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At radius $a$ in a ring rotating at rate $\omega$, a walker with signed tangential speed $u$ relative to the ring has inertial speed $a\omega +u$. The normal reaction per mass is $(a\omega +u{)}^2/a$; the curvature of the relative path matters as well as the centrifugal force and the Coriolis force.

A stationary mass density disturbance can be balanced by a changed azimuthal velocity in a shearing sheet. The continuity and azimuthal equations give zero radial velocity, while the radial equation balances Coriolis force against pressure and gravity. This mode is separate from the density-wave pair in the softened Toomre dispersion relation and is lost if one divides every equation by frequency.

An unwanted ellipse is an experimental warning: a finite-amplitude pendulum can develop additional orbit-axis precession even without Earth's rotation. A symmetrical, firm support and careful launch reduce this effect.^[ref] Distinguish that nonlinear effect from the tiny ellipse already present in the Exact linear Foucault pendulum trajectory. The linear calculation establishes a controlled approximation, not the complete motion of an arbitrary large swing.

Support anisotropy creates a second competing mechanism. As an illustrative modification, let the two horizontal restoring angular frequencies be ${\omega }_x$ and ${\omega }_y$. Even with $\alpha =0$, a diagonal release from rest givesWhen ${\omega }_x\ne {\omega }_y$, the relative phase changes on the scale $1/|{\omega }_x-{\omega }_y|$: the trace changes shape without any terrestrial Coriolis force. It need not be uniform precession of a Foucault pendulum. To cleanly resolve the terrestrial effect over times of order $1/|\alpha |$, a useful design target is $|{\omega }_x-{\omega }_y|\ll |\alpha |$. Near the equator that target becomes particularly demanding. Compare fast periods along perpendicular launch directions to look for this frequency splitting; repeat at several launch orientations to detect preferred suspension axes.

Other controls address distinct alternatives: shelter the apparatus from air currents, track the support against the building, measure both orbit axes, and compare early and late fitted directions as the amplitude decays. A sustaining drive, if used in a different apparatus, must supply mechanical energy without imposing an azimuthal bias; an undriven run provides a useful comparison. A change in swing direction alone cannot exclude these effects. Agreement of the signed rate with independently known geographical latitude and Earth's rotation, together with stable repeat launches and small competing effects, is the inference the experiment is designed to support.

The Coriolis acceleration is the Coriolis force divided by the particle's mass:It is therefore independent of mass for a specified relative velocity and angular velocity. This is one contribution to the acceleration measured in a rotating reference frame. For a fixed common origin with an inertial frame and constant $\mathbf{\Omega }$, the full equation of motion in a rotating frame isAll vectors here are expressed in the rotating axes; ${\mathbf{F}}_{\mathrm{real}}$ is the sum of actual forces. An accelerating origin adds a translational apparent force, and time-dependent angular velocity also adds the Euler acceleration.

The factor two follows directly from the rotating-frame derivative formula. First ${\mathbf{v}}_{\mathrm{lab}}={\dot{\mathbf{r}}}_{\mathrm{rot}}+\mathbf{\Omega }\times \mathbf{r}$. Differentiate again in the inertial frame: differentiating ${\dot{\mathbf{r}}}_{\mathrm{rot}}$ contributes one $\mathbf{\Omega }\times {\dot{\mathbf{r}}}_{\mathrm{rot}}$, and differentiating $\mathbf{\Omega }\times \mathbf{r}$ contributes another. Move both terms to the rotating side of Newton's second law to obtain the minus sign and the factor two. The centrifugal acceleration comes from the remaining double cross product.

The Coriolis force is the apparent force proportional to a particle's velocity relative to a rotating reference frame. For constant angular velocity $\mathbf{\Omega }$ and particle mass $m$,Here $\mathbf{v}$ in the title formula means ${\mathbf{v}}_{\mathrm{rot}}$, not the velocity measured in an inertial frame. The Coriolis force belongs to the chosen reference frame: a particle stationary in that rotating reference frame has no Coriolis force, even though it can have a nonzero velocity in the laboratory. Other reference frames assign different angular velocities and relative velocities, and hence different Coriolis forces to the same motion.

