Analytical mechanics: the Lagrange exam

Analytical Mechanics (Meccanica Razionale in Italian universities, or simply Dynamics, depending on your degree programme) is the exam where high-school physics (forces, accelerations, $F = ma$ ) stops being enough and you need a new language: generalised coordinates, the kinetic energy of a rigid body, Lagrange’s equations. Many students see it as a huge jump, but the method is much more mechanical (in a good way) than it looks: once you can write $T$ and $V$ for a system, the equations of motion come out almost by themselves, just by differentiating.

Here is the path I follow with my students: what the exam asks, where the marks are lost, the method step by step, and two classic exercises worked right through to the end. They are a pendulum on a moving cart and a rolling disc, the two examples that come up most often in exams.

What the exam asks

The typical structure, in almost every engineering course that includes Analytical Mechanics, is a written paper with a mechanical system (one or two degrees of freedom: a pendulum, a disc, a rod, a cart, often combined). You are asked to:

Identify the degrees of freedom and choose the generalised coordinates.
Write the kinetic energy $T$ and the potential energy $V$ of the system.
Derive Lagrange’s equations of motion.
Find the equilibrium positions and discuss their stability.
Linearise around a stable equilibrium and find the frequency (or frequencies) of the small oscillations.
Sometimes, calculate the constraint reactions with the method of Lagrange multipliers or with the equations for linear momentum and angular momentum.

The common thread: you go from the physical system to the final numbers (equilibrium position, natural frequency) always through the same sequence: coordinates, $T$ and $V$ , equations, linearisation. Learning it as a fixed sequence is what takes away the fear of the blank page.

Where marks are lost, in practice

Incomplete kinetic energy of a rigid body. For a rigid body, König’s theorem gives $T = \frac{1}{2} m v_{G}^{2} + \frac{1}{2} I_{G} ω^{2}$ : the translation term of the centre of mass and the rotation term about the centre of mass. Forgetting one of the two (almost always the rotational one, when the body “seems” to move only by translation) is by far the most common mistake.
Badly written pure-rolling constraints. If a disc of radius $R$ rolls without slipping, the constraint is $v = R \dot{θ}$ (or $\overset{x}{˙} = R \dot{θ}$ ). You must write it before calculating $T$ , so that everything is expressed in a single generalised coordinate. Treating $x$ and $θ$ as independent when they are in fact linked by the constraint gives wrong equations from the very start.
The sign of gravitational potential energy. $V = m g h$ , where $h$ is the height relative to a reference level of your choice. But you must choose it once and keep it consistent throughout the exercise. Changing convention halfway (positive upwards in one place, downwards in another) is the number one cause of Lagrange equations with the wrong sign for gravity.
Equilibrium: looking for it in the equations of motion instead of in $V$ . For conservative systems, the equilibrium positions are the critical points of the potential, $\partial V / \partial q_{i} = 0$ for every coordinate. You do not need to write out the full Lagrange equations and then set $\overset{q}{˙} = \overset{q}{¨} = 0$ . That is a waste of time (even though the result is the same, if you did everything right).
Stability judged “by eye” instead of from the sign of the second derivative. The Lagrange-Dirichlet criterion is precise: an equilibrium is stable if it is a local minimum of $V$ , that is, if $\partial^{2} V / \partial q^{2} > 0$ (for a system with one degree of freedom; with more degrees of freedom the Hessian matrix must be positive definite). “It looks stable because it is at the bottom” is not a proof.
Small oscillations: linearising the equations but forgetting to linearise the kinetic energy too. $T$ must be evaluated (the coefficients that multiply $\overset{q}{˙}^{2}$ ) in the equilibrium configuration, not left with generic $cos θ$ and $sin θ$ . Otherwise the “linearised” system you get is no longer linear, which is a contradiction.
Mixing up the order of truncation. For the equilibrium you need $V$ up to second order (the second derivative decides stability). For the linearised equations of motion you need $sin θ \approx θ$ , $cos θ \approx 1$ to first order in the equations, but the generalised mass must be taken at the value it has exactly at equilibrium (order zero). Mixing truncation orders produces wrong linearised equations.

The method, step by step

1. Generalised coordinates

Count the degrees of freedom: the number of independent coordinates needed to describe the configuration, after imposing all the constraints (rigidity, rolling, guides). Choose the coordinates that make $T$ and $V$ easiest to write: usually angles for rotations, linear positions for free translations.

