ME3120 · Dynamic Modelling

Theme 2 · Newton's method

The pendulum by Newton

Our first turning machine. We swing a mass on a rod, take torques about the pivot, and out comes a famous equation — with a twist that the spring never had.

Source: course notes, after a standard simple-pendulum treatment (cf. Spong, Robot Modeling & Control).

Before you start

What you need first

Newton for turning — $\sum \tau = I\,\ddot\theta$ (twist = rotational inertia × angular acceleration).
Free-body diagrams & torque = force × lever arm (the perpendicular distance).
The model recipe — draw it, pick a coordinate, sum forces/torques, write the balance (from the spring topic).

What you'll be able to do

Build the equation of motion of a pendulum with Newton.
See why it is nonlinear — the $\sin\theta$ term.
Get it ready for small-angle linearization (the next topic).

Step 1 · the picture

The system: a mass swinging on a rod

A small bob of mass $m$ hangs from a fixed pivot on a light, rigid rod of length $L$ (we treat the rod as massless and the bob as a point mass).

The bob can only swing — it moves on a circle of radius $L$. So the system has one degree of freedom, and we pick one coordinate: the angle $\theta$, measured from the straight-down position.

One coordinate $\theta$ fixes everything. We take the swing shown — to the left — as the positive direction for $\theta$.

A bob of mass $m$ on a rod of length $L$; $\theta$ is measured from the straight-down dashed line. Gravity $mg$ pulls straight down.

Step 2 · the rotational inertia

A point mass that turns: $I = mL^2$

Because the pendulum turns, we use the turning law $\sum\tau = I\ddot\theta$. That needs $I$, the moment of inertia — the "rotational mass," i.e. how hard the body is to spin up about the pivot.

$$I = m\,L^{2}$$

where:
Symbol	Meaning	SI unit
I	moment of inertia of the bob about the pivot	kg·m²
m	mass of the bob	kg
L	length of the rod (pivot to bob)	m

For a single point mass $m$ sitting a distance $L$ from the axis, $I = mL^2$. The farther out the mass (bigger $L$), the much harder it is to swing — $L$ is squared.

🔭 Looking ahead: real bodies (rods, disks, whole robot links) are not points — finding their $I$ needs the moment-of-inertia & parallel-axis topic later in the course. Here the bob is a point, so $I=mL^2$ is all we need.

Step 3 · isolate the bob

The forces on the bob

Isolate the bob: only two outside forces — its weight $mg$ (down) and the rod tension $T$ (along the rod, toward the pivot).

Cut the bob free and draw every outside force. There are just two:

Weight $mg$ — straight down, always.
Rod tension $T$ — along the rod, pulling toward the pivot.

We will take torques about the pivot. The tension $T$ points straight at the pivot, so its lever arm is zero — $T$ makes no torque. That leaves gravity as the only twist.

Step 3 · the twist

The torque that swings it back

Torque about the pivot = force × lever arm (the perpendicular distance from the pivot to the line of the force). Gravity $mg$ acts straight down, so its line is the vertical through the bob. The perpendicular distance from the pivot to that line is the horizontal offset of the bob, $L\sin\theta$:

The lever arm of gravity is the horizontal distance $L\sin\theta$ from the pivot to the vertical line through the bob.

So the size of the gravity torque is $mg \times L\sin\theta$. Which way does it turn the pendulum? Always back toward straight-down — it fights $\theta$. A torque that opposes the coordinate gets a minus sign:

$$\tau_{\text{grav}} = -\,m\,g\,L\sin\theta$$

This is a restoring torque — exactly like the spring's $-kx$. Pulled out to one side, gravity always twists the bob back.

