Rotating Frames: Centrifugal and Coriolis Forces

Derivation

Rotating Frames: Centrifugal and Coriolis Forces

D-037 Home PU-101 Threads force · symmetry Depends on Galilean Invariance of Newton's Laws

Statement

If \(S\) is an inertial frame and \(R\) is a frame rotating relative to \(S\) with angular velocity \(\vec{\omega}(t)\) about a common origin, then a particle of mass \(m\) obeys, in the rotating frame, \(m\,\vec{a}_R = \vec{F} - 2m\,\vec{\omega}\times\vec{v}_R - m\,\vec{\omega}\times(\vec{\omega}\times\vec{r}) - m\,\dot{\vec{\omega}}\times\vec{r}\), where \(\vec{F}\) is the real (inertial) force and the three added terms are the Coriolis, centrifugal, and Euler fictitious forces.

Why it matters

Almost every laboratory sits on a rotating planet, and almost every convenient frame for describing turntables, centrifuges, gyroscopes, cyclones, and orbiting spacecraft is non-inertial. Rather than abandon Newton's second law, we keep its familiar form \(m\vec{a}=\vec{F}_{\text{eff}}\) by absorbing the frame's rotation into three velocity- and position-dependent inertial forces. This is the workhorse identity of geophysical fluid dynamics, rotordynamics, and ballistics.

The derivation also sharpens a conceptual point: fictitious forces are not "unreal" so much as frame-artifacts of insisting on \(F=ma\) in a frame where \(a\) is not the true inertial acceleration. Their appearance is dictated entirely by kinematics, not by any new physical interaction.

Assumptions

Newton's second law holds in the inertial frame \(S\): Without an inertial frame in which \(m\vec{a}_S=\vec{F}\) is exact, there is no anchor to transform from, and the fictitious forces have no reference against which to be defined.

The two frames share a common origin (pure rotation, no relative translation): If the origins accelerate relative to each other, an additional translational inertial force \(-m\vec{A}_0\) appears and must be added by hand; the pure-rotation result below is then incomplete.

Mass \(m\) is constant and speeds are non-relativistic: If \(m\) varies or \(v\sim c\), \(\vec{F}=d\vec{p}/dt\) no longer reduces to \(m\vec{a}\), and the clean split into three named forces breaks down.

The angular velocity \(\vec{\omega}(t)\) is a well-defined axial vector at each instant (the frame rotates rigidly): If different parts of the frame rotate about different axes (a non-rigid or shearing frame), a single \(\vec{\omega}\) cannot represent the motion and the operator identity \((d/dt)_S=(d/dt)_R+\vec{\omega}\times\) fails.

Derivation

\[ \left(\frac{d\hat{e}_i}{dt}\right)_S = \vec{\omega}\times\hat{e}_i \]

The orthonormal basis \(\{\hat{e}_i\}\) of \(R\) is carried rigidly by the rotation; each basis vector has fixed length, so its inertial rate of change is a pure rotation at angular velocity \(\vec{\omega}\). B

\[ \left(\frac{d\vec{A}}{dt}\right)_S = \left(\frac{d\vec{A}}{dt}\right)_R + \vec{\omega}\times\vec{A} \]

Write any vector as \(\vec{A}=\sum_i A_i\hat{e}_i\) and differentiate with the product rule: \(\sum_i \dot{A}_i\hat{e}_i\) is the rate an observer in \(R\) sees, and \(\sum_i A_i(\vec{\omega}\times\hat{e}_i)=\vec{\omega}\times\vec{A}\) uses step 1. This operator identity is the crux. B

\[ \vec{v}_S \equiv \left(\frac{d\vec{r}}{dt}\right)_S = \vec{v}_R + \vec{\omega}\times\vec{r} \]

Apply the operator identity to the position vector \(\vec{A}=\vec{r}\), with \(\vec{v}_R\equiv(d\vec{r}/dt)_R\) the velocity measured in the rotating frame. A

\[ \vec{a}_S = \left(\frac{d\vec{v}_S}{dt}\right)_S = \left(\frac{d}{dt}\right)_R\vec{v}_S + \vec{\omega}\times\vec{v}_S \]

