Visual walkthrough — Circular motion — centripetal acceleration derivation

This page is the visual companion to the parent derivation. If a word here feels new, we build it from zero right where it appears.

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Step 1 — What "moving in a circle" even means

WHAT. Picture a single dot travelling around the edge of a circle. The dot never slows down or speeds up — its speed (how many metres it covers each second) is fixed. We call that speed $v$ (units: metres per second, $\text{m/s}$). The circle has a fixed radius $r$ — the distance from the centre to the edge (units: metres, $\text{m}$).

WHY. We must pin down "constant speed circular motion" precisely, because the whole surprise of this topic is that constant speed still means accelerating. To see that, we first need to separate two ideas the everyday word "speed" mashes together:

• Speed = the length of the velocity arrow (a plain number).
• Velocity = an arrow: it carries both length AND a direction.

PICTURE. Below, the dot sits on the circle. The straight arrow leaving the dot is the velocity $\vec{v}$. The little half-arrow on top of $v$, written $\vec{v}$, is our reminder that this is an arrow-quantity, not just a number. Notice the arrow is tangent — it just grazes the circle, pointing the way the dot is about to go.

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Step 2 — Two snapshots, a tiny time apart

WHAT. Freeze the film twice. At the first freeze the dot is at point A; a very short time later — we name that gap $\Delta t$ — it is at point B. The symbol $\Delta$ (Greek "delta") means "a small change in." So $\Delta t$ = "a small amount of elapsed time."

WHY. Acceleration is change of velocity over time. To measure a change we need a before and an after. Two snapshots give us exactly that: velocity at A versus velocity at B.

PICTURE. From the centre we draw two spokes: ${\vec{r}}_A$ to point A and ${\vec{r}}_B$ to point B. Both spokes have the same length $r$ (it's the same circle). The wedge between them is the small angle $\Delta \theta$ ("delta theta") — the slice of turn the dot swept during $\Delta t$.

Term-by-term, the arc from A to B: $\underset{\text{curved distance travelled}}{\underbrace{\text{arc AB}}}=\underset{\text{radius}}{\underbrace{r}}\underset{\text{angle swept, in radians}}{\underbrace{\Delta \theta }}$

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Step 3 — For a tiny wedge, the chord ≈ the arc

WHAT. The dot really travels along the curved arc, but the straight line AB (the chord) is almost the same length when the wedge is thin.

WHY. We want to compare straight arrows (velocities are straight arrows), so it helps to replace the curve with a straight segment. This is legal only in the limit of a tiny angle — and taking that limit is exactly what acceleration demands (an instantaneous rate).

PICTURE. Watch the chord (deep-teal) hug the arc (burnt-orange) closer and closer as the wedge narrows. That shrinking gap is the whole trick.

$\underset{\text{straight chord length}}{\underbrace{|\text{AB}|}}\approx \underset{\text{arc length}}{\underbrace{r\Delta \theta }}(\text{exact as }\Delta \theta \to 0)$

The $\approx$ ("approximately equals") becomes a true $=$ once we let $\Delta \theta$ shrink to nothing.

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Step 4 — The velocity arrows make the SAME wedge

WHAT. Take the two velocity arrows, ${\vec{v}}_A$ (velocity at A) and ${\vec{v}}_B$ (velocity at B). Both have length $v$. Slide them so their tails touch. The angle between them is also $\Delta \theta$ — the very same wedge as between the spokes.

WHY — the key insight. Each velocity is tangent, so it sits at a right angle ($90°$) to its own spoke: ${\vec{v}}_A\perp {\vec{r}}_A$ and ${\vec{v}}_B\perp {\vec{r}}_B$. If you rotate two arrows by the same $90°$, the angle between them is untouched. So the velocity pair is just the spoke pair spun a quarter turn — same enclosed angle $\Delta \theta$.

PICTURE. Left: the spoke wedge. Right: the velocity wedge, visibly the same opening angle, just rotated. This is why the two triangles we're about to compare are similar (same shape, different size).

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Step 5 — Similar triangles hand us $|\Delta \vec{v}|$

WHAT. Put the two triangles side by side. In each, one long side and a short "opening" side:

triangle	two long (equal) sides	short opening side
position	$r$ and $r$	chord $
velocity	$v$ and $v$	$

Here $\Delta \vec{v}={\vec{v}}_B-{\vec{v}}_A$ is the change in velocity — the arrow you add to ${\vec{v}}_A$ to get ${\vec{v}}_B$. Its length is $|\Delta \vec{v}|$.

WHY. Similar triangles mean short-side ÷ long-side is identical in both. That single equation is the pivot of the whole derivation — it converts a fact about positions into a fact about velocities.

PICTURE. The plum arrow is $\Delta \vec{v}$, the missing third side that closes the velocity triangle.

