Euler force

Everyone knows the lurch of a subway car pulling away: the floor accelerates forward, and in the car's frame something seems to shove you backward — the inertial force $-m\mathbf{A}$ of a linearly accelerating frame. Now trade the subway for a playground merry-go-round. The moment your friends start pushing, before the platform is even turning fast, you feel the rotational twin of that shove: a tangential push opposite to the way the spin is picking up. No motion of yours triggered it — you were standing perfectly still on the platform. That push is the Euler force, $-m\,\dot{\boldsymbol{\omega}}\times\mathbf{r}$ : the fictitious force that appears in a rotating frame whenever the rotation itself is speeding up, slowing down, or tilting its axis. It is the least famous member of the rotating-frame trio, and for a good reason: the frame we all live in is astonishingly steady. Earth's spin rate wanders by only parts in $10^8$ over a day, so the terrestrial Euler force ( $\sim6\times10^{-11}\ \mathrm{m/s^2}$ ) is some nine orders of magnitude below the centrifugal term $\omega_\oplus^2R\approx3.4\times10^{-2}\ \mathrm{m/s^2}$ — and still about seven orders below even the feeble Coriolis force on a body moving at $10\ \mathrm{m/s}$ , and every geophysics textbook silently sets $\dot{\boldsymbol{\omega}}=0$ . But the moment engineers build frames whose spin genuinely changes — centrifuges screaming up to speed, reaction wheels torquing a space telescope, robot arms whipping through a pick-and-place cycle, a gyroscope forced to precess — the Euler force returns from exile and must be paid for in motor torque, bearing loads, and structural design. Conceptually it completes a satisfying pattern: the centrifugal force reads your position, the Coriolis force reads your velocity, and the Euler force reads the frame's own angular acceleration. Three forces, three inputs, one common origin in the kinematics of rotation.

Key takeaways

The Euler force is the fictitious force $-m\,\dot{\boldsymbol\omega}\times\mathbf r$ that appears in a rotating frame whenever the frame's angular velocity $\boldsymbol\omega$ changes in time.
It is one of three fictitious forces produced by rotation — alongside the centrifugal force $-m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)$ and the Coriolis force $-2m\,\boldsymbol\omega\times\mathbf v_{\mathrm{rot}}$ — but it is the only one tied to angular acceleration.
For a body fixed in the frame it is purely tangential (azimuthal), with magnitude $m\,\dot\omega\,r_\perp$ : the inertial reaction to angular spin-up or spin-down.
It vanishes identically for steady rotation ( $\dot{\boldsymbol\omega}=0$ ), which is why rotating-Earth treatments ignore it — Earth's $\dot\omega$ makes it about $10^{-11}\ \mathrm{m/s^2}$ at the equator.
Summed over a rigid body it delivers the inertial torque $-I\dot{\boldsymbol\omega}$ , the rotational analogue of the linear inertial force $-m\mathbf A$ .
Key idea: In a frame rotating with time-dependent angular velocity $\boldsymbol\omega(t)$ , the rotating-frame equation of motion is $m\,\mathbf a_{\mathrm{rot}} = \mathbf F_{\mathrm{real}} - m\,\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r) - 2m\,\boldsymbol\omega\times\mathbf v_{\mathrm{rot}} - m\,\dot{\boldsymbol\omega}\times\mathbf r$ , and the last term — proportional to the angular acceleration — is the Euler force.

