📐

Topic 2 of 11

Lagrangian Mechanics

The Lagrangian formulation replaces Newton's vector equations with a single scalar function $L=T-V$. By choosing generalized coordinates that automatically satisfy the constraints, you reduce any system to as many equations as degrees of freedom — and the method is coordinate-independent.

Key Concepts

Generalized Coordinates

A minimal set of independent variables

q_i

that fully specifies the configuration of a system while automatically satisfying all constraints. For a pendulum, the angle

\theta

is a generalized coordinate;

x

and

y

separately are not minimal.

Lagrangian

L(q,\dot{q},t)=T-V

, the difference of kinetic and potential energies. Written in generalized coordinates,

L

captures the full dynamics without explicit constraint forces.

Euler-Lagrange Equation

For each generalized coordinate:

\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i}-\frac{\partial L}{\partial q_i}=0

. This yields the equation of motion directly.

Noether's Theorem

L

does not depend on

q_i

(cyclic coordinate), then

\partial L/\partial\dot{q}_i=

const. Translational symmetry → linear momentum; rotational symmetry → angular momentum; time symmetry → energy.

Key Equations

Lagrangian

L = T - V

Kinetic minus potential energy, expressed in generalized coordinates.

Euler-Lagrange

\frac{d}{dt}\frac{\partial L}{\partial \dot{q}_i} - \frac{\partial L}{\partial q_i} = 0

One equation per degree of freedom; yields the equation of motion for qᵢ.

Generalized momentum

p_i = \frac{\partial L}{\partial \dot{q}_i}

Conjugate momentum to qᵢ; conserved when qᵢ is cyclic.

Worked Example

Pendulum via Euler-Lagrange

Problem

Derive the equation of motion for a simple pendulum (mass $m$ , length $l$ ) and find its small-angle angular frequency.

Solution

Generalized coordinate: $q=\theta$ . Kinetic energy: $T=\frac{1}{2}ml^2\dot\theta^2$ . Potential: $V=mgl(1-\cos\theta)$ .

L = \frac{1}{2}ml^2\dot\theta^2 - mgl(1-\cos\theta)

Apply E-L: $\frac{d}{dt}(ml^2\dot\theta)-(-mgl\sin\theta)=0$

ml^2\ddot\theta + mgl\sin\theta = 0 \implies \ddot\theta = -\frac{g}{l}\sin\theta

Small angle: $\sin\theta\approx\theta$ , giving $\ddot\theta=-\omega_0^2\theta$ with $\omega_0=\sqrt{g/l}$ .

Answer

\omega_0 = \sqrt{g/l}

; for

l=1.0

\omega_0=3.13

rad/s.

Practice

Exercises

7 problems

1 of 7

A simple pendulum has length $l=0.50$ m. Using the Lagrangian, find its small-angle angular frequency $\omega_0$ (in rad/s).

Your answer

rad/s

2 of 7

A spring-mass system has $k=200$ N/m and $m=2.0$ kg. Find $\omega_0$ (in rad/s) from the Lagrangian.

Your answer

rad/s

3 of 7

A solid cylinder ( $m=2.0$ kg, $R=0.10$ m, $I=\frac{1}{2}mR^2$ ) rolls without slipping down a $30°$ incline. Find its linear acceleration (in m/s²).

Unlock Exercise 3