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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 mechanism in which rotating weights move outward as angular speed increases, allowing their displacement to indicate or regulate a machine's speed. The steady configuration follows the balance of actual applied forces and centripetal acceleration, or equivalently of applied forces and centrifugal force in a rotating reference frame. Geometry and the restoring mechanism determine the speed-displacement relation.

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 frame rotating with constant angular velocity $\mathbf{\omega }$,The extra terms produce the Coriolis and centrifugal forces in the rotating-frame equation.

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 $m{u}^2/2$.

Two positive point masses $M_1,M_2$ separated by $D$ have a unique intervening gravitational field zero. At distance $x$ from $M_1$, Newton's law of universal gravitation gives $M_1/x^2=M_2/(D-x{)}^2$, hence the displayed value. The signed offset towards $M_2$ from their centre of mass is $x-D{M}_2/(M_1+M_2)$. This instantaneous cancellation is distinct from a Lagrange point, which includes the centrifugal force of a co-rotating description. Along the joining line the zero is unstable: the derivative of the signed acceleration is $2G{M}_1/x^3+2G{M}_2/(D-x{)}^3>0$.

A Lagrange point is an equilibrium of the circular restricted three-body problem in the frame rotating with the two massive bodies.

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]

Start in an inertial frame with the inviscid magnetohydrodynamic momentum equation,Here $\rho$ is mass density, $P$ is the isotropic fluid pressure, $\mathbf{v}$ is inertial-frame velocity, $\mathbf{J}$ is electric current density, $\mathbf{B}$ is the magnetic field, and $\Phi$ is gravitational potential per unit mass. In the low-frequency approximation to Ampère's law, $\mathbf{J}=\nabla \times \mathbf{B}/{\mu }_0$. The vector calculus identityseparates the Lorentz force density into magnetic tension and a magnetic pressure gradient. Transform the material acceleration to a rigid frame rotating with constant angular velocity ${\mathbf{\Omega }}_0$, using the stated frame transformation. The rotating-frame magnetohydrodynamic momentum equation isThe final terms are the Coriolis force and centrifugal force densities; $\mathbf{u}$ is the velocity relative to the rotating frame, and $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ uses that frame.

This requires a smooth conducting-fluid continuum with nonrelativistic motion, a constant magnetic permeability ${\mu }_0$, the solenoidal magnetic-field constraint, a negligible displacement current and negligible net electric-charge force. Viscous forces and anisotropic fluid stresses are omitted. A time-dependent frame would add an Euler force. Incompressible flow is not needed for this force equation. Finite electrical resistivity does not change its displayed instantaneous magnetic force; the later frozen-field calculation additionally uses ideal magnetohydrodynamics. A differentially rotating fluid does not define a single rigid frame with spatially varying ${\mathbf{\Omega }}_0$: choose a fixed local reference rate and retain the fluid's shear in $\mathbf{u}$.

In an inertial frame, a circular orbit at radius $R$ has radial acceleration $-R\Omega (R{)}^2{\mathbf{e}}_R$. Neglecting fluid and magnetic forces givesIn a local rotating frame with ${\Omega }_0=\Omega (R_0)$, the fluid is instantaneously at rest at the reference radius $R_0$. The radial centrifugal force $+\rho R_0{\Omega }_0^2$ then balances the inward gravitational force. Away from $R_0$, the fluid's nonzero rotating-frame velocity must also be retained; the entire differentially rotating disk is not at rest in that one frame.

All time derivatives in this part are taken relative to the rotating reference frame, so its basis is held fixed during differentiation. Put $f(\varphi )=p(\varphi )\sin \varphi$ and $q(\varphi )=p(\varphi )\cos \varphi$. The chain rule and product rule giveExplicitly,The separate Coriolis force and centrifugal force terms in the stated equation account for the motion of the basis relative to an inertial observer.