Where this lives today

Spacecraft attitude control is applied Euler force. A reaction wheel changes a satellite's orientation precisely by commanding $\dot{\boldsymbol\omega}$ on a flywheel: the wheel's inertial torque $-I\dot{\boldsymbol\omega}$ reacts on the spacecraft body, slewing a space telescope to its next target without a gram of propellant. The engineering limits are Euler-force limits — motor torque, bearing loads during wheel spin-up, and the momentum-dump budget. The same $I\dot\omega$ accounting sizes every high-cycle industrial robot joint (pick-and-place arms routinely command $\dot\omega$ transients that dwarf their steady loads), every hard-braking flywheel energy-storage unit, and the emergency-stop loads on wind-turbine rotors, where halting a multi-tonne blade set puts enormous transient tangential forces on the hub. At the planetary scale the Euler force is a measurement rather than a load. Earth's length-of-day wobbles by milliseconds — atmospheric winds, ocean currents, and core flows trading angular momentum with the crust — and ring-laser gyroscopes plus VLBI radio astronomy now track those $\dot\omega_\oplus$ fluctuations to exquisite precision. The implied Euler acceleration at the equator, of order $10^{-11}\ \mathrm{m/s^2}$ , is far too small to feel but is a live diagnostic of the planet's interior. And pulsars make the opposite point: what is extreme in a neutron star is $\omega$ (about $190\ \mathrm{rad/s}$ for the Crab), not $\dot\omega$ — the Crab's steady spin-DOWN rate is a mere $\sim2\times10^{-9}\ \mathrm{rad/s^2}$ , and even the abrupt spin-UP "glitches" that punctuate it reach at most $\sim10^{-4}\ \mathrm{rad/s^2}$ , far below this article's playground merry-go-round. The glitches are feeble Euler events by laboratory standards, yet a precious probe of superfluid neutron-star interiors.

The physical picture

Picture yourself standing motionless at the edge of a large turntable that is suddenly switched on. As the platform accelerates its spin, you feel a push opposite to the direction of increasing rotation — you lurch backward. In the platform's frame nothing is moving you, so you must invent a force: that is the Euler force. It is the rotational analogue of being thrown back into your seat when a car accelerates in a straight line.

Figure

Spin-up on a turntable — the Euler force pushes backward against the growing rotation

The defining feature is that the Euler force cares only about how fast the spin is changing, not about how fast you are moving. A particle bolted to the rotating frame and perfectly at rest within it still feels the Euler force the moment $\boldsymbol\omega$ starts to change — unlike the Coriolis force, which switches off whenever the particle stops moving relative to the frame. Geometrically, for a co-rotating point the Euler force is purely azimuthal (tangent to circles about the axis). If $\dot{\boldsymbol\omega}$ is parallel to $\boldsymbol\omega$ — the axis direction fixed, only the rate changing — the force is a clean tangential push or drag. If $\dot{\boldsymbol\omega}$ tilts the axis, the Euler force can point in more elaborate directions — this is what makes gyroscopic dynamics interesting.

Figure

Two ways to have a nonzero ω̇ — changing the spin rate, or swinging the spin axis

Think of $-m\,\dot{\boldsymbol\omega}\times\mathbf r$ as inertia resisting angular acceleration, the rotational cousin of $-m\mathbf A$ in a linearly accelerating frame. Both are reactions to the frame's own change of motion; both vanish the instant the frame stops accelerating.

The core equations

Start with the definition. In a frame whose angular velocity $\boldsymbol\omega(t)$ is changing, every mass $m$ at position $\mathbf r$ is assigned the fictitious force

\mathbf F_{\mathrm{Euler}} = -\,m\,\dot{\boldsymbol\omega}\times\mathbf r \qquad (1)

where: $\dot{\boldsymbol\omega}=\mathrm d\boldsymbol\omega/\mathrm dt$ — the angular acceleration of the rotating frame; $\mathbf r$ — the particle's position relative to an origin on the rotation axis. Note what is absent: the particle's velocity. Like the centrifugal force and unlike the Coriolis force, the Euler force acts in full on a body sitting perfectly still in the frame. All the frame bookkeeping flows through one operator identity — the master key derived in the article on time derivatives in rotating frames. For any vector $\mathbf A$ ,

\left(\frac{\mathrm d\mathbf A}{\mathrm dt}\right)_{\!\mathrm{in}} = \left(\frac{\mathrm d\mathbf A}{\mathrm dt}\right)_{\!\mathrm{rot}} + \boldsymbol\omega\times\mathbf A \qquad (2)

where: the subscripts mark the frame in which the derivative is evaluated. One immediate corollary matters enormously here: feed the identity $\mathbf A=\boldsymbol\omega$ itself and the cross term dies, $\boldsymbol\omega\times\boldsymbol\omega=0$ , so $\dot{\boldsymbol\omega}_{\mathrm{in}}=\dot{\boldsymbol\omega}_{\mathrm{rot}}\equiv\dot{\boldsymbol\omega}$ . The angular acceleration needs no frame label — both observers agree on it, which is why equation (1) is unambiguous as written. Applying the identity twice to the position vector and sorting terms — the full derivation, factor-of-2 bookkeeping included, lives in the article on the transformation of acceleration in rotating frames — produces Newton's law as a rotating observer must write it:

m\,\mathbf a_{\mathrm{rot}} = \mathbf F_{\mathrm{real}} - 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 (3)

where: $\mathbf F_{\mathrm{real}}$ — the sum of genuine interaction forces; the three correction terms are the centrifugal, Coriolis, and Euler forces respectively. Each reads a different input: position, velocity, angular acceleration. The Coriolis force article and the centrifugal force article each own their term's story; here we take the last one apart.

Derivation AWhere the angular-acceleration term is born

The goal is to see exactly which step of the frame transformation creates the Euler term $-m\,\dot{\boldsymbol\omega}\times\mathbf r$ — we will not re-derive the whole acceleration transformation (that is the business of the acceleration-transformation article), but trace the single lineage that matters here. The strategy: apply the operator identity (2) to the position vector to get the velocity relation, then watch what the rotating-frame time derivative does to the sweep term $\boldsymbol\omega\times\mathbf r$ when $\boldsymbol\omega$ is allowed to depend on time. The point to watch is that the Euler term appears exactly once — unlike the Coriolis contribution, which arises twice and earns its factor of 2 — because only one of the two derivative applications ever differentiates $\boldsymbol\omega$ itself.

Step 1. Apply (2) to $\mathbf r$ : the inertial velocity is the frame velocity plus the sweep,

\mathbf v_{\mathrm{in}} = \mathbf v_{\mathrm{rot}} + \boldsymbol\omega\times\mathbf r

Why no $\dot{\boldsymbol\omega}$ yet: the identity involves $\boldsymbol\omega$ at one instant; differentiating position once cannot probe how the rotation is changing.

Step 2. Differentiate the sweep term in the rotating frame, now letting $\boldsymbol\omega=\boldsymbol\omega(t)$ :

\left(\frac{\mathrm d}{\mathrm dt}\right)_{\!\mathrm{rot}}\big(\boldsymbol\omega\times\mathbf r\big) = \dot{\boldsymbol\omega}\times\mathbf r + \boldsymbol\omega\times\mathbf v_{\mathrm{rot}}

Why $\dot{\boldsymbol\omega}$ carries no frame label: by the corollary above, $\boldsymbol\omega\times\boldsymbol\omega=0$ makes the angular acceleration identical in both frames — the one vector whose rate everyone agrees on.

Step 3. Assemble the full second derivative (the remaining terms are exactly the ones sorted in the acceleration-transformation article):

\mathbf a_{\mathrm{in}} = \mathbf a_{\mathrm{rot}} + \dot{\boldsymbol\omega}\times\mathbf r + 2\,\boldsymbol\omega\times\mathbf v_{\mathrm{rot}} + \boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)

Why the Euler term appears only once: $\dot{\boldsymbol\omega}\times\mathbf r$ comes solely from Step 2's differentiation of the sweep; the second, cross-product half of the operator acts on vectors, not on $\boldsymbol\omega$ , so it can never manufacture another one.

Step 4. Impose $m\mathbf a_{\mathrm{in}}=\mathbf F_{\mathrm{real}}$ and solve for $m\mathbf a_{\mathrm{rot}}$ : each kinematic term crosses to the force side with a minus sign, and the angular-acceleration piece becomes the Euler force

\mathbf F_{\mathrm{Euler}} = -m\,\dot{\boldsymbol\omega}\times\mathbf r

Why the sign matters physically: during spin-up ( $\dot{\boldsymbol\omega}$ along $+\boldsymbol\omega$ ) the force on a co-rotating body points against the direction of increasing motion — the backward lurch — just as $-m\mathbf A$ points backward in an accelerating train.

