Lagrangian Mechanics: An Alternative Approach

A block slides down the face of a wedge, and the wedge is free to slide on the floor. Try that with a free body diagram. The block pushes on the wedge, the wedge pushes back, the floor pushes up, and the block's acceleration is measured in the ground frame while its motion is naturally described along a face that is itself accelerating. Three unknown forces, two frames, a lot of sign errors.

Now notice something. The normal force between block and wedge is the only thing coupling them, it is the thing you had to solve for, and it does not appear anywhere in the answer. That is the pattern. In constrained problems most of the forces you work hardest to find are bookkeeping. Lagrangian mechanics is a way of writing the equations of motion in which those forces never get written down at all. You pay for it with a little calculus, and the calculus is the same every time.

Generalized coordinates

Start by counting. A system's degrees of freedom is the number of numbers you need to pin down where everything is. A pendulum on a rigid rod: one number, the angle $\theta$. A bead threaded on a fixed parabolic wire: one, because $x$ determines $y = cx^2$. The block-and-wedge: two: how far the wedge has slid, and how far the block has slid along its face.

Any set of independent numbers that does the job is a set of generalized coordinates, written $q_1, q_2, \dots$ They need not be lengths. An angle is fine, an arc length is fine, a mixture is fine.

The whole trick is in the choice. If you describe the bead on the wire by $x$ and $y$ separately, you need an extra equation, $y = cx^2$, and the wire has to push on the bead to enforce it, so a normal force enters your equations. If you describe it by $x$ alone, the constraint is already true by construction and there is nothing left for the wire to enforce. The constraint force has been absorbed into the choice of coordinate.

This works because ideal constraint forces do no work. A normal force is perpendicular to the surface, and the allowed motion is along the surface. A rod's tension is along the rod, and the bob moves perpendicular to it. A rolling contact point is instantaneously at rest. None of them transfers energy, so none of them shows up in an energy-based formulation.

The Lagrangian and the equation that uses it

Write the total kinetic energy $T$ and the total potential energy $V$ of the whole system, both expressed in your generalized coordinates and their time derivatives. The Lagrangian is their difference:

$L = T - V$

Not the sum. The sum is the energy, which is a number you already know how to use; the difference is the thing whose derivatives give you motion. Then for every coordinate $q$ in your list, one equation:

$\frac{d}{dt}\left(\frac{\partial L}{\partial \dot q}\right) = \frac{\partial L}{\partial q}$

That is the Euler–Lagrange equation. The dot means a time derivative, $\dot q = dq/dt$. The symbol $\partial$ means a partial derivative: differentiate with respect to that one slot and treat all the others, including $q$ versus $\dot q$, as independent constants while you do it.

Read the two sides physically. For a free particle in one dimension, $L = \tfrac12 m\dot x^2 - V(x)$, so $\partial L/\partial \dot x = m\dot x$, which is momentum, and $\partial L/\partial x = -dV/dx$, which is force. The equation says $\dot p = F$. Newton's second law is the special case; Euler–Lagrange is what that law looks like when you are allowed to use any coordinates you like, including ones that are angles or that ride on moving objects.

The recipe, once and for all:

1. Count degrees of freedom and pick that many coordinates.
2. Write every mass's position in fixed ground axes in terms of those coordinates.
3. Differentiate those positions to get velocities, square them, and sum to get $T$.
4. Write $V$ in the same coordinates.
5. Form $L = T - V$, take the two derivatives for each coordinate, and set them equal.

Step 2 is the one people skip, and it is the one that goes wrong.

Example 1: a pendulum, from scratch

A bob of mass $m$ on a massless rod of length $\ell = 1.20$ m, pivot fixed, $\theta$ measured from straight down. One degree of freedom.

The bob sits at $x = \ell\sin\theta$, $y = -\ell\cos\theta$. Differentiating, $\dot x = \ell\dot\theta\cos\theta$ and $\dot y = \ell\dot\theta\sin\theta$, so

$v^2 = \ell^2\dot\theta^2(\cos^2\theta + \sin^2\theta) = \ell^2\dot\theta^2$

Height above the lowest point is $\ell(1-\cos\theta)$, so $V = mg\ell(1-\cos\theta)$ and

$L = \tfrac12 m\ell^2\dot\theta^2 - mg\ell(1-\cos\theta)$

Now the two derivatives. Treating $\dot\theta$ as the variable, $\partial L/\partial\dot\theta = m\ell^2\dot\theta$. Treating $\theta$ as the variable, only the potential term depends on it, and $d(-\cos\theta)/d\theta = \sin\theta$, so $\partial L/\partial\theta = -mg\ell\sin\theta$. Taking $d/dt$ of the first gives $m\ell^2\ddot\theta$. Setting them equal:

$m\ell^2\ddot\theta = -mg\ell\sin\theta \quad\Longrightarrow\quad \ddot\theta = -\frac{g}{\ell}\sin\theta$

For small $\theta$, $\sin\theta \approx \theta$, so $\ddot\theta = -(g/\ell)\theta$, which is simple harmonic motion with $\omega = \sqrt{g/\ell} = \sqrt{9.80/1.20} = \sqrt{8.167} = 2.86$ rad/s. The rod tension never appeared, which is the point.

Example 2: the free wedge

Block of mass $m = 2.0$ kg on the $\alpha = 30.0^\circ$ face of a wedge of mass $M = 5.0$ kg, everything frictionless. Coordinates: $X$ for the wedge's position along the floor, $s$ for the block's distance down the face, measured relative to the wedge.

The wedge's velocity is $\dot X$ horizontally. The block's velocity in the ground frame is the wedge's velocity plus the block's velocity relative to the wedge, which has magnitude $\dot s$ pointing down the face: horizontal component $\dot X + \dot s\cos\alpha$, vertical component $-\dot s\sin\alpha$. Square and add:

$v_m^2 = (\dot X + \dot s\cos\alpha)^2 + \dot s^2\sin^2\alpha = \dot X^2 + 2\dot X\dot s\cos\alpha + \dot s^2$

There is the cross term. The kinetic energy is $T = \tfrac12 M\dot X^2 + \tfrac12 m(\dot X^2 + 2\dot X\dot s\cos\alpha + \dot s^2)$. The block drops by $s\sin\alpha$, so $V = -mgs\sin\alpha$, and

$L = \tfrac12(M+m)\dot X^2 + m\dot X\dot s\cos\alpha + \tfrac12 m\dot s^2 + mgs\sin\alpha$

Coordinate $X$ does not appear in $L$, only $\dot X$ does. So $\partial L/\partial X = 0$ and the Euler–Lagrange equation says $(M+m)\dot X + m\dot s\cos\alpha$ stays constant. That is just horizontal momentum conservation, handed to you for free. Differentiating once,

$(M+m)\ddot X + m\ddot s\cos\alpha = 0$

For $s$: $\partial L/\partial\dot s = m\dot X\cos\alpha + m\dot s$ and $\partial L/\partial s = mg\sin\alpha$, giving

$\ddot X\cos\alpha + \ddot s = g\sin\alpha$

Two linear equations, two unknowns. From the first, $\ddot s = -(M+m)\ddot X/(m\cos\alpha)$. Substitute:

$\ddot X\cos\alpha - \frac{(M+m)\ddot X}{m\cos\alpha} = g\sin\alpha$

Multiply through by $m\cos\alpha$ and collect:

$\ddot X\left(m\cos^2\alpha - M - m\right) = mg\sin\alpha\cos\alpha$

Since $m\cos^2\alpha - m = -m\sin^2\alpha$, the magnitude is

$|\ddot X| = \frac{mg\sin\alpha\cos\alpha}{M + m\sin^2\alpha}$

Numbers: $\sin 30^\circ = 0.500$, $\cos 30^\circ = 0.866$. Top: $2.0 \times 9.80 \times 0.500 \times 0.866 = 8.49$. Bottom: $5.0 + 2.0(0.250) = 5.50$. So $|\ddot X| = 8.49/5.50 = 1.54$ m/s². Figure 1 lets you check the limits: make $M$ huge and the wedge acceleration goes to zero, as it must.

Rigid bodies and rolling

Nothing changes except that a rigid body's kinetic energy has two pieces: $\tfrac12 Mv_{\text{cm}}^2$ for the centre of mass plus $\tfrac12 I_{\text{cm}}\omega^2$ for the spin about it. Rolling without slipping is a constraint, so it should be built into the coordinate, not enforced by a friction force.

