Classical-Mechanics · Unit 23 · Video 4 · Interactive Practice

Angular Momentum of Many Particles: Only External Torque Changes the Total

IKey Formulas

Formula	Name	What it says
$\vec{\mathbf{L}}_S^{\,\text{sys}} = \sum_j \vec{\mathbf{r}}_{S,j}\times\vec{\mathbf{p}}_j$	System angular momentum	A vector sum over the particles, all about the same point $S$
$\vec{\boldsymbol{\tau}}_S^{\,\text{ext}} = \dfrac{d\vec{\mathbf{L}}_S^{\,\text{sys}}}{dt}$	Torque law for a system	Only the external torque about $S$ changes the total about $S$
$\vec{\mathbf{L}}_S = 2mR^2\omega_z\hat{\mathbf{k}} = I_S\vec{\boldsymbol{\omega}}$	Symmetric pair on a circle	The radial parts $\mp hmR\omega_z\hat{\mathbf{r}}$ cancel for every height $h$
$\vec{\mathbf{L}}_A = \vec{\mathbf{L}}_B + \vec{\mathbf{r}}_{A,B}\times\vec{\mathbf{p}}_{\text{sys}}$	Change of reference point	If $\vec{\mathbf{p}}_{\text{sys}} = \vec{\mathbf{0}}$ , $\vec{\mathbf{L}}$ is the same about every point

Key Insight: Internal torques cancel in pairs because each internal pair of forces is equal and opposite and directed along the line joining the two particles. Equal and opposite alone is not enough: two such forces off that line form a couple, whose torque is the same nonzero vector about every point.

IIVisualization 1 — Two Cones, One Fixed Total

Each particle's angular momentum about $S$ sweeps a cone, yet the symmetric pair's total stays fixed on the axis.

System Height h/R 1 Position φ 30°

IIIVisualization 2 — Why Internal Torques Cancel

Equal and opposite internal forces give zero total torque only when they also act along the joining line.

Internal pair

IVVisualization 3 — Changing the Reference Point

Moving the reference point from $B$ to $A$ adds $\vec{\mathbf{r}}_{A,B}\times\vec{\mathbf{p}}_{\text{sys}}$ , which vanishes when $\vec{\mathbf{p}}_{\text{sys}} = \vec{\mathbf{0}}$ .

System

💡 In three dimensions, moving the reference point from the centre of the circle down the axis to $S$ gives $\vec{\mathbf{r}}_{A,B} = h\hat{\mathbf{k}}$ , and for one particle the extra term $h\hat{\mathbf{k}}\times mR\omega_z\hat{\boldsymbol{\theta}} = -hmR\omega_z\hat{\mathbf{r}}$ is exactly its tilt; for the pair $\vec{\mathbf{p}}_{\text{sys}} = \vec{\mathbf{0}}$ , so no tilt appears.

VQuiz Questions

Problem 1 · The Symmetric Pair

Given: Two particles, each of mass $m = 2\ \text{kg}$ , move on a circle of radius $R = 0.5\ \text{m}$ about the $z$ -axis with $\omega_z = 3\ \text{rad/s}$ , at opposite ends of a diameter. The circle lies $h = 0.4\ \text{m}$ above the origin $S$ . Find the magnitude of the pair's angular momentum about $S$ .

✅ Correct! The radial parts cancel, leaving

\vec{\mathbf{L}}_S = 2mR^2\omega_z\hat{\mathbf{k}} = 3.0\,\hat{\mathbf{k}}\ \text{kg}\cdot\text{m}^2/\text{s}

, independent of

h

❌ Not quite.

mR^2\omega_z = 1.5

is the vertical part of one particle's angular momentum; the pair has two.

❌ Close, but that adds magnitudes.

|\vec{\mathbf{L}}_{S,1}| + |\vec{\mathbf{L}}_{S,2}| = 3.84

ignores that the two tilted vectors point in different directions; their horizontal parts cancel in the vector sum.

❌ Not quite.

5.4 = 3.0 + 2(1.2)

adds the two radial terms to the vertical part as plain numbers. The radial terms

-hmR\omega_z\hat{\mathbf{r}}_1

and

-hmR\omega_z\hat{\mathbf{r}}_2

point opposite ways because

\hat{\mathbf{r}}_1 = -\hat{\mathbf{r}}_2

, so they cancel rather than add — and a horizontal part could never add straight onto a vertical one anyway.

Show solution

For each particle, $\vec{\mathbf{r}}_{S,j} = R\hat{\mathbf{r}}_j + h\hat{\mathbf{k}}$ and $\vec{\mathbf{p}}_j = mR\omega_z\hat{\boldsymbol{\theta}}_j$ , so

\vec{\mathbf{L}}_{S,j} = mR^2\omega_z\hat{\mathbf{k}} - hmR\omega_z\hat{\mathbf{r}}_j.