Apply the same operator identity a second time, now to \(\vec{A}=\vec{v}_S\), to obtain the inertial acceleration. A

\[ \left(\frac{d}{dt}\right)_R\vec{v}_S = \vec{a}_R + \dot{\vec{\omega}}\times\vec{r} + \vec{\omega}\times\vec{v}_R \]

Substitute \(\vec{v}_S=\vec{v}_R+\vec{\omega}\times\vec{r}\) and differentiate term by term in \(R\): \((d\vec{v}_R/dt)_R=\vec{a}_R\), and \((d/dt)_R(\vec{\omega}\times\vec{r})=\dot{\vec{\omega}}\times\vec{r}+\vec{\omega}\times\vec{v}_R\). Note \(\dot{\vec{\omega}}\) has the same value in both frames since \(\vec{\omega}\times\vec{\omega}=0\). C

\[ \vec{\omega}\times\vec{v}_S = \vec{\omega}\times\vec{v}_R + \vec{\omega}\times(\vec{\omega}\times\vec{r}) \]

Substitute \(\vec{v}_S=\vec{v}_R+\vec{\omega}\times\vec{r}\) into the second term of step 4 and distribute the cross product. B

\[ \vec{a}_S = \vec{a}_R + 2\,\vec{\omega}\times\vec{v}_R + \vec{\omega}\times(\vec{\omega}\times\vec{r}) + \dot{\vec{\omega}}\times\vec{r} \]

Add steps 5 and 6; the two \(\vec{\omega}\times\vec{v}_R\) contributions combine into the factor of 2 that becomes the Coriolis term. A

\[ m\,\vec{a}_R = \vec{F} - 2m\,\vec{\omega}\times\vec{v}_R - m\,\vec{\omega}\times(\vec{\omega}\times\vec{r}) - m\,\dot{\vec{\omega}}\times\vec{r} \]

Impose Newton's law in the inertial frame, \(m\vec{a}_S=\vec{F}\) (assumption 1), solve step 7 for \(m\vec{a}_R\), and move the kinematic terms to the right-hand side where they read as forces. A

Result

\[ m\,\vec{a}_R = \vec{F} \;\underbrace{-\,2m\,\vec{\omega}\times\vec{v}_R}_{\text{Coriolis}} \;\underbrace{-\,m\,\vec{\omega}\times(\vec{\omega}\times\vec{r})}_{\text{centrifugal}} \;\underbrace{-\,m\,\dot{\vec{\omega}}\times\vec{r}}_{\text{Euler}} \]

Reading. In the rotating frame, Newton's second law keeps its form provided we add three fictitious forces. The Coriolis force \(-2m\vec{\omega}\times\vec{v}_R\) acts only on moving bodies and is perpendicular to their velocity, deflecting motion sideways. The centrifugal force \(-m\vec{\omega}\times(\vec{\omega}\times\vec{r})=m\omega^2\vec{\rho}\) points radially outward from the rotation axis (\(\vec{\rho}\) is the perpendicular vector from axis to particle) and depends only on position. The Euler force \(-m\dot{\vec{\omega}}\times\vec{r}\) appears only while the rotation rate changes.

Units check. Each term is a force: \(m[\text{kg}]\cdot\omega[\text{s}^{-1}]\cdot v[\text{m s}^{-1}] = \text{kg m s}^{-2}=\text{N}\) for Coriolis; \(m\cdot\omega^2[\text{s}^{-2}]\cdot r[\text{m}]=\text{N}\) for centrifugal; \(m\cdot\dot{\omega}[\text{s}^{-2}]\cdot r[\text{m}]=\text{N}\) for Euler. All balance the real force \(\vec{F}\) in newtons.

