Worked examples — Derivation of Euler-Lagrange equations from D'Alembert's principle

Before we touch numbers, one reminder of what the symbols mean, so nothing is used unearned:

Recall The three quantities you must be able to read off any problem

• $T$ ::: the kinetic energy, ${\sum }_k\frac{1}{2}{m}_k{\dot{\vec{r}}}_k^2$ — "how much motion energy the system has." Depends on the generalized velocities ${\dot{q}}_i$ (and sometimes $q_i,t$).
• $V$ ::: the potential energy — stored energy that depends on where things are, the $q_i$ only.
• $Q_i$ ::: the generalized force on coordinate $q_i$ — every applied (non-constraint) force, projected onto the direction that changing $q_i$ moves the system. When forces come from a potential, $Q_i=-\partial V/\partial q_i$.

Here $q_i$ is any generalized coordinate (see Constraints and generalized coordinates): a number that pins down the configuration while automatically respecting the constraints (an angle, a distance, an arc-length). A dot means $d/dt$ — rate of change in time.

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The scenario matrix

Cell	What it stress-tests	Example
A — trivial / sanity	$V=0$, straight line, recover Newton	Ex 1
B — sign of restoring force	$\partial V/\partial q$ changes sign across equilibrium	Ex 2 (spring)
C — $\partial T/\partial q\ne 0$	kinetic energy itself depends on position	Ex 3 (pendulum, full)
D — coupled coordinates ($n>1$)	two brackets, cross terms	Ex 4 (2D polar orbit)
E — velocity-dependent / time-dependent $T$	driven / rotating constraint, "fictitious" terms	Ex 5 (bead on rotating wire, general $\omega$)
F — limiting / degenerate input	$\ell \to \infty$, $m\to 0$, small-angle limit	Ex 6 (small-angle pendulum)
G — real-world word problem	translate prose → coordinates → EL	Ex 7 (block sliding on a wedge, both free)
H — exam twist	non-obvious coordinate, an ignorable (cyclic) coordinate	Ex 8 (Atwood-on-a-cone / conserved momentum)

Every worked example below is tagged with its cell(s).

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Ex 1 — Cell A · the sanity check

Forecast: guess the answer before reading — what does a particle with no force do? Write it down.

1. Write $T,V$. $T=\frac{1}{2}m{\dot{x}}^2$, $V=0$, so $L=\frac{1}{2}m{\dot{x}}^2$. Why this step? Every EL problem begins by naming the energies; nothing else can be computed until they exist.
2. Momentum slot. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}$. Why this step? EL needs $\partial L/\partial \dot{q}$ — the "generalized momentum" — as its first ingredient.
3. Time-derivative. $\frac{d}{dt}(m\dot{x})=m\ddot{x}$.
4. Position slot. $\frac{\partial L}{\partial x}=0$ (nothing depends on $x$). Why this step? $L$ has no $x$ in it, so $x$ is cyclic — a preview of conservation (Ex 8).
5. Assemble EL: $m\ddot{x}-0=0\Rightarrow \ddot{x}=0$.

Verify: $\ddot{x}=0$ means constant velocity — Newton's first law. Units: $[m\ddot{x}]=\text{kg}\cdot {\text{m/s}}^2=\text{N}$, a force, balancing $0$. ✓

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Ex 2 — Cell B · sign of the restoring force

Forecast: which direction does the force point when $x>0$? When $x<0$? Guess the sign pattern.

1. Energies. $T=\frac{1}{2}m{\dot{x}}^2$, $V=\frac{1}{2}k{x}^2$. So $L=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2$. Why this step? Spring potential is the archetype of "position energy" — see Generalized forces and potentials.
2. Momentum slot & derivative. $\frac{\partial L}{\partial \dot{x}}=m\dot{x}\Rightarrow \frac{d}{dt}=m\ddot{x}$.
3. Position slot. $\frac{\partial L}{\partial x}=-kx$. Why this step? This is $-\partial V/\partial x=Q_x$, the generalized force. Its sign is the whole point: for $x>0$ it is negative (pulls back to $0$); for $x<0$ it is positive (pushes back to $0$). Restoring in both signs.
4. EL: $m\ddot{x}-(-kx)=0\Rightarrow m\ddot{x}+kx=0$.
5. Read the frequency. $\ddot{x}=-\frac{k}{m}x$, so $\omega =\sqrt{k/m}=\sqrt{4/1}=2\text{rad/s}$.