Adding, with $\hat{\mathbf{r}}_1 = -\hat{\mathbf{r}}_2$ , the radial terms cancel:

\vec{\mathbf{L}}_S = 2mR^2\omega_z\hat{\mathbf{k}} = 2(2)(0.5)^2(3)\,\hat{\mathbf{k}} = 3.0\,\hat{\mathbf{k}}\ \text{kg}\cdot\text{m}^2/\text{s}.

The height $h = 0.4\ \text{m}$ appears only in the cancelled terms, so it does not affect the answer. This is $I_S\omega_z$ with $I_S = 2mR^2 = 1.0\ \text{kg}\cdot\text{m}^2$ .

Problem 2 · Equal and Opposite Is Not Enough

Given: Particle 1 is at $(1, 0)\ \text{m}$ and particle 2 at $(3, 0)\ \text{m}$ . Suppose (hypothetically) the internal forces were equal and opposite but perpendicular to the line joining them: $\vec{\mathbf{F}}_{2,1} = 2\,\hat{\mathbf{j}}\ \text{N}$ on particle 1 and $\vec{\mathbf{F}}_{1,2} = -2\,\hat{\mathbf{j}}\ \text{N}$ on particle 2. Find the total internal torque about the origin.

✅ Correct! The pair is a couple:

(\vec{\mathbf{r}}_1 - \vec{\mathbf{r}}_2)\times\vec{\mathbf{F}}_{2,1} = -4\,\hat{\mathbf{k}}\ \text{N}\cdot\text{m}

about every point. Internal torques cancel only for forces along the joining line.

❌ Not quite. Equal and opposite is not enough. The torques cancel only when the two forces also lie along the line joining the particles, so that they share one line of action.

❌ Check the sign.

\tau_{1,z} = x_1F_y = (1)(2) = 2

and

\tau_{2,z} = x_2F_y = (3)(-2) = -6

; the far particle's torque wins.

❌ Not quite. The two forces point in opposite directions, so their torques have opposite signs:

2 + (-6)

, not

-2 - 6

Show solution

In the plane, $\tau_z = xF_y - yF_x$ . For the two particles:

\tau_{1,z} = (1)(2) - (0)(0) = 2\ \text{N}\cdot\text{m},\qquad \tau_{2,z} = (3)(-2) - (0)(0) = -6\ \text{N}\cdot\text{m}.

\vec{\boldsymbol{\tau}}^{\,\text{int}} = (2 - 6)\,\hat{\mathbf{k}} = -4\,\hat{\mathbf{k}}\ \text{N}\cdot\text{m}.

In general $\vec{\boldsymbol{\tau}}^{\,\text{int}} = (\vec{\mathbf{r}}_1 - \vec{\mathbf{r}}_2)\times\vec{\mathbf{F}}_{2,1} = (-2\,\hat{\mathbf{i}})\times(2\,\hat{\mathbf{j}}) = -4\,\hat{\mathbf{k}}$ , which does not depend on the reference point. It vanishes only when $\vec{\mathbf{F}}_{2,1}$ is parallel to $\vec{\mathbf{r}}_1 - \vec{\mathbf{r}}_2$ , which is why the torque law needs central internal forces.

Problem 3 · The Tilt from the Reference Point

Given: A single particle of mass $m = 0.5\ \text{kg}$ moves on a circle of radius $R = 0.4\ \text{m}$ about the $z$ -axis with $\omega_z = 5\ \text{rad/s}$ , in a plane $h = 0.3\ \text{m}$ above the point $S$ on the axis. Find the horizontal part of $\vec{\mathbf{L}}_S$ (the extra term from moving the reference point from the circle's centre down to $S$ ) and the magnitude $|\vec{\mathbf{L}}_S|$ .

Magnitude of the horizontal part of $\vec{\mathbf{L}}_S$ ?

Magnitude $|\vec{\mathbf{L}}_S|$ ?

✅ Correct!

\vec{\mathbf{L}}_S = 0.4\,\hat{\mathbf{k}} - 0.3\,\hat{\mathbf{r}}

, a 3-4-5 triangle with magnitude

0.5\ \text{kg}\cdot\text{m}^2/\text{s}

❌ Check the extra term. It is

h\hat{\mathbf{k}}\times mR\omega_z\hat{\boldsymbol{\theta}} = -hmR\omega_z\hat{\mathbf{r}}

; a single particle has nonzero momentum, so the term does not vanish.

❌ That is the vertical part.

mR^2\omega_z = 0.4

\vec{\mathbf{L}}_B

about the circle's centre; the horizontal part is the extra term

hmR\omega_z

❌ Close, but that adds magnitudes. The vertical and horizontal parts are perpendicular, so combine them with Pythagoras, not by adding.

