Foundations — Conservation of angular momentum — conditions

This is the toolbox page. Before you can say "angular momentum is conserved when the net external torque is zero", you must know what time, position, velocity, momentum, force, cross product, torque, moment of inertia and angular velocity each mean and each look like. We build them one at a time, in an order where nothing is used before it is defined — no symbol appears in a formula until its own section has drawn it.

0. Time $t$ — "the clock everything is measured against"

We need $t$ first because conservation is a statement about time: "the amount of spin does not change as $t$ advances." Every rate of change on this page — velocity, the changing of momentum, and eventually the changing of the spin quantity itself — is a "per-second" quantity, so the clock must exist before anything else.

1. Position vector $\vec{r}$ — "where the object is, measured from a chosen dot"

Why do we need a chosen dot at all? Because "spin about what?" has no answer until you say about which point. The whole topic depends on this choice — change the origin and $\vec{r}$ changes, and (as we will see once it is defined) so does the spin quantity built from $\vec{r}$.

Figure s01 shows exactly this: a black dot at the origin labelled "chosen dot", a second black dot for the object, and the red arrow $\vec{r}$ running from origin to object. The picture makes clear that $\vec{r}$ is drawn from the origin — no origin, no arrow.

The little arrow from tail (at the origin) to head (at the object) is the whole picture: an arrow is a vector — a thing with both a size and a direction. Plain numbers (like "3 kg") have no direction; those are called scalars.

2. Velocity $\vec{v}$ and momentum $\vec{p}=m\vec{v}$ — "how fast, which way, how hard to stop"

The symbol $\frac{d\vec{r}}{dt}$ just means "how much $\vec{r}$ changes in a tiny sliver of time $dt$, divided by that sliver". Read it as "rate of change of position" — nothing scarier.

3. Force $\vec{F}$ — "the push or pull that changes motion"

Why bring force in now? Because Newton's second law says a force is precisely what changes momentum:

We need this line because torque (§6) is built from force, and the master equation of the whole topic comes from feeding $\vec{F}=\frac{d\vec{p}}{dt}$ into the derivative of the spin quantity we build in §5.

4. The cross product $\vec{a}\times \vec{b}$ — "how much two arrows are at right angles, packaged as a new arrow"

This is the single most important tool on the page, so we build it carefully.

Why this exact tool, and not ordinary multiplication? Because the topic asks "how much does a force twist rather than push?" Twisting is a sideways effect. The factor $\sin \theta$ is the machine that measures sideways-ness:

• when the two arrows are parallel ($\theta =0$), $\sin 0=0$ → the cross product is the zero vector. Nothing sideways, no twist.
• when they are perpendicular ($\theta =90^{\circ }$), $\sin 90^{\circ }=1$ → the cross product is as big as possible. All sideways, maximum twist.

Figure s02 shows both extremes side by side: on the left, two black arrows lying on top of each other (parallel) labelled "sin0 = 0, cross = 0"; on the right, a black arrow $\vec{a}$ and a red arrow $\vec{b}$ meeting at a right angle, labelled "sin90 = 1, cross = biggest". Reading left to right, the sideways-ness grows.

The right-hand rule (direction). Point your right hand's fingers along the first arrow $\vec{a}$, curl them toward the second arrow $\vec{b}$; your thumb points along $\vec{a}\times \vec{b}$. This is why the spin quantity and torque (built next) point along the spin axis, not in the plane of motion.

5. Angular momentum $\vec{L}=\vec{r}\times \vec{p}$ — "the amount of spin about the chosen dot"

Now every piece exists, so we can assemble the star of the show.

Figure s03 shows a particle out at position $\vec{r}$ (black arrow from the origin) with its momentum $\vec{p}$ as the red arrow. A dashed line marks the part of $\vec{p}$ that is sideways to $\vec{r}$ — the only part that drives spin. If $\vec{p}$ pointed straight along $\vec{r}$, the sideways part (and hence $\vec{L}$) would vanish.

Only the component of $\vec{p}$ perpendicular to $\vec{r}$ contributes to spin — a particle heading straight at the origin ($\vec{p}\parallel \vec{r}$) has $\vec{L}=0$, exactly like a stone dropped straight toward your eye isn't circling you at all. (And a particle at the origin has $\vec{r}=0$, so $\vec{L}=0$ too — §1's edge case, now made precise.)

