Visual walkthrough — Rotating frames — centrifugal force, Coriolis force

We only assume you can:

• add arrows tip-to-tail (vectors),
• and picture a merry-go-round turning.

Everything else — including the cross product — we build as we need it.

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Step 1 — Two observers, one arrow

WHAT. Picture a flat spinning disc (a merry-go-round). One observer floats above, not spinning — call her the inertial observer. Another sits on the disc, turning with it — the rotating observer. On the disc we paint two arrows, ${\hat{e}}_1$ and ${\hat{e}}_2$, glued to the floor. These are our basis vectors: little rulers of length 1 that tell us "this way is x, that way is y."

WHY. Every position, velocity, or force is just a combination of these two rulers. If we understand how the rulers themselves move, we understand how everything moves. The catch: to the rotating observer the rulers sit still; to the inertial observer the rulers sweep around. That disagreement is the entire subject.

PICTURE. In the figure, the black arrows ${\hat{e}}_1,{\hat{e}}_2$ point along the disc. The curved gray arrow shows the disc turning at rate $\omega$ (radians per second). The floating observer (top) sees the arrows rotate; the seated observer (on the disc) sees them frozen.

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Step 2 — How a glued-on ruler moves: the cross product appears

WHAT. Watch the tip of one ruler, ${\hat{e}}_1$, over a tiny time. It doesn't stretch (still length 1); it only swings sideways. We want a formula for that sideways velocity of the tip.

WHY this tool — the cross product. We need an operation that takes the spin arrow $\vec{\omega }$ and the ruler ${\hat{e}}_1$ and returns a third arrow perpendicular to both, whose length is "spin rate × how far the tip is from the axis." That is exactly what the cross product $\vec{\omega }\times {\hat{e}}_1$ does — no other simple product gives a perpendicular direction. We pick it because the tip must move perpendicular to both the axis and the ruler (it circles the axis).

PICTURE. The tip of ${\hat{e}}_1$ traces a circle. The green arrow $\vec{\omega }\times {\hat{e}}_1$ is tangent to that circle — it's the instantaneous velocity of the tip. Longer $\vec{\omega }$ (faster spin) ⇒ longer green arrow.

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Step 3 — The Transport Theorem: translating between the two observers

WHAT. Take any arrow $\vec{A}$ (position, velocity, whatever). We need a full set of three rulers to write any arrow, so alongside the in-disc rulers ${\hat{e}}_1,{\hat{e}}_2$ we add a third one:

With all three rulers, any arrow $\vec{A}$ has three numbers: $\vec{A}=A_1{\hat{e}}_1+A_2{\hat{e}}_2+A_3{\hat{e}}_3.$ Now ask how fast $\vec{A}$ changes to the inertial observer.

WHY. An arrow can change for two independent reasons: (1) its numbers $A_1,A_2,A_3$ change — the rotating observer sees this — or (2) the rulers underneath it swing around — only the inertial observer sees this. We must add both. This is just the product rule for differentiation, applied to "number × ruler."

PICTURE. The figure splits one blue arrow's motion into two red pieces: a straight piece (numbers changing, seen on the disc) plus a curved sideways piece (rulers rotating). Their tip-to-tail sum is what the floating observer sees.

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Step 4 — First application: velocities

WHAT. Feed the position arrow $\vec{r}$ (from the axis to the object) into the transport theorem.

WHY. Velocity is the rate of change of position. So plugging $\vec{A}=\vec{r}$ tells us how the two observers' velocities relate — the first rung of the ladder to acceleration.

PICTURE. A person walks outward (orange ${\vec{v}}_{\mathrm{rot}}$) while the floor sweeps them sideways (green $\vec{\omega }\times \vec{r}$). The true inertial velocity (blue) is the diagonal sum.

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Step 5 — Second application: acceleration (where the two forces are born)

WHAT. Apply the operator again, this time to ${\vec{v}}_{\mathrm{in}}={\vec{v}}_{\mathrm{rot}}+\vec{\omega }\times \vec{r}$. We hold $\vec{\omega }$ constant (steady spin).

WHY. Acceleration is the rate of change of velocity, and Newton only speaks in acceleration ($\vec{F}=m\vec{a}$, see Newton's Second Law). So we must differentiate once more to reach the language of forces.