$\underset{\begin{array}{c}\text{short/long in}\\ \text{velocity triangle}\end{array}}{\underbrace{\frac{|\Delta \vec{v}|}{v}}}=\underset{\begin{array}{c}\text{short/long in}\\ \text{position triangle}\end{array}}{\underbrace{\frac{|\text{AB}|}{r}}}=\frac{r\Delta \theta }{r}=\Delta \theta$

Multiply both sides by $v$: $\boxed{|\Delta \vec{v}|=\underset{\text{speed}}{\underbrace{v}}\underset{\text{angle swept}}{\underbrace{\Delta \theta }}}$

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Step 6 — Divide by time, then shrink it: the limit

WHAT. Acceleration magnitude is "how fast the velocity arrow changes," i.e. $|\Delta \vec{v}|$ per unit time. Divide by $\Delta t$: $\frac{|\Delta \vec{v}|}{\Delta t}=\frac{v\Delta \theta }{\Delta t}=v\frac{\Delta \theta }{\Delta t}$

WHY the limit. The ratio $\frac{\Delta \theta }{\Delta t}$ is "angle swept per second" — but only averaged over $\Delta t$. To get the instantaneous acceleration we squeeze $\Delta t\to 0$. That squeezing is the limit, written $\lim$. As the gap vanishes, $\frac{\Delta \theta }{\Delta t}$ becomes the exact rate of turning, called angular velocity $\omega$ (see Angular velocity and period).

PICTURE. The bar chart shows $\frac{|\Delta \vec{v}|}{\Delta t}$ settling onto a single value as $\Delta t$ shrinks — it stops wobbling and locks in.

$a_c={\lim }_{\Delta t\to 0}\frac{|\Delta \vec{v}|}{\Delta t}=v\underset{=\omega }{\underbrace{\underset{\Delta t\to 0}{\lim }\frac{\Delta \theta }{\Delta t}}}=v\omega$

Now use the bridge between speed and turning-rate, $v=\omega r$ (a dot on radius $r$ turning at $\omega$ radians/s sweeps $v=\omega r$ metres/s): $\boxed{a_c=v\omega =\frac{v^2}{r}={\omega }^{2}r}$

Each swap uses only $v=\omega r$: put $\omega =v/r$ to get $v^2/r$, or $v=\omega r$ to get ${\omega }^{2}r$.

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Step 7 — Which way does it point? (direction)

WHAT. As $\Delta t\to 0$, the arrow $\Delta \vec{v}$ turns until it is perpendicular to $\vec{v}$ and points to the centre.

WHY. In the velocity triangle the two equal sides are both length $v$. When the opening angle $\Delta \theta$ goes to zero, an isosceles triangle's base becomes perpendicular to its equal sides. Geometrically that base ($\Delta \vec{v}$) ends up aiming straight inward — hence centre-seeking, "centripetal." An inward-bending velocity is precisely what Newton's second law then ties to a real inward force.

PICTURE. As the wedge closes, watch $\Delta \vec{v}$ swing until it points dead at the centre.

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Step 8 — Edge & degenerate cases (never leave a gap)

WHAT / WHY / PICTURE — each limit checked.

• $v=0$ (dot not moving). Then $a_c=0^2/r=0$. A parked dot needs no inward pull. ✔
• $r\to \infty$ (circle straightens into a line). $a_c=v^2/r\to 0$. A giant circle is locally a straight road — no bending, no centripetal acceleration. This is why a gentle bend feels easy. ✔
• $r\to 0$ (spin on the spot). $a_c=v^2/r\to \infty$ for any $v>0$: an impossibly tight turn demands infinite acceleration. That's why real turns can't be perfectly sharp. ✔
• Quadrant check (all around the circle). The calculus form $\vec{a}=-{\omega }^2\vec{r}$ works at every point: top, bottom, left, right. The minus sign guarantees $\vec{a}$ always points opposite the outward spoke — i.e. inward — in all four quadrants, no special cases. ✔

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The one-picture summary

Everything at once: the position wedge → rotate $90°$ → the velocity wedge → similar-triangle ratio → divide by time → limit → $a_c=v^2/r$, pointing inward.

Recall Feynman retelling — the whole film in plain words

A dot races round a circle at a steady pace. I take two photos a heartbeat apart: at A and at B. The two spokes to A and B open up a thin slice of angle. Because the dot's velocity arrow always sits at a right angle to its spoke, the two velocity arrows open up the exact same thin slice — just turned a quarter-turn. So the little triangle made by the velocities is a shrunken twin of the triangle made by the spokes. Twins share side-ratios, and that ratio tells me the length of the velocity-change arrow: $|\Delta \vec{v}|=v\Delta \theta$. Divide by the heartbeat and shrink the heartbeat to nothing, and I get $a_c=v\omega =v^2/r$. Finally, as the slice closes, that velocity-change arrow swings around to point straight at the centre. So: constant speed, yet always accelerating — inward, forever, at $v^2/r$.

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Recall

Why do the velocity vectors enclose the same angle $\Delta \theta$ as the radius vectors? ::: Each velocity is perpendicular to its radius; rotating both radii by $90°$ to get the velocities preserves the angle between them. In the limit $\Delta t\to 0$, what does $\Delta \theta /\Delta t$ become? ::: The angular velocity $\omega$, the instantaneous turning rate in radians per second. As $r\to \infty$ with $v$ fixed, what happens to $a_c$, and why physically? ::: $a_c\to 0$; a huge circle is locally straight, so the velocity barely bends. In $\vec{a}=-{\omega }^2\vec{r}$, what does the minus sign guarantee? ::: The acceleration points opposite the outward radius — i.e. toward the centre — in every quadrant.