Centrifugal force

Your washing machine's spin cycle turns the drum at 1200 revolutions per minute. At the drum wall, 24 centimetres from the axis, the clothes are pressed outward with an apparent gravity of roughly 390 times g — water is wrung straight through the fabric and out the perforations. Ask a physicist what force does the pressing and you get a curious answer: officially, none. In the inertial frame the drum wall pushes the clothes inward, forcing them around a circle they would otherwise leave along a straight tangent line. Yet every centrifuge designer, every astronaut trainer, every engineer who has ever sized a flywheel does the calculation with an outward force $m\omega^2\rho$ — and gets the right answer every time. Both descriptions are correct, because they are bookkeeping for the same physics in two different frames. The moment you insist on writing Newton's $\mathbf{F}=m\mathbf{a}$ inside the rotating drum — where the clothes sit still — you must add the position-dependent fictitious force $-m\,\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ : the centrifugal force. It points directly away from the rotation axis, grows linearly with distance from it, acts on moving and stationary bodies alike, and — unlike its velocity-dependent sibling, the Coriolis force — it is conservative, flowing downhill from the tidy potential $-\tfrac12 m\omega^2\rho^2$ . That last property is what makes it a workhorse. The same potential that wrings your laundry holds spinning buckets of water in parabolic sheets, sets the five Lagrange equilibrium points of the restricted three-body problem, subtracts $0.34\%$ from your weight at the equator, and — over geological time — squashed the Earth itself into its oblate shape.

Key takeaways

The centrifugal force is the fictitious force $-m\,\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ that appears in Newton's second law when it is written in a uniformly rotating frame.
It points radially outward from the rotation axis with magnitude $m\omega^2\rho$ , where $\rho$ is the perpendicular distance to the axis — position-dependent, unlike the velocity-dependent Coriolis force.
It is one of three frame-fictitious terms — alongside the Coriolis force and the Euler force — generated by transforming the acceleration into a rotating frame.
It acts on stationary as well as moving bodies and is derivable from the potential $U_{\mathrm{cf}}=-\tfrac12 m\omega^2\rho^2$ , so it folds neatly into effective-potential-energy arguments.
On Earth it reduces measured gravity at the equator by about $0.34\%$ and tilts the local vertical; in centrifuges it manufactures artificial gravity thousands of times stronger than $g$ .
Key idea: In a frame rotating at constant $\boldsymbol{\omega}$ , every body feels an outward force $-m\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ of magnitude $m\omega^2\rho$ directed away from the rotation axis — present even at rest, and derivable from the potential $-\tfrac12 m\omega^2\rho^2$ .

Where this lives today

The centrifugal force is quietly load-bearing across modern technology. Every biochemistry lab benches an ultracentrifuge — up to $10^6\,g$ for protein and nucleic-acid separation — and uranium enrichment worldwide runs on cascades of gas centrifuges in which the $\omega^2\rho$ field, acting on a 1% mass difference between $^{235}\mathrm{UF_6}$ and $^{238}\mathrm{UF_6}$ molecules, does what chemistry cannot. At planetary scale, spacecraft mission design lives inside the co-rotating effective potential of Derivation B's Jacobi integral: the James Webb Space Telescope orbits the Sun–Earth $L_2$ point, a saddle of the combined gravitational-plus-centrifugal landscape, and every low-energy transfer trajectory to the Moon threads the same potential's ridgelines. The one place the force is being engineered from scratch is artificial gravity. With multi-year crewed Mars mission studies and commercial space stations in active development, rotating-habitat design has returned from science fiction to trade-study spreadsheets — and the numbers of Worked Example 1 are the actual design tension: large radius for a tolerable gravity gradient and gentle Coriolis effects, against the structural cost of a 100-m ring whose rim carries permanent 1-g hoop loading. Meanwhile every human-rated centrifuge at every space agency still runs crew adaptation studies on the same $m\omega^2\rho$ that Huygens wrote down in 1659.