Applying ${\left(\frac{d}{dt}\right)}_{\mathrm{in}}={\left(\frac{d}{dt}\right)}_{\mathrm{rot}}+\vec{\omega }\times$ to each piece:

${\vec{a}}_{\mathrm{in}}=\underset{={\vec{a}}_{\mathrm{rot}}}{\underbrace{(\frac{d}{dt}{)}_{\mathrm{rot}}{\vec{v}}_{\mathrm{rot}}}}+\underset{\text{from operator on }{\vec{v}}_{\mathrm{rot}}}{\underbrace{\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}}+\underset{=\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}{\underbrace{(\frac{d}{dt}{)}_{\mathrm{rot}}(\vec{\omega }\times \vec{r})}}+\underset{\text{from operator on }\vec{\omega }\times \vec{r}}{\underbrace{\vec{\omega }\times (\vec{\omega }\times \vec{r})}}$

Watch the factor of 2 appear. Two separate terms each gave $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$: one from the $\vec{\omega }\times$ operator hitting ${\vec{v}}_{\mathrm{rot}}$, one from differentiating $\vec{\omega }\times \vec{r}$ inside the rotating frame (since $(d\vec{r}/dt{)}_{\mathrm{rot}}={\vec{v}}_{\mathrm{rot}}$). They add:

PICTURE. The figure tracks the two clones of $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ merging into one doubled green arrow — that is why the 2 is there, not a typo.

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Step 6 — Insert Newton, read off the fictitious forces

WHAT. Real forces obey ${\vec{F}}_{\mathrm{real}}=m{\vec{a}}_{\mathrm{in}}$. Substitute the boxed line and solve for what the seated observer needs, $m{\vec{a}}_{\mathrm{rot}}$.

WHY. The seated observer wants to keep using the simple rule "force = mass × the acceleration I measure." We rearrange so the leftover $\vec{\omega }$-terms move to the force side and masquerade as forces.

$m{\vec{a}}_{\mathrm{rot}}=m{\vec{a}}_{\mathrm{in}}-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}-m\vec{\omega }\times (\vec{\omega }\times \vec{r})$

$\boxed{m{\vec{a}}_{\mathrm{rot}}={\vec{F}}_{\mathrm{real}}\underset{{\vec{F}}_{\mathrm{Coriolis}}}{\underbrace{-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}}\underset{{\vec{F}}_{\mathrm{centrifugal}}}{\underbrace{-m\vec{\omega }\times (\vec{\omega }\times \vec{r})}}}$

Term by term:

• ${\vec{F}}_{\mathrm{real}}$ — genuine pushes/pulls (ropes, gravity, the seat), same for both observers.
• $-2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ — the Coriolis pseudo-force: needs motion ${\vec{v}}_{\mathrm{rot}}$, sits $\perp$ to it, flips if you reverse direction.
• $-m\vec{\omega }\times (\vec{\omega }\times \vec{r})$ — the centrifugal pseudo-force: depends only on where you are, always present.

PICTURE. The figure shows the seated observer's free-body: real force inward, plus the two invented arrows, balancing to give the acceleration she actually measures.

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Step 7 — Which way does centrifugal point? (the double cross product)

WHAT. Decode $-\vec{\omega }\times (\vec{\omega }\times \vec{r})$ into a plain direction.

WHY. A cross product inside a cross product is scary. We unpack it geometrically so you can see "outward from the axis" without algebra every time.

• Inner: $\vec{\omega }\times \vec{r}$ points along the circle (tangent), length $\omega r_{\perp }$, where $r_{\perp }$ is the distance from the axis.
• Outer: crossing $\vec{\omega }$ with that tangent arrow rotates it another 90° — landing it pointing inward toward the axis, length ${\omega }^2{r}_{\perp }$.
• The minus sign flips it: outward, away from the axis.

PICTURE. Two nested right-angle turns: $\vec{r}$ → (cross $\vec{\omega }$) → tangent → (cross $\vec{\omega }$) → inward → (minus) → outward. Each 90° turn is drawn.

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Step 8 — Edge & degenerate cases (never get surprised)

WHAT. Sweep every special input so no scenario ambushes you.