The familiar result follows if the force laws have Galilean invariance; the three abstract Newton's laws of motion do not guarantee that invariance. Isaac Newton's Corollary V correctly identifies a central feature of ordinary Newtonian mechanics, but its general physical scope requires assumptions about interactions as well as about the relation between force and acceleration.^[ref]

Take a constant velocity $\mathbf{U}$ and a Galilean transformationThen ${\ddot{\mathbf{x}}}_i^{\prime }={\ddot{\mathbf{x}}}_i$, ${\mathbf{x}}_i^{\prime }-{\mathbf{x}}_j^{\prime }={\mathbf{x}}_i-{\mathbf{x}}_j$, and ${\dot{\mathbf{x}}}_i^{\prime }-{\dot{\mathbf{x}}}_j^{\prime }={\dot{\mathbf{x}}}_i-{\dot{\mathbf{x}}}_j$. Any interaction depending only on these relative quantities, with unchanged parameters, supplies the same forces for the transformed initial data. Newton's second law consequently gives the same relative accelerations; uniqueness of the resulting initial value problem supplies the same relative histories. An inverse-square central interaction is a standard example. Newton's third law helps establish the common centre of mass result, but the boost argument principally needs invariant interactions and the second law.

Here is a counterexample to deduction from the three laws alone. Give two equal masses a preferred inertial frame and write $\mathbf{r}={\mathbf{x}}_1-{\mathbf{x}}_2$, $\mathbf{V}=({\dot{\mathbf{x}}}_1+{\dot{\mathbf{x}}}_2)/2$. For fixed $k>0$ and speed scale $u_{\ast }>0$, specifyThe forces are equal, opposite and central, the centre of mass is inertial, and each body obeys $m{\ddot{\mathbf{x}}}_i={\mathbf{F}}_i$. An elementary body with no force is inertial. Nevertheless,Boosting an isolated pair from $\mathbf{V}=0$ to $\mathbf{V}=\mathbf{U}$ changes its internal oscillation frequency. All three motion laws hold, but the proposed universal boost equivalence fails. The extra preferred structure in the force law is what makes the failure possible.

A passive coordinate transformation of this same model remains possible: transform the preferred frame's description too. That mathematical freedom does not establish an active dynamical symmetry which changes the physical pair's velocity relative to the preferred structure while holding the laws' background data fixed. Confusing those operations would hide the counterexample.

Finally, a change to rotating coordinates introduces centrifugal forces and Coriolis forces; their dependence on position or velocity can change internal experiments. Constant translational velocity has no analogous effect when the interactions have Galilean invariance. Corollary V expresses that physical symmetry, with the interaction assumption made explicit, rather than a consequence of action–reaction alone.

There are two different issues: whether a law follows logically from the other laws, and whether an experiment can test it without assuming anything else. Newton's laws of motion have empirical content when combined with rules for identifying inertial frames, measuring mass and specifying forces. An experiment tests predictions of that combination. Calling every observed acceleration an otherwise unspecified force would remove the empirical constraint rather than verify Newton's second law. The distinction is central to the empirical testing of Newton's laws of motion.

Newton's first law contributes the claim that a common family of inertial frames exists: bodies with zero net external force have uniform straight motion in those frames. It is not enough to choose coordinates which straighten the trajectory of just one body; an admissible frame must accommodate many independent bodies and interactions. Once that frame and the ordinary constant-positive-mass form of Newton's second law are assumed,This conditional deduction does not establish the existence of the required frame, or independently identify the physical circumstances in which the net force vanishes. Defining an inertial frame and then counting a straight trajectory as proof of the definition would not be a test.

Nevertheless, the law of inertia has testable implications in a network of assumptions. Consider low-friction motion along an air track: determine positions at measured times, reduce the known residual influences, and ask whether the remaining acceleration approaches zero in the same laboratory frame for different bodies. Rotating the frame changes the description through Coriolis forces and centrifugal forces; this dependence must be accounted for. No finite experiment establishes that every unmeasured influence is absent, but varying the apparatus can discriminate an interaction from a frame effect. A failed prediction is evidence against some part of the package, not automatically the isolated verbal statement of Newton's first law.

