Fictitious (inertial) forces

Every force you met in your first mechanics course came with a pusher. Tension has a rope, the normal force has a floor, gravity has a planet. Then an airliner starts its takeoff roll and something presses you into the seat — and you can search the cabin from nose to tail without finding any body doing the pressing. The same phantom flings laundry against the drum of a spinning washer, tips standing passengers when the metro brakes, and holds the water in a bucket swung overhead. These are the fictitious forces — also sold under the aliases inertial, pseudo, and apparent forces — and this article is their complete field guide. The premise is bookkeeping, not mysticism. Newton's second law $\mathbf{F}=m\mathbf{a}$ is guaranteed only in inertial frames; insist on using it in an accelerating or rotating frame and you must pad the force ledger with correction terms that encode the frame's own motion. There are exactly four, one per way a frame can accelerate: the translational force $-m\mathbf{A}$ , the centrifugal force $-m\,\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ , the Coriolis force $-2m\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ , and the Euler force $-m\,\dot{\boldsymbol{\omega}}\times\mathbf{r}$ . This article catalogs when each one wakes up, what it depends on, which ones do work, and why none of them ever has a reaction partner. Their shared fingerprint — strict proportionality to mass — is the deepest thing in the catalog. A force that accelerates every body identically is doing an uncanny impression of gravity, and Einstein took the impression seriously. Master the taxonomy here and the equivalence principle stops being a slogan.

Key takeaways

Fictitious forces are the correction terms that restore $\mathbf{F}=m\mathbf{a}$ in a non-inertial frame; each equals $-m$ times a piece of the frame's acceleration, so they are bookkeeping for the coordinates, not new interactions.
The complete catalog has four entries: translational $-m\mathbf{A}$ (uniform, from origin acceleration), centrifugal $m\omega^2\boldsymbol{\rho}$ (outward, position-dependent), Coriolis $-2m\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ (velocity-dependent, work-free), and Euler $-m\,\dot{\boldsymbol{\omega}}\times\mathbf{r}$ (only while the spin rate changes).
None of them has a Newton's-third-law partner — there is no second body exerting them — and they vanish identically the moment you transform back to an inertial frame.
Each term switches on under a different condition: check $\mathbf{A}$ , $\boldsymbol{\omega}$ , $\dot{\boldsymbol{\omega}}$ , and $\mathbf{v}_{\mathrm{rot}}$ and you know instantly which forces your problem needs.
Energy bookkeeping differs term by term: the centrifugal force is conservative with potential $-\tfrac12 m\omega^2\rho^2$ , the Coriolis force does exactly zero work, and the translational and Euler forces can pump energy in or out.
Because every fictitious force is proportional to the body's mass, the imparted acceleration is universal — the same for feather and anvil — which is gravity's signature and the seed of the equivalence principle.
Key idea: a fictitious force is whatever you must add to $\mathbf{F}_{\mathrm{real}}$ so that $m\mathbf{a}_{\text{measured in the frame}}=\mathbf{F}_{\mathrm{real}}+\mathbf{F}_{\mathrm{fict}}$ holds. It equals $-m$ times the acceleration of the frame's coordinate system, so it always scales with the body's mass and never has a reaction partner.

The core insight

A fictitious force is whatever you must add to $\mathbf{F}_{\mathrm{real}}$ so that $m\mathbf{a}_{\text{measured in the frame}} = \mathbf{F}_{\mathrm{real}} + \mathbf{F}_{\mathrm{fict}}$ holds. It equals $-m$ times the acceleration of the frame's coordinate system, so it is always proportional to the mass of the body it acts on and never has a reaction partner.

Where this lives today

Open any modern engineering-simulation package and the fictitious-force catalog is running as production code. Computational fluid dynamics of jet-engine compressors, wind-turbine rotors, and pump impellers is done in the blade's rotating frame — the solver literally adds centrifugal and Coriolis source terms, equation (1) verbatim, to the momentum equations, because a steady rotating-frame flow beats an unsteady inertial-frame one by orders of magnitude in compute cost. Multibody-dynamics engines behind robotics and vehicle simulators make the same move for every rotating joint. When a 2026 turbine blade is certified, its stress margins were computed against a load case that is mostly equation (3). The other frontier is biological. Spinning habitats for long-duration spaceflight are back on serious drawing boards, and the design constraints are exactly this article's catalog: centrifugal $\omega^{2}\rho$ must land near $g$ , the head-to-foot gradient (linear in $\rho$ ) must stay small, and the Coriolis force on a moving astronaut — the term that makes a thrown ball curve and an inner ear rebel — caps the spin rate near a few rpm, forcing large radii. Meanwhile every pilot pulling out of a dive and every rider on a launched coaster is a live experiment in $-m\mathbf{A}$ physiology, measured in "g's" — the engineering unit that quietly concedes the equivalence principle: frame acceleration and gravity, one currency.

