Intro to Lagrangian Mechanics (IPhO)

IPhO mechanics lesson introducing the Lagrangian method, generalized coordinates, and when it simplifies problems.

International Physics Olympiad preparation

On this page

The Lagrangian approach is a tool for problems where forces are awkward but energy is simple. In IPhO-style mechanics, it often turns a messy free-body diagram (with many constraint forces) into one clean differential equation in the right coordinate.

The official IPhO syllabus does not name Lagrangian mechanics or calculus of variations. Use this lesson as an optional problem-solving extension. Any result here should also be interpretable through the official mechanics content: Newton’s laws, energy, momentum, angular momentum, constraints and symmetry.

Learning objectives

By the end, you should be able to:

IPHO-M-LAG-01: reduce a constrained configuration to a minimal set of generalized coordinates;
IPHO-M-LAG-02: construct L = T-V and derive the corresponding equation of motion; and
connect a cyclic coordinate to a physical symmetry and a conserved quantity.

Scroll diagram horizontally to read all labels.

Choose the smallest coordinate set that describes every allowed configuration. Here, θ replaces x and y together with the constraint x² + y² = R².

1. Definitions

Generalized coordinates qᵢ: a minimal set of variables that uniquely specifies the configuration (given the constraints).
Degrees of freedom: the number of independent qᵢ.
Holonomic constraint: a relation f(q,t) = 0 that reduces degrees of freedom.
Lagrangian: for conservative systems, L(q,q dot,t) = T-V.
Action: S = ∫_t₁^t₂ L dt (stationary for the true path).
Euler-Lagrange equations:
(d/dt)((∂ L)/(∂ q dot ᵢ))-(∂ L)/(∂ qᵢ) = 0
(or = Qᵢ if you include nonconservative generalized forces).
Conjugate momentum: pᵢ = ∂ L/∂ q dot ᵢ.
Cyclic (ignorable) coordinate: if ∂ L/∂ qᵢ = 0, then pᵢ is conserved.
Energy function: E = ∑ᵢ q dot ᵢ pᵢ - L; if ∂ L/∂ t = 0, then E is conserved (often the mechanical energy).
Small oscillations: expand L to quadratic order about equilibrium to get linear equations and normal-mode frequencies.

2. Key ideas

Choose qᵢ that build in constraints (you then do not need to solve for tensions/normal forces explicitly).
Compute T carefully from geometry: write positions as functions of qᵢ, then differentiate to get velocities.
Look for symmetries: a cyclic coordinate gives an immediate conservation law.
If L has no explicit time dependence, use energy conservation as a fast check (or as the solution method).
For small oscillations, expand to second order about the correct equilibrium and read off an effective “mass” and “spring constant”.

Pick generalized coordinate(s) qᵢ consistent with constraints.
Express all positions in terms of qᵢ.
Differentiate to get velocities and build T.
Write V from heights/springs/fields.
Form L = T-V and apply Euler-Lagrange.
Use cyclic coordinates and limiting cases to simplify/check.

3. Detailed explanation

3.1 Why Lagrangians help in olympiad mechanics

Newton’s laws are always valid, but they make you solve for constraint forces you do not ultimately care about. Lagrangians let you:

eliminate constraint forces by choosing coordinates that satisfy the constraints,
use energy expressions directly (often far simpler than force components),
exploit symmetries (cyclic coordinates) to get constants of motion quickly.

3.2 The core equation

For each generalized coordinate qᵢ:

(d/dt)((∂ L)/(∂ q dot ᵢ))-(∂ L)/(∂ qᵢ) = 0.

If a nonconservative force does virtual work δ W = ∑ᵢ Qᵢ δ qᵢ, then:

(d/dt)((∂ L)/(∂ q dot ᵢ))-(∂ L)/(∂ qᵢ) = Qᵢ.

3.3 Handling constraints (two standard options)

Best option (most common in IPhO): choose coordinates that already satisfy the constraints (e.g. use an angle, or use x = Rθ for rolling).
Lagrange multipliers: if a holonomic constraint is f(q,t) = 0, you can solve in the full space by using
L' = L + λ f(q,t),
and treating λ as an extra variable. This is useful when you actually want the constraint force.

3.4 Cyclic coordinates and conserved quantities

If ∂ L/∂ q = 0, then

(d/dt)((∂ L)/(∂ q dot)) = 0 ⇒ p = (∂ L)/(∂ q dot) = constant.

This is the Lagrangian way to see conservation laws (linear momentum, angular momentum, etc).

3.5 Small oscillations (the mark-rich shortcut)

For one coordinate q near stable equilibrium q₀:

Expand the potential: V(q) ≈ V(q₀) + 1/2 k (q-q₀)².
Expand the kinetic energy: T ≈ 1/2 M q dot ² (where M is the effective inertia for that coordinate).

Then the equation is Mq double dot + k(q-q₀) = 0 and

ω = square root of (k/M) .

For multiple coordinates, the same idea becomes matrix algebra: quadratic T defines a mass matrix and quadratic V defines a stiffness matrix; normal-mode frequencies come from an eigenvalue problem.

4. Common mistakes

Writing L = T + V (wrong sign).
Forgetting that T must be expressed in terms of qᵢ and q dot ᵢ (not in mixed coordinates).
Missing cross terms in T when velocities depend on multiple coordinates.
Using a constraint for positions but forgetting to differentiate it for velocities (e.g. using x = Rθ but not x dot = Rθ dot).
Expanding about the wrong equilibrium when doing small oscillations.
Declaring a coordinate cyclic when T or V actually depends on it.

