Visual walkthrough — Torque = dL - dt

---

Step 0 — The three arrows we are allowed to use

Before any formula, let us fix the cast of characters as pictures. Everything on this page is built from just three arrows and one point.

The little "$\times$" between arrows we will meet in Step 2 — we do not use it until we have drawn what it means. (If cross products are new, see Cross Product.)

Recall Check the cast

Which arrow starts at the pivot? ::: $\vec{r}$, the position arrow — it always runs from $O$ to the particle. How do $\vec{p}$ and $\vec{v}$ differ as pictures? ::: Same direction; $\vec{p}$ is just $\vec{v}$ scaled longer (or shorter) by the mass $m$. Which way does $\vec{r}\times \vec{p}$ point when both lie in the board? ::: Out of the board (right-hand rule, curling $\vec{r}$ into $\vec{p}$).

---

Step 1 — What "angular momentum" $\vec{L}$ looks like

We want a single number-with-direction that captures how much this moving particle is circling around $O$. Two things should matter: how fast it moves ($\vec{p}$) and how far off to the side it swings ($\vec{r}$). The tool that combines two arrows into "how much they sweep sideways" is the cross product — that is why it enters here, and no other product would do (a plain multiply loses the direction; a dot product measures alignment, the opposite of what we want).

$\vec{L}=\underset{\text{from }O\text{ to particle}}{\underbrace{\vec{r}}}\times \underset{\text{momentum }m\vec{v}}{\underbrace{\vec{p}}}$

WHAT: we defined $\vec{L}$ as the cross product of position and momentum. WHY: the cross product's $\sin {\theta }_{rp}$ automatically throws away the part of the motion aimed straight at $O$ (that part cannot circle anything) and keeps only the sideways sweep. PICTURE: the shaded parallelogram in s02 — its area is exactly $|\vec{L}|$. Wider swing ⇒ bigger area ⇒ more angular momentum. The green dot with a ⊙ shows $\vec{L}$ pointing out of the board by the right-hand rule from Step 0. See Angular Momentum for the full story of this arrow.

---

Step 2 — Watch $\vec{L}$ change: differentiate the picture

Angular momentum changes when the picture changes. Both arrows $\vec{r}$ and $\vec{p}$ can move as time ticks, so the rate of change of $\vec{L}$ must account for both moving. The tool for "rate of change" is the derivative $d/dt$ — that is why it enters: we are asking "how fast is the parallelogram morphing per second?"

$\frac{d\vec{L}}{dt}=\underset{\text{term A: }\vec{r}\text{ moved}}{\underbrace{\frac{d\vec{r}}{dt}\times \vec{p}}}+\underset{\text{term B: }\vec{p}\text{ changed}}{\underbrace{\vec{r}\times \frac{d\vec{p}}{dt}}}$

• $\frac{d\vec{r}}{dt}$ — how the position arrow moves per second. But that is the velocity: $\frac{d\vec{r}}{dt}=\vec{v}$.
• $\frac{d\vec{p}}{dt}$ — how the momentum arrow changes per second (we deal with this in Step 4).

---

Step 3 — Term A vanishes (the self-cross-product picture)

Look hard at term A. We just said $d\vec{r}/dt=\vec{v}$, and $\vec{p}=m\vec{v}$. So term A is $\vec{v}\times m\vec{v}$ — the cross product of an arrow with a scaled copy of itself.

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

WHAT: we deleted term A. WHY: velocity and momentum always point the same way, so they can never "sweep" against each other. PICTURE: the flattened, area-zero parallelogram in s04.

Only term B survives: $\frac{d\vec{L}}{dt}=\vec{r}\times \frac{d\vec{p}}{dt}$

---

Step 4 — Plug in Newton: $d\vec{p}/dt$ is the force

We still have $d\vec{p}/dt$ hiding in term B. What is the rate of change of momentum? That is precisely what Newton's Second Law tells us:

$\frac{d\vec{p}}{dt}=\vec{F}$

Substitute it straight into the picture:

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

• $\vec{F}$ — the push on the particle (the momentum arrow's rate of change).
• $\vec{r}\times \vec{F}$ — position crossed with force: this is a new parallelogram, made of the lever arm and the push.

WHAT: we replaced $d\vec{p}/dt$ with $\vec{F}$. WHY: Newton's law is the one fact that turns "how momentum changes" into "the force acting" (with $m$ held constant). PICTURE: s05 redraws the parallelogram with $\vec{F}$ as the second edge instead of $\vec{p}$.