Limiting cases

\(\vec{\omega}=0\): all three terms vanish and \(m\vec{a}_R=\vec{F}\) recovers the inertial law — consistent with Galilean invariance.
\(\dot{\vec{\omega}}=0\) (steady rotation): the Euler force drops out, leaving only centrifugal and Coriolis; the usual turntable and planetary case.
\(\vec{v}_R=0\) (particle at rest in \(R\)): Coriolis vanishes; a body clamped to a steadily spinning platform feels only the outward centrifugal force.
Motion parallel to the axis (\(\vec{v}_R\parallel\vec{\omega}\)): \(\vec{\omega}\times\vec{v}_R=0\), so there is no Coriolis deflection for purely axial motion.
On the rotation axis (\(\vec{\rho}=0\)): centrifugal force vanishes; only Coriolis and Euler survive.

Breaks when

The origins accelerate relative to one another. If \(R\)'s origin has inertial acceleration \(\vec{A}_0\neq 0\) (e.g. a turntable mounted on an accelerating vehicle), an extra translational fictitious force \(-m\vec{A}_0\) must be added; the pure-rotation formula above is then incomplete.
Speeds approach \(c\) or \(m\) is not constant. The step \(\vec{F}=m\vec{a}_S\) fails; one must return to \(\vec{F}=d\vec{p}/dt\) with relativistic momentum, and the tidy three-force decomposition no longer holds.
The frame does not rotate rigidly about a single instantaneous axis. A shearing or deforming "frame" has no well-defined \(\vec{\omega}(t)\), so the operator identity of step 2 — the entire basis of the derivation — is invalid.

Failure modes

Dropping the factor of 2 in the Coriolis term by forgetting that \(\vec{\omega}\times\vec{v}_R\) appears in both step 5 and step 6.
Using inertial-frame velocity \(\vec{v}_S\) in the Coriolis term instead of the rotating-frame velocity \(\vec{v}_R\); the fictitious forces are written for the observer inside \(R\).
Treating centrifugal force as radial from the origin rather than from the rotation axis; only the component \(\vec{\rho}\) perpendicular to \(\vec{\omega}\) enters, since \(\vec{\omega}\times(\vec{\omega}\times\vec{r})=\vec{\omega}(\vec{\omega}\cdot\vec{r})-\omega^2\vec{r}\).
Applying centrifugal force in the inertial frame. Adding centrifugal force to a free-body diagram drawn in \(S\); there it does not exist — only the real centripetal force acts.
Forgetting the Euler term whenever \(\dot{\vec{\omega}}\neq 0\), e.g. during spin-up or spin-down of a centrifuge.
Sign errors in the cross products from mis-ordering factors; recall \(\vec{\omega}\times\vec{v}_R=-\vec{v}_R\times\vec{\omega}\).

Discussion

The three fictitious forces have sharply different physical characters. The centrifugal and Euler forces depend only on position (and on \(\vec{\omega}\), \(\dot{\vec{\omega}}\)) and act on every body whether moving or not; they behave like static force fields. The Coriolis force is qualitatively different: proportional to velocity and always perpendicular to it, it does no work and cannot change a particle's speed in the rotating frame — it only curves the trajectory. This velocity-dependence makes the Coriolis force the direct mechanical analogue of the magnetic Lorentz force \(q\vec{v}\times\vec{B}\), with \(2m\vec{\omega}\) playing the role of \(q\vec{B}\).

Geophysically this single term organizes the large-scale atmosphere and ocean. In the Northern Hemisphere the vertical component of Earth's \(\vec{\omega}\) deflects horizontal winds to the right; balanced against the pressure-gradient force it produces geostrophic flow, cyclonic (counter-clockwise) circulation around lows, and the precession of the Foucault pendulum. The relevant deflecting factor is \(2\omega\sin\lambda\) at latitude \(\lambda\), vanishing at the equator and maximal at the poles.

The centrifugal force reshapes effective gravity. On Earth it reduces the measured \(g\) by up to \(\omega^2 R\approx 0.034\ \text{m s}^{-2}\) at the equator and, integrated over geological time, flattens the planet into an oblate spheroid whose equatorial bulge is precisely the surface on which effective gravity is everywhere normal. The combination \(\vec{g}_{\text{eff}}=\vec{g}-\vec{\omega}\times(\vec{\omega}\times\vec{r})\) is what a plumb line actually points along.