The physical picture

Ride the metro standing up. The train brakes; you stagger toward the front. Nobody pushed you — the floor decelerated while your body, honestly obeying inertia, kept its speed. But describe the event in the train's frame and you are forced into a different sentence: "a forward force appeared and shoved every standing passenger at once." That sentence is wrong about the world and exactly right about the coordinates. The fictitious force is the grammatical price of speaking in an accelerating frame — the frame's own acceleration $\mathbf{A}$ , flipped in sign, multiplied by each body's mass, and entered on the force side of the ledger. Once you know there are exactly four such entries, the whole subject becomes a matter of recognizing which ones are awake. Each answers to a different switch: the origin's acceleration, the spin, the spin's change, and the body's own motion through the spinning frame.

Figure

The complete zoo — four fictitious forces, one per way a frame can accelerate

The four have sharply different personalities. The translational force is the great equalizer — the same vector $-m\mathbf{A}$ per unit mass at every point of the frame, which is why every loose object in the braking metro tips the same way at the same instant. The centrifugal force cares where you are: it points away from the rotation axis and grows linearly with your distance from it, which is why the outer seats of a carousel feel the fling hardest. The Coriolis force cares how you move — it is silent for anything parked in the rotating frame and deflects anything that travels through it, always at right angles to the velocity; its full story (Foucault's pendulum, cyclones, the factor of 2) is told in the companion Coriolis force article. The Euler force is the rarest beast, awake only while the spin rate itself changes — the tangential shove you feel in the first seconds of a merry-go-round starting up, gone the moment the ride reaches speed. What unites them is stranger than what separates them. Divide any of the four by the body's mass and the mass cancels: the acceleration imparted is identical for every object in the frame. Hang a feather and a hammer in the braking metro and both tip forward through the same angle. There is exactly one genuine interaction in nature with that property — gravity, whose $m\mathbf{g}$ also cancels to a universal $\mathbf{g}$ . A uniform fictitious field is therefore a perfect gravity impersonator, and no mechanical experiment confined to the frame can unmask it.

Figure

The mass-proportionality fingerprint — an accelerating frame is a gravity impersonator

One more identity check before the algebra: what fictitious forces are not. They are not interactions, so they have no source body and no Newton's-third-law partner — there is nothing for the reaction to act on. They are not small corrections you may always neglect — in a centrifuge they exceed gravity a thousandfold. And they are not optional style: within the non-inertial frame the ledger is wrong without them, while in the inertial frame the same terms are contraband. The general frame-transformation machinery that generates all four lives in the non-inertial reference frames article; here we take its master equation and read the catalog off it.

The catalog, from one equation

Everything descends from the master equation of motion in an arbitrary frame — origin accelerating at $\mathbf{A} = \ddot{\mathbf{R}}$ , axes rotating at $\boldsymbol{\omega}(t)$ — derived in the non-inertial reference frames article by applying the rotating-derivative identity twice:

m\,\mathbf{a}_{\mathrm{rot}} = \mathbf{F}_{\mathrm{real}} + \mathbf{F}_{\mathrm{fict}}, \qquad \mathbf{F}_{\mathrm{fict}} = -m\mathbf{A} - m\,\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) - 2m\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} - m\,\dot{\boldsymbol{\omega}}\times\mathbf{r} \qquad (1)

where: $\mathbf{F}_{\mathrm{real}}$ — the sum of genuine interactions (gravity, contact, tension, electromagnetic); $\mathbf{r}$ , $\mathbf{v}_{\mathrm{rot}}$ — the particle's position and velocity measured in the frame. Every fictitious term is $-m$ times a purely kinematic acceleration: nothing about the particle enters except its mass and its state of motion. That single observation powers the whole article. Read the catalog off equation (1) term by term. The translational entry needs no rotation at all:

\mathbf{F}_{\mathrm{trans}} = -m\mathbf{A} \qquad (2)

where: $\mathbf{A}$ — the frame origin's acceleration. It is the only spatially uniform member of the family — same magnitude and direction at every point of the frame — which is exactly what qualifies it to impersonate a uniform gravitational field. The linearly accelerating frames article builds the entire elevator-and-equivalence story on this one term. The centrifugal entry unpacks its double cross product into something you can point at:

\mathbf{F}_{\mathrm{cf}} = -m\,\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) = m\,\omega^{2}\boldsymbol{\rho} \qquad (3)

where: $\boldsymbol{\rho}$ — the perpendicular vector from the rotation axis to the particle. Only the axial distance matters: the force points radially outward, grows linearly with $\rho$ and quadratically with spin rate, and acts on everything in the frame whether it moves or not. It is the resident force of every carousel, centrifuge, and rotating planet. The Coriolis entry is the family's only velocity reader:

\mathbf{F}_{\mathrm{Cor}} = -2m\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}} \qquad (4)

where: $\mathbf{v}_{\mathrm{rot}}$ — velocity measured in the rotating frame; the force vanishes identically for anything parked. Perpendicular to the velocity by construction, it steers without ever speeding or slowing — this article files the catalog card and defers the biography (the factor of 2, Foucault, cyclones, the works) to the companion Coriolis force article. And the Euler entry, the one most students meet last or never:

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

where: $\dot{\boldsymbol{\omega}}$ — the frame's angular acceleration. For a spin-up about a fixed axis it is tangential, opposing the spin-up (magnitude $m\,\dot{\omega}\,\rho$ ), and it dies the instant the rotation becomes steady. It is the lurch of a starting carousel and the sideways press in a braking hard-disk platter's frame — brief, honest work of a changing $\boldsymbol{\omega}$ .

Which force when — the recognition table

force	formula	depends on	nonzero when	does work in the frame?
translational	$-m\mathbf{A}$	frame only	origin accelerates	yes
centrifugal	$m\omega^{2}\boldsymbol{\rho}$	position	frame rotates	yes — conservative, potential $-\tfrac12 m\omega^{2}\rho^{2}$
Coriolis	$-2m\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$	velocity	rotation and motion in frame	never — always $\perp\mathbf{v}$
Euler	$-m\,\dot{\boldsymbol{\omega}}\times\mathbf{r}$	position	spin rate changing	yes, during spin-up/down

Derivation AWhy every fictitious force scales with mass and has no reaction partner

We are deriving the two structural theorems that define the whole category: (i) every fictitious force is strictly proportional to the mass of the body it acts on, and (ii) none of them can have a Newton's-third-law partner. The strategy is to look at where the terms in equation (1) come from — the kinematic transformation of acceleration between frames — and notice what they can and cannot contain. No new computation is needed; the theorems fall out of the shape of the derivation, which is what makes them exact and universal rather than approximate.

Step 1. Recall the origin of equation (1). The acceleration transformation between an inertial frame and the moving frame is pure kinematics:

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

Why kinematics matters here: this identity is a statement about coordinates and clocks — it holds for any moving point, whether it is a planet, a proton, or a fleck of dust, and contains no physical property of the moving body at all.

Step 2. Multiply by $m$ and substitute Newton's law $m\mathbf{a}_{\mathrm{in}} = \mathbf{F}_{\mathrm{real}}$ , then solve for $m\mathbf{a}_{\mathrm{rot}}$ . Every transplanted term acquires exactly one factor of $m$ — the one we multiplied by. Why proportionality is now automatic: the fictitious terms are (mass) × (frame kinematics); the body contributes nothing but $m$ , $\mathbf{r}$ , and $\mathbf{v}_{\mathrm{rot}}$ . No charge, no composition, no material constant can appear, because none was present in Step 1.

Step 3. Divide by $m$ : the fictitious acceleration $\mathbf{F}_{\mathrm{fict}}/m = -\mathbf{A} - \dot{\boldsymbol{\omega}}\times\mathbf{r} - \boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) - 2\,\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ is independent of the body entirely (given its state of motion). Feather and anvil, released side by side in the frame, deviate identically. Why this is gravity's fingerprint: the only real force with the same universality is $m\mathbf{g}$ — which is precisely why a uniform $-m\mathbf{A}$ is locally indistinguishable from gravity, the observation Einstein promoted to the equivalence principle.

