ME3120 · Dynamic Modelling
Theme 3 · Energy / Lagrange

Kinetic & potential energy

A second way to model machines: instead of chasing forces, we count energy. First, meet the only two kinds we ever need.

Source: course notes, Week 2 (the energy method); standard mechanics.

Before you start

What you need first

  • Coordinates & velocities \(q,\dot q\) — one per degree of freedom.
  • The spring (\(k\)) and moment of inertia \(I\) (the "rotational mass") from Theme 2.

What you'll be able to do

  • Write the kinetic energy \(T\) of any moving system.
  • Write the potential energy \(V\) (gravity and springs).
  • Add up parts and pick a sensible height reference.

Why a second method?

To use \(\sum F = m\ddot x\) you must know every force — including ones you do not care about: the tension in a pendulum rod, the reaction at a pin, the normal force from a surface.

These "constraint forces" hold parts together but do no useful work. With Newton you must find them anyway, then watch them cancel. For a robot with many joints, that becomes a nightmare.
The energy method's big idea: energy does not care about constraint forces. If we build the equations from energy, the tensions and reactions never appear. (Remember the pendulum rod tension dropping out? Energy makes that happen automatically, every time.)

Only two kinds of energy matter here

Kinetic energy \(T\)

the energy of motion — anything moving or spinning has it.

Potential energy \(V\)

stored energy of position — height (gravity) or a stretched spring.

Get good at writing these two scalars and you can model almost anything. Everything in this theme is built from \(T\) and \(V\).

Energy of motion

Kinetic energy: moving and spinning

A mass moving in a line, and a body spinning, each carry kinetic energy:

$$T = \tfrac12 m v^{2} \qquad\quad T = \tfrac12 I \omega^{2}$$
Same shape twice: spinning just swaps mass \(m\) for inertia \(I\), and speed \(v\) for angular speed \(\omega=\dot\theta\).
m v sliding I ω spinning
Kinetic energy comes in two forms — sliding (\(\tfrac12 mv^2\)) and spinning (\(\tfrac12 I\omega^2\)).
where:
SymbolMeaningSI unit
Tkinetic energyJ (joule)
mmasskg
vspeed of the massm/s
Imoment of inertiakg·m²
\(\omega\)angular speed (\(=\dot\theta\))rad/s

Many moving parts? Just add them up

$$T = \tfrac12 m_1 v_1^{2} + \tfrac12 m_2 v_2^{2} + \dots + \tfrac12 I_1 \omega_1^{2} + \dots$$
The kinetic energy of a whole machine is the sum of \(\tfrac12 m v^2\) for every moving mass plus \(\tfrac12 I\omega^2\) for every spinning part.
🔭 Looking ahead: a robot arm is several links, each sliding and spinning. We will simply add up the energy of every link.

Energy of position

Potential energy: gravity and springs

h = 0 m h gravity m k x spring
Potential energy is stored by position: a raised mass (\(mgh\)) and a stretched spring (\(\tfrac12 kx^2\)).

Two stores of potential energy show up in our machines:

$$V_{\text{grav}} = m g h \qquad\quad V_{\text{spring}} = \tfrac12 k x^{2}$$
where:
SymbolMeaningSI unit
hheight above a chosen referencem
kspring stiffnessN/m
xstretch from natural lengthm

Where is \(h=0\)? You choose

Only changes in potential energy matter, so you are free to put the reference level \(h=0\) anywhere convenient. For the pendulum, two common choices:

Reference levelPotential energy \(V\)
at the pivot\(V = -mgL\cos\theta\)
at the lowest point\(V = mgL(1-\cos\theta)\)
Both give the same equation of motion, because they differ only by a constant and \(\dfrac{\partial V}{\partial\theta}=mgL\sin\theta\) is identical. The constant washes out.
📐 Worked example

Write \(T\) and \(V\) for a hanging block

A block of mass \(m = 2\ \text{kg}\) hangs on a vertical spring (\(k = 300\ \text{N/m}\)). At one instant it moves at \(v = 1.5\ \text{m/s}\), the spring is stretched \(x = 0.08\ \text{m}\), and the block is \(h = 0.5\ \text{m}\) above the floor. Take the floor as \(h=0\) and \(g = 9.81\ \text{m/s}^2\). Find \(T\) and \(V\).

x = 0.08 m 2 kg v h=0 h = 0.5 m
The hanging block, with the numbers from the problem.
1Kinetic energy (the block slides, no spin):
$$T = \tfrac12 m v^{2} = \tfrac12 (2)(1.5)^{2} = 2.25\ \text{J}$$
2Potential energy — add gravity and the spring:
$$V = mgh + \tfrac12 k x^{2} = (2)(9.81)(0.5) + \tfrac12 (300)(0.08)^{2} = 9.81 + 0.96 = 10.8\ \text{J}$$
Notice the recipe: one \(\tfrac12 mv^2\) for the motion, then a sum of every store of potential energy (here gravity + spring). That is all \(T\) and \(V\) ever are.

✏️ Try it yourself

A disk of moment of inertia \(I = 0.2\ \text{kg·m}^2\) spins at \(\omega = 8\ \text{rad/s}\) while its centre (mass \(m = 3\ \text{kg}\)) moves at \(v = 1.2\ \text{m/s}\). Find the total kinetic energy.

Sliding part. \(\tfrac12 m v^2 = \tfrac12 (3)(1.2)^2 = 2.16\ \text{J}\) Spinning part. \(\tfrac12 I \omega^2 = \tfrac12 (0.2)(8)^2 = 6.4\ \text{J}\) Total. \(T = 2.16 + 6.4 = 8.56\ \text{J}\) — add both forms of motion.

Common mistakes to avoid

MistakeFix
Forgetting a moving part in \(T\)Every mass that moves needs its \(\tfrac12 mv^2\); every spinning part its \(\tfrac12 I\omega^2\).
Using \(v\) instead of \(v^2\)Kinetic energy grows with the square of speed.
Mixing up \(m\) and \(I\)\(m\) (kg) goes with speed \(v\); \(I\) (kg·m²) goes with angular speed \(\omega\).
Worrying about where \(h=0\) isAny reference works — only changes in \(V\) matter.
Putting a constraint force into \(V\)\(V\) is only gravity and springs; tensions/reactions never enter the energy.

Recap — the whole topic on one screen

$$T = \sum \tfrac12 m v^{2} + \sum \tfrac12 I\omega^{2} \qquad V = \sum mgh + \sum \tfrac12 k x^{2}$$
IdeaWhat you own now
Why energyIt ignores constraint forces — no tension-chasing
Kinetic energy\(\tfrac12 mv^2\) (sliding) + \(\tfrac12 I\omega^2\) (spinning), added up
Potential energy\(mgh\) (gravity) + \(\tfrac12 kx^2\) (spring), added up
ReferencePut \(h=0\) anywhere — only changes matter

Next topic

Velocity from position

To write \(\tfrac12 mv^2\) we need the speed \(v\) of each part — and that comes from where the part is. Next we turn a position into a velocity (a quick chain-rule step), and find the speed of a swinging pendulum bob.

→ Velocity from position