Case	${\vec{v}}_{\mathrm{rot}}$	Position	Coriolis	Centrifugal
Standing still on disc	$\vec{0}$	off-axis	$\vec{0}$ (needs velocity)	outward, $m{\omega }^2{r}_{\perp }$
Sitting on the axis	$\vec{0}$	$r_{\perp }=0$	$\vec{0}$	$\vec{0}$
Moving parallel to $\vec{\omega }$	$\parallel \vec{\omega }$	any	$\vec{0}$ ($\sin \varphi =0$)	outward as usual
Moving radially (in/out)	$\perp \vec{\omega }$	any	full, sideways	outward
Not spinning ($\omega =0$)	any	any	$\vec{0}$	$\vec{0}$ (frame is inertial!)

WHY each vanishes.

• Coriolis needs the cross product to be nonzero: $|\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}|=\omega v\sin \varphi$. If ${\vec{v}}_{\mathrm{rot}}=0$, or ${\vec{v}}_{\mathrm{rot}}\parallel \vec{\omega }$ (so $\varphi =0$), it's zero.
• Centrifugal needs $r_{\perp }\ne 0$: on the axis there's nothing to be flung from.
• If $\omega =0$ the frame doesn't spin — it's an inertial frame — and both fictitious forces disappear, as they must.

PICTURE. Four mini-panels: still rider (only outward arrow), axis rider (no arrows), radial thrower (sideways Coriolis), and axis-parallel mover (no Coriolis). Each labelled with which force survives.

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The one-picture summary

This final panel compresses the whole ladder:

$\vec{r}\underset{\text{transport}}{\overset{d/dt}{\to }}{\vec{v}}_{\mathrm{in}}={\vec{v}}_{\mathrm{rot}}+\vec{\omega }\times \vec{r}\underset{\text{transport}}{\overset{d/dt}{\to }}{\vec{a}}_{\mathrm{in}}={\vec{a}}_{\mathrm{rot}}+2\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}+\vec{\omega }\times (\vec{\omega }\times \vec{r})$

$\Rightarrow m{\vec{a}}_{\mathrm{rot}}={\vec{F}}_{\mathrm{real}}-\underset{\text{Coriolis: turns motion}}{\underbrace{2m\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}}}-\underset{\text{centrifugal: flings outward}}{\underbrace{m\vec{\omega }\times (\vec{\omega }\times \vec{r})}}$

Take these numbers to D3 Worked Examples, then test yourself in D4 Exercises. See the accumulation effect in action in Coriolis Effect in Weather and Foucault Pendulum.

Recall Feynman retelling — the whole walkthrough in plain words

Two friends watch a spinning merry-go-round. One floats above (not spinning), one rides on it. They paint two arrows on the floor as rulers, plus a third pointing straight up out of the floor. To the rider the rulers sit still; to the floater they sweep around — and the speed of that sweep is "spin arrow crossed with ruler," a perpendicular nudge (Step 2). Since everything is built from those rulers, the floater always sees the rider's change plus the rulers' sweep — that's the transport theorem (Step 3). Apply it once to position → velocities differ by "the floor drags you" (Step 4). Apply it again to velocity → accelerations differ by two leftover terms, and one of them shows up twice, giving the famous factor of 2 (Step 5). Newton only trusts the floater's acceleration; when the rider insists on using force = mass × her acceleration, those two leftovers sneak onto the force side as make-believe forces (Step 6): centrifugal, which always flings you straight out from the axis (Step 7), and Coriolis, which only bites when you move and always shoves you sideways. Stand still → no Coriolis. Sit on the axis → no forces at all. Stop the spin → the whole ghost show vanishes and Newton is plain again (Step 8). If the spin changes rate, one more ghost — the Euler force — joins the party, but we kept the spin steady so it stayed home.

Recall Quick self-test

The factor of 2 in Coriolis comes from ::: two identical $\vec{\omega }\times {\vec{v}}_{\mathrm{rot}}$ terms appearing when we differentiate a second time. Centrifugal magnitude equals ::: $m{\omega }^2{r}_{\perp }$, pointing outward from the axis. Coriolis is zero when ::: you are at rest in the rotating frame, or move parallel to $\vec{\omega }$, or $\omega =0$. Centrifugal is zero when ::: you sit on the rotation axis ($r_{\perp }=0$) or $\omega =0$. The extra pseudo-force present only when the spin rate changes is ::: the Euler force $-m\dot{\vec{\omega }}\times \vec{r}$.