At a deeper level the fictitious forces are the connection (Christoffel-symbol) terms that appear when Newton's law is written in a non-inertial coordinate system; they are geometry, not interaction. This is the Newtonian shadow of the equivalence principle: locally, a uniform gravitational field and a uniformly accelerated frame are indistinguishable, and the centrifugal potential \(-\tfrac{1}{2}m\omega^2\rho^2\) enters the Lagrangian on exactly the same footing as a genuine potential energy — which is why one can define effective gravity and rotating-frame "geopotential" surfaces at all.

Common misconceptions. Fictitious forces are not "fake" in the rotating frame — they produce real, measurable accelerations (cyclones, deflected projectiles, the bulge of the Earth) — but they have no reaction partner and no source body, and they vanish the instant one transforms back to an inertial frame. "Centrifugal" is the outward inertial force felt in \(R\); "centripetal" is the inward real force seen in \(S\). They are not an action-reaction pair.

Worked examples

Centrifugal reduction of \(g\) at the equator

A mass rests on the equator; find the outward centrifugal acceleration due to Earth's spin. Symbols first: \(a_{\text{cf}}=\omega^2 R\), since at the equator \(\vec{\rho}=\vec{R}\) is perpendicular to \(\vec{\omega}\). A

\[ \omega = \frac{2\pi}{86164\ \text{s}} = 7.292\times10^{-5}\ \text{s}^{-1},\qquad R = 6.371\times10^{6}\ \text{m} \]

Use the sidereal day (86164 s), not the solar day, because \(\omega\) is the rotation rate relative to the fixed stars — the inertial frame. B

\[ a_{\text{cf}} = (7.292\times10^{-5})^2 (6.371\times10^{6}) \]

Substitute numbers into \(a_{\text{cf}}=\omega^2 R\). A

\[ a_{\text{cf}} = 3.39\times10^{-2}\ \text{m s}^{-2} \]

Reading. The spin lightens an equatorial object by about \(0.034\ \text{m s}^{-2}\), roughly \(0.34\%\) of \(g\). Units check. \((\text{s}^{-1})^2\cdot\text{m}=\text{m s}^{-2}\), an acceleration.

Coriolis force on a northbound train

A train of mass \(m=1.0\times10^{4}\ \text{kg}\) moves due north at \(v=30\ \text{m s}^{-1}\) at latitude \(\lambda=45^\circ\)N. Find the horizontal Coriolis force. Symbols first: for horizontal motion the deflecting force has magnitude \(F_C=2m\omega v\sin\lambda\), directed to the right of \(\vec{v}_R\). B

\[ F_C = 2m\,\omega\, v\,\sin\lambda \]

Only the local-vertical component \(\omega\sin\lambda\) of \(\vec{\omega}\) drives horizontal deflection; the horizontal component gives a vertical (Eötvös) force, neglected here. C

\[ F_C = 2(1.0\times10^{4})(7.292\times10^{-5})(30)\sin 45^\circ \]

Substitute; \(\sin 45^\circ = 0.7071\). A

\[ F_C = 30.9\ \text{N}\quad\text{(directed eastward)} \]

Reading. A 10-tonne train pressing north is pushed sideways with only \(\sim 31\ \text{N}\) — tiny per unit mass (\(3.1\times10^{-3}\ \text{m s}^{-2}\)), but over continental scales this is what curves winds and ocean currents. Units check. \(\text{kg}\cdot\text{s}^{-1}\cdot\text{m s}^{-1}=\text{kg m s}^{-2}=\text{N}\).

