straight up ($\theta =\pi$). $\sin \pi =0$ so again $\ddot{\theta }=0$ — an equilibrium — but the tiniest nudge makes $\sin \theta$ push away, not back. This is the unstable equilibrium, exactly what real experience says. ✔
• Longer rod (bigger $\ell$). $\frac{g}{\ell }$ shrinks, so the restoring push weakens and the swing slows — long pendulums tick slower. ✔

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The one-picture summary

Everything above compressed: the picture defines $\theta ,\ell ,m$; the arc gives speed $\ell \dot{\theta }$; that builds $T$; the geometry gives height $\ell (1-\cos \theta )$; that builds $V$; $L=T-V$; two derivatives; the E–L rule; the boxed law — all on one canvas.

Recall Feynman retelling — the whole walk in plain words

Picture a ball on a stiff stick, swinging. First we agree on names: how heavy ($m$), how long the stick ($\ell$), how far it's swung ($\theta$). Because the ball rides a circle, its speed is just "how fast the angle turns" times "how long the stick is" — that's $\ell \dot{\theta }$. From speed we get its moving energy, $T$. Then we notice that swinging up lifts the ball; the lift is $\ell (1-\cos \theta )$, and gravity turns lift into height energy, $V$. Nature keeps a running score of moving-minus-height energy, called $L$. There's a magic rule that turns this score into the law of motion: take how the score responds to the swing's speed, watch how that changes in time, and subtract how the score responds to the angle itself. Doing that bookkeeping — and the mass politely cancels out — leaves one clean sentence: the swing accelerates back toward the bottom, and the harder it leans out, the stronger the pull, following $\sin \theta$. Check it: tiny swings behave like a spring, resting at the bottom stays put, balanced on top is a knife-edge. And never once did we mention the tug of the stick — good coordinates made that force disappear.

Recall Quick self-test

Speed of the bob in terms of $\theta$ ::: $v=\ell \dot{\theta }$ Kinetic energy $T$ ::: $\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$ Height of bob above lowest point ::: $\ell (1-\cos \theta )$ Potential energy $V$ ::: $mg\ell (1-\cos \theta )$ Final equation of motion ::: $\ddot{\theta }=-\frac{g}{\ell }\sin \theta$ Why does mass $m$ disappear? ::: it divides out — gravity accelerates all masses equally

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Related: 2.1.05-Deriving-the-Euler-Lagrange-equation · 2.1.07-Cyclic-coordinates-and-conservation-laws · 2.1.08-Generalized-momenta-and-the-Hamiltonian · Constraints-and-degrees-of-freedom · Noethers-theorem