The physical picture

Sit on a spinning merry-go-round holding a ball on a string at arm's length. Your friend on the ground sees textbook circular motion: the ball perpetually accelerates toward the axis, and the string supplies the inward pull that makes it happen. Nothing pushes outward on anything. But you see the ball hanging motionless beside you — and your string is undeniably taut. For your rotating-frame ledger to balance, $\mathbf{F} = m\mathbf{a} = 0$ demands a second force, equal and opposite to the string tension, pointing away from the axis. That entry is the centrifugal force: not a new interaction, but the outward-flying tendency of inertia, written into the books of an observer who refuses to admit they are turning.

Figure

One ball, two ledgers — the inertial frame needs only the inward pull, the rotating frame must invent the outward push

The direction and strength of the force come straight out of the double cross product. $\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ takes the position vector, and crossing with $\boldsymbol{\omega}$ twice nets a vector of magnitude $\omega^2\rho$ pointing inward toward the axis; the minus sign in the force flips it outward. Crucially, only the component of $\mathbf{r}$ perpendicular to the axis survives the operation — a body sitting exactly on the rotation axis feels nothing, and sliding a body parallel to the axis changes nothing.

Figure

The geometry of the double cross product — only the perpendicular component of position matters

Because the force depends only on position — not velocity — it behaves inside the rotating frame like a bona-fide conservative field. It has equipotential surfaces (nested cylinders about the axis), it stores and releases energy, and a free particle "rolls downhill" outward along $-\nabla U_{\mathrm{cf}}$ . That is exactly why a centrifuge separates by density: everything is pushed outward, but in a fluid the push per unit volume wins for the denser component, which migrates to the rim while the lighter fraction is displaced inward. The Coriolis force, by contrast, is always perpendicular to the velocity, does no work, and can never be written as the gradient of anything — the two forces divide the labour of rotating-frame dynamics between them.

The mathematics of the outward push

Everything starts from one definition. In a frame rotating at constant angular velocity $\boldsymbol{\omega}$ , a particle of mass $m$ at position $\mathbf{r}$ (measured from any origin on the axis) is assigned the fictitious force

\mathbf{F}_{\mathrm{cf}}=-m\,\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r}) \qquad (1)

where: $\boldsymbol{\omega}$ — the angular velocity vector of the rotating frame (constant here); $\mathbf{r}$ — the particle's position from an origin on the rotation axis. The nested cross product looks opaque, but it is doing one geometric job: isolating how far the particle sits from the axis and pushing it further out. Working the double cross product through (Derivation A below) collapses equation (1) to its useful everyday form. Split $\mathbf{r} = \mathbf{r}_\parallel + \boldsymbol{\rho}$ into pieces along and perpendicular to the axis; only the perpendicular piece survives:

\mathbf{F}_{\mathrm{cf}}=m\omega^2\boldsymbol{\rho},\qquad |\mathbf{F}_{\mathrm{cf}}|=m\omega^2\rho \qquad (2)

where: $\boldsymbol{\rho}$ — the component of $\mathbf{r}$ perpendicular to the axis, pointing radially outward; $\rho$ — the perpendicular distance from the rotation axis. Read it physically: double the spin rate and the force quadruples; double the distance from the axis and it doubles. A centrifuge earns its thousands of $g$ from the $\omega^2$ , not from its radius. Where does equation (1) come from in the first place? It is one term of the master acceleration transformation between an inertial frame and a rotating one — derived in full, factor-of-2 bookkeeping and all, in the article on the transformation of acceleration in rotating frames:

\mathbf{a}_{\mathrm{in}}=\mathbf{a}_{\mathrm{rot}}+2\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}+\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})+\dot{\boldsymbol{\omega}}\times\mathbf{r} \qquad (3)

where: $\mathbf{a}_{\mathrm{in}}$ , $\mathbf{a}_{\mathrm{rot}}$ — acceleration in the inertial and rotating frames; $\mathbf{v}_{\mathrm{rot}}$ — velocity measured in the rotating frame; $\dot{\boldsymbol{\omega}}$ — angular acceleration of the frame, zero for the uniform rotation assumed in this article. Moving every kinematic term to the force side of Newton's law turns each into a fictitious force with its own name and its own physics:

m\mathbf{a}_{\mathrm{rot}}=\mathbf{F}_{\mathrm{real}}\underbrace{-2m\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}}_{\text{Coriolis}}\underbrace{-m\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})}_{\text{centrifugal}}\underbrace{-m\dot{\boldsymbol{\omega}}\times\mathbf{r}}_{\text{Euler}} \qquad (4)

where: $\mathbf{F}_{\mathrm{real}}$ — the sum of genuine forces (gravity, contact, tension, ...). The Coriolis force needs motion within the frame ( $\mathbf{v}_{\mathrm{rot}}\neq0$ ) — the Coriolis force article owns that story; the Euler force needs a changing spin rate — see the Euler force article. For a body simply held at rest in a steadily rotating frame, the centrifugal term is the only fictitious force in play, which is why it dominates the everyday experience of rotation.

Derivation AExtracting the outward push with the BAC-CAB identity

The goal is to reduce the abstract double cross product of equation (1) to the working form (2), $\mathbf{F}_{\mathrm{cf}} = m\omega^2\boldsymbol{\rho}$ , and in doing so prove the two structural facts quoted everywhere else in this article: the force points radially away from the axis (not away from the origin), and only the perpendicular part of the position vector contributes. The strategy is pure vector algebra — one application of the BAC-CAB triple-product identity followed by a decomposition of $\mathbf{r}$ along and across the axis. The step to watch is the sign at the end: the kinematic term $\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})$ points inward (it is the centripetal acceleration of a co-rotating point), and it is the minus sign in the force definition that turns it into the outward centrifugal push.

Step 1. Expand the triple product with the BAC-CAB rule:

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

Why this identity: for any three vectors, $\mathbf{A}\times(\mathbf{B}\times\mathbf{C})=\mathbf{B}(\mathbf{A}\cdot\mathbf{C})-\mathbf{C}(\mathbf{A}\cdot\mathbf{B})$ — the result must lie in the plane of $\mathbf{B}$ and $\mathbf{C}$ , here the plane of $\boldsymbol{\omega}$ and $\mathbf{r}$ .

Step 2. Decompose the position vector along and across the axis, $\mathbf{r}=\mathbf{r}_\parallel+\boldsymbol{\rho}$ with $\mathbf{r}_\parallel=(\hat{\boldsymbol{\omega}}\cdot\mathbf{r})\,\hat{\boldsymbol{\omega}}$ . Then $\boldsymbol{\omega}(\boldsymbol{\omega}\cdot\mathbf{r})=\omega^2\mathbf{r}_\parallel$ , and

\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})=\omega^2\mathbf{r}_\parallel-\omega^2(\mathbf{r}_\parallel+\boldsymbol{\rho})=-\omega^2\boldsymbol{\rho}

Why the parallel parts cancel: the dot product $\boldsymbol{\omega}\cdot\mathbf{r}$ sees only $\mathbf{r}_\parallel$ , and the $\omega^2\mathbf{r}$ term contains it once more with the opposite sign — the axis-parallel information is systematically erased. What survives is the inward-pointing vector $-\omega^2\boldsymbol{\rho}$ : exactly the centripetal acceleration of the co-rotating point at P.

Step 3. Insert into the definition (1) and let the minus sign do its work:

\mathbf{F}_{\mathrm{cf}}=-m\left(-\omega^2\boldsymbol{\rho}\right)=m\omega^2\boldsymbol{\rho}

Why the flip matters physically: in the inertial frame, $\omega^2\rho$ inward is the acceleration the real forces must supply to keep a body co-rotating. In the rotating frame the same magnitude reappears with the opposite sign as the push the body seems to exert outward against whatever holds it. *Two quick checks confirm the result: on the axis $\boldsymbol{\rho}=0$ gives zero force, and for $\mathbf{r}\perp\boldsymbol{\omega}$ the full $m\omega^2 r$ is recovered.*

The position-dependence of equation (2) has a payoff the Coriolis force can never offer: a potential energy. Since $\mathbf{F}_{\mathrm{cf}}$ points along $\boldsymbol{\rho}$ and grows linearly with $\rho$ , it is the gradient of a quadratic bowl — inverted:

U_{\mathrm{cf}}=-\tfrac12 m\omega^2\rho^2,\qquad \mathbf{F}_{\mathrm{cf}}=-\nabla U_{\mathrm{cf}} \qquad (5)

where: $U_{\mathrm{cf}}$ — the centrifugal potential energy, defined up to a constant. The equipotentials are cylinders around the axis, and the potential decreases outward — the rotating frame is a hill whose summit ridge is the axis itself, and everything not nailed down rolls off. Folding $U_{\mathrm{cf}}$ into energy bookkeeping is what makes rotating-frame statics tractable.

Go deeper · Derivation B · 1 minThe centrifugal potential and why the force is conservative

Derivation BThe centrifugal potential and why the force is conservative

We want to establish equation (5): that the centrifugal force, alone among the rotational fictitious forces, comes from a scalar potential. The strategy is constructive — guess the quadratic form suggested by the linear force law, verify by direct differentiation that its negative gradient reproduces $m\omega^2\boldsymbol{\rho}$ , and then note why the same construction is impossible for the Coriolis force. Conservativeness is what licenses every effective-potential argument downstream, from spinning buckets to Lagrange points, so it deserves an explicit check rather than a wave of the hand.

Step 1. Set the axis along $\hat{\mathbf{z}}$ so that $\rho^2=x^2+y^2$ , and propose

U_{\mathrm{cf}}(x,y)=-\tfrac12 m\omega^2\left(x^2+y^2\right)

Why a quadratic: a force linear in displacement is always the gradient of a quadratic — this is the spring potential with the sign of the spring constant reversed, an inverted parabola rather than a bowl.

Step 2. Differentiate:

-\nabla U_{\mathrm{cf}}=m\omega^2\left(x\,\hat{\mathbf{x}}+y\,\hat{\mathbf{y}}\right)=m\omega^2\boldsymbol{\rho}

Why this settles it: the gradient reproduces equation (2) exactly, at every point; and since $\mathbf{F}_{\mathrm{cf}}$ is a gradient field its curl vanishes and its work integral is path-independent.

Step 3. Contrast with Coriolis. The Coriolis force $-2m\boldsymbol{\omega}\times\mathbf{v}_{\mathrm{rot}}$ depends on velocity, not position, so no function of position can generate it; moreover it is always perpendicular to $\mathbf{v}_{\mathrm{rot}}$ and therefore does zero work along any path. Why this division of labour is convenient: in the rotating frame, energy conservation reads $\tfrac12 m v_{\mathrm{rot}}^2 + V + U_{\mathrm{cf}} = \text{const}$ — the Coriolis force steers trajectories without ever appearing in the energy ledger. *As a direct check: $\nabla\times\mathbf{F}_{\mathrm{cf}}=0$ by inspection of $m\omega^2(x,y,0)$ , confirming the force is curl-free; the rotating-frame energy integral is the Jacobi integral of celestial mechanics.*

Two more forms of the same physics are worth displaying, because they are how the centrifugal potential is met in practice. Write the Lagrangian of a particle in the rotating frame — kinetic energy plus the frame terms — and the centrifugal potential appears automatically with the opposite sign, as an effective potential-energy credit:

L=\tfrac12 m\,|\dot{\mathbf{r}}|^2+m\dot{\mathbf{r}}\cdot(\boldsymbol{\omega}\times\mathbf{r})+\tfrac12 m|\boldsymbol{\omega}\times\mathbf{r}|^2-V \qquad (6)

where: $V$ — the potential of the real forces; $\dot{\mathbf{r}}$ — the velocity in the rotating frame. The middle term generates the Coriolis force in the equations of motion; the third term is $+\tfrac12 m\omega^2\rho^2 = -U_{\mathrm{cf}}$ , the centrifugal contribution. And in any central-force problem, transforming to the frame co-rotating with the radius vector converts the angular-momentum barrier into precisely such a term:

V_{\mathrm{eff}}(r)=\frac{\ell^2}{2mr^2}+V(r) \qquad (7)

where: $\ell$ — the conserved angular momentum; $r$ — the radial distance. The $\ell^2/2mr^2$ barrier is a centrifugal potential in disguise — the minimum of $V_{\mathrm{eff}}$ marks circular orbits, and its curvature sets the radial oscillation frequency. One concept, three costumes. The cleanest laboratory demonstration of the centrifugal potential is a bucket of water on a turntable. Once the water co-rotates, its free surface must be an equipotential of gravity plus centrifugal — and that surface is a parabola:

z(\rho)=z_0+\frac{\omega^2\rho^2}{2g} \qquad (8)

where: $z_0$ — the height of the surface on the axis; $g$ — gravitational acceleration. Spin a 30-cm-wide bucket at one revolution per second and the rim water stands about $4.5\ \mathrm{cm}$ above the centre. Liquid-mirror telescopes exploit exactly this: a slowly spun dish of mercury is a perfect parabolic mirror, refigured live by physics.

Go deeper · Derivation C · 1 minThe parabolic surface of a spinning bucket

Derivation CThe parabolic surface of a spinning bucket

The goal is equation (8): the equilibrium free surface of a liquid co-rotating at constant $\omega$ is a paraboloid of revolution, with rim rise $\omega^2\rho^2/2g$ . The strategy is the effective-potential argument licensed by Derivation B — in the co-rotating frame the liquid is in statics, and a fluid at rest can have no force component along its own surface, so the surface must be an equipotential of the combined gravitational-plus-centrifugal potential. Watch the signs: the centrifugal potential is negative and deepens outward, so the equipotential must climb outward to compensate.

Step 1. In the co-rotating frame the effective potential energy per unit mass is

\Phi_{\mathrm{eff}}(\rho,z)=gz-\tfrac12\omega^2\rho^2

Why per unit mass: every fluid element feels both forces in proportion to its mass, so the shape cannot depend on $m$ — dividing it out keeps the geometry clean.

Step 2. Demand the free surface be the equipotential $\Phi_{\mathrm{eff}}=\text{const}$ :

gz-\tfrac12\omega^2\rho^2=gz_0 \quad\Longrightarrow\quad z(\rho)=z_0+\frac{\omega^2\rho^2}{2g}

Why an equipotential: if it were not, $-\nabla\Phi_{\mathrm{eff}}$ would have a component tangent to the surface, and surface fluid would flow along it — contradicting equilibrium. (The pressure force acts only perpendicular to the free surface.)

Step 3. Put in numbers. For $\omega=2\pi\ \mathrm{rad/s}$ (one revolution per second) and bucket radius $\rho=0.15\ \mathrm{m}$ :

\Delta z=\frac{(2\pi)^2(0.15)^2}{2\times9.81}=\frac{39.48\times0.0225}{19.62}\approx0.045\ \mathrm{m}

Why the shape is $\omega$ -tunable but always parabolic: the exponent 2 in $\rho^2$ is inherited directly from the linearity of the centrifugal force — a deep-seated consequence of Derivation A, not an accident of water. *Two sanity checks: the dimensions $[\omega^2\rho^2/g]=\mathrm{m}$ are correct, and the limit $\omega\to0$ returns a flat surface — both as expected. This paraboloid shape is the operating principle of liquid-mirror telescopes.*

Limiting cases

Case	Result	Meaning
Object on the rotation axis	$\rho=0\;\Rightarrow\;\mathbf{F}_{\mathrm{cf}}=0$	A point on the axis has no perpendicular distance, so no centrifugal force; only positions off-axis feel it.
Position along the axis	$\mathbf{r}\parallel\boldsymbol{\omega}\;\Rightarrow\;\boldsymbol{\omega}\times\mathbf{r}=0$	The double cross product vanishes; centrifugal force acts only through the component of $\mathbf{r}$ perpendicular to $\boldsymbol{\omega}$ .
Non-rotating frame	$\omega\to 0\;\Rightarrow\;\mathbf{F}_{\mathrm{cf}}\to 0$	All fictitious forces disappear; the frame becomes inertial and Newton's law holds in its bare form.
Equatorial surface of Earth	$a_{\mathrm{cf}}=\omega^2 R_E\approx 0.0339\ \mathrm{m/s^2}$	With $\omega=7.292\times10^{-5}\,\mathrm{rad/s}$ and $R_E=6.378\times10^6\,\mathrm{m}$ , this reduces measured $g$ by about $0.34\%$ .
Fast lab rotation	$\omega^2\rho\gg g$ for modest $\omega$	At 4000 rpm and $\rho=10\ \mathrm{cm}$ , $\omega^2\rho\approx1.75\times10^4\ \mathrm{m/s^2}\approx1790\,g$ — the $\omega^2$ scaling makes artificial gravity cheap.