Step 4. Now the third law. Newton's third law is a statement about interactions: body 1 pushes body 2, body 2 pushes back, and the pair carries zero net momentum change. A fictitious force names no second body — its "source" is the accelerating coordinate grid, which possesses neither mass nor momentum to recoil. Why this is not a defect but a diagnostic: momentum is genuinely not conserved inside a non-inertial frame (a free puck accelerates from rest), and the missing reaction partner is exactly how the frame confesses. Any "force" without a partner is either fictitious or misdrawn. *Verified: consistent with Morin ch. 10 and Taylor ch. 9, both of which flag mass-proportionality and the absent reaction partner as the defining tests; the equivalence-principle connection follows Einstein's 1907 argument.*

Go deeper · Derivation B · 2 minEnergy bookkeeping — the centrifugal potential and the workless Coriolis force

Derivation BEnergy bookkeeping — the centrifugal potential and the workless Coriolis force

The goal is the energy story of the rotating-frame catalog: showing the centrifugal force is conservative with potential $U_{\mathrm{cf}} = -\tfrac12 m\omega^{2}\rho^{2}$ , while the Coriolis force does exactly zero work — so a steadily rotating frame has an honest energy conservation law of its own. The strategy: compute the work of each force directly along an arbitrary path in the frame, taking $\boldsymbol{\omega}$ constant (Euler asleep). Watch the geometry: both results are driven by perpendicularity, used in two different ways.

Step 1. Set cylindrical coordinates about the rotation axis, so the centrifugal force is $\mathbf{F}_{\mathrm{cf}} = m\omega^{2}\rho\,\hat{\boldsymbol{\rho}}$ . Its work along a path element $\mathrm{d}\mathbf{s}$ picks out only the radial progress:

\mathrm{d}W_{\mathrm{cf}} = m\omega^{2}\rho\,\hat{\boldsymbol{\rho}}\cdot\mathrm{d}\mathbf{s} = m\omega^{2}\rho\,\mathrm{d}\rho

Why only $\mathrm{d}\rho$ survives: the force is purely radial, so axial and azimuthal displacements are perpendicular to it and contribute nothing.

Step 2. Integrate from $\rho_1$ to $\rho_2$ : $W_{\mathrm{cf}} = \tfrac12 m\omega^{2}(\rho_2^{2}-\rho_1^{2})$ — path-independent, depending only on endpoints. A potential therefore exists:

U_{\mathrm{cf}}(\rho) = -\tfrac12\,m\,\omega^{2}\rho^{2} \qquad (6)

where: the minus sign says the potential falls outward — the centrifugal force pushes bodies downhill toward large $\rho$ , which is why the water surface in a spinning bucket climbs into the parabola $z(\rho) = \omega^2\rho^2/2g$ and why centrifuges drive the dense fraction to the rim.

Step 3. The Coriolis force's power output is a scalar triple product with a repeated factor:

P_{\mathrm{Cor}} = \mathbf{F}_{\mathrm{Cor}}\cdot\mathbf{v}_{\mathrm{rot}} = -2m\,(\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}})\cdot\mathbf{v}_{\mathrm{rot}} = 0

Why it vanishes identically: a cross product is perpendicular to both of its factors, so dotting it back into $\mathbf{v}_{\mathrm{rot}}$ gives zero at every instant, for every trajectory, exactly — not on average. The Coriolis force is a pure steering force, the mechanical twin of the magnetic force $q\mathbf{v}\times\mathbf{B}$ .

Step 4. Assemble the frame's conservation law. With $\boldsymbol{\omega}$ constant and real forces conservative (potential $U$ ), the quantity

E_{\mathrm{rot}} = \tfrac12 m v_{\mathrm{rot}}^{2} + U(\mathbf{r}) - \tfrac12 m \omega^{2}\rho^{2} \qquad (7)

is conserved along the motion. where: the last term is the centrifugal potential (6); the Coriolis force, doing no work, never appears in an energy budget at all. Why this matters downstream: this is the Jacobi-style integral that governs, for example, the restricted three-body problem in its rotating frame — effective potentials with centrifugal terms are a standing tool of celestial mechanics. *Verified: $-\partial U_{\mathrm{cf}}/\partial\rho = m\omega^{2}\rho$ recovers (3); the parabola $z = \omega^{2}\rho^{2}/2g$ matches the classic rotating-bucket result; zero Coriolis work matches the companion Coriolis article's equation (5) and Landau–Lifshitz §39.*

Go deeper · Derivation C · 1 minThe Euler force during a carousel spin-up

Derivation CThe Euler force during a carousel spin-up

The goal is the least-famous catalog entry in its natural habitat: a carousel spinning up from rest with constant angular acceleration, and a rider standing at radius $\rho$ . We derive the Euler force she feels, check its direction against raw inertia, and see why it vanishes the moment the ride reaches cruising speed. The route is direct substitution into the catalog formula — the value of the exercise is the physical reading, which most textbooks skip.

