Transformation of acceleration in rotating frames

Roll a ball gently across a moving merry-go-round and watch it from on board: it swerves, drifts outward, and — if someone is still pushing the ride faster — picks up a sideways kick that wasn't there a moment ago. Nothing is touching it. An observer on the ground sees a free ball rolling straight. Who is doing the pushing in your frame? Nobody is. The pushes are the bill the rotating frame charges you for insisting on Newton's second law aboard a turning platform, and the remarkable thing is that the bill is itemized, exact, and short. One kinematic identity — no forces, no masses, no assumptions about what causes the rotation — relates the acceleration you measure on the platform to the true inertial acceleration: the difference is precisely three terms. The centrifugal term knows only where you are; the Coriolis term knows only how fast you move; the Euler term switches on only while the spin rate itself changes. Gaspard-Gustave de Coriolis sorted this bookkeeping for rotating machines in 1835, factor of two and all. That factor of two is the derivation's one genuine subtlety — a velocity hides in two different places, and rotation charges for both — and this article's centrepiece is the derivation that exposes it honestly. Once you have the identity, the entire strange dynamics of rotating platforms becomes ordinary Newtonian mechanics plus three known corrections: it runs beneath every weather model, every inertial navigation system, every helicopter-blade stress analysis, and all the rotating-Earth physics — effective gravity, the equatorial bulge, the deflection of falling bodies — that follows in this track.

Key takeaways

The inertial acceleration of a particle, expressed through rotating-frame measurements, splits into four kinematic pieces: relative, Coriolis, centripetal (centrifugal), and Euler.
The master identity is $\mathbf{a}_{\mathrm{in}} = \mathbf{a}_{\mathrm{rot}} + 2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} + \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) + \dot{\boldsymbol{\omega}}\times\mathbf{r}$ — exact, purely kinematic, valid for any motion and any rotation.
It follows from applying the operator identity $\left(\tfrac{\mathrm{d}}{\mathrm{d}t}\right)_{\mathrm{in}} = \left(\tfrac{\mathrm{d}}{\mathrm{d}t}\right)_{\mathrm{rot}} + \boldsymbol{\omega}\times$ twice to the position vector.
Moving every term except $\mathbf{a}_{\mathrm{rot}}$ to the force side of Newton's law produces the fictitious forces of the rotating frame: Coriolis, centrifugal, and Euler.
The Coriolis term carries the factor 2 because rotation acts both on the explicit velocity and on the velocity hidden inside the entrainment term $\boldsymbol{\omega}\times\mathbf{r}$ — two equal contributions.
Key idea: Apply the rotating-frame derivative rule $(\mathrm{d}/\mathrm{d}t)_{\mathrm{in}} = (\mathrm{d}/\mathrm{d}t)_{\mathrm{rot}} + \boldsymbol{\omega}\times$ twice to $\mathbf{r}$ : the inertial acceleration becomes relative + Coriolis + centripetal + Euler, all evaluated at one instant.

Where this lives today

Every aircraft, submarine, and smartphone that navigates without GPS is integrating this article numerically. A strapdown inertial navigation system reads accelerometers and gyroscopes hundreds of times per second and runs the mechanization equations — which are exactly equation (5) rearranged: subtract the Coriolis correction $2\boldsymbol{\omega}_\oplus\times\mathbf{v}$ (navigation texts call it the Coriolis compensation), subtract the centrifugal piece (folded into the local gravity model), and integrate what remains to track velocity and position in the rotating Earth frame. Get the factor of 2 wrong in flight software and a transatlantic flight misses its destination by tens of kilometres. The other daily customer is rotating machinery. Helicopter and wind-turbine engineers write blade dynamics in the rotor frame, where the Coriolis term couples a blade's out-of-plane flapping to in-plane lead-lag oscillations — a coupling that has to be damped by design, because it can destroy a rotor hub. Robotic arms on rotating platforms, spacecraft with spinning sections, and centrifuge-based materials labs all budget the same three terms. The 2026 twist is scale: MEMS gyroscopes costing under a dollar now run the same mechanization arithmetic that once required a room of gimbals — equation (5), a million units a day.

The physical picture

Put a bead on a turntable and let it slide outward while the operator brakes the spin. In the rotating frame the bead feels three phantom pushes at once, and each one answers a different question. The centrifugal push asks only "where are you?" — it points away from the axis and grows with distance, acting even on beads glued in place. The Coriolis push asks only "how fast are you moving across the frame?" — it acts at right angles to the bead's platform velocity and vanishes the moment the bead stops. The Euler push asks only "is the spin changing?" — it is the tangential jolt you feel when a carousel starts or brakes, and it disappears entirely at steady rotation.