A solid cylinder rolling down a $30.0^\circ$ incline, with $s$ measured along the slope: rolling means $\omega = \dot s/R$, and $I_{\text{cm}} = \tfrac12 MR^2$, so

$T = \tfrac12 M\dot s^2 + \tfrac12\left(\tfrac12 MR^2\right)\frac{\dot s^2}{R^2} = \tfrac34 M\dot s^2$

with $V = -Mgs\sin\theta$. Then $\partial L/\partial\dot s = \tfrac32 M\dot s$ and $\partial L/\partial s = Mg\sin\theta$, so $\tfrac32 M\ddot s = Mg\sin\theta$ and

$\ddot s = \tfrac23 g\sin\theta = \tfrac23(9.80)(0.500) = 3.27\ \text{m/s}^2$

Friction never appeared. Same move for a massive pulley: the string's speed is $\dot x$, the pulley spins at $\dot x/R$, and its contribution to $T$ is $\tfrac12(\tfrac12 MR^2)(\dot x/R)^2 = \tfrac14 M\dot x^2$, so the pulley behaves exactly like an extra $M/2$ of mass being dragged along. The $R$ always cancels.

Small oscillations about equilibrium

Most of the hard problems in this unit ask for an oscillation frequency, and there is a shortcut. Suppose one coordinate, and suppose near the equilibrium value $q_0$ you can write the Lagrangian in the form

$L \approx \tfrac12 a\,\dot u^2 - \tfrac12 b\,u^2, \qquad u = q - q_0$

where $a$ and $b$ are constants. Euler–Lagrange gives $a\ddot u = -bu$, so

$\omega = \sqrt{b/a}$

So the job is to expand $T$ to second order in the velocity and $V$ to second order in the displacement, and read off the two coefficients. Use $\cos\varepsilon \approx 1 - \varepsilon^2/2$ and $\sin\varepsilon\approx\varepsilon$, and keep the quadratic terms, because the constant and linear ones contribute nothing.

For a bead on the parabola $y = cx^2$: $\dot y = 2cx\dot x$, so $v^2 = \dot x^2(1 + 4c^2x^2) \approx \dot x^2$ near the bottom, giving $a = m$. And $V = mgcx^2$, so $b = 2mgc$ and $\omega = \sqrt{2gc}$. With $c = 0.80$ m⁻¹ that is $\sqrt{2(9.80)(0.80)} = \sqrt{15.68} = 3.96$ rad/s.

Example 3: the bead on a spinning hoop

A bead slides on a hoop of radius $R = 0.40$ m that is forced to spin about its vertical diameter at $\omega = 8.00$ rad/s. Let $\theta$ be the angle from the bottom. One degree of freedom, since the spin is imposed rather than free.

The bead is a distance $R\sin\theta$ from the axis, so it is carried around at speed $\omega R\sin\theta$, perpendicular to the hoop plane. Within the plane it moves at $R\dot\theta$. Those are perpendicular, so

$T = \tfrac12 mR^2\dot\theta^2 + \tfrac12 mR^2\omega^2\sin^2\theta, \qquad V = -mgR\cos\theta$

The second term of $T$ has no $\dot\theta$ in it, so as far as the $\theta$ equation is concerned it acts like a negative potential. Collect everything that depends on $\theta$ alone into an effective potential

$U_{\text{eff}}(\theta) = -mgR\cos\theta - \tfrac12 mR^2\omega^2\sin^2\theta$

and the problem becomes a particle of "mass" $mR^2$ in that potential. Equilibrium is where $dU_{\text{eff}}/d\theta = 0$:

$mgR\sin\theta - mR^2\omega^2\sin\theta\cos\theta = 0 \quad\Longrightarrow\quad \sin\theta\left(g - R\omega^2\cos\theta\right) = 0$

One root is $\theta = 0$, the bottom. The other exists only if $\cos\theta = g/(R\omega^2)$ is at most 1, that is if $\omega \ge \sqrt{g/R}$. That threshold is the critical spin rate: $\sqrt{9.80/0.40} = 4.95$ rad/s. Below it the bead sits at the bottom; above it the bottom turns into a hump and two tilted minima appear.

Here $\omega = 8.00$ exceeds 4.95, so $\cos\theta_0 = 9.80/(0.40 \times 64.0) = 9.80/25.6 = 0.3828$, giving $\theta_0 = 67.5^\circ$.

For the oscillation frequency about that tilted spot, differentiate again:

$\frac{d^2U_{\text{eff}}}{d\theta^2} = mgR\cos\theta - mR^2\omega^2\cos 2\theta$

Evaluate at $\cos\theta_0 = g/(R\omega^2)$, using $\cos 2\theta = 2\cos^2\theta - 1$. The first term is $mR^2\omega^2\cos^2\theta_0$ and the second is $-mR^2\omega^2(2\cos^2\theta_0 - 1)$, so the sum is $mR^2\omega^2(1 - \cos^2\theta_0) = mR^2\omega^2\sin^2\theta_0$. Dividing by the coefficient $mR^2$ of $\tfrac12\dot\theta^2$:

$\omega_{\text{osc}} = \omega\sin\theta_0 = 8.00\sin 67.5^\circ = 8.00(0.9239) = 7.39\ \text{rad/s}$

Below critical the same second derivative at $\theta = 0$ gives $mgR - mR^2\omega^2$, so $\omega_{\text{osc}} = \sqrt{g/R - \omega^2}$, which goes to zero exactly at the critical rate and tells you the bottom has gone unstable.

