Applying E-L equations to various systems: Question bank

Intuition What this page is for

The parent note taught you the recipe. This page attacks the places where the recipe feels like it should say one thing but actually says another. Each item is a one-line reveal: read the prompt, cover the answer, say your reasoning out loud, then uncover.

Before we start, one convention used everywhere below, spelled out from zero so nothing is assumed:

$q$ = a generalized coordinate — any single number (an angle, a distance, anything) that helps locate the system. One per degree of freedom.
$\overset{q}{˙}$ = its time-rate-of-change, "how fast that number is changing right now".
$L = T - V$ where $T$ is kinetic (moving) energy and $V$ is potential (stored) energy.
$\frac{\partial L}{\partial q ˙}$ = the generalized momentum conjugate to $q$ ; freeze everything except $\overset{q}{˙}$ and see how $L$ responds. (More in 2.1.08-Generalized-momenta-and-the-Hamiltonian.)
A cyclic coordinate is one that does not appear by name in $L$ (only its dot appears). See 2.1.07-Cyclic-coordinates-and-conservation-laws.

True or false — justify

Every answer starts with True/False, then why — never bare.

The Lagrangian is always the total energy

T + V

of the system.

False.

L = T - V

, a difference. The sum

T + V

is the energy (often the Hamiltonian);

L

is a different bookkeeping object whose stationary integral picks the real path.

If a coordinate

q

is cyclic, then

q

itself stays constant in time.

False. Cyclic means

q

is absent from $L$ , so its momentum

\partial L / \partial \overset{q}{˙}

is constant — but

q

can still change (e.g.

θ

in central force grows steadily while

L_{z}

is fixed).

Choosing coordinates that respect the constraint means the constraint force does no work.

True — that is exactly why we can drop it. An ideal constraint force is perpendicular to the allowed motion, so

F \cdot d r = 0

, and it never enters

T

V

The Euler–Lagrange equation only works in Cartesian coordinates.

False. Its whole selling point is that the same form

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

holds in any coordinates — angles, arc-length, polar — because it comes from a coordinate-free variational principle.

For the pendulum,

\ddot{θ} = - \frac{g}{ℓ} sin θ

is simple harmonic motion.

False in general. It is only approximately SHM for small

θ

where

sin θ \approx θ

. For large swings

sin θ

makes the period grow with amplitude — genuine anharmonic motion.

Inside the partial derivatives of

L

, we may substitute

\overset{q}{˙} = d q / d t

before differentiating.

False. When taking

\partial / \partial q

\partial / \partial \overset{q}{˙}

, treat

q

and

\overset{q}{˙}

as independent slots. The link

\overset{q}{˙} = d q / d t

is restored only by the outer

d / d t

Adding a constant to

V

changes the equations of motion.

False. Only derivatives of

V

enter E–L (

\partial L / \partial q

), so a constant shift cancels. Only the shape of

V

matters, never its zero point.

Spot the error

Each line states a flawed move; the reveal names the flaw and the fix.

"In polar coordinates

T = \frac{1}{2} m \overset{r}{˙}^{2}

Wrong: the tangential motion is missing. Speed is

v^{2} = \overset{r}{˙}^{2} + r^{2} \dot{θ}^{2}

, so

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

. Always build

v^{2}

from

\dot{r}

first.

"For the Atwood machine I'll include the string tension

T_{string}

as a force in the equation."

Wrong: the inextensible string is an ideal constraint, already baked into using one coordinate

x

. Adding tension double-counts it. Drop it entirely.

"The pendulum has 2 DOF because the bob lives in a plane."

Wrong: the rigid rod fixes the radius, killing one coordinate. Only the angle

θ

is free ⇒ 1 DOF. Count independent freedoms after constraints.

\frac{d}{d t} \frac{\partial L}{\partial θ ˙}

for

L = \frac{1}{2} m ℓ^{2} \dot{θ}^{2}

equals

m ℓ^{2} \dot{θ}

Wrong: that is only

\partial L / \partial \dot{θ}

. You still owe the outer

d / d t

, giving

m ℓ^{2} \ddot{θ}

. The E–L rule is "partial then total-time-derivative".

"On the incline I'll add the normal force to

V

so energy is conserved."

Wrong: the normal force does no work (perpendicular to the slope), so it never belongs in

V

. Using

s

along the slope already removes it.

"Since gravity acts on the incline block,

V = m g s

Wrong sign and geometry: the block descends by

s sin α

, losing PE, so

V = - m g s sin α

. Track the actual height change, not the path length.

\partial L / \partial θ = - m g ℓ sin θ

so the pendulum EOM is

\ddot{θ} = - g ℓ sin θ

Wrong: you forgot to divide by the

m ℓ^{2}

coming from

\frac{d}{d t} \frac{\partial L}{\partial θ ˙}

. Correct EOM is

\ddot{θ} = - \frac{g}{ℓ} sin θ

"The centrifugal term

m r \dot{θ}^{2}

is a real applied force I added to the central-force problem."