Problems

A child of mass \(m=40\ \text{kg}\) sits \(r=2.0\ \text{m}\) from the axis of a merry-go-round spinning steadily at \(\omega=1.5\ \text{rad s}^{-1}\). What centrifugal force does the child feel in the rotating frame?
Solution
Steady rotation, so only centrifugal acts on the seated (\(\vec{v}_R=0\)) child: \(F_{\text{cf}}=m\omega^2 r = (40)(1.5)^2(2.0)=(40)(2.25)(2.0)=180\ \text{N}\), directed radially outward. This equals the inward (centripetal) real force the seat must supply.
A baseball of mass \(m=0.145\ \text{kg}\) is thrown horizontally at \(v=40\ \text{m s}^{-1}\) at latitude \(\lambda=51^\circ\)N. Find the horizontal Coriolis force and the resulting horizontal acceleration.
Solution
\(F_C=2m\omega v\sin\lambda\). With \(\omega=7.292\times10^{-5}\ \text{s}^{-1}\) and \(\sin 51^\circ=0.777\): \(F_C=2(0.145)(7.292\times10^{-5})(40)(0.777)=6.57\times10^{-4}\ \text{N}\). Acceleration \(a=F_C/m=2\omega v\sin\lambda=4.53\times10^{-3}\ \text{m s}^{-2}\), deflecting the ball to the right. Over a \(0.5\ \text{s}\) flight the sideways displacement is \(\tfrac12 a t^2\approx 5.7\times10^{-4}\ \text{m}\) — under a millimetre, hence negligible for baseball.
A turntable of radius \(0.50\ \text{m}\) starts from rest with constant angular acceleration \(\dot{\omega}=3.0\ \text{rad s}^{-2}\). A puck of mass \(m=0.20\ \text{kg}\) sits at the rim, momentarily at rest in the rotating frame at \(t=0\). What is the Euler force on it then, and how does it compare with the centrifugal force at that instant?
Solution
Euler force \(F_E=m\dot{\omega} r=(0.20)(3.0)(0.50)=0.30\ \text{N}\), directed tangentially opposite to the angular acceleration. At \(t=0\), \(\omega=0\), so the centrifugal force \(m\omega^2 r=0\). The Euler force dominates entirely during spin-up; as \(\omega\) grows the centrifugal term \(m\omega^2 r\) rises and overtakes it once \(\omega^2 r>\dot\omega r\), i.e. \(\omega>\sqrt{\dot\omega}=1.73\ \text{rad s}^{-1}\).
Estimate the reduction in effective gravity due to the centrifugal force at latitude \(\lambda=45^\circ\). Give the radially-outward (vertical) component of the centrifugal acceleration.
Solution
At latitude \(\lambda\) the perpendicular distance from the axis is \(\rho=R\cos\lambda\), so the centrifugal acceleration has magnitude \(\omega^2\rho=\omega^2 R\cos\lambda\), directed outward from the axis. Its component along the local vertical is \(\omega^2 R\cos^2\lambda\). With \(\omega^2 R=3.39\times10^{-2}\ \text{m s}^{-2}\) and \(\cos^2 45^\circ=0.5\): reduction \(=3.39\times10^{-2}\times0.5=1.7\times10^{-2}\ \text{m s}^{-2}\). So \(g_{\text{eff}}\approx 9.81-0.017=9.79\ \text{m s}^{-2}\) from this effect alone (the full latitude variation of \(g\) also includes the Earth's oblateness).
A Foucault pendulum precesses because of the Coriolis force, at angular rate \(\Omega=\omega\sin\lambda\). Derive the precession period and evaluate it at latitude \(\lambda=30^\circ\)N. Why does the pendulum not precess at the equator?
Solution
The horizontal Coriolis deflection makes the swing plane rotate slowly at rate \(\Omega=\omega\sin\lambda\) (the local-vertical component of \(\vec\omega\)). The precession period is \(T=\dfrac{2\pi}{\Omega}=\dfrac{2\pi}{\omega\sin\lambda}=\dfrac{T_{\text{day}}}{\sin\lambda}\), where \(T_{\text{day}}=2\pi/\omega=23.93\ \text{h}\) (sidereal). At \(\lambda=30^\circ\), \(\sin 30^\circ=0.5\), so \(T=23.93/0.5=47.9\ \text{h}\approx 48\ \text{h}\). At the equator \(\lambda=0\Rightarrow\sin\lambda=0\), so \(\Omega=0\): the axial component of \(\vec\omega\) is horizontal there, giving no rotation of the swing plane, and the precession period diverges — the pendulum does not precess.