---

Step 5 — Name the new arrow: that's torque $\vec{\tau }$

The quantity $\vec{r}\times \vec{F}$ deserves its own name because it answers "how much does this force twist the particle around $O$?" We call it torque.

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

PICTURE: s06 splits $\vec{F}$ into a pink piece along $\vec{r}$ (useless for spinning) and a blue piece perpendicular to $\vec{r}$ (the only piece that twists). The $\sin {\theta }_{rF}$ keeps only the blue.

---

Step 6 — Edge cases: does the law survive every scenario?

A law you can trust must handle the weird inputs. Here are all of them, each as a picture.

---

Step 7 — From one particle to a whole rigid body

Real objects are billions of particles. Add up each one's angular momentum: ${\vec{L}}_{\text{tot}}={\sum }_i{\vec{L}}_i$, so $\frac{d{\vec{L}}_{\text{tot}}}{dt}={\sum }_i{\vec{\tau }}_i$.

For a rigid body spinning about a fixed axis, $\vec{L}=I\vec{\omega }$ (with $I$ the moment of inertia and $\vec{\omega }$ the angular velocity). If $I$ is constant: ${\vec{\tau }}_{\text{ext}}=\frac{d(I\vec{\omega })}{dt}=I\frac{d\vec{\omega }}{dt}=I\vec{\alpha }$

This special case is $\tau = I\alpha$. When $I$ changes (a skater pulling arms in), you must keep the full $d\vec{L}/dt$ — that is the door to Conservation of Angular Momentum.

---

The one-picture summary

This single figure compresses all seven steps: the position arrow $\vec{r}$, the momentum arrow $\vec{p}$ sweeping the $\vec{L}$-parallelogram, the force $\vec{F}$ twisting it, the deleted self-cross-product, and the final boxed law $\vec{\tau }=d\vec{L}/dt$.

Recall Feynman: retell the whole walkthrough in plain words

Pick a dot to watch from — call it $O$. Draw an arrow from that dot to a little moving ball; that's $\vec{r}$. Draw where the ball is heading; scale it by its heft; that's $\vec{p}$. Glue these two arrows into a slanted box — the area of that box is the ball's "circling-amount," its angular momentum $\vec{L}$, and which way it points (out of or into the board) you get from the right-hand rule. Now let time tick. The box can change two ways: the position edge slides, or the momentum edge stretches. The sliding edge is useless — it lines up with the momentum, so it squashes the box to nothing. Only the stretching edge matters, and what stretches the momentum is a push — a force (as long as the ball's mass doesn't change). Position-arrow crossed with that push is a brand-new box we name torque: the twist. So the whole thing collapses to one sentence you can see: the twist you apply equals how fast the circling-amount grows. $\vec{\tau }=d\vec{L}/dt$. Straight-line ball with no push? The box never changes — angular momentum sits still, even with nothing going in a circle.

---

Active Recall

Recall Rebuild it yourself

Why does the derivative of $\vec{L}$ have two terms? ::: Product rule: both $\vec{r}$ and $\vec{p}$ depend on time, so each contributes a term (the two edges of the parallelogram can each change). Which term vanishes and why (as a picture)? ::: $\frac{d\vec{r}}{dt}\times \vec{p}=\vec{v}\times m\vec{v}$; parallel arrows, $\sin 0=0$, flattened zero-area box. What replaces $d\vec{p}/dt$, and by which law (and what hidden assumption)? ::: $\vec{F}$, by Newton's second law $\vec{F}=d\vec{p}/dt$, valid in this form only when mass $m$ is constant. What is $\vec{\alpha }$ and how does $\tau =I\alpha$ appear? ::: $\vec{\alpha }=d\vec{\omega }/dt$, angular acceleration; with $I$ constant, $d(I\vec{\omega })/dt=Id\vec{\omega }/dt=I\vec{\alpha }$. A particle drifts in a straight line, no force, passing 3 m to the side of $O$ at $v=4$ m/s, $m=2$ kg. Its $\vec{L}$? ::: $L=mvd=2\cdot 4\cdot 3=24$ kg·m²/s, constant. When does $\vec{\tau }=\vec{r}\times \vec{F}$ equal zero even with a real force? ::: When $\vec{F}$ is parallel to $\vec{r}$ (${\theta }_{rF}=0$), a central push straight at (or from) $O$.