Step 4 · derive

Derive the equation of motion

1Start from Newton's law for turning, about the pivot:

$$\sum \tau = I\,\ddot\theta$$

2Put in the inertia of the point-mass bob, $I = mL^2$, and the only torque, gravity $-mgL\sin\theta$:

$$-\,m\,g\,L\sin\theta = m\,L^{2}\,\ddot\theta$$

3Gather both terms on one side:

$$m\,L^{2}\,\ddot\theta + m\,g\,L\sin\theta = 0$$

4Divide through by $mL^{2}$ (the $m$ cancels, one $L$ cancels):

$$\ddot\theta + \frac{g}{L}\sin\theta = 0$$

The equation of motion

$$\ddot\theta + \frac{g}{L}\sin\theta = 0$$

where:
Symbol	Meaning	SI unit
$\theta$	angle from straight-down	rad
$\ddot\theta$	angular acceleration	rad/s²
g	acceleration due to gravity ($9.81$)	m/s²
L	rod length	m

This single line is the model of the pendulum — the mass $m$ cancelled out, so a heavy bob and a light bob on the same rod swing the same way.

Notice what is missing: there is no spring, yet gravity gives the same restoring action. The $\dfrac{g}{L}\sin\theta$ term is the pendulum's "stiffness," and $\ddot\theta$ is its inertia. They fight → it swings back and forth.

Why this one is different: it is nonlinear

Compare the pendulum with the spring we built last topic:

System	Equation of motion	Restoring term
Mass–spring	$m\ddot x + kx = 0$	$kx$ — straight-line in $x$ ✓ linear
Pendulum	$\ddot\theta + \tfrac{g}{L}\sin\theta = 0$	$\sin\theta$ — curved in $\theta$ ✗ nonlinear

The twist: the unknown $\theta$ is stuck inside $\sin\theta$. That one $\sin$ makes the equation nonlinear — we cannot simply read off a natural frequency, and the neat solving tools from the spring topic do not apply directly.

🔭 Looking ahead — two payoffs. (1) Next topic we tame the $\sin\theta$ for small swings (small-angle linearization) and recover a clean $\omega_n=\sqrt{g/L}$. (2) That same gravity torque $mgL\sin\theta$ is the seed of the robot gravity term $G(q)$ we build much later.

📐 Worked example

Restoring torque and angular acceleration

A simple pendulum has a bob of mass $m = 0.5\ \text{kg}$ on a light rod of length $L = 1.0\ \text{m}$. It is held at $\theta_0 = 20^\circ$ from straight-down and released from rest. Take $g = 9.81\ \text{m/s}^2$. Find (a) its moment of inertia, (b) the restoring torque at release, and (c) the initial angular acceleration.

The worked-example pendulum, with the numbers from the problem.

1Moment of inertia of the point-mass bob, $I = mL^2$:

$$I = (0.5)(1.0)^2 = 0.5\ \text{kg·m}^2$$

2Restoring torque $\tau = mgL\sin\theta$ at $\theta_0=20^\circ$ ($\sin 20^\circ = 0.342$):

$$\tau = (0.5)(9.81)(1.0)(0.342) = 1.68\ \text{N·m}$$

3Angular acceleration $\ddot\theta = -\dfrac{g}{L}\sin\theta$ (it points back toward straight-down):

$$\ddot\theta = -\frac{9.81}{1.0}(0.342) = -3.36\ \text{rad/s}^2$$

Cross-check: $\ddot\theta = \tau/I = 1.68/0.5 = 3.36\ \text{rad/s}^2$ ✓, and the minus sign just says it accelerates back toward the bottom.

✏️ Try it yourself

A simple pendulum has a bob of mass $m = 0.8\ \text{kg}$ on a rod of length $L = 0.5\ \text{m}$, held at $\theta_0 = 30^\circ$ and released from rest. Take $g = 9.81\ \text{m/s}^2$. Find (a) $I$, (b) the restoring torque, and (c) the initial angular acceleration.

(a) $I = mL^2 = (0.8)(0.5)^2 = 0.20\ \text{kg·m}^2$ (b) $\tau = mgL\sin\theta = (0.8)(9.81)(0.5)(0.5) = 1.96\ \text{N·m}$ ($\sin 30^\circ = 0.5$) (c) $\ddot\theta = -\dfrac{g}{L}\sin\theta = -\dfrac{9.81}{0.5}(0.5) = -9.81\ \text{rad/s}^2$ Check: $\ddot\theta = \tau/I = 1.96/0.20 = 9.81\ \text{rad/s}^2$ ✓ — a short rod swings back harder (bigger $g/L$) than the worked example's longer one.