Two coordinates at once: normal modes

When two coordinates are both small, expand to quadratic order and you get

$L \approx \tfrac12\left(a_{11}\dot u_1^2 + 2a_{12}\dot u_1\dot u_2 + a_{22}\dot u_2^2\right) - \tfrac12\left(b_{11}u_1^2 + 2b_{12}u_1u_2 + b_{22}u_2^2\right)$

Assume both coordinates oscillate at the same frequency, $u_i \propto \cos\omega t$, substitute, and you are left with a pair of linear equations that have a non-zero solution only when the determinant vanishes. The condition is $\det(B - \omega^2 A) = 0$: a quadratic in $\omega^2$ giving two roots, the two normal mode frequencies.

Take a pendulum of length $L = 0.80$ m, bob $m = 0.50$ kg, hanging from a cart of mass $M = 2.0$ kg that is free to slide. The bob's position is $x + L\sin\theta$ and $-L\cos\theta$, so

$T = \tfrac12 M\dot x^2 + \tfrac12 m\left(\dot x^2 + 2L\dot x\dot\theta\cos\theta + L^2\dot\theta^2\right)$

$x$ is ignorable, so $(M+m)\dot x + mL\dot\theta\cos\theta$ is conserved; starting from rest it is zero, so $\dot x = -mL\dot\theta\cos\theta/(M+m) \approx -mL\dot\theta/(M+m)$ for small $\theta$. Put that back into the $\theta$ equation, $mL\ddot x + mL^2\ddot\theta = -mgL\theta$, and the $\ddot x$ term shrinks the effective inertia:

$mL^2\ddot\theta\left(1 - \frac{m}{M+m}\right) = -mgL\theta \quad\Longrightarrow\quad \omega^2 = \frac{g(M+m)}{ML}$

Numbers: $9.80(2.5)/(2.0 \times 0.80) = 24.5/1.60 = 15.3$, so $\omega = 3.91$ rad/s: faster than the $3.50$ rad/s a fixed pivot would give, because the recoiling cart lets the bob take a shorter path. One of the two modes here is the free sliding of the whole system, at $\omega = 0$. Put a spring on the cart and that zero mode becomes a real frequency, and you must solve the full determinant.

The double pendulum works the same way, with $\theta_1$ and $\theta_2$; the only extra care is writing the lower bob's position as the sum of two rod vectors before differentiating. For two equal masses on two equal rods of length $L$ the determinant gives $\omega^2 = (2 \mp \sqrt2)g/L$, which for $L = 1.00$ m is $2.40$ rad/s and $5.78$ rad/s.

Recognising when to reach for this

Four signals, any one of which is enough:

More than one body, tied together. Carts with pendulums, wedges with blocks, double Atwood machines, a chain sliding off a table. Each internal tension or normal force you would otherwise have to name is a reason to switch.

A support that is free to move. The moment the pivot, the rail, or the ramp can recoil, Newtonian setup gets frame-confusing while the Lagrangian just picks up a cross term.

The question asks for a frequency of small oscillations. Expand $L$ to quadratic order and read off $\omega = \sqrt{b/a}$. This is faster than any torque analysis and it never asks you to identify a restoring force.

Something rotates or is forced at a fixed rate. A spinning hoop, a rotating rod, a turntable groove. Fold the imposed motion into $T$, sweep the $\theta$-only terms into an effective potential, and read equilibria off its minima.

The counter-signal matters too. If a problem asks for a constraint force (the tension in the rod, the normal force from the sphere, the angle where a block leaves a surface) the Lagrangian will not hand it to you directly, because you deliberately threw it away. Get the motion from the Lagrangian, then go back to $F = ma$ in the radial direction for that one force. That is exactly how the block-leaving-a-sphere problem goes: energy conservation gives $v^2$ as a function of angle, then $mg\cos\theta - N = mv^2/R$ with $N = 0$ gives $\cos\theta = 2/3$, so $\theta = 48.2^\circ$.

The effective-potential idea you met with the spinning hoop is the same machinery that governs orbits, and the two-coordinate determinant is the engine behind coupled oscillators and normal modes, which is where this goes next.