Visual walkthrough — Relation to linear quantities - v = rω, a_t = rα, a_c = rω²

We are translating between two languages of motion: the angular one (how much you turn) and the linear one (how far and how fast you move along your circle). The whole bridge is one word long: the radius $r$.

---

Step 1 — Draw the arc: what "$r$", "$\theta$", "$s$" even are

WHAT. Put a dot at the center of a spinning disk — call it the axis. Pick one point $P$ on the disk sitting a distance $r$ from the axis. As the disk turns, $P$ sweeps out a slice of a circle.

WHY these three. Everything on this page is built from just these. Speed will come from "how fast does $s$ grow?"; the angular quantities will come from "how fast does $\theta$ grow?". So we must nail down the picture first.

PICTURE. The blue wedge below: the radius $r$ (blue), the swept angle $\theta$ (orange), and the curved arc $s$ (green) riding along the rim.

---

Step 2 — Why radians: the equation $s=r\theta$

WHAT. We claim the arc length obeys $s=r\theta$ Read term-by-term: $s$ (the curved distance) equals $r$ (how far out you are) times $\theta$ (how much angle you swept).

WHY radians and not degrees. A radian is defined as the angle for which the arc equals the radius. Turn by exactly $1$ radian and the point walks a curved distance of exactly one $r$. So "angle in radians" literally means "how many radius-lengths of arc did I trace." That is the whole trick — it makes $s=r\theta$ clean with no conversion factor. In degrees you'd be forced to write $s=r\theta \cdot \frac{\pi }{180}$, an ugly tax we refuse to pay. (More in Radian measure and arc length.)

PICTURE. Two dots on the same spoke: an inner one ($r$ small) and an outer one ($r$ big). Same $\theta$, but the outer arc is visibly longer — because $s$ scales with $r$. This is the reason the rim moves faster; we're just about to make it a formula.

---

Step 3 — From arc to speed: $v=r\omega$

WHAT. We now ask: how fast does the point move along its circle? That means: how fast is $s$ growing? "Rate of change of a quantity" is exactly what a derivative measures — that's why the tool enters here. Writing $\frac{ds}{dt}$ means "how much $s$ changes per tiny tick of time $t$."

WHY a derivative and not just $s/t$. If the spin were perfectly steady, $s/t$ would do. But we want a rule that works every instant, even while speeding up. The derivative is the "at this exact moment" version of $s/t$. Apply it to $s=r\theta$: $\frac{ds}{dt}=\frac{d}{dt}\left(r\theta \right)=r\frac{d\theta }{dt}$ We pulled $r$ out front because $r$ is constant (rigid disk). If $r$ could change we'd get an extra piece — see Step 6.

PICTURE. The velocity arrow $\vec{v}$ drawn tangent to the circle (a right angle to the spoke $r$). A twin arrow at double the radius is exactly twice as long — same $\omega$, double $r$, double $v$.

---

Step 4 — Speeding up the spin: $a_t=r\alpha$

WHAT. Now let the disk not just spin but spin faster and faster. Then $\omega$ itself changes with time. The rate at which $\omega$ changes is the angular acceleration $\alpha$ (alpha): $\alpha =\frac{d\omega }{dt}({\text{rad/s}}^2).$

WHY differentiate again. Speed $v$ now changes because $\omega$ changes. "Rate of change of speed" is the tangential acceleration $a_t$ — again a derivative, for the same reason as before: we want the instantaneous rule. Differentiate $v=r\omega$ (with $r$ still constant): $\frac{dv}{dt}=r\frac{d\omega }{dt}\Rightarrow a_t=r\alpha$

Edge case here: if the disk spins steadily, $\omega$ is constant, so $\alpha =0$ and therefore $a_t=0$. No tangential acceleration at all — this is exactly uniform circular motion.

PICTURE. The orange ${\vec{a}}_t$ arrow pointing the same way as $\vec{v}$ (disk speeding up), sitting tangent to the rim.

---

Step 5 — The sneaky one: why a circle accelerates even at constant speed

WHAT. Here's the surprise. Even if the point moves at perfectly constant speed, it is still accelerating — because acceleration cares about velocity the vector, and the velocity's direction is constantly turning.

WHY vectors, not just speed. Speed is a single number; velocity $\vec{v}$ is an arrow that has both a length and a direction. Acceleration means "how the velocity arrow changes." On a circle the arrow's length can stay fixed while its heading swings around — that change is real and must be caused by an acceleration. This is why we now differentiate the position vector, not just $s$.

PICTURE. Four snapshots of $\vec{v}$ at four points around the circle — the arrow keeps the same length but points in four different directions. That rotating heading is the thing we're about to differentiate.