Verify: plug the guessed solution $x=\cos (\omega t)$: $\ddot{x}=-{\omega }^2\cos \omega t=-4x$, and indeed $-k/m=-4$. ✓ Units of $\omega$: $\sqrt{(\text{N/m})/\text{kg}}=\sqrt{1/{\text{s}}^2}={\text{s}}^{-1}$. ✓

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Ex 3 — Cell C · when $\partial T/\partial q\ne 0$

The pendulum's danger is that the mapping $x=\ell \sin \theta$ makes velocity depend on $\theta$, so $T$ itself carries $\theta$. Look at the figure: the bob traces a circle, and the velocity vector is always tangent.

Forecast: will gravity's torque be $\propto \theta$ or $\propto \sin \theta$? Guess.

1. Position → velocity. $x=\ell \sin \theta ,y=-\ell \cos \theta$, so $\dot{x}=\ell \cos \theta \dot{\theta },\dot{y}=\ell \sin \theta \dot{\theta }$, and ${\dot{x}}^2+{\dot{y}}^2={\ell }^2{\dot{\theta }}^2$. Why this step? We must express $T$ in the chosen coordinate; the ${\cos }^2+{\sin }^2=1$ collapse is why $\theta$ is the natural choice.
2. Energies. $T=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2$, $V=mgy=-mg\ell \cos \theta$. $L=\frac{1}{2}m{\ell }^2{\dot{\theta }}^2+mg\ell \cos \theta$.
3. Momentum slot. $\frac{\partial L}{\partial \dot{\theta }}=m{\ell }^2\dot{\theta }\Rightarrow \frac{d}{dt}=m{\ell }^2\ddot{\theta }$.
4. Position slot. $\frac{\partial L}{\partial \theta }=-mg\ell \sin \theta$. Why this step? This is the "$\partial T/\partial q\ne 0$"-style term (here it lives in $V$), the restoring torque — and it is $\sin \theta$, not $\theta$.
5. EL: $m{\ell }^2\ddot{\theta }+mg\ell \sin \theta =0\Rightarrow \ddot{\theta }=-\frac{g}{\ell }\sin \theta$.

Verify: the unknown tension in the rod never appeared — exactly the D'Alembert payoff. Dimensionally $g/\ell$ has units ${\text{s}}^{-2}$, matching $\ddot{\theta }$. ✓

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Ex 4 — Cell D · coupled coordinates ($n=2$)

Forecast: two coordinates → two EL equations. Which of them will have a "${\dot{\varphi }}^2$" term appear from nowhere?

1. Velocity in polar. ${\dot{\vec{r}}}^2={\dot{r}}^2+r^2{\dot{\varphi }}^2$ (radial part + tangential part). Why this step? See Kinetic energy in generalized coordinates — the $r^2{\dot{\varphi }}^2$ term is what couples the two coordinates.
2. Energies. $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)$, $V=0$.
3. $r$-equation. $\frac{\partial L}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r}$; $\frac{\partial L}{\partial r}=mr{\dot{\varphi }}^2$. EL: $m\ddot{r}-mr{\dot{\varphi }}^2=0\Rightarrow \ddot{r}=r{\dot{\varphi }}^2$. Why this step? The $r{\dot{\varphi }}^2$ came from $\partial T/\partial r$ — the centripetal term, born of geometry not of any real force.
4. $\varphi$-equation. $\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }$; $\frac{\partial L}{\partial \varphi }=0$. EL: $\frac{d}{dt}(m{r}^2\dot{\varphi })=0$. Why this step? $\varphi$ is cyclic (absent from $L$), so its momentum is conserved.
5. Conserved quantity. $m{r}^2\dot{\varphi }=\text{const}$ — angular momentum. (This is Noether's theorem in miniature: rotational symmetry → conserved angular momentum.)

Verify: a free particle moves in a straight line; in polar coordinates a straight line indeed has $m{r}^2\dot{\varphi }=$ const (equal-areas / Kepler's 2nd law) and $\ddot{r}=r{\dot{\varphi }}^2$. Check the special case $\dot{\varphi }=0$: then $\ddot{r}=0$, pure radial straight-line motion. ✓

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Ex 5 — Cell E · time/velocity structure in $T$ (rotating constraint)

Forecast: with no applied force, will the bead sit still, or fly outward? Guess.

