ME3120 · Dynamic Modelling
Theme 3 · Energy / Lagrange
Coupled 2-DOF systems
Two coordinates, one recipe. Apply Lagrange twice and the equations stack into a mass matrix — your first look at the shape of the robot equation.
Source: course notes, Week 2 worked example C.
Before you start
What you need first
- Lagrange's equation, applied once per coordinate (Topic 12).
- Spring potential energy \(\tfrac12 kx^2\), including a spring between two moving masses.
What you'll be able to do
- Write \(T\) and \(V\) for a two-mass system.
- Get two coupled equations and stack them into matrices.
- Read the mass matrix \(M\), stiffness \(K\), and the coupling.
The recipe does not change
The real power of Lagrange shows up with more than one degree of freedom. Nothing new is needed — you simply apply the master formula once for each coordinate, giving one equation per coordinate.
🔭 Looking ahead: a 2-link robot arm also has two coordinates. Here we
use a simpler two-mass system so the algebra stays clean — but the shape of the
answer is the same.
Step 1 · the picture
Two masses, two springs
Wall — spring \(k_1\) — mass \(m_1\) — spring \(k_2\) — mass \(m_2\),
on a frictionless line. Coordinates \(x_1,x_2\) from each block's rest position.
Two degrees of freedom → two coordinates \(x_1,x_2\) → we will get
two equations of motion.
Step 2 · the energies
Kinetic and potential energy
1Kinetic energy — the two blocks move independently:
$$T = \tfrac12 m_1\dot x_1^{2} + \tfrac12 m_2\dot x_2^{2}$$
2Potential energy — spring \(k_1\) stretches by \(x_1\); spring \(k_2\) stretches by the difference \((x_2-x_1)\):
$$V = \tfrac12 k_1 x_1^{2} + \tfrac12 k_2 (x_2-x_1)^{2}$$
The key modelling step is \(k_2\)'s stretch: the middle spring feels the
relative motion \((x_2-x_1)\), not \(x_2\) alone. That difference is what ties
the two masses together.
Step 3 · Lagrange for \(x_1\)
First equation (coordinate \(x_1\))
1Momentum term:
$$\frac{\partial \mathcal{L}}{\partial \dot x_1}=m_1\dot x_1 \;\Longrightarrow\; \frac{d}{dt}(m_1\dot x_1)=m_1\ddot x_1$$
2Position term (chain rule on \((x_2-x_1)^2\) gives a \(+\) sign):
$$\frac{\partial \mathcal{L}}{\partial x_1}=-k_1 x_1 + k_2 (x_2-x_1)$$
3Assemble \(m_1\ddot x_1 - \partial\mathcal{L}/\partial x_1 = 0\):
$$m_1\ddot x_1 + (k_1+k_2)x_1 - k_2 x_2 = 0$$
Step 4 · Lagrange for \(x_2\)
Second equation (coordinate \(x_2\))
1Momentum term:
$$\frac{\partial \mathcal{L}}{\partial \dot x_2}=m_2\dot x_2 \;\Longrightarrow\; \frac{d}{dt}(m_2\dot x_2)=m_2\ddot x_2$$
2Position term (only \(k_2\)'s spring depends on \(x_2\)):
$$\frac{\partial \mathcal{L}}{\partial x_2}=-k_2 (x_2-x_1)$$
3Assemble:
$$m_2\ddot x_2 + k_2 x_2 - k_2 x_1 = 0$$
Step 5 · stack them
Stack into matrices: \(M\) and \(K\)
$$\underbrace{\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix}}_{M}\begin{bmatrix}\ddot x_1 \\ \ddot x_2\end{bmatrix} + \underbrace{\begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix}}_{K}\begin{bmatrix}x_1 \\ x_2\end{bmatrix} = \begin{bmatrix}0\\0\end{bmatrix}$$
In short, \(M\ddot{\mathbf{x}} + K\mathbf{x} = \mathbf 0\): a mass matrix \(M\) and a stiffness matrix \(K\).