Figure

Three phantom pushes on a sliding bead — each fictitious force answers a different question

Why does the Coriolis term alone carry a factor of 2? Because the particle's velocity hides in two places in the inertial description. The inertial velocity is $\mathbf{v}_{\mathrm{rot}}+\boldsymbol{\omega}\times\mathbf{r}$ : the part you measure plus the part the frame carries. Differentiate again and the rotation operator acts on both pieces — rotating the measured velocity gives one $\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ , and the changing position inside the entrainment term $\boldsymbol{\omega}\times\mathbf{r}$ gives another, identical one. Two equal contributions, one famous factor. The centrifugal term arises only once, which is why it has no 2. The structure exactly mirrors the linearly accelerating frame, where $\mathbf{a}_{\mathrm{in}}=\mathbf{a}_{\mathrm{rot}}+\mathbf{A}$ with a single frame-acceleration correction $\mathbf{A}$ . Rotation simply makes the correction position- and velocity-dependent — three specialized terms instead of one constant — and that dependence is what makes rotating-frame physics rich instead of trivial.

The core equations

The whole calculation runs on one rule, established in the time-derivatives article: differentiating a vector in the inertial frame equals differentiating it in the rotating frame plus one cross product for the turning axes. Because we are about to apply it twice in succession, keep its operator character in view — it acts on whatever vector you feed it:

\left(\frac{\mathrm{d}\mathbf{Q}}{\mathrm{d}t}\right)_{\mathrm{in}} = \left(\frac{\mathrm{d}\mathbf{Q}}{\mathrm{d}t}\right)_{\mathrm{rot}} + \boldsymbol{\omega}\times\mathbf{Q} \qquad (1)

where: $\mathbf{Q}$ — any vector quantity; $\boldsymbol{\omega}\times\mathbf{Q}$ — rate of change of $\mathbf{Q}$ due solely to the frame's rotation. For a vector frozen into the frame the first term vanishes and the whole inertial change is the rigid sweep $\boldsymbol{\omega}\times\mathbf{Q}$ . Feeding the identity the position vector once yields the velocity transformation — the subject of its own article, quoted here as the intermediate rung of the ladder:

\mathbf{v}_{\mathrm{in}} = \mathbf{v}_{\mathrm{rot}} + \boldsymbol{\omega}\times\mathbf{r} \qquad (2)

where: $\mathbf{v}_{\mathrm{in}}$ — velocity in the inertial frame, $(\mathrm{d}\mathbf{r}/\mathrm{d}t)_{\mathrm{in}}$ ; $\mathbf{v}_{\mathrm{rot}}$ — velocity measured in the rotating frame; $\boldsymbol{\omega}\times\mathbf{r}$ — entrainment velocity of the frame point the particle occupies. Note the two velocity-carrying pieces on the right: both will feel the rotation operator in the next step, and that is where the factor of 2 will come from. Feeding the identity to $\mathbf{v}_{\mathrm{in}}$ itself — the full calculation is Derivation A below — produces the master result of this article:

\mathbf{a}_{\mathrm{in}} = \mathbf{a}_{\mathrm{rot}} + 2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} + \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) + \dot{\boldsymbol{\omega}}\times\mathbf{r} \qquad (3)

where: $\mathbf{a}_{\mathrm{in}}$ — acceleration measured in the inertial (lab) frame; $\mathbf{a}_{\mathrm{rot}}$ — acceleration measured in the rotating frame, $(\mathrm{d}^2\mathbf{r}/\mathrm{d}t^2)_{\mathrm{rot}}$ ; $\dot{\boldsymbol{\omega}}$ — angular acceleration, the same in both frames (Derivation B); $\mathbf{r}$ — position from the common origin on the rotation axis. This equation is exact — no small- $\omega$ , slow-motion, or short-time approximation enters anywhere. Every rotating-frame phenomenon in this track is one of its terms wearing a costume. Dynamics enters only now. Newton's second law holds in the inertial frame, $\mathbf{F}=m\mathbf{a}_{\mathrm{in}}$ ; substitute (3) and solve for what the rotating observer sees. First the kinematic rearrangement:

\mathbf{a}_{\mathrm{rot}} = \mathbf{a}_{\mathrm{in}} - 2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} - \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) - \dot{\boldsymbol{\omega}}\times\mathbf{r} \qquad (4)

where: each term crosses the equation with its sign flipped — the inward centripetal term of (3) becomes the outward centrifugal term here. Multiplying by the mass turns bookkeeping into the working equation of rotating-frame dynamics:

m\,\mathbf{a}_{\mathrm{rot}} = \mathbf{F} - 2m\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} - m\,\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) - m\,\dot{\boldsymbol{\omega}}\times\mathbf{r} \qquad (5)

where: $\mathbf{F}$ — the sum of real (interaction) forces. The three subtracted terms are the Coriolis, centrifugal, and Euler forces: not interactions but the frame's motion billed as forces, mass-proportional exactly like gravity — the observation that later grows into the equivalence principle. Their individual physics lives in the Coriolis force, centrifugal force, and Euler force articles; this equation is their common birth certificate. The centripetal term rewards unpacking once with the BAC–CAB identity (Derivation C), because its geometry is simpler than its double cross product suggests:

\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) = (\boldsymbol{\omega}\cdot\mathbf{r})\,\boldsymbol{\omega} - \omega^2\,\mathbf{r} = -\,\omega^2\,\boldsymbol{\rho} \qquad (6)

where: $\boldsymbol{\rho}$ — the component of $\mathbf{r}$ perpendicular to the axis (the vector from the axis to the particle); $\omega^2=\boldsymbol{\omega}\cdot\boldsymbol{\omega}$ . Whatever direction $\mathbf{r}$ points, the term aims straight at the rotation axis with magnitude $\omega^2\rho$ — the familiar centripetal acceleration of circular motion, recovered as a special case. For the rotating Earth — the arena of the articles downstream — the numbers to carry are:

a_{\mathrm{cf}} = \omega^2 \rho,\qquad \omega_\oplus = 7.292\times10^{-5}\ \mathrm{s^{-1}} \qquad (7)

where: $a_{\mathrm{cf}}$ — centrifugal acceleration magnitude; $\omega_\oplus$ — Earth's sidereal angular velocity ( $2\pi$ per 86164 s). At the equator $\rho=R_\oplus=6.378\times10^6\ \mathrm{m}$ gives $\omega_\oplus^2R_\oplus\approx0.0339\ \mathrm{m/s^2}$ , about $0.35\%$ of $g$ — small, but responsible for the equatorial bulge and the latitude dependence of weight.

Which formula answers which question

you want	use	notes
derivative bookkeeping, any vector	eq. (1)	owned by the time-derivatives article
velocities between frames	eq. (2)	owned by the velocity-transformation article
accelerations between frames	eq. (3)	this article's master identity — exact
equation of motion on the platform	eq. (5)	Newton plus three fictitious forces
direction and size of the centrifugal piece	eq. (6)	$-\omega^2\boldsymbol{\rho}$ , straight at the axis
sorted Coriolis physics on Earth	Coriolis force article	presents eq. (5) with the terms interpreted

Derivation AThe master identity by applying the derivative rule twice

We are deriving equation (3), the exact four-term acceleration transformation, from nothing but the operator identity (1) applied twice to the position vector. The strategy is a ladder: the first application turns position into the velocity transformation (2); the second application must then act on the inertial velocity — both of its pieces — and we expand, watch the same term appear from two different parents, and collect. The step to watch is Step 2: the operator must be fed $\mathbf{v}_{\mathrm{in}}$ , not $\mathbf{v}_{\mathrm{rot}}$ , because acceleration is the inertial derivative of the inertial velocity; feeding it the wrong velocity is the classic way to lose half the Coriolis term. No approximation enters at any point — the result is exact kinematics.

Step 1. Apply (1) to $\mathbf{r}$ to get the velocity transformation:

\mathbf{v}_{\mathrm{in}} = \left(\tfrac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right)_{\mathrm{rot}} + \boldsymbol{\omega}\times\mathbf{r} = \mathbf{v}_{\mathrm{rot}} + \boldsymbol{\omega}\times\mathbf{r}

Why: this is equation (2) — the measured velocity plus the entrainment sweep of the occupied frame point. Note it contains the velocity in two costumes: explicitly as $\mathbf{v}_{\mathrm{rot}}$ , and implicitly inside $\boldsymbol{\omega}\times\mathbf{r}$ , whose $\mathbf{r}$ is about to change.