2. Kinetic energy (König’s theorem)

For each rigid body in the system:

T = \frac{1}{2} m v_{G}^{2} + \frac{1}{2} I_{G} ω^{2}

where $v_{G}$ is the velocity of the centre of mass, $ω$ the angular velocity, and $I_{G}$ the moment of inertia about the axis through the centre of mass perpendicular to the motion. Add up the contributions of all the bodies in the system, expressing everything in the chosen generalised coordinates (using the constraints to eliminate the dependent coordinates).

3. Potential energy

Gravity: $V_{g} = m g h (q)$ , with $h$ a function of the chosen coordinates. Springs: $V_{k} = \frac{1}{2} k δ (q)^{2}$ , with $δ$ the extension relative to the rest length. Add up all the conservative contributions.

4. Lagrange’s equations

With $L = T - V$ :

\frac{d}{d t} (\frac{\partial L}{\partial q ˙ _{i}}) - \frac{\partial L}{\partial q _{i}} = 0, i = 1, \dots, n

If there are non-conservative forces $Q_{i}$ instead (friction, typically), the same equation is written by separating kinetic and potential energy, with the forcing term on the right:

\frac{d}{d t} \frac{\partial T}{\partial q ˙ _{i}} - \frac{\partial T}{\partial q _{i}} + \frac{\partial V}{\partial q _{i}} = Q_{i}

There is one equation for each generalised coordinate. If a coordinate does not appear in $L$ (only its derivative $\overset{q}{˙}_{i}$ does), it is cyclic: the quantity $\partial L / \partial \overset{q}{˙}_{i}$ is conserved, and this is often the quickest shortcut for solving the problem (typically a linear momentum or an angular momentum).

5. Equilibrium and stability

For a conservative system, the equilibria are the critical points of $V (q)$ :

\frac{\partial V}{\partial q _{i}} = 0 \forall i

Stability (Lagrange-Dirichlet): the equilibrium is stable if $V$ has a local minimum there. With one degree of freedom, this means $V^{''} (q_{e q}) > 0$ .

6. Small oscillations

You linearise around a stable equilibrium $q_{e q}$ : write $q = q_{e q} + ε$ , expand $T$ and $V$ to second order in $ε, \overset{ε}{˙}$ (with the coefficients of $T$ evaluated at $q_{e q}$ ), and you get an equation of the form $m_{e q} \overset{ε}{¨} + k_{e q} ε = 0$ , with:

ω^{2} = \frac{k _{e q}}{m _{e q}}, k_{e q} = \frac{\partial ^{2} V}{\partial q ^{2}}_{q_{e q}}, m_{e q} = coefficient of \overset{q}{˙}^{2} in T at q = q_{e q}

Two classic worked exercises

Exercise 1: pendulum on a moving cart

A cart of mass $M$ slides without friction on a horizontal plane. A pendulum of mass $m$ and length $l$ is hinged to the cart, with $θ$ measured from the downward vertical. Generalised coordinates: $X$ (position of the cart), $θ$ (angle of the pendulum).

Position of the pendulum mass: $x_{p} = X + l sin θ$ , $y_{p} = - l cos θ$ . Differentiating:

\overset{x}{˙}_{p} = \dot{X} + l cos θ \dot{θ}, \overset{y}{˙}_{p} = l sin θ \dot{θ}

T = \frac{1}{2} M \dot{X}^{2} + \frac{1}{2} m (\dot{X}^{2} + 2 l cos θ \dot{X} \dot{θ} + l^{2} \dot{θ}^{2}), V = - m g l cos θ

$X$ appears neither in $T$ nor in $V$ : it is a cyclic coordinate. Its Lagrange equation gives a conservation law directly:

\frac{d}{d t} [(M + m) \dot{X} + m l cos θ \dot{θ}] = 0

(the total horizontal momentum of the system is conserved, which makes sense because there are no external horizontal forces). The equation for $θ$ , after differentiating and simplifying (the terms in $\dot{θ} \dot{X}$ cancel exactly), reduces to:

l \ddot{θ} + \ddot{X} cos θ + g sin θ = 0

Equilibrium: $θ = 0$ (pendulum vertical, hanging down, $V$ at a minimum, $V^{''} (0) = m g l > 0$ : stable) and $θ = π$ (vertical, pointing up, $V^{''} (π) = - m g l < 0$ : unstable). This is exactly what you would expect physically.