Common mistakes to avoid

Mistake	Fix
Using torque $= mgL$ (forgetting the $\sin\theta$)	The lever arm is the horizontal offset $L\sin\theta$, not $L$. Torque $= mgL\sin\theta$.
Dropping the restoring minus sign	Gravity always twists $\theta$ back to zero, so the torque is $-mgL\sin\theta$.
Writing $I = mL$	Moment of inertia of a point mass is $I = mL^2$ (units kg·m²).
Reading $\omega_n=\sqrt{g/L}$ at any angle	That only holds after the small-angle step (next topic). At large $\theta$ the $\sin\theta$ rules.
Measuring $\theta$ from the horizontal	Here $\theta$ is from straight-down; that is why the restoring term is $\sin\theta$ (not $\cos\theta$).

Recap — the whole topic on one screen

$$m L^{2}\ddot\theta + m g L\sin\theta = 0\qquad\Longrightarrow\qquad \ddot\theta + \frac{g}{L}\sin\theta = 0$$

Idea	What you own now
Build the model	Torque about the pivot: $\sum\tau = I\ddot\theta$ with $I=mL^2$
The torque	Gravity gives a restoring $-mgL\sin\theta$ (lever arm $L\sin\theta$)
The equation	$\ddot\theta + \tfrac{g}{L}\sin\theta = 0$ — mass cancels out
The twist	It is nonlinear because of $\sin\theta$

Next topic

Small-angle linearization

That $\sin\theta$ is awkward. For small swings, $\sin\theta \approx \theta$ — and the pendulum's equation snaps into the same clean shape as the spring, $\ddot\theta + \tfrac{g}{L}\theta = 0$, with natural frequency $\omega_n=\sqrt{g/L}$. Next we make that approximation honestly and see exactly when it is allowed.

→ Small-angle linearization

Mistake	Fix
Using torque \(= mgL\) (forgetting the \(\sin\theta\))	The lever arm is the horizontal offset \(L\sin\theta\), not \(L\). Torque \(= mgL\sin\theta\).
Dropping the restoring minus sign	Gravity always twists \(\theta\) back to zero, so the torque is \(-mgL\sin\theta\).
Writing \(I = mL\)	Moment of inertia of a point mass is \(I = mL^2\) (units kg·m²).
Reading \(\omega_n=\sqrt{g/L}\) at any angle	That only holds after the small-angle step (next topic). At large \(\theta\) the \(\sin\theta\) rules.
Measuring \(\theta\) from the horizontal	Here \(\theta\) is from straight-down; that is why the restoring term is \(\sin\theta\) (not \(\cos\theta\)).

Symbol	Meaning	SI unit
\(\theta\)	angle from straight-down	rad
\(\ddot\theta\)	angular acceleration	rad/s²
g	acceleration due to gravity (\(9.81\))	m/s²
L	rod length	m

System	Equation of motion	Restoring term
Mass–spring	\(m\ddot x + kx = 0\)	\(kx\) — straight-line in \(x\) ✓ linear
Pendulum	\(\ddot\theta + \tfrac{g}{L}\sin\theta = 0\)	\(\sin\theta\) — curved in \(\theta\) ✗ nonlinear

Idea	What you own now
Build the model	Torque about the pivot: \(\sum\tau = I\ddot\theta\) with \(I=mL^2\)
The torque	Gravity gives a restoring \(-mgL\sin\theta\) (lever arm \(L\sin\theta\))
The equation	\(\ddot\theta + \tfrac{g}{L}\sin\theta = 0\) — mass cancels out
The twist	It is nonlinear because of \(\sin\theta\)

The pendulum by Newton

Before you start

What you need first

What you'll be able to do

The system: a mass swinging on a rod

A point mass that turns: \(I = mL^2\)

The forces on the bob

The torque that swings it back

Derive the equation of motion

The equation of motion

Why this one is different: it is nonlinear

Restoring torque and angular acceleration

Common mistakes to avoid

Recap — the whole topic on one screen

Small-angle linearization