5. Problem-solving tips

Start by writing your coordinate choice and the constraint relation in one line.
Build T from geometry: write x(q),y(q), then compute x dot,y dot and v² = x dot ² + y dot ².
Look for a cyclic coordinate before you differentiate anything.
If L has no explicit t, compute E and use it as a check (often it is the cleanest route to v(q)).
For small oscillations: expand to second order and stop. Higher-order terms rarely earn marks unless asked.

6. Worked examples

1) Simple pendulum via Lagrangian (and small-angle limit)

Let the generalized coordinate be the angle θ from the downward vertical, with length ℓ.

Speed: v = ℓθ dot, so

T = 1/2 mℓ²θ dot ².

Height change: Δ h = ℓ(1- cos θ), so

V = mgℓ(1- cos θ).

Lagrangian:

L = 1/2 mℓ²θ dot ² - mgℓ(1- cos θ).

Euler-Lagrange gives

mℓ²θ double dot + mgℓ sin θ = 0 ⇒ θ double dot + g/ℓ sin θ = 0.

Small-angle (θ≪ 1): sin θ ≈ θ, so θ double dot + (g/ℓ)θ = 0 and ω = square root of (g/ℓ).

2) Rolling down an incline (constraint built in)

A cylinder of mass m, radius R, and moment of inertia I rolls without slipping down an incline of angle α.

Use θ as the coordinate. The no-slip constraint is x = Rθ, so x dot = Rθ dot.

Kinetic energy:

T = 1/2 mx dot ² + 1/2 Iθ dot ² = (1/2)(mR² + I)θ dot ².

Potential energy (down the slope): V = -mgx sin α = -mgRθ sin α.

Lagrangian: L = T-V = (1/2)(mR² + I)θ dot ² + mgRθ sin α.

Euler-Lagrange:

(mR² + I)θ double dot = mgR sin α.

Therefore the translational acceleration is

a = x double dot = Rθ double dot = (g sin α)/(1 + I/(mR²)).

3) Two equal masses and a spring (cyclic coordinate shows up automatically)

Two equal masses m move on a frictionless line, connected by a spring of constant k and natural length ℓ₀.

Use the centre-of-mass coordinate X = (x₁ + x₂)/2 and separation r = x₁-x₂. Then x₁ = X + r/2 and x₂ = X-r/2.

Velocities: x dot ₁ = X dot + r dot/2 and x dot ₂ = X dot -r dot/2, so

T = 1/2 m(x dot ₁² + x dot ₂²) = mX dot ² + m/4r dot ².

Potential: V = 1/2 k(r-ℓ₀)².

The Lagrangian is L = mX dot ² + m/4r dot ²-1/2 k(r-ℓ₀)².

Since X is cyclic, ∂ L/∂ X = 0 and p_X = ∂ L/∂X dot = 2mX dot is constant: centre-of-mass motion is uniform.

For the internal coordinate q = r-ℓ₀, Euler-Lagrange gives

q double dot + (2k/m)q = 0,

so ω = square root of (2k/m) for the relative oscillation.

7. Extensions

1) Central forces: reduce 2D motion to 1D using a cyclic coordinate

For a particle in a central potential V(r), use plane polar coordinates (r,φ):

L = 1/2 m(r dot ² + r²φ dot ²)-V(r).

The coordinate φ is cyclic, so

p_φ = (∂ L)/(∂ φ dot) = mr²φ dot = ℓ

is conserved (angular momentum).

Substitute φ dot = ℓ/(mr²) into the energy to get an effective 1D radial motion:

E = 1/2 mr dot ² + V_eff(r), V_eff(r) = V(r) + ℓ²/2mr².

Task: show that circular orbits satisfy dV_eff/dr = 0, and stability requires d²V_eff/dr² > 0 at the orbit radius.

2) Bead on a rotating hoop (effective potential and bifurcation)

A bead of mass m slides without friction on a circular hoop of radius R. The hoop rotates about its vertical diameter with constant angular speed Ω.

Let θ be the bead’s angle from the lowest point. The kinetic energy includes the bead’s motion along the hoop and the hoop’s rotation:

T = 1/2 mR²(θ dot ² + Ω² sin² θ).

The gravitational potential is V = mgR(1- cos θ), so

L = 1/2 mR²(θ dot ² + Ω² sin² θ)-mgR(1- cos θ).

The equilibrium condition is θ dot = θ double dot = 0, giving

sin θ(Ω² cos θ-g/R) = 0.

So θ = 0 is always an equilibrium, and for Ω² > g/R there are two additional equilibria with

cos θ₀ = g/RΩ².

Small oscillations about θ = 0 have frequency ω² = g/R-Ω² (stable only if Ω² < g/R). About the tilted equilibria, ω² = Ω² sin² θ₀.

8. Practice and evidence

Solve two constraint-heavy problems twice: once with Newton or energy, then with L = T-V. Treat IPHO-M-LAG-01/02 as passed only if both methods agree and you can explain which constraint force disappeared from the Lagrangian calculation. Then return to the IPhO Mechanics Hub.

Syllabus and review details

No official syllabus alignment is listed for this lesson.

Last reviewed: 2026-07-20