For the workhorse case — axis fixed, rate changing, $\boldsymbol\omega=\omega(t)\hat{\mathbf z}$ — the geometry collapses to something you can feel in your hands. The cross product of $\dot\omega\hat{\mathbf z}$ with the position singles out the azimuthal direction:

\mathbf F_{\mathrm{Euler}} = -m\,\dot\omega\,r_\perp\,\hat{\boldsymbol\varphi},\qquad |\mathbf F_{\mathrm{Euler}}| = m\,\dot\omega\,r_\perp \qquad (4)

where: $r_\perp$ — perpendicular distance from the rotation axis; $\hat{\boldsymbol\varphi}$ — the azimuthal (tangential) unit vector in the direction of increasing rotation angle. During spin-up the force points backward along $-\hat{\boldsymbol\varphi}$ ; during braking it reverses and throws you forward. It is largest at the rim and zero on the axis — riders at the edge of the carousel get the biggest jolt. Summing the Euler force over an extended rigid body upgrades the force statement to a torque statement, and lands on something familiar:

\boldsymbol\tau_{\mathrm{Euler}} = \int \mathbf r\times(-\dot{\boldsymbol\omega}\times\mathbf r)\,\mathrm dm = -\,I\,\dot{\boldsymbol\omega}\;\big|_{\text{about the axis}} \qquad (5)

where: $I$ — the moment of inertia about the rotation axis. This is the rotational twin of the statement that a frame acceleration $\mathbf A$ applies a net inertial force $-M\mathbf A$ to a body of mass $M$ : an angular acceleration $\dot{\boldsymbol\omega}$ applies a net inertial torque $-I\dot{\boldsymbol\omega}$ . It is precisely the torque a motor must overcome to spin the body up — the term $I\dot\omega$ in every machine-design calculation is the integrated Euler force wearing its inertial-frame costume.

Go deeper · Derivation B · 1 minThe Euler force summed over a rigid body gives the inertial torque

Derivation BThe Euler force summed over a rigid body gives the inertial torque

The goal is equation (5): integrating the Euler force's moment over a rigid body spinning up about a fixed axis yields the torque $-I\dot\omega$ , connecting the fictitious-force picture to the familiar $\tau=I\dot\omega$ of rigid-body dynamics. The strategy is direct integration — write the moment $\mathbf r\times\mathbf F_{\mathrm{Euler}}$ of the force on each mass element, expand the resulting triple product with the BAC-CAB identity, and project on the axis. The satisfying payoff is consistency: the "fictitious" bookkeeping of the rotating frame reproduces, term for term, the real torque budget an inertial engineer would write down.

Step 1. Sum the moment of the Euler force over all mass elements:

\boldsymbol\tau=\int \mathbf r\times\mathrm d\mathbf F_{\mathrm{Euler}} = -\int \mathbf r\times(\dot{\boldsymbol\omega}\times\mathbf r)\,\mathrm dm

Why a torque and not just a force: the Euler force grows linearly with distance from the axis, so its distributed effect on an extended body is fundamentally a twisting one.

Step 2. Expand the triple product:

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

Why BAC-CAB again: the same identity that reduced the centrifugal double cross product — it is the standard tool for unpacking nested cross products into dot products.

Step 3. Take $\dot{\boldsymbol\omega}=\dot\omega\,\hat{\mathbf z}$ and project on the axis. With $\mathbf r\cdot\hat{\mathbf z}=z$ and $r^2 = r_\perp^2+z^2$ , the $z$ -component of the integrand is $\dot\omega(r^2 - z^2) = \dot\omega\,r_\perp^2$ :

\tau_z = -\dot\omega\int r_\perp^2\,\mathrm dm = -I\,\dot\omega

Why only $r_\perp$ survives: the axial coordinate drops out exactly as it does for the moment of inertia itself — twisting about $\hat{\mathbf z}$ cares only about mass distribution around, not along, the axis. *As a check: the dimensions $[\dot\omega\, r_\perp^2\,\mathrm{kg}]=\mathrm{N\,m}$ confirm the result is a torque, and this inertial torque is precisely the $I\dot\omega$ term at the heart of Euler's rigid-body equations.*

Go deeper · Derivation C · 1 minFriction budget for a bead on a spinning-up turntable

Derivation CFriction budget for a bead on a spinning-up turntable

The goal is the cleanest experimental signature of the Euler force: a bead held by friction at radius $r_\perp$ on a turntable that accelerates from rest at constant $\alpha=\dot\omega$ demands a friction force $F=m r_\perp\sqrt{\alpha^2+\omega^4}$ , and at the very first instant that demand is purely the Euler term. The strategy is rotating-frame statics: the bead is at rest in the frame ( $\mathbf v_{\mathrm{rot}}=0$ , $\mathbf a_{\mathrm{rot}}=0$ ), so the real friction must exactly cancel the fictitious forces present. Watch how the two demands evolve at different rates — the Euler piece is constant in time while the centrifugal piece grows as $\omega^4$ — because that competition decides when the bead finally slips.