Step 2. Apply (1) a second time, to the vector $\mathbf{v}_{\mathrm{in}}$ :

\mathbf{a}_{\mathrm{in}} = \left(\tfrac{\mathrm{d}\mathbf{v}_{\mathrm{in}}}{\mathrm{d}t}\right)_{\mathrm{rot}} + \boldsymbol{\omega}\times\mathbf{v}_{\mathrm{in}}

Why $\mathbf{v}_{\mathrm{in}}$ and not $\mathbf{v}_{\mathrm{rot}}$ : acceleration is defined as the inertial derivative of the inertial velocity. The operator happily differentiates any vector you hand it — hand it the right one.

Step 3. Expand the first piece, the rotating-frame derivative of $\mathbf{v}_{\mathrm{rot}}+\boldsymbol{\omega}\times\mathbf{r}$ , term by term:

\left(\tfrac{\mathrm{d}\mathbf{v}_{\mathrm{in}}}{\mathrm{d}t}\right)_{\mathrm{rot}} = \mathbf{a}_{\mathrm{rot}} + \dot{\boldsymbol{\omega}}\times\mathbf{r} + \boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}

Why: $(\mathrm{d}\mathbf{v}_{\mathrm{rot}}/\mathrm{d}t)_{\mathrm{rot}}$ is by definition $\mathbf{a}_{\mathrm{rot}}$ ; the product rule on $\boldsymbol{\omega}\times\mathbf{r}$ gives $\dot{\boldsymbol{\omega}}\times\mathbf{r}$ plus $\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ , using $(\mathrm{d}\mathbf{r}/\mathrm{d}t)_{\mathrm{rot}}=\mathbf{v}_{\mathrm{rot}}$ and the frame-independence of $\dot{\boldsymbol{\omega}}$ (Derivation B). This is the first Coriolis twin: the particle keeps arriving at frame points with different sweep velocities.

Step 4. Expand the second piece by substituting (2) for $\mathbf{v}_{\mathrm{in}}$ :

\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{in}} = \boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} + \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})

Why: this is the rotation acting on the velocity itself — the measured velocity's basis vectors turn, giving the second Coriolis twin, while the entrainment part contributes the double cross product that will become the centripetal term. The two twins have genuinely different physical parents; they merely happen to be equal.

Step 5. Add the two pieces and collect the twins into the factor of 2:

\mathbf{a}_{\mathrm{in}} = \mathbf{a}_{\mathrm{rot}} + 2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} + \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) + \dot{\boldsymbol{\omega}}\times\mathbf{r}

Why this ends the kinematics: every term is now expressed in rotating-frame measurables. Impose $\mathbf{F}=m\mathbf{a}_{\mathrm{in}}$ and solve for $m\mathbf{a}_{\mathrm{rot}}$ to obtain the equation of motion (5), where the three transferred terms — signs flipped — become the Coriolis, centrifugal, and Euler forces. *Verified: Result matches Morin Eq. (10.11), Goldstein Sec. 4.10, and Taylor Eq. (9.34); setting $\boldsymbol{\omega}=0$ recovers $\mathbf{a}_{\mathrm{in}}=\mathbf{a}_{\mathrm{rot}}$ , and a particle glued to the frame recovers pure centripetal acceleration $-\omega^2\boldsymbol{\rho}$ .*

Figure

The family tree of the four terms — where each piece of the master identity is born

Go deeper · Derivation B · 1 minWhy the angular acceleration is the same in both frames

Derivation BWhy the angular acceleration is the same in both frames

The master identity writes a single, unsubscripted $\dot{\boldsymbol{\omega}}$ in the Euler term — a quiet claim that inertial and rotating observers agree on how the angular velocity is changing, even though they disagree about almost every other rate of change. The goal here is to justify that claim in two lines, because Derivation A silently used it in Step 3. The strategy is self-referential and rather elegant: feed the operator identity its own angular velocity. The thing to notice is where the agreement comes from — the antisymmetry of the cross product, not any special property of rotation.

Step 1. Set $\mathbf{Q}=\boldsymbol{\omega}$ in the operator identity (1):

\left(\tfrac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t}\right)_{\mathrm{in}} = \left(\tfrac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t}\right)_{\mathrm{rot}} + \boldsymbol{\omega}\times\boldsymbol{\omega}

Why this is allowed: the identity holds for any vector, including the very vector that defines the rotation.