---

Step 6 — Differentiate the vector twice: $a_c=r{\omega }^2$

WHAT. Differentiate the position once to get velocity, twice to get acceleration. (Recall the calculus facts: the slope of $\cos \theta$ is $-\sin \theta$, the slope of $\sin \theta$ is $\cos \theta$, and $\theta =\omega t$ contributes a factor $\omega$ each time by the chain rule.)

Velocity (first derivative): $\vec{v}=r\omega (-\sin \theta ,\cos \theta ).$ Its length is $r\omega$ — matching Step 3 — and it points at a right angle to $\vec{r}$ (tangent). Good, consistent.

Acceleration (second derivative): $\vec{a}=r{\omega }^2(-\cos \theta ,-\sin \theta )=-{\omega }^2\vec{r}.$

• The bracket $(-\cos \theta ,-\sin \theta )$ is exactly $\vec{r}$ flipped — it points from $P$ back toward the center.
• The minus sign is the headline: acceleration aims inward. That's why it's called centripetal ("center-seeking").

WHY $r$ was pulled out earlier — the payoff: in Steps 3–4 we assumed $r$ constant. Step 6 shows what "constant $r$" buys us: a pure inward acceleration with no messy in/out drift term. If $r$ changed (a bead sliding on a spinning rod), extra terms appear — that lives in advanced motion problems, beyond this page.

PICTURE. The red ${\vec{a}}_c$ arrow pointing dead at the center, drawn together with the tangential arrows so you see them at right angles.

---

Step 7 — Putting the two accelerations together

WHAT. A general point has both accelerations at once: $a_t$ (tangent, changes speed) and $a_c$ (inward, changes direction). Because they point at 90° to each other, the total magnitude uses Pythagoras: $a=\sqrt{a_t^2+a_c^2}.$

WHY Pythagoras. Two perpendicular arrows add like the legs of a right triangle; the resultant is the hypotenuse. There's no cross-term because $\cos 90°=0$.

PICTURE. The right triangle of ${\vec{a}}_t$ (leg), ${\vec{a}}_c$ (leg), and total $\vec{a}$ (hypotenuse), drawn at the point.

---

Step 8 — Every case checked

Recall Walk through all the corner cases

• Point on the axis ($r=0$): $v=0$, $a_t=0$, $a_c=0$. The center doesn't move — correct.
• Steady spin ($\alpha =0$): $a_t=0$, but $a_c=r{\omega }^2\ne 0$. Constant speed, still turning. Uniform circular motion.
• Momentarily at rest but winding up ($\omega =0,\alpha \ne 0$): $a_c=0$ (no speed to redirect yet) but $a_t=r\alpha \ne 0$. All acceleration is tangential at that instant.
• Slowing down ($\alpha <0$): $a_t$ flips to point against $\vec{v}$; $a_c$ still points inward. The formulas already carry the sign.
• Double $\omega$, fix $r$ and $\alpha$: $v$ doubles, $a_t$ unchanged (it never contained $\omega$!), $a_c$ quadruples. The $a_t$-trap from the parent note.

---

The one-picture summary

Everything on this page grew from one arc. The figure below compresses it: arc $\to$ speed $\to$ two accelerations, each labelled with its formula and the tool that produced it.

Recall Feynman retelling — the whole walkthrough in plain words

Start by drawing a slice of pizza: the tip is the axis, one point on the crust is your traveller. The radius $r$ is how far out the traveller sits; the angle $\theta$ (in radians, chosen so that turning by one radian walks you exactly one radius along the crust) says how much you turned; the arc $s=r\theta$ is the curved distance walked.

Ask "how fast along the crust?" — that's speed, the rate $s$ grows, and since $r$ never changes it's just $r$ times the rate the angle grows, i.e. $v=r\omega$. Farther out means longer arc for the same turn, so the rim is faster.

Now spin the pizza faster and faster. The angle's growth-rate itself changes; call that $\alpha$. The traveller's speed changes at rate $a_t=r\alpha$ — a forward (or backward) shove along the crust.

Finally, the sneaky part: even at a steady spin, the traveller is being yanked toward the tip the whole time, because the direction it's facing keeps swinging around. Track its position as $(r\cos \theta ,r\sin \theta )$, differentiate twice, and out pops an inward pull of size $a_c=r{\omega }^2$. Double the spin and this pull is four times as strong — which is exactly why fast-spinning things try to fling themselves apart. The two accelerations sit at a right angle, so the total is the hypotenuse $\sqrt{a_t^2+a_c^2}$. One arc, three formulas, one dictionary word: $r$.