Step 1. List the fictitious forces on the co-rotating bead with $\omega(t)=\alpha t$ :

\mathbf F_{\mathrm{cent}}=m\omega^2 r_\perp\,\hat{\mathbf r}_\perp,\qquad \mathbf F_{\mathrm{Cor}}=0,\qquad \mathbf F_{\mathrm{Euler}}=-m\alpha\,r_\perp\,\hat{\boldsymbol\varphi}

Why Coriolis is absent: the bead does not move relative to the turntable, and the Coriolis force reads velocity in the frame — zero in, zero out.

Step 2. Impose rotating-frame equilibrium, $0=\mathbf F_{\mathrm{friction}}+\mathbf F_{\mathrm{cent}}+\mathbf F_{\mathrm{Euler}}$ , and solve:

\mathbf F_{\mathrm{friction}} = -m\omega^2 r_\perp\,\hat{\mathbf r}_\perp + m\alpha\,r_\perp\,\hat{\boldsymbol\varphi}

Why friction must lead the bead: the tangential piece points forward, in $+\hat{\boldsymbol\varphi}$ — the floor drags the bead up to speed, and the backward Euler force is the bead's inertia resisting exactly that.

Step 3. The two components are perpendicular, so the magnitude is

F = m\,r_\perp\sqrt{\alpha^2+\omega^4}

Why the odd pairing of powers: the tangential demand $\alpha$ is constant while the radial demand $\omega^2=(\alpha t)^2$ grows quadratically in time — at switch-on the friction budget is pure Euler; seconds later it is overwhelmingly centrifugal.

Step 4. Slipping begins when $F$ exhausts static friction: $m r_\perp\sqrt{\alpha^2+\omega^4}=\mu_s m g$ , i.e. at

\omega_{\mathrm{slip}}=\left[\left(\frac{\mu_s g}{r_\perp}\right)^{2}-\alpha^2\right]^{1/4}

Why the Euler term barely moves the slip point in slow spin-ups: for small $\alpha$ the $\alpha^2$ inside the bracket is negligible — but it is the only term at $t=0$ , which is why the initial jolt, not the steady spin, is what you feel first on the carousel. *As a check: $[\alpha]=[\omega^2]=\mathrm{s^{-2}}$ so the two terms under the square root share the same dimensions, and taking $\alpha\to0$ recovers the pure centripetal requirement $m\omega^2 r_\perp$ of steady rotation.*

Limiting cases

The Euler force is governed entirely by $\dot{\boldsymbol\omega}$ , so its presence or absence follows simple conditions on how the rotation evolves.

Case	Result	Meaning
Steady rotation	$\dot{\boldsymbol\omega}=0 \;\Rightarrow\; \mathbf F_{\mathrm{Euler}}=0$	Constant angular velocity (magnitude and direction) — the rotating-Earth case. Only centrifugal and Coriolis survive; the Euler force vanishes exactly.
Spin-up about a fixed axis	$\boldsymbol\omega=\omega(t)\,\hat{\mathbf z},\quad \mathbf F_{\mathrm{Euler}}=-m\,\dot\omega\,r_\perp\,\hat{\boldsymbol\varphi}$	Axis direction fixed, rate changing. The force is purely azimuthal with magnitude $m\,\dot\omega\,r_\perp$ , opposing the increase in rotation.
On the axis	$r_\perp=0 \;\Rightarrow\; \mathbf F_{\mathrm{Euler}}=0$ (fixed-axis spin-up)	A particle on the rotation axis feels no Euler force during fixed-axis spin-up, since $\dot{\boldsymbol\omega}\times\mathbf r$ vanishes — exactly like the centrifugal force.
Precessing axis	$\dot{\boldsymbol\omega}=\boldsymbol\Omega\times\boldsymbol\omega \;(\vert\boldsymbol\omega\vert=\text{const})$	If the spin rate is constant but the axis sweeps around at rate $\boldsymbol\Omega$ , $\dot{\boldsymbol\omega}$ is nonzero and the Euler force persists — central to gyroscope dynamics.
Rotating Earth	$\dot\omega_\oplus/\omega_\oplus\sim10^{-8}\ \text{per day} \;\Rightarrow\; a_{\mathrm{Euler}}\sim10^{-11}\ \mathrm{m/s^2}$	Length-of-day fluctuations of milliseconds make the terrestrial Euler force about a trillionth of $g$ — utterly negligible next to Coriolis and centrifugal effects.