The cross product makes the Coriolis force perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}_{\mathrm{rot}}$, with magnitude $2m\Omega |{\mathbf{v}}_{\mathrm{rot}}|\sin \alpha$, where $\alpha$ is the angle between them. Apply the right-hand rule to $\mathbf{\Omega }\times {\mathbf{v}}_{\mathrm{rot}}$ and then reverse its direction. Motion parallel to the rotation axis gives zero Coriolis force; reversing the relative velocity reverses the Coriolis force.

The appeal of this description is practical: axes attached to the Earth or a turning apparatus can be kept fixed on the page, provided Newton's second law includes the extra terms. A free particle in rotating coordinates shows exactly how a straight laboratory trajectory can become curved, while Coriolis acceleration in local Earth coordinates gives a useful terrestrial scale. The rotating-table demonstration of Coriolis deflection makes the two viewpoints visible in one experiment.

The instantaneous power of the Coriolis force, evaluated with the velocity in the same rotating reference frame, vanishes:This follows from the perpendicularity of the cross product and the dot product definition of power. The Coriolis force alone does no work and cannot change the rotating-frame kinetic energy $m|{\mathbf{v}}_{\mathrm{rot}}{|}^2/2$. It can change the direction of the relative velocity. Actual forces and the centrifugal force can still change that kinetic energy, as the free particle in rotating coordinates illustrates. Conservation of an appropriate total energy has the additional conditions stated under energy balance in a uniformly rotating frame.

For a zero-mean magnetic shearing wave, the real wavevector satisfies $k_x=k_{x0}+2A{k}_{y}t$, with $k_y,k_z$ constant. Dotting the velocity-amplitude equation with its complex conjugate eliminates Coriolis force and pressure work, giving $d|\tilde{\mathbf{v}}{|}^2/dt=4A\mathrm{Re}({\tilde{v}}_x{\tilde{v}}_y^{\ast })-2\nu k^2|\tilde{\mathbf{v}}{|}^2$. The inequality $2|v_x{v}_y|\le |\mathbf{v}{|}^2$ proves the displayed bound; the magnetic bound has the same form with $\eta$ replacing $\nu$. NowFor $A{k}_y\ne 0$ and positive kinematic viscosity and magnetic diffusivity, the negative cubic exponent dominates the linear shear-work bound, so both amplitudes tend to zero. Transient growth is allowed. Positivity matters: an inviscid vertical velocity mode with $k_z=0$ can remain constant, and an ideal vertical magnetic mode with $k_z=0$ can also remain constant.

For a rotating conducting sphere, a geostrophic azimuthal velocity $U=rV$ couples to the cylinder-integrated magnetic stress $T=\int B_r{B}_{\varphi }dz$. Put $H=\int B_r^{2}dz$ and let $2z_b$ be the cylinder height. With an insulating boundary and no flow through it, cylindrical averaging removes the Coriolis force and gives $U_t=(2z_b{\mu }_0\rho r^2{)}^{-1}{\partial }_r(r^{2}T)$, $T_t=Hr{\partial }_r(U/r)$. The local squared wave speed is $H/(2z_b{\mu }_0\rho )$, the cylinder average of $B_r^2/({\mu }_0\rho )$.

The rotating-frame effective potential of the circular restricted three-body problem has negative singular wells at both primaries. Its three Collinear Lagrange points are planar saddles, and its two Triangular Lagrange points are planar local maxima for nonzero primary masses. These geometric labels do not replace a dynamical stability analysis including the Coriolis force.

The ratio of viscous force to Coriolis force in a rotating fluid at length scale $L$. This convention uses rotation frequency $2\Omega$; conventions omitting the factor two are also common. For small $E$, bulk viscosity is weak but no-slip boundaries can form Ekman layers of thickness of order $L\sqrt{E}$, with relatively larger boundary shear stress.

For uniform angular velocity $\omega$, potential $\varphi$, and additional forces doing no work in the rotating frame,The Coriolis force does no work; the centrifugal potential accounts for the remaining apparent force. This energy is conserved precisely when the potential has no explicit time dependence in the rotating coordinates. An inertial time-independent conservative force need not meet that condition unless its potential is also invariant under the frame's rotation. The balance prevents applying a rotating-frame conservation law to a nonaxisymmetric stationary inertial potential.