Newton's second law is more than a definition if the force law and mass assignments constrain predictions before the new acceleration is measured. For instance, independently characterize a spring's reproducible extension response, specify the same interaction in several configurations, and compare the resulting accelerations for different bodies. If its strength and every mass are freely refitted to each outcome, agreement is empty. If they are fixed by an independently checked model and calibration, the law predicts held-out trajectories. Such a test remains conditional on the spring model, initial conditions and frame identification; this is an instance of the Duhem problem, not a reason to deny all empirical content.

Newton's third law supplies a different constraint. For two bodies exchanging forces, Newton's second law giveswhen there is no net external force. Equal-and-opposite interaction forces imply conservation of momentum for the pair. The two second-law equations alone do not require that cancellation: assigning ${\mathbf{F}}_{12}=\mathbf{f}$ and ${\mathbf{F}}_{21}=-2\mathbf{f}$ is algebraically consistent with both second-law equations while violating the third law. This is a logical countermodel, not a proposed isolated physical interaction. Measuring paired impulses or checking the pair's total momentum therefore tests an additional constraint, conditional on the assumed absence of momentum exchange with anything else.

The system boundary matters. In electromagnetism, matter can exchange momentum with the electromagnetic field. Instantaneous mechanical forces between two particles need not satisfy the simple pairwise statement, although total matter-plus-field momentum is conserved under the appropriate closed-system conditions. Likewise, Noether theorem can derive conservation of momentum from a translationally invariant action; that uses an extra symmetry assumption and does not derive the third law from the second alone.

Thus the first law is not an additional differential consequence once the full second-law framework is fixed, and the third law is not a consequence of that framework. All three can contribute to empirical tests, but none is tested in total isolation from frame, interaction and measurement assumptions. This preserves the difference between logical dependence and experimental testability.^[ref]

For the uniform unloaded space-elevator cable, let the constant linear mass density be $\lambda$ and the outer endpoint be $b={\eta }^{\prime }R$. In the co-rotating frame the outward effective force per unit length is $\lambda ({\omega }^{2}r-G{M}_E/r^2)$. The cable is just clear of the surface, with neither a ground load nor an endpoint counterweight, so both ends have zero applied tension. The integrated effective force must vanish:Using $G{M}_E={\omega }^2{\eta }^3{R}^3$, this becomesThe root ${\eta }^{\prime }=1$ is a zero-length cable. The physical nonzero cable hasIt extends well beyond the geostationary orbit so its outer centrifugal force balances the inner segment's weight. If a cable is anchored with a nonzero endpoint load or has a counterweight, that load changes the integrated balance; its length is not fixed by the zero-end-load formula.

Let $x$ be the signed distance from the Earth's centre along the polar tunnel, with $x(0)=r_0$ and $\dot{x}(0)=0$. The shell theorem leaves only the enclosed mass $4\pi \rho |x{|}^3/3$ contributing to the gravitational field. ConsequentlyThis is simple harmonic motion with $\omega =\sqrt{g/r_0}$ and $x=r_0\cos (\omega t)$. Crossing the centre takes a quarter-period; reaching the opposite surface takes a half-period:The body's mass cancels. The polar diameter lies on the rotation axis, so the tunnel motion has no additional centrifugal force or Coriolis force from the Earth's rotation. A realistic radial density profile would change the harmonic equation.

Let a frame's origin have inertial acceleration ${\mathbf{a}}_0$ and its axes rotate with angular velocity $\mathbf{\Omega }(t)$. For position $\mathbf{r}$ and velocity ${\mathbf{v}}^{\prime }$ measured in that frame, Newton's law becomesThe extra terms are translational, Coriolis, Euler and centrifugal inertial forces. Centrifugal force points away from the rotation axis; Coriolis force deflects moving bodies sideways and does no instantaneous work on their relative motion, since it is perpendicular to ${\mathbf{v}}^{\prime }$. The Euler term appears when the rotation rate or axis changes.

For constant $\mathbf{\Omega }$, centrifugal force has potential $U_{\mathrm{cf}}=-m|\mathbf{\Omega }\times \mathbf{r}{|}^2/2$. It can be combined with a conservative real potential to find rotating-frame equilibria. The Coriolis term remains velocity dependent. These forces arise from describing acceleration in a noninertial frame; they are not additional interactions between bodies. They are essential for trajectories on rotating apparatus and for large-scale terrestrial motions. Frame-dependent terms must be included consistently with the chosen coordinates.