Step 2. The cross product of any vector with itself vanishes:

\boldsymbol{\omega}\times\boldsymbol{\omega} = 0 \quad\Longrightarrow\quad \left(\tfrac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t}\right)_{\mathrm{in}} = \left(\tfrac{\mathrm{d}\boldsymbol{\omega}}{\mathrm{d}t}\right)_{\mathrm{rot}} \equiv \dot{\boldsymbol{\omega}}

Why it matters: $\boldsymbol{\omega}$ is the one vector whose rate of change is frame-agnostic — the rotation cannot rotate itself. That is what licenses the single unambiguous $\dot{\boldsymbol{\omega}}$ in the Euler term of (3) and (5). *Verified: Standard identity used in Goldstein and Landau-Lifshitz; geometrically, a vector along the rotation axis is unmoved by the rotation it generates.*

Go deeper · Derivation C · 1 minUnpacking the centripetal term with BAC-CAB

Derivation CUnpacking the centripetal term with BAC-CAB

The double cross product $\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ looks opaque, but the goal of this short derivation is to show it is the most familiar vector in circular motion: the centripetal acceleration, aimed at the axis, of magnitude $\omega^2\rho$ . The strategy is the BAC-CAB expansion followed by a decomposition of $\mathbf{r}$ into components along and perpendicular to the axis. Watch how the axial component of $\mathbf{r}$ cancels out entirely — the term knows only about distance from the axis, which is why centrifugal effects on Earth depend on latitude through $\rho=R\cos\lambda$ and not on $R$ alone.

Step 1. Apply the BAC-CAB identity $\mathbf{A}\times(\mathbf{B}\times\mathbf{C}) = \mathbf{B}(\mathbf{A}\cdot\mathbf{C}) - \mathbf{C}(\mathbf{A}\cdot\mathbf{B})$ with $\mathbf{A}=\mathbf{B}=\boldsymbol{\omega}$ , $\mathbf{C}=\mathbf{r}$ :

\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) = \boldsymbol{\omega}\,(\boldsymbol{\omega}\cdot\mathbf{r}) - \mathbf{r}\,\omega^2

Step 2. Decompose the position into axial and perpendicular parts, $\mathbf{r} = \mathbf{r}_\parallel + \boldsymbol{\rho}$ with $\mathbf{r}_\parallel = (\hat{\boldsymbol{\omega}}\cdot\mathbf{r})\,\hat{\boldsymbol{\omega}}$ :

\boldsymbol{\omega}\,(\boldsymbol{\omega}\cdot\mathbf{r}) = \omega^2\,\mathbf{r}_\parallel

Why: only the axial part of $\mathbf{r}$ survives the dot product with $\boldsymbol{\omega}$ .

Step 3. Subtract:

\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) = \omega^2\,\mathbf{r}_\parallel - \omega^2(\mathbf{r}_\parallel + \boldsymbol{\rho}) = -\,\omega^2\,\boldsymbol{\rho}

Why the geometry is now transparent: the term points from the particle straight at the rotation axis (the direction of $-\boldsymbol{\rho}$ ), with magnitude $\omega^2\rho$ — the centripetal acceleration of the circle the particle would trace if glued to the frame. Moved to the force side of (5) it flips into the outward centrifugal force $+m\omega^2\boldsymbol{\rho}$ , whose physics is the centrifugal-force article. *Verified: For a glued particle ( $\mathbf{v}_{\mathrm{rot}}=0$ , $\mathbf{a}_{\mathrm{rot}}=0$ , steady $\boldsymbol{\omega}$ ), (3) collapses to $\mathbf{a}_{\mathrm{in}}=-\omega^2\boldsymbol{\rho}$ — exactly uniform circular motion, as it must.*

Figure

Geometry of the centripetal term — only the distance from the axis survives

Limiting cases

The four-term expansion collapses in instructive special cases:

Case	Result	Meaning
Steady rotation	$\dot{\boldsymbol{\omega}}=0 \;\Rightarrow\; \mathbf{a}_{\mathrm{in}}=\mathbf{a}_{\mathrm{rot}}+2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}+\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$	The Euler term drops; this is the case for the rotating Earth on short timescales.
Object at rest in rotating frame	$\mathbf{v}_{\mathrm{rot}}=0,\ \mathbf{a}_{\mathrm{rot}}=0 \;\Rightarrow\; \mathbf{a}_{\mathrm{in}}=\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$	Pure centripetal acceleration $-\omega^2\boldsymbol{\rho}$ toward the axis, magnitude $\omega^2\rho$ with $\rho$ the distance from the axis.
Non-rotating frame	$\boldsymbol{\omega}=0 \;\Rightarrow\; \mathbf{a}_{\mathrm{in}}=\mathbf{a}_{\mathrm{rot}}$	Recovers the inertial (or linearly accelerating, after adding $\mathbf{A}$ ) result.
Motion parallel to the axis	$\mathbf{v}_{\mathrm{rot}}\parallel\boldsymbol{\omega} \;\Rightarrow\; 2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}=0$	No Coriolis deflection for motion along the rotation axis; deflection requires a velocity component perpendicular to $\boldsymbol{\omega}$ .