6. Torque $\vec{\tau }=\vec{r}\times \vec{F}$ and the master equation

Now we can derive the changer of $\vec{L}$, using every tool built above. Start from $\vec{L}=\vec{r}\times \vec{p}$ and differentiate with respect to time:

$\frac{d\vec{L}}{dt}=\frac{d\vec{r}}{dt}\times \vec{p}+\vec{r}\times \frac{d\vec{p}}{dt}$

What we did: applied the cross-product product rule (§4, line three). Why: $\vec{L}$ is a product of two arrows that both change with $t$.

$\frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times (m\vec{v})=m(\vec{v}\times \vec{v})=0$

What we did: used $\frac{d\vec{r}}{dt}=\vec{v}$ (§2) and $\vec{p}=m\vec{v}$; then $\vec{v}\times \vec{v}=0$ because a vector is parallel to itself (§4, line one). So the first term vanishes.

$\vec{r}\times \frac{d\vec{p}}{dt}=\vec{r}\times \vec{F}=\vec{\tau }$

What we did: used Newton's second law $\vec{F}=\frac{d\vec{p}}{dt}$ (§3). What survives is exactly the torque.

A force pointing straight toward the origin (a central force, like gravity toward the Sun) has $\vec{r}\parallel \vec{F}$, so $\sin \theta =0$, so $\vec{\tau }=0$. That single fact is Kepler's equal-area law.

7. From $\vec{L}=\sum \vec{r}\times \vec{p}$ to $L=I\omega$ — building the rigid-body shortcut

For a rigid spinning body (like the skater), we add up $\vec{r}\times \vec{p}$ over every mass element. Here is how the tidy shortcut appears.

The build, one honest step at a time. Label the mass elements $m_i$ at positions ${\vec{r}}_i$ from the axis. Total angular momentum is the sum of each particle's: $\vec{L}={\sum }_i{\vec{r}}_i\times (m_i{\vec{v}}_i).$

For a rigid body spinning about a fixed axis, each particle moves in a circle of radius $r_i$ (its distance from the axis) at speed $v_i=\omega r_i$, in the sideways direction. So its little angular momentum along the axis has size $m_i{r}_i{v}_i=m_i{r}_i(\omega r_i)=m_i{r}_i^2\omega$. Adding them: $L={\sum }_i{m}_i{r}_i^2\omega =\left(\underset{\text{call this }I}{\underbrace{\sum _i{m}_i{r}_i^2}}\right)\omega .$

That bracketed sum — a property of shape and mass alone, independent of how fast it spins — is what we name the moment of inertia $I$. It falls out naturally from the $r^2$ that the cross product plus $v=\omega r$ produce.

Figure s04 shows why $r^2$ dominates: a black spin axis with a near mass (small $r$, black) and a red far mass (large $r$). The same mass moved twice as far from the axis quadruples its contribution to $I$.

Why does this let the skater speed up? With $\vec{\tau }=0$, $L=I\omega$ stays fixed. Pull arms in → the mass moves to smaller $r$ → $I=\sum m_i{r}_i^2$ shrinks → $\omega$ must grow to keep the product constant. No outside twist needed; the $r^2$ inside $I$ does the trick.

Prerequisite map

Read it top to bottom: set a clock and an origin; build $\vec{r}$, then $\vec{v}$, then $\vec{p}$; bring in $\vec{F}$ and Newton's law; feed $\vec{r}$ and $\vec{p}$ through the cross product to get $\vec{L}$, and $\vec{r}$ and $\vec{F}$ to get $\vec{\tau }$; differentiate $\vec{L}$ to reach the master equation; and summing $\vec{r}\times \vec{p}$ over a rigid body gives $L=I\omega$.

Equipment checklist

Connections

• Yeh note Hinglish mein →
• Moment of inertia — the $I$ built in §7.
• Torque — the twist built in §6.
• Newton's third law — why internal torques cancel.
• Central forces — the classic $\vec{\tau }=0$ case.
• Conservation of linear momentum — the straight-line cousin ($p=mv$).
• Kepler's laws — where zero torque becomes equal areas.
• Rotational kinetic energy — uses $I$ and $\omega$ built here.