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A stationary person relative to the rim has centripetal acceleration $a{\omega }^2=g$. If $u$ is a signed walking speed relative to the rim, the inertial tangential speed is $a\omega +u$, so the inward normal reaction satisfiesConsequently the support force per mass, and hence apparent weight, is $4g$ for $u=+a\omega$ and zero for $u=-a\omega$. In the latter case the walker has zero inertial speed and needs no centripetal support; ordinary weight-bearing contact cannot be assumed to persist without a restraint. For artificial gravity in a rotating ring, the centrifugal force and the Coriolis force alone are insufficient: the walker also has relative-path centripetal acceleration $u^2/a$. Omitting it would give the wrong apparent weights at these large walking speeds.

A horizontal pressure gradient force accelerates air from higher toward lower pressure. In a rotating frame the Coriolis force deflects the flow. Away from strong friction, at large spatial scales and small Rossby number, a slowly changing flow can reach geostrophic balance:where $f=2\Omega \sin \phi$ is the Coriolis parameter. The geostrophic wind runs along atmospheric isobars, with speed $|{\nabla }_{h}p|/(\rho |f|)$. Closely spaced atmospheric isobars therefore indicate stronger geostrophic wind when density and latitude are comparable. In the northern hemisphere lower pressure lies to the left of this wind, producing counterclockwise circulation around a low; the sense reverses in the southern hemisphere.

A curved flow also has centripetal acceleration. For cyclonic circular flow in the northern hemisphere, the gradient wind relation isNear the ground, friction reduces wind speed and weakens its Coriolis contribution, so air crosses atmospheric isobars toward lower pressure. The simple geostrophic formula fails near the equator, where $f$ becomes small, and where rapid changes or tight curvature make neglected acceleration important. Vertically, large-scale motion is often approximated by hydrostatic balance, $dp/dz=-\rho g$.

Bernoulli's principle can relate speed changes to pressure changes along an appropriate inviscid streamline, but its simple dynamic-pressure relation is not a replacement for the rotating force balance governing a broad atmospheric pressure map.

Changing basis and acceleration. Attach orthonormal unit vectors ${\mathbf{e}}_i(t)$ to the rotating frame, with a common origin fixed in the inertial frame. An infinitesimal rotation $d\mathbf{\theta }=\mathbf{\omega }dt$ changes a vector fixed in those axes by $d\mathbf{A}=d\mathbf{\theta }\times \mathbf{A}$: its tip moves tangentially to the circle about the rotation axis. More generally, write $\mathbf{A}={\sum }_i{A}_i{\mathbf{e}}_i$. Differentiating the components and the rotating basis separately proves the rotating-frame derivative formula,

Apply the formula first to position: ${\dot{\mathbf{r}}}_0=\mathbf{v}+\mathbf{\omega }\times \mathbf{r}$, where $\mathbf{v}=(d\mathbf{r}/dt{)}_S$. Apply it again to both terms. For constant $\mathbf{\omega }$,One cross term comes from differentiating $\mathbf{v}$ in the inertial frame and another from differentiating $\mathbf{r}$ in $\mathbf{\omega }\times \mathbf{r}$. The displayed source formula implicitly requires constant angular velocity and an unaccelerated common origin. For changing rotation rate there is also $\dot{\mathbf{\omega }}\times \mathbf{r}$; an accelerating origin supplies its own translational acceleration. Moving the rotation terms to the force side gives the apparent Coriolis force and centrifugal force with negative signs.

Southward motion on Earth. Use local east, north and up axes, with latitude $\lambda$ positive northward. Then $\mathbf{\Omega }=\Omega (0,\cos \lambda ,\sin \lambda )$ and $\mathbf{v}=(0,-v,0)$. The Coriolis acceleration in local Earth coordinates givesIt is westward in the northern hemisphere, eastward in the southern hemisphere and zero for exactly southward motion at the equator. There is no vertical Coriolis component for this particular velocity.

Correct steering. On the desired meridional track, drag has no eastward component because it is antiparallel to the velocity. If the thrust makes signed angle $\varphi$ east of south, its eastward component is $T\sin \varphi$. The cross-track force balance is thereforesoThe argument must have magnitude at most one for this simple thrust-only correction to be feasible. The signed angle changes sign at the equator: correction is east of south in the north and west of south in the south. Thus the phrase “east of south” cannot describe a positive correction throughout the entire flight.