Common trap“The factor of 2 is just a stronger centrifugal effect”

The Coriolis and centrifugal terms are structurally different animals, and merging them is the most reliable way to wreck a rotating-frame calculation. The centrifugal term $-m\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ depends on position and acts on everything, moving or not; the Coriolis term $-2m\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ depends on velocity and vanishes for anything at rest in the frame. The factor of 2 is not adjustable and not centrifugal in origin: it counts two genuinely distinct effects that happen to be equal — the rotation of the measured velocity's basis, and the changing sweep velocity of the occupied position. Two related traps: feeding the second differentiation $\mathbf{v}_{\mathrm{rot}}$ instead of $\mathbf{v}_{\mathrm{in}}$ silently discards one twin (you get 1 instead of 2); and using the inertial velocity inside the Coriolis force when applying (5) double-counts the entrainment. The velocity in the Coriolis term is always the velocity measured in the rotating frame.

Worked examples

1 · A bead on a spinning-up turntable — all three forces at once (confidence builder)

A turntable rotates at $\omega=2.0\ \mathrm{rad/s}$ and is being spun up at $\dot{\omega}=0.50\ \mathrm{rad/s^2}$ . A bead slides along a radial groove at $r=0.50\ \mathrm{m}$ , moving outward at $v=0.30\ \mathrm{m/s}$ . Find the magnitude of each fictitious acceleration acting on the bead in the turntable frame.

Step 1. Centrifugal, from (6): $a_{\mathrm{cf}} = \omega^2 r = (2.0)^2\times0.50 = \boxed{2.0\ \mathrm{m/s^2}}$ , radially outward — it acts whether or not the bead moves.

Step 2. Coriolis: the groove velocity is perpendicular to $\boldsymbol{\omega}$ , so $a_{\mathrm{Cor}} = 2\omega v = 2\times2.0\times0.30 = \boxed{1.2\ \mathrm{m/s^2}}$ , tangential — pressing the bead against the trailing wall of the groove (the groove wall, in turn, supplies the real force that makes the bead follow the groove).

Step 3. Euler: $a_{\mathrm{Eul}} = \dot{\omega}\, r = 0.50\times0.50 = \boxed{0.25\ \mathrm{m/s^2}}$ , tangential, opposing the spin-up — the bead tends to lag as the table accelerates beneath it.

Step 4. Hierarchy check: $2.0 : 1.2 : 0.25$ . All three are comparable here because $\omega$ , $v$ , and $\dot\omega$ were chosen so — on the rotating Earth, by contrast, the Euler term is utterly negligible ( $\dot\omega_\oplus\sim10^{-22}\ \mathrm{rad/s^2}$ from tidal braking) and the hierarchy is centrifugal $\gg$ Coriolis for everyday speeds.

2 · Inside an ultracentrifuge — who cares about which term (exam level)

A laboratory ultracentrifuge spins at $60{,}000\ \mathrm{rpm}$ with samples at $r=8.0\ \mathrm{cm}$ . (a) Find the centrifugal acceleration in units of $g$ . (b) The rotor reaches full speed from rest in $120\ \mathrm{s}$ at constant $\dot\omega$ ; find the Euler acceleration during spin-up. (c) A particle sedimenting radially outward at $1.0\ \mathrm{mm/s}$ feels what Coriolis acceleration? Rank the three.

Step 1. Convert: $\omega = 60{,}000\times\dfrac{2\pi}{60} = 6283\ \mathrm{rad/s}$ .

Step 2. (a) Centrifugal: $a_{\mathrm{cf}} = \omega^2 r = (6283)^2\times0.080 = 3.16\times10^{6}\ \mathrm{m/s^2} = \boxed{3.2\times10^{5}\ g}$ . This is the working number biochemists quote as "320,000 × g" — the master identity's centripetal term is literally printed on the machine.

Step 3. (b) Euler: $\dot\omega = 6283/120 = 52.4\ \mathrm{rad/s^2}$ , so $a_{\mathrm{Eul}} = \dot\omega\,r = 52.4\times0.080 = \boxed{4.2\ \mathrm{m/s^2}}$ — under half a $g$ , tangential, present only during the two minutes of spin-up.