Small oscillations about $θ = 0$ : linearising ( $cos θ \approx 1$ , $sin θ \approx θ$ , $\dot{θ}^{2} \approx 0$ ), the momentum conservation equation gives $\ddot{X} \approx - \frac{m l}{M + m} \ddot{θ}$ . Substituting into the equation for $θ$ :

l \ddot{θ} - \frac{m l}{M + m} \ddot{θ} + g θ = 0 ⟹ \ddot{θ} + \frac{g ( M + m )}{l M} θ = 0

ω^{2} = \frac{g ( M + m )}{l M}

If the cart is much heavier than the pendulum ( $M \to \infty$ ), $ω^{2} \to g / l$ : you get the simple pendulum back, as you should. With a light cart, the frequency is higher. The system oscillates faster because the cart itself takes part in the motion, shortening the “effective length” of the pendulum.

Exercise 2: a disc rolling against a spring

A uniform disc of mass $m$ and radius $R$ rolls without slipping on a horizontal plane. Its centre is connected to a fixed wall by a spring of stiffness $k$ (at rest when the centre is at the origin). Generalised coordinate: $x$ , the position of the centre of the disc.

Rolling constraint: $\dot{θ} = \overset{x}{˙} / R$ . Moment of inertia of the disc about its own axis: $I_{G} = \frac{1}{2} m R^{2}$ .

T = \frac{1}{2} m \overset{x}{˙}^{2} + \frac{1}{2} I_{G} \dot{θ}^{2} = \frac{1}{2} m \overset{x}{˙}^{2} + \frac{1}{2} (\frac{1}{2} m R^{2}) \frac{x ˙ ^{2}}{R ^{2}} = \frac{1}{2} m \overset{x}{˙}^{2} + \frac{1}{4} m \overset{x}{˙}^{2} = \frac{3}{4} m \overset{x}{˙}^{2}

V = \frac{1}{2} k x^{2}

The Lagrangian is $L = \frac{3}{4} m \overset{x}{˙}^{2} - \frac{1}{2} k x^{2}$ . Lagrange’s equation:

\frac{d}{d t} (\frac{3}{2} m \overset{x}{˙}) - (- k x) = 0 ⟹ \frac{3}{2} m \overset{x}{¨} + k x = 0 ⟹ \overset{x}{¨} + \frac{2 k}{3 m} x = 0

ω^{2} = \frac{2 k}{3 m}

Compare this with a point mass $m$ attached to the same spring without rolling: it would be $ω^{2} = k / m$ , which is faster. The disc oscillates more slowly because part of the kinetic energy goes into rotation, not translation. It is as if the spring had to move an “effective mass” of $\frac{3}{2} m$ instead of $m$ . Once you have seen this result, you recognise it instantly in every exercise with rolling.

How we work in lessons

On Analytical Mechanics I always start from a hand-drawn picture of the system: where the coordinates are, where the constraints are, what is connected to what. Most of the mistakes I see come from a badly written constraint, long before you reach Lagrange’s equations. Only after fixing the coordinates, $T$ and $V$ without ambiguity do we move on to the derivation, which by then is almost mechanical.

We redo together the exercises from your problem classes and from past exam papers, paying particular attention to equilibrium and small oscillations. In my experience they are the part worth the most marks and the one that is understood worst, because it requires you to connect an algebraic result (the sign of a second derivative) with a physical intuition (stable or not).

If you are missing the basics, point mechanics is in Physics 1: kinematics and dynamics, electromagnetism from the same first two years is in Physics 2: Gauss’s law, and the statics of constraints is in constraint reactions. The rigid body in space, with Euler angles, is in attitude dynamics.

For the overall picture of how I work on university physics, see the physics page; if your analytical mechanics course also requires a solid grounding in calculus (derivatives, Taylor expansions for linearisation), have a look at the maths page.

Before booking, try the university exam practice tests: the correct answer is free, and the guided solution is unlocked with your first lesson. If you have a specific exercise, perhaps a system with two degrees of freedom that does not add up, write to me through the contact page for a free 30-minute introductory call.

Analytical mechanics: passing the Lagrange exam