An equatorial inertia--gravity wave is a high-frequency equatorially trapped mode restored jointly by buoyancy and the Coriolis force. Its dispersion relation has eastward and westward branches lying outside the low-frequency equatorial Rossby wave branch.

For release from rest relative to the building, take $x(0)=A>0$, $y(0)=\dot{x}(0)=\dot{y}(0)=0$. The exact solution of the stated coupled linear differential equations isIndeed $u(0)=A$ and $\dot{u}(0)=i\alpha A$, so $\dot{z}(0)=0$. In the slowly rotating coordinates, the ellipse has semiaxes $A$ and $A|\alpha |/\beta$. A bob released from rest therefore does not describe a perfectly straight line even in this ideal linear model. Replacing the bracket by $\cos {\omega }_{0}t$ gives the familiar leading approximation but introduces a spurious initial transverse velocity $-i\alpha A$. The exact expression keeps the initial condition and separates that tiny orbit width from the slow rotation of its long axis.

For the illustrative $\ell =20\mathrm{m}$ and $g=9.81m/{\mathrm{s}}^2$, the fast period $2\pi /\beta$ is approximately $8.9714\mathrm{s}$. At $45^{\circ }$ north, $|\alpha |/\beta \simeq 7.3624\times 10^{-5}$: the transverse semiaxis is too small to distinguish at the figure's scale. Nevertheless the long-axis orientation changes by about $-63.8138^{\circ }$ in six hours. Changing the sign of geographical latitude conjugates the entire solution, reflecting the trajectory across the eastward axis; setting $\alpha =0$ gives $z=A\cos {\omega }_{0}t$ exactly. These are stronger checks than merely drawing a rotating line.

The coupled equations give an additional independent constraint. Multiplying them by $\dot{x}$ and $\dot{y}$ and adding cancels the Coriolis force's contributions:This is mechanical energy per unit mass in the reduced model; precession does not require a loss of mechanical energy. Also, direct differentiation givesFor this release the constant is $\alpha A^2$. The exact trajectory satisfies both identities. They test the coupled dynamics separately from the visible slow rotation and express the fact that the Coriolis force does no work.

An inertial force is an additional term used to express Newton's second law in an accelerating reference frame. For origin acceleration $\mathbf{A}$ and axes with angular velocity $\mathbf{\Omega }$, the apparent force on a particle isHere $\mathbf{r}$ and ${\mathbf{v}}^{\prime }$ are measured relative to that frame. The rotation terms are the Coriolis force, Euler force and centrifugal force. Their values depend on the frame rather than on an additional physical interaction.

Consider a ball moving horizontally with no net horizontal force in the laboratory. Treat its horizontal motion as that of a free particle: ${\mathbf{r}}_{\mathrm{lab}}=(ut,0)$ with constant $u>0$. Axes rotating counterclockwise at constant angular velocity $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$ about the launch point assign it the positionThis is an exact rotation matrix transformation, with the axes coinciding at $t=0$. Taking $u=0.25\mathrm{m}{\mathrm{s}}^{-1}$, $\Omega =0.8\mathrm{rad}{\mathrm{s}}^{-1}$, and $0\le t\le 3\mathrm{s}$ gives the two views below. At $t=3\mathrm{s}$, the laboratory position is $(0.75,0)\mathrm{m}$, but the rotating position is approximately $(-0.5530,-0.5066)\mathrm{m}$. Both assign the same distance $0.75\mathrm{m}$ from the origin.

The entire rotating acceleration must include the centrifugal acceleration as well as the Coriolis acceleration. To see the balance explicitly, write $c=\cos \Omega t$ and $s=\sin \Omega t$. Direct differentiation givesThe two apparent contributions areAt launch $\mathbf{r}=0$, so only the Coriolis acceleration contributes to the initial bending. Later the centrifugal acceleration also contributes; the full curved trajectory cannot be explained by keeping only the Coriolis force. In particular, $|{\mathbf{v}}_{\mathrm{rot}}{|}^2=u^2(1+{\Omega }^2{t}^2)$ grows even though the Coriolis force does no work: the rotating-frame kinetic energy increase is supplied by work from the centrifugal force. The laboratory kinetic energy remains $m{u}^2/2$.

The geostrophic wind is the horizontal flow for which the pressure gradient force balances the Coriolis force. It runs along atmospheric isobars and has speed proportional to their pressure gradient, at fixed density and nonzero Coriolis parameter. Curvature, friction and small spatial scales can invalidate that approximation.