Common trap“The centrifugal force is the reaction partner of the centripetal force”

It is not — and the misreading breaks Newton's third law. A third-law pair always acts on two different bodies: the string pulls the ball inward, and the ball pulls the string outward; that outward pull on the string is real and inertial-frame-legal, but it is not the centrifugal force. The centrifugal force acts on the ball itself, exists only in the rotating frame's ledger, and has no reaction partner because it is not an interaction at all. Two further traps: never invoke centrifugal force while working in an inertial frame (there, the real centripetal force alone supplies the $m\omega^2\rho$ of inward acceleration — adding an outward force would wrongly cancel it); and never hand it a velocity dependence — a body moving in the rotating frame feels the centrifugal force appropriate to its instantaneous position, plus a separate Coriolis force for its motion. Keeping the two ledgers strictly apart is the entire discipline of this subject.

Worked examples

Figure

Artificial gravity by rotation — in the habitat frame the centrifugal force is the floor pressing up

1 · Spinning up artificial gravity (confidence builder)

A rotating space habitat is a ring of radius $r = 100\ \mathrm{m}$ . What spin rate gives the rim crew an apparent gravity of $1g$ , and what is the rim speed? How different is "gravity" at the head of a $1.8$ -m-tall astronaut?

Step 1. Set the centrifugal acceleration equal to $g$ : $\omega^2 r = 9.81\ \mathrm{m/s^2}$ , so $\omega=\sqrt{9.81/100}=0.313\ \mathrm{rad/s}$ .

Step 2. Convert to revolutions per minute: $\omega\times\dfrac{60}{2\pi} = 0.313\times9.549 = \boxed{3.0\ \mathrm{rpm}}$ — a leisurely one turn every 20 seconds.

Step 3. Rim speed: $v=\omega r=0.313\times100=\boxed{31.3\ \mathrm{m/s}}$ (about 113 km/h) — the speed the structure must sustain forever, and the speed at which arriving spacecraft must match the dock.

Step 4. Head-to-toe gradient: apparent gravity scales linearly with $\rho$ , so at the head, $\rho = 98.2\ \mathrm{m}$ , gravity is $98.2/100 = 98.2\%$ of the floor value — a 1.8% gradient, imperceptible. In a small 10-m centrifuge the same astronaut's head would sit at $82\%$ of foot gravity, one reason habitat designs favour large radii (the other is the Coriolis force on moving crew — see the Coriolis force article).

2 · The bench-top ultracentrifuge (exam level)

A laboratory centrifuge spins samples at $4000\ \mathrm{rpm}$ at an effective radius of $10\ \mathrm{cm}$ . Find the centrifugal acceleration in units of $g$ , the outward force on a $2$ -g sample, and comment on why the rotor walls — not the sample tubes — are the engineering constraint.

Step 1. Convert the spin rate: $\omega = 4000\times\dfrac{2\pi}{60} = 418.9\ \mathrm{rad/s}$ .

Step 2. Centrifugal acceleration: $a=\omega^2\rho=(418.9)^2\times0.10=1.75\times10^4\ \mathrm{m/s^2}$ . In units of $g$ : $a/g = 1.75\times10^4/9.81 = \boxed{\approx1790\,g}$ .

Step 3. Force on the sample: $F=ma=0.002\times1.75\times10^4=\boxed{35\ \mathrm{N}}$ — the 2-gram sample presses on its tube bottom like a $3.6$ -kg mass under ordinary gravity.