Step 4. (c) Coriolis: $a_{\mathrm{Cor}} = 2\omega v = 2\times6283\times1.0\times10^{-3} = \boxed{12.6\ \mathrm{m/s^2}}$ — about $1.3\,g$ sideways on any particle that dares to move radially, which is why sedimenting particles drift slightly azimuthally rather than settling along a perfect radius.

Step 5. Ranking: centrifugal $3.16\times10^{6}$ : Coriolis $12.6$ : Euler $4.2$ (all in $\mathrm{m/s^2}$ ) — the centrifugal term wins by five orders of magnitude, which is the entire design point of a centrifuge: make one term of equation (5) overwhelm everything else, including gravity.

Origins · 1 minGaspard-Gustave de Coriolis and how this was discovered

Gaspard-Gustave de Coriolis

1792–1843 · École Polytechnique, Paris

Coriolis reached the acceleration transformation not from planets but from payroll: as an engineer-professor at the École Polytechnique he wanted the work and efficiency of rotating machines — waterwheels, mill gearing — computed honestly in the machine's own turning frame. His 1832 memoir on the principle of forces vives (kinetic energy) in relative motion set up the bookkeeping; the 1835 sequel, Sur les équations du mouvement relatif des systèmes de corps, delivered the complete transformation, where he split the fictitious corrections into the ordinary centrifugal force and a second, velocity-dependent piece he called the compound centrifugal force — factor of 2 attached. He never saw a hurricane in it: the name "Coriolis force" was pinned on the term by meteorologists only in the early twentieth century, decades after his death. Along the way he gave physics its modern definition of work and the $\tfrac12$ in $\tfrac12mv^2$ — so both sides of equation (5), the forces and the energy accounting they feed, carry his fingerprints.

History

1673

— Huygens' Horologium Oscillatorium quantifies the centrifugal tendency of circular motion, the first of the four terms to be tamed.

1749

— Euler, attacking the precession of the equinoxes, writes the equations of motion in rotating coordinates for the first time.

1835

— Coriolis publishes the complete acceleration transformation for machines; his "compound centrifugal force" carries the factor of 2.

1851

— Foucault's pendulum at the Panthéon makes the Coriolis term visible to the public, and the transformation becomes geophysics.

1953

— The SPIRE system flies a transcontinental route on pure inertial navigation — equation (5) integrated continuously by instruments, no stars and no radio.

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 10 (Accelerating frames of reference) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Sec. 4.9–4.10 · Landau & Lifshitz, Mechanics, Vol. 1, §39 · Taylor, Classical Mechanics, Ch. 9

Provenance. Master identity and derivation verified against Morin Ch. 10 (Eq. 10.11), Goldstein 3rd ed. Sec. 4.9–4.10, Landau-Lifshitz Vol. 1 §39, and Taylor Ch. 9. A verifier should recompute the double application of the operator identity to confirm the factor-of-2 Coriolis term, and check the numerical equatorial centrifugal acceleration: $\omega_\oplus=7.292\times10^{-5}\,\mathrm{s^{-1}}$ , $R_\oplus=6.378\times10^6\,\mathrm{m}$ give $\omega_\oplus^2 R_\oplus\approx 0.0339\,\mathrm{m/s^2}$ (≈0.34% of g=9.81), all standard textbook values.

Builds on: Time derivatives in rotating frames · Angular velocity vector of a rotating frame · Transformation of velocity between rotating frames · Non-inertial reference frames Leads to: Fictitious (inertial) forces · Coriolis force · Centrifugal force · Euler force · Effective gravity on the rotating Earth

Are the fictitious forces fake? They press beads against real groove walls

They are real effects with an unreal cause. In the rotating frame the bead genuinely accelerates relative to the frame, groove walls genuinely bend, and strain gauges genuinely read forces — none of that is illusion. What is missing is an interaction: no object exerts the centrifugal or Coriolis force, so there is no reaction partner and Newton's third law does not apply to them. They are the kinematics of your turning coordinates billed as forces. The tell-tale signature is mass-proportionality: like gravity, they accelerate everything equally — the observation that eventually led Einstein to the equivalence principle.

Why is a_rot not just the inertial acceleration written in rotating components?

