ME3120 · Dynamic Modelling
Theme 3 · Energy / Lagrange
The Lagrangian & Lagrange's equation
Combine the two energies into one quantity, turn a fixed crank, and the equation of motion drops out — no forces, no free-body diagram.
Source: course notes, Week 2 (the energy method); Spong, Lynch & Park.
Before you start
What you need first
- Kinetic & potential energy \(T\), \(V\) (Topic 10).
- Velocity from position — getting \(\dot q\) into \(T\) (Topic 11).
- Generalized coordinates \(q\) — one per degree of freedom.
What you'll be able to do
- Form the Lagrangian \(\mathcal{L}=T-V\).
- Apply Lagrange's equation — the master formula.
- Run the partial-derivative and \(d/dt\) steps, and re-derive the spring.
A note on the symbol \(\mathcal{L}\)
We write the Lagrangian as the script \(\mathcal{L}\), on
purpose. Plain \(L\) already means the rod length in this course (the
pendulum), so to avoid two meanings in one equation, the Lagrangian gets its own symbol
\(\mathcal{L}\). Read \(\mathcal{L}\) as "the Lagrangian," \(L\) as "a length."
Step 1 · combine the energies
The Lagrangian: \(\mathcal{L} = T - V\)
$$\mathcal{L} = T - V$$
| Symbol | Meaning | SI unit |
|---|---|---|
| \(\mathcal{L}\) | the Lagrangian | J |
| T | kinetic energy (of motion) | J |
| V | potential energy (of position) | J |
It is a minus, not a plus. \(\mathcal{L}\) is not
the total energy \(T+V\). It is a special combination \(T-V\) that makes the next formula
produce the equation of motion.
Step 2 · the master formula
Lagrange's equation
$$\frac{d}{dt}\!\left(\frac{\partial \mathcal{L}}{\partial \dot q}\right) - \frac{\partial \mathcal{L}}{\partial q} = Q$$
| Symbol | Meaning | SI unit |
|---|---|---|
| q | a generalized coordinate | m or rad |
| \(\dot q\) | its velocity | m/s or rad/s |
| Q | generalized force (applied push/torque, damping) | N or N·m |
Apply it once for each coordinate \(q\) — each application
gives one equation of motion. If nothing external pushes the system, \(Q=0\) (we cover
\(Q\) fully next topic).
Reading the master formula
| Piece | What it does |
|---|---|
| \(\dfrac{\partial \mathcal{L}}{\partial \dot q}\) | differentiate \(\mathcal{L}\) treating the velocity \(\dot q\) as the variable → the "momentum" term |
| \(\dfrac{d}{dt}(\ \cdot\ )\) | then differentiate that in time → produces the acceleration \(\ddot q\) |
| \(\dfrac{\partial \mathcal{L}}{\partial q}\) | differentiate \(\mathcal{L}\) treating the position \(q\) as the variable → the spring/gravity term |
| \(Q\) | applied forces / torques / damping (else \(0\)) |
Quick refresher: partial derivatives
A partial derivative differentiates with respect to one variable, holding the others fixed. The key trick here: treat the position \(q\) and the velocity \(\dot q\) as two separate, independent symbols. For \(\mathcal{L}=\tfrac12 m\dot x^{2}-\tfrac12 k x^{2}\):
Hold \(x\) fixed, differentiate by \(\dot x\)
$$\frac{\partial \mathcal{L}}{\partial \dot x}=m\dot x$$
Hold \(\dot x\) fixed, differentiate by \(x\)
$$\frac{\partial \mathcal{L}}{\partial x}=-k x$$
Quick refresher: the \(d/dt\) step
After the partial \(\partial\mathcal{L}/\partial\dot q\), the \(d/dt\) lets everything depend on time again and differentiates in time — turning a velocity into an acceleration:
$$\frac{\partial \mathcal{L}}{\partial \dot x}=m\dot x \quad\xrightarrow{\ d/dt\ }\quad \frac{d}{dt}(m\dot x)=m\ddot x$$
That step is where the acceleration \(\ddot q\) in every equation of motion
comes from.