Common trap“If I'm standing still on the platform, only the centrifugal force can act on me”

Standing still in the frame kills the Coriolis force, not the Euler force. The full checklist for a body at rest in a rotating frame is: centrifugal force always (off-axis), Euler force whenever $\dot{\boldsymbol\omega}\neq0$ — and the merry-go-round's starting jolt is the proof your body already knows. Two more traps. First, do not conflate Euler with Coriolis by their formulas: Coriolis reads your velocity at constant spin, Euler reads the spin's own change even at zero velocity; each can exist without the other. Second, mind the sign and the object: the fictitious Euler force on you is $-m\,\dot{\boldsymbol\omega}\times\mathbf r$ , opposite to the genuine tangential acceleration $\dot{\boldsymbol\omega}\times\mathbf r$ of the platform point beneath your feet — the platform surges forward, and the frame books your lag as a backward force on you.

Worked examples

1 · The starting jolt of a merry-go-round (confidence builder)

A playground merry-go-round accelerates uniformly from rest to $\omega = 0.5\ \mathrm{rad/s}$ in $2\ \mathrm{s}$ . A 30-kg child stands at $r_\perp = 2\ \mathrm{m}$ from the axis. Find the Euler force on the child during the spin-up, and compare it with the centrifugal force at the moment full speed is reached.

Step 1. Angular acceleration: $\dot\omega = 0.5/2 = 0.25\ \mathrm{rad/s^2}$ , directed along the spin axis.

Step 2. Euler acceleration at the child's position: $a_{\mathrm{E}} = \dot\omega\,r_\perp = 0.25\times2 = 0.5\ \mathrm{m/s^2}$ , tangential and backward — against the direction the platform is picking up speed.

Step 3. Euler force: $F_{\mathrm{E}} = m\,a_{\mathrm{E}} = 30\times0.5 = \boxed{15\ \mathrm{N}}$ — about the weight of a 1.5-kg school bag pulling the child backward for the full two seconds. This is what the child's grip must resist at the very first instant, when the centrifugal force is still zero.

Step 4. Centrifugal force at full speed: $F_{\mathrm{cf}} = m\omega^2 r_\perp = 30\times(0.5)^2\times2 = \boxed{15\ \mathrm{N}}$ — by coincidence of these numbers, exactly equal. The hand-off is characteristic: the Euler force owns the start of the ride, then vanishes the instant $\omega$ steadies, leaving the centrifugal force holding the child outward. (Once the child walks on the platform, the Coriolis force joins in — see the Coriolis force article.)

2 · When does the child slip? (exam level)

Suppose the same merry-go-round keeps accelerating at $\dot\omega = 0.25\ \mathrm{rad/s^2}$ indefinitely, and the coefficient of static friction between shoes and platform is $\mu_s = 0.5$ . At what angular speed, and at what time, does a child standing at $r_\perp = 2\ \mathrm{m}$ begin to slide?

Step 1. From the friction-budget derivation, the total fictitious demand on the shoes has perpendicular components — tangential $m\dot\omega r_\perp$ (constant) and radial $m\omega^2 r_\perp$ (growing) — so slipping starts when $m r_\perp\sqrt{\dot\omega^2+\omega^4} = \mu_s m g$ .

Step 2. Solve for the slip speed: $\omega_{\mathrm{slip}} = \left[\left(\dfrac{\mu_s g}{r_\perp}\right)^2 - \dot\omega^2\right]^{1/4} = \left[\left(\dfrac{0.5\times9.81}{2}\right)^2 - (0.25)^2\right]^{1/4} = \left[6.015 - 0.0625\right]^{1/4} = \boxed{1.56\ \mathrm{rad/s}}$ .