Because changing basis never changes a vector, while $\mathbf{a}_{\mathrm{rot}}$ is a genuinely different vector. $\mathbf{a}_{\mathrm{rot}}$ means: differentiate the position twice while holding the rotating basis fixed — it is what a platform-riding observer measures with their own rulers and clocks. A particle glued to the turntable has $\mathbf{a}_{\mathrm{rot}}=0$ , but its inertial acceleration $-\omega^2\boldsymbol{\rho}$ is nonzero in every basis you could possibly write it in. The gap between the two vectors is exactly the three extra terms of equation (3).

Does it matter where I put the origin?

Yes, in one specific way: the derivation assumes both frames share an origin on the rotation axis, so that $\mathbf{r}$ in the centrifugal and Euler terms is measured from the axis. The Coriolis term is origin-independent (it involves only the velocity), and the centrifugal term really depends only on $\boldsymbol{\rho}$ , the perpendicular offset from the axis — Derivation C shows the axial component of $\mathbf{r}$ cancels. If the rotating frame's origin also translates (a turntable on an accelerating truck), you must add the origin's inertial acceleration $\mathbf{A}_O$ as a fourth correction, exactly as in linearly accelerating frames.

Which of the four terms actually matter on Earth?

All are present; their sizes differ wildly. The centrifugal term is the biggest: $\omega_\oplus^2R_\oplus\approx0.034\ \mathrm{m/s^2}$ , a steady 0.35% of $g$ , absorbed into "effective gravity" (its own article). The Coriolis term on a $v=10\ \mathrm{m/s}$ object is $2\omega v\approx1.5\times10^{-3}\ \mathrm{m/s^2}$ — tiny per second, decisive over hours and kilometres (weather, artillery, Foucault). The Euler term is essentially zero: Earth's spin changes on tidal-braking timescales, $\dot\omega\sim10^{-22}\ \mathrm{rad/s^2}$ , giving accelerations twenty orders below gravity. That hierarchy — centrifugal absorbed, Coriolis kept, Euler dropped — defines practical geophysics.

Why do the signs flip when the terms move to the force side?

Equation (3) states pure kinematics: $\mathbf{a}_{\mathrm{in}}$ equals $\mathbf{a}_{\mathrm{rot}}$ plus three corrections. Newton's law constrains $\mathbf{a}_{\mathrm{in}}$ , not $\mathbf{a}_{\mathrm{rot}}$ ; solving for the rotating observer's $m\mathbf{a}_{\mathrm{rot}}$ means subtracting the three corrections from $\mathbf{F}$ . Each subtracted term is then experienced as a push in the opposite direction: the inward centripetal term becomes the outward centrifugal force, and the sign of the Coriolis force is opposite the raw $2\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ . Nothing physical flips — only which side of the ledger the bookkeeping sits on.

Could I lose the factor of 2 by doing the derivation some plausible-looking way?

Yes, and it is the classic exam disaster. The standard mistake is applying the operator identity the second time to $\mathbf{v}_{\mathrm{rot}}$ instead of $\mathbf{v}_{\mathrm{in}}$ : that produces $\mathbf{a}_{\mathrm{rot}}+\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ — one twin only — and silently drops the piece coming from differentiating the entrainment term $\boldsymbol{\omega}\times\mathbf{r}$ . The discipline that prevents it: acceleration is the inertial derivative of the inertial velocity, so the operator must be fed the full $\mathbf{v}_{\mathrm{rot}}+\boldsymbol{\omega}\times\mathbf{r}$ . If your result has a bare $\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ , a twin is missing.

Is there also a term with the angular acceleration crossed with velocity, like ω̇×v?

No, and the absence is informative. Run Derivation A and watch what each differentiation can produce: $\dot{\boldsymbol{\omega}}$ appears only from differentiating the explicit $\boldsymbol{\omega}$ inside the entrainment term $\boldsymbol{\omega}\times\mathbf{r}$ , and it arrives already married to $\mathbf{r}$ . A hypothetical $\dot{\boldsymbol{\omega}}\times\mathbf{v}$ term would require differentiating $\boldsymbol{\omega}$ and $\mathbf{r}$ in the same single product rule application — a second-order combination that simply never arises in a first-and-second derivative calculation. It would show up only in the third derivative (the rotating-frame "jerk"), which mechanics rarely needs.

Transformation of acceleration in rotating frames

Where this lives today

The physical picture

The core equations

Which formula answers which question

Limiting cases

Worked examples

1 · A bead on a spinning-up turntable — all three forces at once (confidence builder)

2 · Inside an ultracentrifuge — who cares about which term (exam level)

Going further

Before this

Where this leads