The gradient wind retains the centripetal acceleration of a curved flow as well as the Coriolis force and horizontal pressure gradient force. For cyclonic circular flow in the northern hemisphere, $v^2/r+fv={\rho }^{-1}dp/dr$, with $r$ increasing away from the low-pressure centre.

The free-drift velocity of an iceberg follows from air and water drag, the Coriolis force, pressure gradients, currents and other external forces. Large submerged area and mass make its wind response weak compared with that of a thin floe.

For particles with the same charge-to-mass ratio, rotationally invariant mutual central forces, and a weak uniform magnetic field, a frame with the displayed angular velocity cancels the velocity-dependent magnetic Lorentz force against the Coriolis force. The remaining rotational and magnetic terms are quadratic in $\mathbf{B}$. Thus, to first order in the field, the rotating-frame equations coincide with the zero-field equations. This is the classical dynamical basis of Larmor precession; equality is an approximation rather than an exact removal of the magnetic field.

Use local east, north, and up unit vectors, with $x$ eastward, $y$ northward, and positive orientation $\psi$ anticlockwise as viewed from above. Let $\phi$ be signed geographical latitude, positive north, and $\Omega >0$ the magnitude of Earth's angular velocity. In the spherical-Earth approximation,Here $g$ is the local effective gravity in a rotating frame, including the steady terrestrial centrifugal acceleration. For small horizontal displacements compared with $\ell$, the horizontal restoring acceleration from tension is $-{\omega }_0^2(x,y)$. Vertical displacement is second order in the swing amplitude, so vertical velocity can be omitted at this order. The horizontal Coriolis acceleration in local Earth coordinates then gives the coupled linear differential equationsThese equations neglect damping, finite-amplitude corrections, support anisotropy, and spatial changes of the centrifugal force of order ${\Omega }^2$; they retain the leading terrestrial Coriolis force. A related derivation uses south/east axes instead.^[ref]

Introduce the complex number $z=x+iy$. Multiplying the second equation by $i$ and adding givesSet $z=e^{-i\alpha t}u$. Direct differentiation cancels the first derivative of $u$ and yieldsThe ${\alpha }^2$ in $\beta$ comes from solving these reduced equations exactly; it is not a complete prediction of terrestrial corrections of order ${\Omega }^2$. Thus the horizontal trajectory is an ellipse in coordinates rotating at angular velocity $-\alpha$ relative to the building. When $|\alpha |\ll {\omega }_0$, each rapid swing samples an almost fixed long axis; its slow precession isThe sidereal day $T_{\mathrm{sid}}=2\pi /\Omega$ is a rotation relative to distant directions. $T_{\mathrm{p}}$ denotes a full $2\pi$ turn of a continuously followed orientation; an unlabelled swing line repeats after $T_{\mathrm{p}}/2$. At the equator there is no finite precession period in this model. The Coriolis parameter is $f=2\alpha$, so the plane's rate is $-f/2$, not $-f$.

The limiting cases check the sign and scale. At the north pole $\alpha =\Omega$: the swing direction turns clockwise once per sidereal day. At the equator $\alpha =0$: both horizontal equations reduce to independent simple harmonic motions and a line launch stays on its original line. At the south pole $\alpha =-\Omega$: the direction turns anticlockwise once per sidereal day. Vanishing horizontal precession at the equator does not mean that every component of Coriolis acceleration vanishes there.

For a stipulated $\Omega =7.292115\times 10^{-5}{\mathrm{s}}^{-1}$, the sidereal day is approximately $23.9345\mathrm{h}$. At $45^{\circ }$ north the predicted rate is $-10.6356^{\circ }/\mathrm{h}$ and $T_{\mathrm{p}}\simeq 33.8485\mathrm{h}$; at $45^{\circ }$ south the rate changes sign. At the illustrative geographical latitude $34^{\circ }$ south, the prediction is $+8.4109^{\circ }/\mathrm{h}$. These calculations are model predictions, not new observations of the UNSW apparatus.

A swing plane fixed among the stars is generally only a polar picture. A north–south swing at the equator turns with Earth in an inertial frame, while having no precession against the building.^[ref]