1. Constraint kills $\varphi$. The wire forces $\varphi =\omega t$, so $\dot{\varphi }=\omega$ is not a coordinate — only $r$ is free. Why this step? A moving (time-dependent) constraint reduces the freedom; $\omega$ enters $T$ as a fixed number.
2. Energies. $T=\frac{1}{2}m({\dot{r}}^2+r^2{\omega }^2)$, $V=0$.
3. Momentum slot. $\frac{\partial T}{\partial \dot{r}}=m\dot{r}\Rightarrow \frac{d}{dt}=m\ddot{r}$.
4. Position slot. $\frac{\partial T}{\partial r}=mr{\omega }^2$. Why this step? This is the term the parent note flagged: kinetic energy depends on position, so $\partial T/\partial r\ne 0$ produces a real dynamical push.
5. EL: $m\ddot{r}-mr{\omega }^2=0\Rightarrow \ddot{r}={\omega }^{2}r$.

Verify: solution $r=r_0\cosh (\omega t)$ (for release from rest at $r_0$) gives $\ddot{r}={\omega }^2{r}_0\cosh (\omega t)={\omega }^{2}r$. ✓ The bead accelerates outward — the "centrifugal" effect, produced with no constraint force in sight. Degenerate check: $\omega =0$ ⇒ $\ddot{r}=0$, a free bead on a fixed wire. ✓

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Ex 6 — Cell F · limiting / degenerate input

Forecast: near the bottom, does the pendulum look like a spring (Ex 2)? Guess the period.

1. Small-angle expansion. For $|\theta |\ll 1$, $\sin \theta \approx \theta$ (leading term of the series). Why this step? We test the EL result in a degenerate corner of input space, where geometry linearizes.
2. Reduced equation. $\ddot{\theta }=-\frac{g}{\ell }\theta$ — identical form to Ex 2's $\ddot{x}=-\frac{k}{m}x$.
3. Frequency & period. $\omega =\sqrt{g/\ell }$, so $T_{\text{period}}=2\pi \sqrt{\ell /g}$. Numeric: $\ell =1\text{m},g=9.8{\text{m/s}}^2\Rightarrow T_{\text{period}}=2\pi \sqrt{1/9.8}\approx 2.007\text{s}$.
4. $\ell \to \infty$ limit. $g/\ell \to 0\Rightarrow \ddot{\theta }\to 0$: an infinitely long pendulum feels no restoring pull — locally the arc looks flat, motion becomes uniform. Consistent with Ex 1 (free motion). ✓

Verify: period formula gives $\approx 2.007\text{s}$ for a 1 m pendulum (the classic "seconds pendulum" is $\ell \approx 0.994$ m). Units: $\sqrt{\text{m}/({\text{m/s}}^2)}=\text{s}$. ✓

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Ex 7 — Cell G · real-world word problem (block on a free wedge)

Forecast: as the block slides down-right, which way does the wedge recoil? Momentum says…?

1. Coordinates → positions. Wedge horizontal position $X$. Block position: $x_m=X+s\cos \alpha ,y_m=-s\sin \alpha$. Why this step? Two moving objects, two free coordinates $(X,s)$; the incline constraint is baked into these formulas so the normal force disappears.
2. Velocities. ${\dot{x}}_m=\dot{X}+\dot{s}\cos \alpha ,{\dot{y}}_m=-\dot{s}\sin \alpha$.
3. Energies. $T=\frac{1}{2}M{\dot{X}}^2+\frac{1}{2}m[(\dot{X}+\dot{s}\cos \alpha {)}^2+{\dot{s}}^2{\sin }^2\alpha ]$, and $V=mg{y}_m=-mgs\sin \alpha$.
4. $X$-equation (cyclic — $X$ absent from $L$): $\frac{\partial L}{\partial \dot{X}}=M\dot{X}+m(\dot{X}+\dot{s}\cos \alpha )=\text{const}$. Why this step? Horizontal momentum of the whole system is conserved (floor is frictionless). Differentiate: $(M+m)\ddot{X}+m\ddot{s}\cos \alpha =0$.
5. $s$-equation. $\frac{\partial L}{\partial \dot{s}}=m(\dot{X}+\dot{s}\cos \alpha )\cos \alpha +m\dot{s}{\sin }^2\alpha$, and $\frac{\partial L}{\partial s}=mg\sin \alpha$. EL: $m\ddot{X}\cos \alpha +m\ddot{s}=mg\sin \alpha$.
6. Solve the pair. From (4): $\ddot{X}=-\frac{m\cos \alpha }{M+m}\ddot{s}$. Substitute: $\ddot{s}=\frac{g\sin \alpha }{1-\frac{m{\cos }^2\alpha }{M+m}}=\frac{(M+m)g\sin \alpha }{M+m{\sin }^2\alpha }.$ And $\ddot{X}=-\frac{m\cos \alpha g\sin \alpha }{M+m{\sin }^2\alpha }$ (negative: wedge recoils opposite to the block's horizontal motion — matches the momentum forecast).