A fictitious force is a term required when writing Newton's second law in an accelerating frame; it comes from the coordinate choice rather than a new interaction. For axes rotating at constant $\mathbf{\Omega }$ about a fixed inertial origin,Apply this identity twice to position to get ${\mathbf{a}}_I={\mathbf{a}}_R+2\mathbf{\Omega }\times {\mathbf{v}}_R+\mathbf{\Omega }\times (\mathbf{\Omega }\times \mathbf{r})$. Moving the last two terms to the right of $m{\mathbf{a}}_I=\mathbf{F}$ gives the stated Coriolis and centrifugal forces. An accelerating origin or changing rotation rate would add further terms.

A slowly rotating fluid planet follows an equipotential of gravitational plus centrifugal potential. The centrifugal potential is $-{\Omega }^2{r}^2{\cos }^2\lambda /2$, most negative at the equator. An outward equatorial shift raises gravitational potential and restores the balance, producing an oblate figure. Self-gravity also changes with the shape; no flattening coefficient follows from rotation alone without a density model.

Using local north and radial-up unit vectors, the centrifugal acceleration on a stationary mass isIt points away from the rotation axis: upward and equatorward, at angle $\lambda$ to radial up in the northern hemisphere. Its vertical fraction of gravitational force isusing a $24\mathrm{h}$ day for this approximate number. The resultant effective weight tilts equatorward relative to radial down. A plumb-defined local vertical already follows effective gravity; the component calculation uses the spherical radial vertical. A supported stationary mass of course has zero net acceleration.

For the Coriolis deflection of vertical launch and drop, retain only first order in $\Omega$, use effective constant $g$, and neglect horizontal velocities in the Coriolis term as instructed. With east positive, vertical velocity gives $\ddot{x}=-2\Omega \cos \lambda \dot{z}$ and no northward acceleration. For launch, $z=ut-g{t}^2/2$ and the flight time is $2u/g$. Integrating with zero initial horizontal velocity givesThe landing is westward. The westward velocity accumulated on ascent is only recovered on descent, so the two displacement contributions do not cancel.

For release from rest at $h_0=u^2/(2g)$, $z=h_0-g{t}^2/2$, $t_{\mathrm{fall}}=u/g$, and $\ddot{x}=2\Omega gt\cos \lambda$. ThusThe drop lands eastward with one quarter the launch-return displacement magnitude. North–south corrections from horizontal Coriolis coupling are beyond the retained approximation.

A fictitious force accounts for acceleration of the chosen reference frame rather than an additional interaction. For axes rotating at constant $\mathbf{\Omega }$, differentiating a vector uses $(d/dt{)}_{\mathrm{lab}}=(d/dt{)}_{\mathrm{rot}}+\mathbf{\Omega }\times$. Twice applying this givesso moving the last two terms to the force side produces the Coriolis and centrifugal forces. Here $\mathbf{v}$ is measured in rotating axes.

The rotating saddle trap has potential force $(-m{\omega }^{2}x,+m{\omega }^{2}y)$. ThereforeThe trial solution givesBoth modes are bounded for arbitrary sufficiently small initial displacements and velocities precisely when $|\Omega |>\omega$. At equality there is a zero-frequency secular mode; the exactly stationary solution does not establish stability.

For rapidly rotating saddle confinement with $\Omega \gg \omega >0$, real mode ellipses can be writtenThey are nearly circular clockwise motions at rates ${ϵ}_{\pm }\simeq \Omega \pm {\omega }^2/(2\Omega )$. Rotating these axes back to the laboratory cancels the large angular rate: the leading laboratory motion is a slow approximately harmonic ellipse with frequency ${\omega }^2/(2\Omega )$, plus small rapid micromotion. It results from the changing saddle force, not a static potential minimum. At next order the ellipse precesses prograde at ${\omega }^4/(8{\Omega }^3)$. The rotating-potential model and this guiding-centre distinction are analysed in the primary work cited here.^[ref]

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.