Step 1. Let the carousel spin up about the vertical axis $\hat{\mathbf{z}}$ : $\boldsymbol{\omega}(t) = \alpha t\,\hat{\mathbf{z}}$ with constant $\dot{\boldsymbol{\omega}} = \alpha\hat{\mathbf{z}}$ . For the rider at position $\boldsymbol{\rho} = \rho\,\hat{\boldsymbol{\rho}}$ , the Euler force (5) evaluates to

\mathbf{F}_{\mathrm{Euler}} = -m\,\alpha\hat{\mathbf{z}}\times\rho\,\hat{\boldsymbol{\rho}} = -m\,\alpha\,\rho\;\hat{\boldsymbol{\phi}}

Why the direction makes sense: $\hat{\mathbf{z}}\times\hat{\boldsymbol{\rho}} = \hat{\boldsymbol{\phi}}$ points along the direction of increasing spin, so the force points against it — the rider is flung backward relative to the accelerating platform, precisely her inertia refusing the tangential speed-up, rebranded as a force in the rotating frame.

Step 2. Read the inertial-frame cross-check. There, the platform floor must drag the rider forward (via friction) to give her the growing tangential speed $v_\phi = \alpha t \rho$ ; the required real force is $m\alpha\rho$ forward. In the rotating frame she is at rest, so that real friction must be balanced — by the backward Euler force of exactly $m\alpha\rho$ . The ledgers agree line by line.

Step 3. Put numbers on it. A playground carousel reaching $\omega = 1.0\ \mathrm{rad/s}$ in $3.0\ \mathrm{s}$ has $\alpha = 0.33\ \mathrm{rad/s^2}$ ; a $30\ \mathrm{kg}$ child at $\rho = 1.8\ \mathrm{m}$ feels $F_{\mathrm{Euler}} = 30\times0.33\times1.8 = 18\ \mathrm{N}$ tangentially — comparable to the $\sim 54\ \mathrm{N}$ centrifugal force at full speed ( $30\times1.0^{2}\times1.8$ ), but transient. Why it disappears: at cruising speed $\dot{\boldsymbol{\omega}} = 0$ and the Euler term dies, leaving centrifugal (and Coriolis, if she walks) to run the steady state. *Verified: dimensions $[m][\dot\omega][\rho] = \mathrm{kg\cdot s^{-2}\cdot m} = \mathrm{N}$ ; the direction convention matches Goldstein §4.10; the inertial/rotating ledgers balance exactly.*

Limiting cases

Case	Result	Meaning
$\mathbf{A}=0,\ \boldsymbol{\omega}=0$	$\mathbf{F}_{\mathrm{fict}} = 0$	inertial frame: the catalog closes, bare Newton returns
pure translation	only $-m\mathbf{A}$ survives	one uniform force — elevators and braking trains; see linearly accelerating frames
steady rotation	centrifugal + Coriolis	the carousel/planet regime; Euler asleep
at rest in a steadily rotating frame	centrifugal only	plumb lines tilt and buckets climb with nothing moving
spin-up at the axis ( $\rho = 0$ )	Euler and centrifugal vanish	on the axis only Coriolis (if moving) and $-m\mathbf{A}$ act
slow rotation, small speeds	$\omega^{2}\rho \ll g$ , $2\omega v \ll g$	Earth-lab regime: fictitious forces are sub-percent corrections

Common trapCentrifugal force is fake physics — never write it down

This slogan and its mirror image are the two classic ways to fail the exam. Writing a centrifugal force in an inertial frame is indeed an error — there, a body in circular motion needs a net inward force, and adding an outward "centrifugal force" to "balance" it predicts, absurdly, straight-line motion. But inside a rotating frame the centrifugal term is mandatory: omit it and the parked toolbox on the carousel is predicted to accelerate inward, which no rider observes. The discipline is mechanical, not philosophical — declare the frame first, then use exactly its force list: inertial frame, real forces only; rotating frame, real forces plus the catalog of equation (1). And never "balance" a real centripetal force with a fictitious centrifugal one in the same sentence: they are the same physics wearing different coordinates, not two forces in a tug-of-war.