Verify — two limits.

• $M\to \infty$ (immovable wedge): $\ddot{s}\to g\sin \alpha$, the textbook block-on-fixed-incline result, and $\ddot{X}\to 0$. ✓
• $\alpha \to 90^{\circ }$: $\ddot{s}\to g$ (free fall) and $\ddot{X}\to 0$. ✓ Numeric check with $M=3,m=1,\alpha =30^{\circ },g=9.8$: $\ddot{s}=\frac{4\cdot 9.8\cdot 0.5}{3+1\cdot 0.25}=\frac{19.6}{3.25}\approx 6.031{\text{m/s}}^2$; $\ddot{X}=-\frac{1\cdot (\sqrt{3}/2)\cdot 9.8\cdot 0.5}{3.25}\approx -1.305{\text{m/s}}^2$. (See VERIFY.)

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Ex 8 — Cell H · exam twist (cyclic coordinate → conservation)

Forecast: which of $r,\varphi$ is cyclic? What does its conservation buy you?

1. Energies. $T=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)$, $V=V(r)$. $L=T-V$.
2. $\varphi$ is cyclic. $\frac{\partial L}{\partial \varphi }=0$ since neither $T$'s $\varphi$-independence nor $V$ mentions $\varphi$. Why this step? A missing coordinate signals a conserved momentum — this is the variational shadow of Noether.
3. Conserved momentum. $p_{\varphi }=\frac{\partial L}{\partial \dot{\varphi }}=m{r}^2\dot{\varphi }={\ell }_0$ (constant angular momentum).
4. $r$-equation. $\frac{d}{dt}(m\dot{r})-(mr{\dot{\varphi }}^2-V^{\prime }(r))=0\Rightarrow m\ddot{r}=mr{\dot{\varphi }}^2-V^{\prime }(r)$.
5. Eliminate $\dot{\varphi }$. Use $\dot{\varphi }={\ell }_0/(m{r}^2)$: $m\ddot{r}=\frac{{\ell }_0^2}{m{r}^3}-V^{\prime }(r).$ Why this step? One conserved quantity turns a 2-D problem into a 1-D one with an effective potential $V_{\text{eff}}=V(r)+\frac{{\ell }_0^2}{2m{r}^2}$.

Verify — recover Ex 4. Set $V=0$: $m\ddot{r}={\ell }_0^2/(m{r}^3)$. With ${\ell }_0=m{r}^2\dot{\varphi }$, the right side is $m{r}^2{\dot{\varphi }}^2/r=mr{\dot{\varphi }}^2$, i.e. $\ddot{r}=r{\dot{\varphi }}^2$ — exactly Ex 4. ✓ Numeric spot-check ($m=1,r=2,\dot{\varphi }=3$): ${\ell }_0=1\cdot 4\cdot 3=12$, and ${\ell }_0^2/(m{r}^3)=144/8=18=r{\dot{\varphi }}^2=2\cdot 9$. ✓

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Recall

Recall Which cell needs which move?

Cyclic coordinate (absent from $L$) ::: its generalized momentum $\partial L/\partial \dot{q}$ is conserved (Ex 4, 8). $\partial T/\partial q\ne 0$ ::: kinetic energy depends on position → "fictitious" (centrifugal/centripetal) terms appear (Ex 3, 4, 5). Restoring force sign check ::: $-\partial V/\partial q$ must point back to equilibrium for both signs of $q$ (Ex 2). Two free bodies ::: two coordinates → two EL equations, solve simultaneously (Ex 7). Immovable / infinite limit ::: send a mass or length to $\infty$ and confirm the textbook special case (Ex 6, 7).