The off-diagonal \(-k_2\) is the coupling — it
is what ties \(x_1\) and \(x_2\) together. If \(k_2=0\) the masses uncouple into two separate
springs. Coupling lives off the diagonal.
📄 Full line-by-line derivation (printable PDF)
— every partial derivative and algebra step, with the setup figure. A pen-and-paper companion;
the page above is already complete on its own.
Why this matters for robots
The robot equation has the very same shape — a mass matrix, plus extra terms:
$$M(q)\,\ddot q + C(q,\dot q)\,\dot q + G(q) = \tau$$
| Today's 2-mass system | Robot arm |
|---|---|
| \(M\) (constant) | \(M(q)\) — depends on the pose |
| \(K\mathbf{x}\) (springs) | \(G(q)\) — gravity on the links |
| — | \(C(q,\dot q)\dot q\) — Coriolis, from moving fast |
🔭 Every one of these comes straight out of Lagrange's equation. You now have
the engine; Themes 4–5 build the pieces \(M(q)\), \(C\), and \(G\) for a real arm.
Write \(M\) and \(K\) for real numbers
Take \(m_1 = 2\ \text{kg}\), \(m_2 = 1\ \text{kg}\), \(k_1 = 100\ \text{N/m}\), \(k_2 = 50\ \text{N/m}\). Write the mass and stiffness matrices and the two equations.
1Mass matrix (the masses on the diagonal):
$$M=\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}\ \text{kg}$$
2Stiffness matrix (\(k_1+k_2=150\) top-left, \(-k_2=-50\) off-diagonal, \(k_2=50\) bottom-right):
$$K=\begin{bmatrix} 150 & -50 \\ -50 & 50 \end{bmatrix}\ \text{N/m}$$
3The two equations of motion:
$$2\ddot x_1 + 150 x_1 - 50 x_2 = 0, \qquad \ddot x_2 + 50 x_2 - 50 x_1 = 0$$
Read the coupling: the \(-50\) terms are how each mass feels the other through
the middle spring \(k_2\).
✏️ Try it yourself
For the same chain with \(m_1 = m_2 = 3\ \text{kg}\) and \(k_1 = k_2 = 200\ \text{N/m}\), write the mass matrix \(M\) and stiffness matrix \(K\).
Mass matrix.
\(M=\begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}\ \text{kg}\)
Stiffness matrix.
\(K=\begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix}=\begin{bmatrix} 400 & -200 \\ -200 & 200 \end{bmatrix}\ \text{N/m}\)
Coupling.
The off-diagonal \(-200\) ties the two masses together through the middle spring.
Common mistakes to avoid
| Mistake | Fix |
|---|---|
| Writing \(k_2\)'s energy as \(\tfrac12 k_2 x_2^2\) | The middle spring stretches by the difference: \(\tfrac12 k_2 (x_2-x_1)^2\). |
| Applying Lagrange only once | Two coordinates → apply it twice, once for \(x_1\) and once for \(x_2\). |
| Sign slip on the coupling term | Differentiating \((x_2-x_1)^2\) by \(x_1\) gives \(-2(x_2-x_1)\); track the signs carefully. |
| Putting coupling on the diagonal | Coupling lives off the diagonal (\(-k_2\)); the diagonal holds each coordinate's own stiffness. |
Recap — the whole topic on one screen
$$M\ddot{\mathbf{x}} + K\mathbf{x} = \mathbf 0, \qquad M=\begin{bmatrix} m_1 & 0 \\ 0 & m_2 \end{bmatrix},\quad K=\begin{bmatrix} k_1+k_2 & -k_2 \\ -k_2 & k_2 \end{bmatrix}$$
| Idea | What you own now |
|---|---|
| Two coordinates | Apply Lagrange once per coordinate → two equations |
| The coupling | Middle spring feels \((x_2-x_1)\); appears off-diagonal |
| Matrix form | \(M\ddot{\mathbf x}+K\mathbf x=\mathbf 0\) — mass & stiffness matrices |
| Robot link | Same shape as \(M(q)\ddot q + C\dot q + G(q)=\tau\) |