Wrong-sign correction and the miss distance. A literally constant velocity vector cannot develop a cross-track displacement. The physically useful interpretation of the last task is constant speed, with an erroneous lateral steering correction. In the usual leading-track model, thrust and drag balance along the actual motion, and the applied transverse correction has the correct magnitude but the wrong sign. It therefore reinforces rather than cancels the Coriolis force. Starting at the equator, use the unperturbed southward track $\lambda \simeq -vt/R$, and let $x_E$ be the eastward departure from that track. To first order in the departure,There is a factor two relative to uncompensated Coriolis deflection because the wrong correction adds an equal lateral force. Integration gives the wrong-sign Coriolis steering error,For southern angular distance $\Lambda =16\pi /180=0.27925$, $R=6.4\times 10^6\mathrm{m}$ and $v=800/3.6=222.2\mathrm{m}{\mathrm{s}}^{-1}$, the nominal flight time after the equator isTaking $\Omega \simeq 2\pi /(86400\mathrm{s})$ givesThe small-latitude approximation $\sin (vt/R)\simeq vt/R$ gives ${\ddot{x}}_E\simeq 4\Omega v^{2}t/R$ and $x_E\simeq 2\Omega v^2{t}^3/(3R)$, or $1.95\times 10^5\mathrm{m}$, confirming the scale and direction.

This is an estimate under an explicit steering model, not an exact trajectory forced by the wording. For example, retaining spherical transport of the local axes while still linearizing in the cross-track motion gives, with $s=-\lambda$, $\dot{u}-(v/R)u\tan s=4\Omega v\sin s$. Consequently $u\cos s=2\Omega R{\sin }^{2}s$ and the longitude departure is $\Delta \ell =2\Omega R(\tan \Lambda -\Lambda )/v$. The distance along the destination parallel is then $R\cos \Lambda \Delta \ell \simeq 1.93\times 10^5\mathrm{m}$. Corrections from the changed track and a more detailed control law are not fixed by the problem. If “keeps the velocity constant” instead means the full vector, the pilot would need an additional compensating force and there would be no miss; if thrust is kept at a specified geographical bearing despite drift, its along-track component and lateral drag must also be modelled.

Let a coordinate frame rotate with angular velocity $\mathbf{\Omega }$ relative to an inertial frame. For any vector $\mathbf{u}$,Applying this twice to position, with coincident nonaccelerating origins, givesNewton's second law written using rotating coordinates therefore contains the Coriolis forceThe other inertial corrections are the centrifugal force and, for changing rotation rate, the Euler force. An accelerating origin introduces an additional translational fictitious force.

The Coriolis force depends on velocity and is perpendicular to it, so ${\mathbf{F}}_C\cdot {\mathbf{v}}_{\mathrm{rotating}}=0$: the Coriolis force does no work in the rotating frame. It bends trajectories without directly changing their rotating-frame speed. It is absent for motion parallel to the rotation axis.

For locally horizontal motion on Earth at latitude $\lambda$, take $x$ east, $y$ north and $z$ up. Keeping the vertical component $\Omega \sin \lambda$ givesHorizontal motion is deflected right in the northern hemisphere and left in the southern hemisphere. In the ideal absence of other horizontal forces these equations give inertial circles with angular frequency $|f|$. The same velocity-dependent correction explains the precession of a Foucault pendulum, whose precession-rate magnitude is $\Omega |\sin \lambda |$. Terrestrial vertical motions also involve Earth's horizontal rotation component, so the reduced horizontal equations are not the full force law.

Let the rotating frame have origin acceleration ${\mathbf{A}}_0$ and angular velocity $\mathbf{\Omega }(t)$ relative to an inertial frame. For a vector referred to the moving axes, the transport rule isApply it twice to position. With rotating-frame velocity $\mathbf{v}$ and acceleration ${\mathbf{a}}_R$,Thus Newton's equation expressed in the rotating frame isThe added terms are translational, Coriolis, Euler and centrifugal fictitious forces. They reflect the frame's acceleration, rather than additional interactions with nearby objects.