Wrong: nobody added it. It falls out of

\partial T / \partial r

automatically because

T

contains

r^{2} \dot{θ}^{2}

. It is a consequence of using rotating polar coordinates, not an external push.

Why questions

Why does the E–L method let us ignore tension and normal forces entirely?

Because those ideal constraint forces are perpendicular to the permitted motion and do zero work, and a good choice of independent coordinates already forbids the forbidden motion — the forces are geometrically absorbed.

Why do we need exactly one generalized coordinate per degree of freedom — no more, no fewer?

Fewer can't locate the system; more introduces redundant variables tied by constraints, breaking the independence that E–L assumes. One-per-DOF is the minimal complete description.

Why does a coordinate absent from

L

produce a conserved quantity?

\partial L / \partial q = 0

, the E–L equation becomes

\frac{d}{d t} (\partial L / \partial \overset{q}{˙}) = 0

, so the momentum

\partial L / \partial \overset{q}{˙}

has zero time-derivative — it is constant. This is Noether's theorem in its simplest guise.

Why is

r^{2} \dot{θ}^{2}

(not

\dot{θ}^{2}

) the tangential kinetic term in polar coordinates?

Because tangential speed is radius times angular rate,

v_{⊥} = r \dot{θ}

. Kinetic energy uses speed squared, so

v_{⊥}^{2} = r^{2} \dot{θ}^{2}

. The

r

converts an angle-rate into a real distance-rate.

Why does the Atwood acceleration depend only on

(m_{1} - m_{2}) / (m_{1} + m_{2})

and not on the tension?

The numerator is the net driving weight difference; the denominator is the total mass being accelerated. Tension is an internal constraint force that never appears once the single coordinate is chosen.

Why can we treat

q

and

\overset{q}{˙}

as independent variables during differentiation even though physically

\overset{q}{˙} = d q / d t

Because

L

is a function of two independent slots along the whole space of possible paths; the physical relation only holds on the actual trajectory, which the outer

d / d t

reinstates. The variational calculus demands this separation.

Why is Hamilton's principle (

δ S = 0

) the deeper statement, with E–L just its consequence?

The principle says the true path makes the action stationary against all wiggles fixed at the endpoints; demanding that for arbitrary wiggles and integrating by parts yields E–L. See 2.1.05-Deriving-the-Euler-Lagrange-equation.

Edge cases

Atwood machine with equal masses

m_{1} = m_{2}

: what does the EOM predict?

\overset{x}{¨} = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2}} = 0

. No net weight difference means no acceleration — the system is in equilibrium at any position. Sanity check passes.

Atwood machine with one mass set to zero, say

m_{2} = 0

\overset{x}{¨} = \frac{m _{1} g}{m _{1}} = g

. The remaining mass is in free fall since nothing opposes it — the limiting case reproduces plain gravity.

Central-force particle with

\dot{θ} = 0

(purely radial launch): what happens to the conserved

L_{z}

L_{z} = m r^{2} \dot{θ} = 0

and stays zero. Motion is a straight radial line; the centrifugal term

m r \dot{θ}^{2}

vanishes, leaving

m \overset{r}{¨} = - V^{'} (r)

, a 1D problem.

Pendulum released from exactly

θ = π

(straight up): what does the EOM say?

\ddot{θ} = - \frac{g}{ℓ} sin π = 0

. It is an equilibrium — but an unstable one; the slightest perturbation grows, unlike the stable

θ = 0

point. Zero acceleration hides the instability.

Incline angle

α \to 0

(flat ground):

\overset{s}{¨} = g sin 0 = 0

. No slope, no gravitational pull along the surface, so the block does not accelerate. The formula degrades gracefully to "nothing happens".

Incline angle

α \to 9 0^{\circ}

(vertical wall):

\overset{s}{¨} = g sin 9 0^{\circ} = g

. The "slope" is now a cliff and the block is in free fall — the tilted-coordinate result correctly reproduces vertical free fall.

Central force with

r \to 0

while

L_{z}

is fixed and nonzero: what dominates the radial equation?

The centrifugal term

m r \dot{θ}^{2} = L_{z}^{2} / (m r^{3})

blows up, acting as an infinite outward barrier. This is why a particle with angular momentum cannot reach the origin — the "centrifugal wall".

A system where

T

depends explicitly on time (e.g. a wire spun at forced rate

Ω t

): is energy still conserved?

Not necessarily. Explicit time-dependence in

L

breaks the time-translation symmetry, so the Hamiltonian (energy) need not be constant even though E–L still applies. See 2.1.08-Generalized-momenta-and-the-Hamiltonian.

Recall One-line summary to carry away

The recipe never changes; the traps are all about what you are allowed to ignore (constraint forces, constant shifts in $V$ ) and what you must not confuse (cyclic ≠ frozen coordinate, partial vs total derivative, energy $T + V$ vs Lagrangian $T - V$ ).