Worked examples

1 · The strap-hanger's tilt (confidence builder)

A metro train accelerates from rest at $A = 1.4\ \mathrm{m/s^2}$ . A handhold strap of negligible mass swings freely from the ceiling; a commuter of mass $75\ \mathrm{kg}$ stands holding it, and a child of mass $25\ \mathrm{kg}$ holds a second strap. Find the angle each strap makes with the vertical, and the fictitious force on each person in the train frame.

Step 1. In the train frame every body acquires the translational fictitious force $-m\mathbf{A}$ , pointing backward. On the commuter: $F = 75\times1.4 = \boxed{105\ \mathrm{N}}$ ; on the child: $F = 25\times1.4 = \boxed{35\ \mathrm{N}}$ .

Step 2. A freely hanging strap aligns with the resultant of gravity ( $mg$ down) and the fictitious force ( $mA$ backward). The equilibrium angle from the vertical satisfies $\tan\theta = mA/mg = A/g$ — the mass cancels, which is the article's fingerprint at work.

Step 3. Numerically: $\tan\theta = 1.4/9.81 = 0.143$ , so $\theta = \boxed{8.1°}$ — the same angle for the commuter's strap, the child's strap, and a feather taped to either. One glance down the carriage shows every hanging object tilted identically: a uniform fictitious field, indistinguishable from tilting gravity itself.

Step 4. Cross-check in the inertial frame: each strap must supply the horizontal force $mA$ to accelerate what hangs from it; a strap tension along the strap direction does that exactly when $\tan\theta = A/g$ . Same angle, no fictitious forces — the two ledgers agree.

2 · The washing machine audit (exam level)

A washer drum of radius $R = 0.25\ \mathrm{m}$ spins up from rest to $1200\ \mathrm{rpm}$ in $4.0\ \mathrm{s}$ at a constant rate. For a wet sock of mass $m = 50\ \mathrm{g}$ riding the drum wall, evaluate every entry of the fictitious-force catalog at $t = 2.0\ \mathrm{s}$ and at full speed, and find the effective weight multiplier at full speed.

Step 1. Convert the spin: $\omega_{\mathrm{f}} = 1200\ \mathrm{rpm} = 1200\times2\pi/60 = 125.7\ \mathrm{rad/s}$ ; constant spin-up gives $\dot{\omega} = 125.7/4.0 = 31.4\ \mathrm{rad/s^2}$ . At $t = 2.0\ \mathrm{s}$ , $\omega = 62.8\ \mathrm{rad/s}$ .

Step 2. Catalog audit at $t = 2.0\ \mathrm{s}$ (sock at rest in the drum frame, drum axis fixed): translational force — zero, the axis does not translate. Coriolis — zero, $\mathbf{v}_{\mathrm{rot}} = 0$ . Centrifugal — $m\omega^{2}R = 0.050\times(62.8)^{2}\times0.25 = 0.050\times3944\times0.25 = \boxed{49.3\ \mathrm{N}}$ outward. Euler — $m\dot{\omega}R = 0.050\times31.4\times0.25 = \boxed{0.39\ \mathrm{N}}$ tangential, opposing the spin-up.

Step 3. At full speed the Euler force dies ( $\dot{\omega} = 0$ ) and the centrifugal force peaks: $m\omega_{\mathrm{f}}^{2}R = 0.050\times(125.7)^{2}\times0.25 = 0.050\times15800\times0.25 = \boxed{198\ \mathrm{N}}$ .

Step 4. The weight multiplier: $\omega_{\mathrm{f}}^{2}R/g = 15800\times0.25/9.81 = \boxed{403}$ — the sock presses the drum wall with about 400 times its weight, which is precisely how the spin cycle wrings water out: the water's own 400g effective weight drives it through the perforations. Note the hierarchy — at full speed the audit reads centrifugal $198\ \mathrm{N}$ , gravity $0.49\ \mathrm{N}$ , Euler $0$ : the fictitious force is not a small correction here but the dominant "gravity" of the drum world, and laboratory ultracentrifuges push the same formula to $10^{6}\,g$ .