Step 3. Time to reach it: $t = \omega_{\mathrm{slip}}/\dot\omega = 1.56/0.25 = \boxed{6.2\ \mathrm{s}}$ — about 15 rpm, a brisk playground spin.

Step 4. Check the anatomy of the demand at slip: centrifugal $\omega^2 r_\perp = (1.56)^2\times2 = 4.88\ \mathrm{m/s^2}$ versus Euler $\dot\omega\,r_\perp = 0.5\ \mathrm{m/s^2}$ — the available budget is $\mu_s g = 4.905\ \mathrm{m/s^2}$ . The centrifugal term does the final shoving, but note the asymmetry in time: at $t=0$ the entire demand was the Euler term. Ignore it and you would predict zero required friction at start-up — yet that is exactly when an unwary rider stumbles.

Origins · 1 minLeonhard Euler and how this was discovered

Leonhard Euler

1707–1783 · Basel → St Petersburg → Berlin

The force is named for Euler not because he wrote it down as a "force," but because the entire mechanics of changing rotation is his. In 1750 he announced what he considered a new principle of mechanics — that $\mathbf F=m\mathbf a$ applies to every element of a continuous body — and used it to crack the motion of rigid bodies, a problem Newton's Principia had left untouched. His 1765 Theoria motus corporum solidorum delivered the equations still bearing his name, in which the $I\dot{\boldsymbol\omega}$ term — the integrated Euler force of this article — plays the lead. Both of those works belong to his Berlin years (1741–1766) and predate his blindness: he lost the sight of his right eye around 1738 but went totally blind only after a failed cataract operation in 1771, by then back in St Petersburg. From there he dictated calculations to assistants with legendary fluency — the 1775 rotation theorem is one of the results — and his output increased after losing his sight. The naming convention is modern textbook shorthand, but the honour is precise: every term in the rotating-frame equation of motion passes through machinery Euler built — the angular velocity vector, the rigid-body equations, and the analysis of frames themselves.

History

1736

— Euler's Mechanica recasts Newtonian mechanics in the language of analysis, replacing geometry with differential equations

1750

— Euler announces the principle that F = ma applies to each element of any body, unlocking rigid-body dynamics

1765

— Theoria motus corporum solidorum publishes the Euler equations of rigid-body motion, with the I dω/dt inertial term at their heart

1775

— Euler proves that any displacement of a rigid body about a fixed point is a single rotation, founding the kinematics of the angular velocity vector

1852

— Léon Foucault builds and names the gyroscope; its analysis runs entirely on Euler's rigid-body equations and their inertial torque

Going further

References. 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 (Vol. 1), §39 (Motion in a non-inertial frame) · Taylor, Classical Mechanics, ch. 9 (Mechanics in noninertial frames) All equations follow from the rotating-frame operator identity $\mathrm d/\mathrm dt|_{\mathrm{in}}=\mathrm d/\mathrm dt|_{\mathrm{rot}}+\boldsymbol\omega\times$ : double application yields the Euler term $\dot{\boldsymbol\omega}\times\mathbf r$ and the factor of 2 in Coriolis; the rigid-body torque $-\mathbf I\dot{\boldsymbol\omega}$ follows from the BAC–CAB expansion; and the turntable friction magnitude $m r_\perp\sqrt{\alpha^2+\omega^4}$ is confirmed dimensionally and in the $\alpha\to0$ limit. The Earth rotation rate $\omega_\oplus\approx7.29\times10^{-5}\,\mathrm{s^{-1}}$ is the standard sidereal value; length-of-day fluctuations of order milliseconds give the quoted relative variation of parts in $10^{8}$ .

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

Is the jolt I feel when a bus brakes an Euler force?

Not quite — it is the Euler force's linear cousin. A braking bus is a linearly decelerating frame, and the fictitious force there is $-m\mathbf A$ : uniform, the same for every passenger regardless of where they sit. The Euler force is the rotational version: it requires a frame that spins at a changing rate, and it grows with distance from the axis, $m\dot\omega r_\perp$ — riders at the rim of a starting carousel lurch harder than riders near the middle. Same physics family (inertia booked as a force in an accelerating frame), different geometry: one reads $\mathbf A$ , the other reads $\dot{\boldsymbol\omega}$ and your position.