The centrifugal force points outward from the rotation axis and, for constant $\mathbf{\Omega }$, has potential $-m|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. The Coriolis force is perpendicular to velocity, so does no work on the particle measured in that rotating frame; it changes direction rather than speed directly. The Euler force is present when the frame's rotation rate or direction changes.

On Earth, gravity and centrifugal acceleration combine into effective gravity. For slow horizontal motion at latitude $\lambda$, the vertical component of Earth's rotation gives Coriolis parameter $f=2\Omega \sin \lambda$: an initially northward motion deflects eastward in the northern hemisphere. This supplies a useful local explanation of rotating flows and Foucault precession. A bucket of rotating liquid illustrates the centrifugal potential: its steady free surface is an equipotential of $gz-{\Omega }^2{r}^2/2$, giving $z={\Omega }^2{r}^2/(2g)+\mathrm{constant}$. These equations require a declared frame; adding a centrifugal force while retaining inertial-frame acceleration would count the same kinematic effect twice.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

Let $x$ be signed displacement along the rails, measured from the centre. The diagram in the original paper shows a guide extending across the centre, so the signed coordinate can pass through zero. Hooke's law gives the radial force $-kx$. In the rotating reference frame, the radial centrifugal force is $+m{\Omega }^{2}x$. The Coriolis force is perpendicular to the rails and is balanced by their reaction; it has no radial component. Therefore Newton's second law givesEquivalently, the radial component of acceleration in polar coordinates in the inertial frame is $\ddot{x}-{\Omega }^{2}x$. The radial spring oscillator on a rotating guide has simple harmonic motion only when $k/m-{\Omega }^2>0$, withThe angular frequency is in $\mathrm{rad}{\mathrm{s}}^{-1}$ and the frequency in $\mathrm{Hz}$. The critical angular speed isStrictly this is an upper bound, not an attained maximum for oscillations: at equality $\ddot{x}=0$, so a displaced object released from rest stays at its displaced position relative to the rails; above it, the central equilibrium is unstable.

A planetary dynamo maintains an intrinsic magnetic field by motion of an electrically conducting fluid. The resistive induction equation balances stretching and transport against magnetic diffusion. A sufficiently large magnetic Reynolds number is necessary but does not ensure that the flow regenerates every required field component. Thermal or compositional convection supplies mechanical power; rotation organizes the flow through the Coriolis force, and the Lorentz force density eventually modifies the flow and limits growth. A fluid core can fail to operate a dynamo if it is stably stratified or inadequately driven.

A point mass constrained to a straight guide through the axis of a uniformly rotating reference frame, with signed central displacement $x$ and Hooke's law force $-kx$, satisfies $m\ddot{x}=-kx+m{\Omega }^{2}x$. The transverse Coriolis force is balanced by the guide. The central equilibrium is stable for ${\Omega }^2<k/m$, neutral at equality and unstable above it. The angular frequency is the positive square root in the title.

For shear ${\mathbf{u}}_0=-2Ax{\mathbf{e}}_y$, the spatially averaged perturbation kinetic energy obeys $dE/dt=2A\rho ⟨u_x^{\prime }{u}_y^{\prime }⟩-\rho \nu ⟨|\nabla {\mathbf{u}}^{\prime }{|}^2⟩$. Pressure and the Coriolis force do no net work. Thus positive radial-azimuthal velocity correlation extracts mean-flow energy. A leading two-dimensional shearing wave has $\mathrm{Re}(v_x{v}_y^{\ast })=-k_x{k}_y|Z{|}^2/k^4>0$.

// Target: astrophysics.bigb

For a uniformly rotating incompressible fluid of constant mass density, pressure and conservative body forces combine as the gradient of this scalar. The centrifugal term has a negative sign. Pressure enforces incompressible flow, whereas the velocity-dependent Coriolis force remains explicit and cannot be included in a scalar position potential. Adding a spatially constant function of time to $\psi$ does not change the motion.