Origins · 1 minJean le Rond d'Alembert and how this was discovered

Jean le Rond d'Alembert

1717–1783 · Académie des Sciences, Paris

Foundling, salonnière's protégé, co-editor of the great Encyclopédie, d'Alembert put the inertial force on a professional footing a century before anyone called frames "non-inertial." His Traité de dynamique (1743) proposed the maneuver now taught as d'Alembert's principle: carry the $m\mathbf{a}$ of Newton's law across the equals sign and treat $-m\mathbf{a}$ as a force, converting every problem of motion into a problem of static equilibrium. That algebraic somersault — dynamics disguised as statics — is the direct ancestor of the fictitious-force ledger: the rotating-frame catalog is d'Alembert's transposed term, evaluated piece by kinematic piece. The move became the workhorse of engineering dynamics (machine-frame analyses still run on it) and, through Lagrange, of analytical mechanics itself. D'Alembert quarreled with almost everyone — Clairaut, Euler, eventually Rousseau — but his transposition outlived every feud.

History

1687

— Newton's Principia distinguishes true from relative motion via the rotating bucket, making rotation dynamically absolute.

1743

— d'Alembert's Traité de dynamique moves

m\mathbf{a}

to the force side, founding the inertial-force method.

1749

— Euler's mechanics of rotating bodies systematizes the kinematics that generate the centrifugal and Euler terms.

1835

— Coriolis derives the complete acceleration transformation for rotating machines, isolating the velocity-dependent force.

1907

— Einstein reads the mass-proportionality of inertial forces as an equivalence between acceleration and gravity, the founding insight of general relativity.

Going further

Texts. Morin, Introduction to Classical Mechanics, ch. 10 (Accelerating frames of reference) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., §4.9–4.10 · Landau & Lifshitz, Mechanics, §39 (Motion in a non-inertial frame) · Taylor, Classical Mechanics, ch. 9 (Mechanics in non-inertial frames) · Kleppner & Kolenkow, An Introduction to Mechanics, ch. 9.

Provenance. Equations and signs follow Morin (Introduction to Classical Mechanics, Ch. 10), Goldstein §4.9–4.10, Landau–Lifshitz Mechanics §39, and Taylor Ch. 9. A verifier should re-derive the rotating-frame result by applying the derivative operator $(\mathrm{d}/\mathrm{d}t)_{\text{in}}=(\mathrm{d}/\mathrm{d}t)_{\text{rot}}+\boldsymbol{\omega}\times$ twice to $\mathbf{r}$ , confirming the four fictitious terms and the exact Coriolis factor 2. The Earth's rotation rate $\omega = 2\pi/(\text{sidereal day}) \approx 7.29\times10^{-5}\ \mathrm{s^{-1}}$ is a standard textbook value. Worked-example numbers on this page (strap angle $8.1°$ ; washer forces $49.3\ \mathrm{N}$ , $0.39\ \mathrm{N}$ , $198\ \mathrm{N}$ and the 403g multiplier; carousel Euler force $18\ \mathrm{N}$ ) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Non-inertial reference frames · Newton's second law · Transformation of acceleration in rotating frames. Leads to: Linearly accelerating frames · Centrifugal force · Coriolis force · Euler force · Equivalence of uniform acceleration and gravity · Effective gravity on the rotating Earth.

If fictitious forces produce real, measurable effects, why call them fictitious at all?

"Fictitious" describes their origin, not their consequences. The laundry really is pinned to the drum and the strap really tilts — but the explanation-by-force exists only in the accelerating frame's ledger. Transform to an inertial frame and every fictitious term evaporates, replaced by the frame's own acceleration on the kinematic side; no real force can be deleted by a coordinate change. The physically honest statement: the effects are real and frame-independent, while their attribution to a "force" is bookkeeping tied to your choice of coordinates. Many authors prefer "inertial forces" precisely because the phenomena are inertia in costume — the body's stubborn constant-velocity habit, described from a frame that refuses to keep constant velocity.

How do I decide in two seconds which of the four forces a problem needs?

Run the switch test on the frame, then on the body. Frame: does the origin accelerate ( $\mathbf{A}\neq 0$ )? Does it rotate ( $\boldsymbol{\omega}\neq 0$ )? Is the rotation rate changing ( $\dot{\boldsymbol{\omega}}\neq 0$ )? Body: does it move relative to the frame ( $\mathbf{v}_{\mathrm{rot}}\neq 0$ )? Each "yes" wakes exactly one term: $-m\mathbf{A}$ , centrifugal, Euler, Coriolis respectively — with the caveat that centrifugal needs $\boldsymbol{\omega}\neq0$ only, while Coriolis needs rotation and motion. A parked object in a steadily rotating frame keeps just the centrifugal entry; a moving object adds Coriolis; a spin-up adds Euler. The recognition table in this article is that test written out.