Why is it called the Euler force if Euler never wrote about fictitious forces?

The name is a 20th-century textbook convention, but a fair one. The term $-m\dot{\boldsymbol\omega}\times\mathbf r$ , integrated over a rigid body, is exactly the $-I\dot{\boldsymbol\omega}$ inertial torque at the heart of Euler's 1765 rigid-body equations — and the whole kinematic apparatus of the angular velocity vector, without which the formula cannot even be stated, is Euler's 1775 work. Naming the angular-acceleration force after the person who built the mechanics of angular acceleration is like naming the Coriolis force after Coriolis: the label marks intellectual ownership of the term, not the first appearance of the phrase "fictitious force."

If I stand still on the carousel, the Euler force acts on me — but does it do any work?

In the rotating frame, no: work requires displacement, and you are not displacing in that frame, so the Euler force (like every force on a frame-stationary body) does zero rotating-frame work. But let something move in the frame — a bead sliding along a spoke during spin-up, say — and the Euler force, which is tangential while the bead's motion is radial, still does no work on it; tilt the motion and it can. Contrast the ledgers: in the inertial frame the platform's friction genuinely accelerates you and does real work, paid for by the motor. Energy bookkeeping is frame-dependent for fictitious forces; each frame's ledger balances internally.

The article says Earth's Euler force is negligible — is it ever measurable at all?

Directly as a force on laboratory objects, no: an acceleration of order $10^{-11}\ \mathrm{m/s^2}$ is millions of times below what the best gravimeters can attribute to it against noise. But the underlying $\dot\omega_\oplus$ is measured superbly — just not through the force. Very-long-baseline interferometry tracks the orientation of the Earth against quasars, and large ring-laser gyroscopes sense the rotation rate optically; both resolve the millisecond wanderings of the length of day that atmospheric and core angular-momentum exchanges produce. So the honest statement is: the terrestrial Euler force is negligible, while the terrestrial $\dot{\boldsymbol\omega}$ is one of geodesy's precision observables.

Precession keeps the spin rate constant, so where does a nonzero ω̇ come from?

Because $\boldsymbol\omega$ is a vector, and a vector can change by turning as well as by stretching. Under precession at rate $\boldsymbol\Omega$ , the tip of $\boldsymbol\omega$ rides around a circle: $\dot{\boldsymbol\omega}=\boldsymbol\Omega\times\boldsymbol\omega$ , which is perpendicular to $\boldsymbol\omega$ and has magnitude $\Omega\,\omega\sin\theta$ — nonzero even though $|\boldsymbol\omega|$ never changes. This is the same logic as circular motion itself, where speed is constant but velocity is not. It is why gyroscopes being steered feel persistent inertial torques, and why the fixed-axis formula $m\dot\omega r_\perp$ is a special case, not the definition.

In the inertial frame there is no Euler force — so what really pushes the rider backward at start-up?

Nothing pushes the rider backward; the platform accelerates forward underneath them. In the inertial frame the story is: the rider's feet are dragged forward by friction, the rest of the body's inertia lags, and the rider tips rearward relative to the platform until their grip and posture supply the tangential force $m\dot\omega r_\perp$ needed to accelerate with it. The rotating frame compresses that whole narrative into a single entry — a backward force $-m\dot{\boldsymbol\omega}\times\mathbf r$ on the rider. Both descriptions predict the identical stumble; only the bookkeeping differs. This is the same two-ledger discipline as for the centrifugal force.

How do I keep the three rotational fictitious forces straight on an exam?

Read each force's input. Centrifugal reads position: $-m\boldsymbol\omega\times(\boldsymbol\omega\times\mathbf r)$ , outward, present for anything off-axis, even at rest, even at steady spin. Coriolis reads velocity: $-2m\boldsymbol\omega\times\mathbf v_{\mathrm{rot}}$ , sideways, present only when the body moves within the frame. Euler reads the frame's angular acceleration: $-m\dot{\boldsymbol\omega}\times\mathbf r$ , tangential (for fixed-axis changes), present only while the spin changes. Then run the two switch tests: is the body moving in the frame? (toggles Coriolis) — is the spin changing? (toggles Euler). The centrifugal force has no off-switch except sitting on the axis.