ME3120 · Dynamic Modelling

Theme 3 · Energy / Lagrange

Lagrange on one DOF

Everything together at last: the spring and the pendulum, derived end to end by energy — and set side by side with the Newton results from Theme 2.

Source: course notes, Week 2 worked examples A & B.

Before you start

What you need first

The energy recipe and Lagrange's equation (Topic 12).
Velocity from position for the pendulum bob (Topic 11).
The Newton results to compare: $m\ddot x+kx=0$ and $\ddot\theta+\tfrac{g}{L}\sin\theta=0$.

What you'll be able to do

Derive the spring equation by energy, end to end.
Derive the pendulum equation by energy — tension never appears.
Say when Lagrange beats Newton.

The recipe, in one line

$$\text{choose } q \;\to\; T \;\to\; V \;\to\; \mathcal{L}=T-V \;\to\; \frac{d}{dt}\!\left(\frac{\partial \mathcal{L}}{\partial \dot q}\right)-\frac{\partial \mathcal{L}}{\partial q}=Q \;\to\; \text{check}$$

We now run this exact recipe twice — on the spring, then the pendulum — and get the Theme 2 answers without drawing a single force.

📐 Worked example A

The mass–spring, by energy

Mass $m$, spring $k$, coordinate $x$ from rest.

1Kinetic energy:

$$T=\tfrac12 m\dot x^{2}$$

2Potential energy (spring):

$$V=\tfrac12 k x^{2}$$

3Lagrangian:

$$\mathcal{L}=\tfrac12 m\dot x^{2}-\tfrac12 k x^{2}$$

4Momentum term — differentiate by $\dot x$, then by time:

$$\frac{\partial \mathcal{L}}{\partial \dot x}=m\dot x \quad\Longrightarrow\quad \frac{d}{dt}\!\left(\frac{\partial \mathcal{L}}{\partial \dot x}\right)=m\ddot x$$

5Position term — differentiate by $x$:

$$\frac{\partial \mathcal{L}}{\partial x}=-k x$$

6Assemble ($Q=0$):

$$m\ddot x + k x = 0$$

Identical to the Newton answer ✓ — built from two scalars, no free-body diagram.

📐 Worked example B

The pendulum, by energy — the pay-off

Bob mass $m$, rod $L$, angle $\theta$. One coordinate.

1Speed of the bob (from Topic 11):

$$v^{2}=L^{2}\dot\theta^{2}$$

2Kinetic energy:

$$T=\tfrac12 m L^{2}\dot\theta^{2}$$

3Potential energy (rise $L(1-\cos\theta)$):

$$V=mgL(1-\cos\theta)$$

4Lagrangian:

$$\mathcal{L}=\tfrac12 m L^{2}\dot\theta^{2}-mgL(1-\cos\theta)$$

5Momentum term:

$$\frac{\partial \mathcal{L}}{\partial \dot\theta}=mL^{2}\dot\theta \quad\Longrightarrow\quad \frac{d}{dt}\!\left(\frac{\partial \mathcal{L}}{\partial \dot\theta}\right)=mL^{2}\ddot\theta$$

6Position term (only $V$ depends on $\theta$):

$$\frac{\partial \mathcal{L}}{\partial \theta}=-mgL\sin\theta$$

7Assemble and divide by $mL^{2}$:

$$mL^{2}\ddot\theta+mgL\sin\theta=0 \;\Longrightarrow\; \ddot\theta+\frac{g}{L}\sin\theta=0$$

Identical to the Newton answer ✓ — and the rod tension never appeared. That is the whole pay-off of the energy method.

Newton vs Lagrange — side by side

	Newton (Theme 2)	Lagrange (this theme)
Start from	forces & free-body diagram	energies $T$ and $V$
Constraint forces	must find them, then cancel	never appear
Geometry	track force vectors carefully	scalars only — no vector balancing
Many DOF	gets messy fast	same recipe, repeated per coordinate

Same equation of motion, far less work. For one simple body it is a tie; for a multi-joint robot, Lagrange wins every time.

✏️ Try it yourself

Re-derive the vertical mass–spring by Lagrange. Let $y$ be measured downward from the spring's natural length, so $T=\tfrac12 m\dot y^2$ and $V=\tfrac12 k y^2 - mgy$ (spring stores energy; gravity lowers it). With $Q=0$, find the equation of motion.

Lagrangian. $\mathcal{L}=\tfrac12 m\dot y^2 - \tfrac12 k y^2 + mgy$ Momentum term. $\dfrac{\partial \mathcal{L}}{\partial \dot y}=m\dot y \Rightarrow \dfrac{d}{dt}(m\dot y)=m\ddot y$ Position term. $\dfrac{\partial \mathcal{L}}{\partial y}=-ky+mg$ Assemble. $m\ddot y-(-ky+mg)=0 \Rightarrow m\ddot y + ky = mg$. It rests at $y_{\text{eq}}=mg/k$, the static stretch. ✓

Common mistakes to avoid

Mistake	Fix
Trying to add the rod tension to $T$ or $V$	Constraint forces never enter energy — that is exactly why Lagrange is clean.
Using $V=mgL\cos\theta$ with the wrong sign	The bob rises $L(1-\cos\theta)$; a sign slip flips the gravity term.
Forgetting the chain rule in $v^2$	$v^2=L^2\dot\theta^2$ — the $\dot\theta$ comes from differentiating the position.
Dropping the gravity term in the vertical spring	With $y$ down, gravity adds $+mgy$ to $\mathcal{L}$ and a $mg$ to the right side.

Recap — the whole topic on one screen

$$\text{spring: } m\ddot x + kx = 0 \qquad\qquad \text{pendulum: } \ddot\theta+\tfrac{g}{L}\sin\theta=0$$

Idea	What you own now
Spring by energy	$T,V \to \mathcal{L} \to$ crank $\to m\ddot x+kx=0$
Pendulum by energy	Same recipe $\to \ddot\theta+\tfrac{g}{L}\sin\theta=0$, no tension
Why it wins	Scalars, no constraint forces, same recipe for any DOF

Next topic

Coupled 2-DOF systems

One coordinate was a warm-up. The real power shows with two: apply Lagrange once per coordinate and the equations stack into a mass matrix — the first glimpse of the robot equation's shape.

→ Coupled 2-DOF systems

Mistake	Fix
Trying to add the rod tension to \(T\) or \(V\)	Constraint forces never enter energy — that is exactly why Lagrange is clean.
Using \(V=mgL\cos\theta\) with the wrong sign	The bob rises \(L(1-\cos\theta)\); a sign slip flips the gravity term.
Forgetting the chain rule in \(v^2\)	\(v^2=L^2\dot\theta^2\) — the \(\dot\theta\) comes from differentiating the position.
Dropping the gravity term in the vertical spring	With \(y\) down, gravity adds \(+mgy\) to \(\mathcal{L}\) and a \(mg\) to the right side.

	Newton (Theme 2)	Lagrange (this theme)
Start from	forces & free-body diagram	energies \(T\) and \(V\)
Constraint forces	must find them, then cancel	never appear
Geometry	track force vectors carefully	scalars only — no vector balancing
Many DOF	gets messy fast	same recipe, repeated per coordinate

Idea	What you own now
Spring by energy	\(T,V \to \mathcal{L} \to\) crank \(\to m\ddot x+kx=0\)
Pendulum by energy	Same recipe \(\to \ddot\theta+\tfrac{g}{L}\sin\theta=0\), no tension
Why it wins	Scalars, no constraint forces, same recipe for any DOF