The energy recipe (replaces the free-body diagram)
1Choose coordinates \(q\) — one per degree of freedom.
2Write \(T\) — add \(\tfrac12 mv^2\) and \(\tfrac12 I\omega^2\) for every part.
3Write \(V\) — gravity \(mgh\) and springs \(\tfrac12 kx^2\).
4Form \(\mathcal{L}=T-V\).
5Apply Lagrange's equation for each \(q\).
6Check — units, limits, equilibrium.
Re-derive the spring by Lagrange
The block-and-spring of Theme 2: mass \(m\), stiffness \(k\), coordinate \(x\) from rest, on a frictionless table. Let us get \(m\ddot x + kx = 0\) again — this time from energy, with no free-body diagram.
The same block-and-spring — now modelled by energy.
1Energies, then the Lagrangian (\(T=\tfrac12 m\dot x^2\), \(V=\tfrac12 kx^2\)):
$$\mathcal{L} = T - V = \tfrac12 m\dot x^{2} - \tfrac12 k x^{2}$$
2Momentum term — differentiate by \(\dot x\), then by time:
$$\frac{\partial \mathcal{L}}{\partial \dot x}=m\dot x \quad\Longrightarrow\quad \frac{d}{dt}\!\left(\frac{\partial \mathcal{L}}{\partial \dot x}\right)=m\ddot x$$
3Position term — differentiate by \(x\):
$$\frac{\partial \mathcal{L}}{\partial x}=-k x$$
4Assemble (no applied force, so \(Q=0\)):
$$m\ddot x - (-k x) = 0 \;\Longrightarrow\; m\ddot x + k x = 0$$
Exactly the Newton answer from Theme 2 ✓ — but we never drew a single force.
Write \(T\) and \(V\), turn the crank, done.
✏️ Try it yourself
A mass \(m\) falls freely under gravity. With \(y\) measured upward, \(T=\tfrac12 m\dot y^2\) and \(V=mgy\). Apply Lagrange's equation (with \(Q=0\)) and find the acceleration \(\ddot y\).
Lagrangian.
\(\mathcal{L}=T-V=\tfrac12 m\dot y^2 - mgy\)
Momentum term.
\(\dfrac{\partial \mathcal{L}}{\partial \dot y}=m\dot y \Rightarrow \dfrac{d}{dt}(m\dot y)=m\ddot y\)
Position term.
\(\dfrac{\partial \mathcal{L}}{\partial y}=-mg\)
Assemble.
\(m\ddot y-(-mg)=0 \Rightarrow m\ddot y + mg = 0 \Rightarrow \ddot y = -g\) — free fall,
accelerating downward at \(g\). ✓
Common mistakes to avoid
| Mistake | Fix |
|---|---|
| Writing \(\mathcal{L}=T+V\) | It is \(T-V\) — always a minus. |
| Mixing \(\partial/\partial q\) and \(\partial/\partial\dot q\) | Treat \(q\) and \(\dot q\) as separate symbols when taking partials. |
| Skipping the \(d/dt\) step | That step makes the acceleration \(\ddot q\) appear. |
| Forgetting the minus on \(-\partial\mathcal{L}/\partial q\) | The formula subtracts the position term; a sign slip flips the spring/gravity term. |
| Applying it once for a 2-DOF system | Apply the formula once per coordinate. |
Recap — the whole topic on one screen
$$\mathcal{L}=T-V \qquad\quad \frac{d}{dt}\!\left(\frac{\partial \mathcal{L}}{\partial \dot q}\right) - \frac{\partial \mathcal{L}}{\partial q} = Q$$
| Idea | What you own now |
|---|---|
| The Lagrangian | \(\mathcal{L}=T-V\) (minus, and \(\mathcal{L}\) not \(L\)) |
| The master formula | The crank, applied once per coordinate |
| The two steps | Partial by \(\dot q\) then \(d/dt\); partial by \(q\) |
| The pay-off | Spring re-derived from energy, no forces drawn |