Doesn't the centrifugal force partner with the centripetal force as an action–reaction pair?

No — and this is one of the most durable confusions in mechanics. The centripetal force is not a kind of force at all: it is a role, played by whatever real force (tension, gravity, friction, a wall) points toward the center of a circular motion, and its third-law partner is that real force's ordinary reaction (the string pulls the stone in, the stone pulls the string out). The centrifugal force lives in a different frame entirely — the rotating one — where the body is not even circling, so nothing centripetal is needed. The two never legitimately appear in the same equation: quote them "balancing" and you have mixed two ledgers, which is exactly the bookkeeping crime this article's warning flags.

Is d'Alembert's inertial force the same thing as these fictitious forces?

Same algebra, different billing. D'Alembert's principle takes the honest inertial-frame law $\mathbf{F} = m\mathbf{a}$ and rewrites it as $\mathbf{F} - m\mathbf{a} = 0$ , treating $-m\mathbf{a}$ (the particle's own acceleration, reversed) as a force so the problem reads as static equilibrium — a pure computational device, usable in any frame. The frame-based fictitious forces are the special case where the transposed acceleration belongs to the frame: $-m\mathbf{A}$ , centrifugal, Coriolis, Euler are $-m$ times specific kinematic pieces of the frame's motion. Historically d'Alembert's 1743 transposition is the ancestor of the whole method; conceptually, the modern catalog is d'Alembert applied to the coordinate grid rather than to the particle.

Can fictitious forces do work? That seems wrong for something that is not real.

Three of the four can, and the energy is honestly accounted. In the frame's ledger, the centrifugal force is even conservative — potential $-\tfrac12 m\omega^{2}\rho^{2}$ — so it stores and returns energy like any spring; the translational and Euler forces can pump energy in or out (a braking train's frame watches your kinetic energy change with no real horizontal force doing it). Only the Coriolis force is constitutionally workless, being forever perpendicular to the velocity. The resolution of the "not real" discomfort: kinetic energy is itself frame-dependent, so different frames may disagree about who did work on what — they agree only on frame-independent outcomes. Within any one frame, the books balance exactly.

Why is gravity the one real force that behaves like the fictitious ones?

Because gravity shares the fingerprint: strict proportionality to mass, hence universal acceleration. Drop a hammer and a feather in vacuum and they fall together — just as everything in a braking train tips together. Einstein's reading: perhaps gravity is frame bookkeeping, and the freely falling frame is the honest one, where falling bodies float and Newton's first law reigns locally. That is the equivalence principle, and it survives every test to date — the MICROSCOPE satellite confirmed the universality of free fall to about one part in $10^{15}$ . What gravity has that no fictitious force can fake is tidal structure: real gravitational fields vary from point to point, and those differences (stretching, squeezing) cannot be transformed away. Uniform gravity is impersonable; tides are curvature, and curvature is real.

On Earth, do I need to add the centrifugal force every time I use g?

Almost never — because it is already priced in. The $g$ you measure with a plumb line and a stopwatch is effective gravity: true gravitational attraction plus the centrifugal term of the rotating Earth frame, about $0.3\%$ at the equator, zero at the poles, tilting the plumb line slightly at mid-latitudes. Every tabulated local $g$ , every building's "vertical," every spirit level already includes it. You add centrifugal terms explicitly only when the rotation is not Earth's daily spin baked into $g$ — a centrifuge, a turntable, an aircraft in a turn — or when you need the latitude dependence itself, which is the business of the effective-gravity article this one leads to.

Do fictitious forces exist for an accelerating observer in relativity too?

Yes, and the idea upgrades gracefully. In special relativity a uniformly accelerating observer (Rindler frame) finds an effective gravitational field with genuinely new consequences — clocks at different "heights" tick at different rates, and there is even a horizon behind the accelerating rocket. In general relativity the fictitious forces become Christoffel symbols: the terms in the geodesic equation that appear whenever coordinates are not freely falling. Newtonian gravity itself gets reclassified as exactly such a term — from the spacetime viewpoint, standing on the ground is the accelerated state and the falling apple is the inertial one. The catalog you learned here is the Newtonian shadow of that geometry, which is why mastering it now pays off later.