❌ Check the magnitude. Combine the vertical part

mR^2\omega_z

and the horizontal part

hmR\omega_z

, which are perpendicular.

Show solution

About the circle's centre $B$ the angular momentum is purely vertical: $\vec{\mathbf{L}}_B = mR^2\omega_z\hat{\mathbf{k}} = (0.5)(0.4)^2(5)\,\hat{\mathbf{k}} = 0.4\,\hat{\mathbf{k}}$ .

Moving the reference point to $S$ ( $A = S$ , $\vec{\mathbf{r}}_{A,B} = h\hat{\mathbf{k}}$ ) adds

h\hat{\mathbf{k}}\times mR\omega_z\hat{\boldsymbol{\theta}} = -hmR\omega_z\hat{\mathbf{r}} = -(0.3)(0.5)(0.4)(5)\,\hat{\mathbf{r}} = -0.3\,\hat{\mathbf{r}}.

So $\vec{\mathbf{L}}_S = 0.4\,\hat{\mathbf{k}} - 0.3\,\hat{\mathbf{r}}$ and

|\vec{\mathbf{L}}_S| = \sqrt{0.4^2 + 0.3^2} = 0.5\ \text{kg}\cdot\text{m}^2/\text{s}.

Check: $|\vec{\mathbf{r}}_S||\vec{\mathbf{p}}| = \sqrt{0.4^2 + 0.3^2}\,(0.5)(0.4)(5) = (0.5)(1.0) = 0.5$ , since $\vec{\mathbf{r}}_S \perp \vec{\mathbf{p}}$ .

Problem 4 · A Spinning Pair That Also Moves

Given: A dumbbell of two $1\ \text{kg}$ masses flies through the air. At one instant its centre of mass $B$ is at $\vec{\mathbf{r}}_{A,B} = 4\,\hat{\mathbf{j}}\ \text{m}$ from a point $A$ , it moves with velocity $3\,\hat{\mathbf{i}}\ \text{m/s}$ , and its angular momentum about $B$ is $\vec{\mathbf{L}}_B = 2\,\hat{\mathbf{k}}\ \text{kg}\cdot\text{m}^2/\text{s}$ . Find $\vec{\mathbf{L}}_A$ .

✅ Correct!

\vec{\mathbf{L}}_A = 2\,\hat{\mathbf{k}} + (4\,\hat{\mathbf{j}})\times(6\,\hat{\mathbf{i}}) = 2\,\hat{\mathbf{k}} - 24\,\hat{\mathbf{k}} = -22\,\hat{\mathbf{k}}

❌ Not quite.

\vec{\mathbf{L}}

is the same about every point only when

\vec{\mathbf{p}}_{\text{sys}} = \vec{\mathbf{0}}

. This dumbbell moves, so

\vec{\mathbf{p}}_{\text{sys}} = 6\,\hat{\mathbf{i}}\ \text{kg}\cdot\text{m/s}

and the extra term survives.

❌ Check the cross product.

\hat{\mathbf{j}}\times\hat{\mathbf{i}} = -\hat{\mathbf{k}}

, not

+\hat{\mathbf{k}}

❌ Check the momentum.

\vec{\mathbf{p}}_{\text{sys}}

uses the total mass,

2\ \text{kg}

\vec{\mathbf{p}}_{\text{sys}} = (2)(3)\,\hat{\mathbf{i}} = 6\,\hat{\mathbf{i}}

Show solution

The system's momentum is $\vec{\mathbf{p}}_{\text{sys}} = M\vec{\mathbf{v}}_{\text{cm}} = (2\ \text{kg})(3\,\hat{\mathbf{i}}\ \text{m/s}) = 6\,\hat{\mathbf{i}}\ \text{kg}\cdot\text{m/s}$ .

\vec{\mathbf{L}}_A = \vec{\mathbf{L}}_B + \vec{\mathbf{r}}_{A,B}\times\vec{\mathbf{p}}_{\text{sys}} = 2\,\hat{\mathbf{k}} + (4\,\hat{\mathbf{j}})\times(6\,\hat{\mathbf{i}})

= 2\,\hat{\mathbf{k}} + 24(\hat{\mathbf{j}}\times\hat{\mathbf{i}}) = 2\,\hat{\mathbf{k}} - 24\,\hat{\mathbf{k}} = -22\,\hat{\mathbf{k}}\ \text{kg}\cdot\text{m}^2/\text{s}.

The spin about the centre of mass is small next to the "orbital" term from the centre of mass moving past $A$ , so $\vec{\mathbf{L}}_A$ even points the opposite way.

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