For uniform rotation, the Newtonian gravitational potential plus the centrifugal potential generates the non-Coriolis part of the equation of motion in a rotating frame. Stationary particles obey $\nabla {\Phi }_{\mathrm{eff}}=0$. The Coriolis force changes perturbation dynamics, so curvature of this potential alone is not a general rotating-frame stability criterion.

A thermally driven fluid layer rotating about the vertical axis experiences a Coriolis force that couples vertical velocity to vertical vorticity. Its onset depends on the Rayleigh number, Prandtl number and Taylor number, as well as thermal and mechanical boundary conditions. The force does no mechanical work, but changes the linear mode balance and nonlinear convection roll competition.

The Technische Universität Darmstadt demonstration uses a steel ball launched outward from the centre of a rotating disc. A fixed lecture-room camera records a straight path, while a camera carried above the disc records a curved path. The instructions specify launching toward the fixed camera so that the laboratory motion is clearly visible. The rotating iPod transmits its view wirelessly through Apple TV; two vertical support rods of length $1\mathrm{m}$ are specified for the slow-motion arrangement. The paired video views are the readout: one physical motion is represented in two reference frames.^[ref]

A useful preparation control appears in the Universität Freiburg instructions: with the disc stationary, check that the ball travels radially toward its edge. A camera is mounted on supports attached to the disc; turn the disc by hand as the ball leaves its centre and observe the camera view.^[ref]

The inference is a dependence of the observed trajectory on the rotating axes; a curved camera trace alone does not isolate the Coriolis force. To compare with the constant-angular velocity free particle in rotating coordinates, measure the disc angular velocity, calibrate lengths in the camera image, and compare launches at rest and with opposite rotation directions. These are checks of the ideal model, not reported measurements from the manuals. A rolling ball can exchange forces with the surface through friction, and hand rotation need not have constant angular velocity; an Euler acceleration then enters too. The ideal free-ball drawing deliberately omits those effects. Neither manual supplies a numerical uncertainty or a precision measurement of the Coriolis acceleration.

The horizontal sea-ice drift momentum balance per unit area includes air and water drag forces, the Coriolis force, depth-integrated internal ice stress and the sea-surface pressure gradient. Here $m$ is ice-plus-snow mass per area and $\Sigma$ is a depth-integrated stress tensor.

// Target: geophysics.bigb

Sea-ice free drift neglects internal stress transmitted from neighboring ice. An isolated floe is governed by air and water drag forces, the Coriolis force and sea-surface tilt; steady drift also neglects acceleration.

// Target: geophysics.bigb

For steady flow of constant-density fluid in a uniformly rotating frame with conservative body forces, the displayed quantity is constant along each relative streamline. The centrifugal potential contributes the negative quadratic term, and the Coriolis force contributes no work because it is perpendicular to the relative velocity. The result follows by dotting the rotating Euler equations for an inviscid fluid with that velocity.

An incompressible Boussinesq fluid with uniform magnetic field, uniform vertical rotation and uniform temperature gradient supports waves combining magnetic tension, Coriolis force and buoyancy. Put $K=|\mathbf{k}|$, $h^2=K^2-k_z^2$, $a^2=(\mathbf{B}\cdot \mathbf{k}{)}^2/({\mu }_0\rho )$ and $H=g\alpha \beta$, with ${\partial }_z{T}_0=-\beta$. Eliminating pressure and the temperature and magnetic perturbations gives the displayed dispersion relation for nonzero frequencies. Stable thermal stratification has $H<0$. Clearing frequency denominators requires checking any resulting zero roots in the original equations; a stationary magnetostatic-buoyancy balance can exist independently of the propagating branches.

For a layer rotating at angular speed $\Omega$, of depth $d$ and kinematic viscosity $\nu$, the Taylor number compares the Coriolis force to viscous damping through the squared ratio $(2\Omega d^2/\nu {)}^2$. Definitions in other geometries may include different geometric factors; this is the plane-layer convention.

A projectile thrown across a rotating disk's diameter at speed $v$ relative to the thrower inherits the thrower's tangential velocity. During its approximate flight time $2a/v$, a receiver at the opposite rim moves tangentially in the opposite direction. The relative sideways displacement is consequently $4\omega a^2/v$ to leading order in $a\omega /v$. Integrating the Coriolis force in the disk's frame gives the same result.