Step 4. Scaling insight: the acceleration grows as $\omega^2$ , so doubling the spin rate to 8000 rpm quadruples the load to $\sim7150\,g$ , while doubling the radius merely doubles it. Rotors therefore fail by hoop stress at the rim — every kilogram of rotor rim at $\rho = 10\ \mathrm{cm}$ demands $17.5\ \mathrm{kN}$ of inward force from the surrounding metal, which is why ultracentrifuge rotors are titanium, inspected for fatigue, and spun inside armored chambers.

Origins · 1 minChristiaan Huygens and how this was discovered

Christiaan Huygens

1629–1695 · The Hague and Paris

Huygens got to the mathematics of circular motion a generation before Newton's Principia. In 1659, aged thirty, he derived the exact law of what he named vis centrifuga — the "centre-fleeing force" — showing the outward tendency of a body swung in a circle grows as $v^2/r$ , equivalently $\omega^2 r$ ; he published the theorems in 1673 as an appendix to his Horologium Oscillatorium, the treatise on pendulum clocks that also gave the world the physics of the isochronous pendulum. The clockmaker's motivation was practical: he wanted to know how spinning and swinging masses load their pivots. Huygens even used his formula to compute how much lighter objects should weigh at the equator of a spinning Earth — decades before anyone could measure it. Newton later inverted the viewpoint, insisting on the inward centripetal force as the physically primary one, and the two names have divided the labour between reference frames ever since: Newton's word for the inertial ledger, Huygens's word for the rotating one.

History

1659

— Christiaan Huygens derives the quantitative law of centrifugal tendency,

v^2/r

, naming it vis centrifuga

1673

— The centrifugal theorems appear in print as an appendix to Huygens's Horologium Oscillatorium

1687

— Newton's Principia recasts circular motion around the inward centripetal force, demoting the outward push to a frame artifact

1883

— Ernst Mach's critique of Newton's rotating-bucket argument asks what rotation is even relative to, seeding ideas that influenced general relativity

1934

— Jesse Beams builds practical ultracentrifuges reaching hundreds of thousands of g, turning the centrifugal force into a precision laboratory tool

Going further

References. Morin, Introduction to Classical Mechanics, Ch. 10 (Accelerated Frames) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., Sec. 4.10 & 5.1 · Landau & Lifshitz, Mechanics, Vol. 1, Sec. 39 · Taylor, Classical Mechanics, Ch. 9 Key numerical benchmarks: the triple-product reduction gives $\boldsymbol{\omega}\times(\boldsymbol{\omega}\times\mathbf{r})=-\omega^2\boldsymbol{\rho}$ ; the equatorial centrifugal acceleration is $a_{\mathrm{cf}}=\omega^2 R_E$ with $\omega=7.292\times10^{-5}\,\mathrm{rad/s}$ and $R_E=6.378\times10^6\,\mathrm{m}$ , giving $\approx0.0339\,\mathrm{m/s^2}$ ( $0.34\%$ of $g_0=9.81$ ); the plumb-line deflection reaches a mid-latitude maximum of $\approx1.7\times10^{-3}$ rad; and the equatorial-polar radius difference is $21\,\mathrm{km}$ (WGS-84: $6378.137$ vs $6356.752$ km).

Builds on: Non-inertial reference frames · Fictitious (inertial) forces · Time derivatives in rotating frames · Angular velocity vector of a rotating frame Leads to: Transformation of acceleration in rotating frames · Coriolis force · Effective gravity on the rotating Earth · Equatorial bulge and the figure of the Earth

If the centrifugal force is fictitious, why does the water actually leave my clothes in the spin cycle?

Because "fictitious" describes the bookkeeping, not the outcome. In the inertial frame the story is: water droplets tend to move in straight lines; the drum wall keeps yanking the clothes inward off those lines, but at the perforations there is nothing to yank the water, so it exits along the tangent. In the rotating frame the same events are booked as an outward force $m\omega^2\rho$ pushing water through the holes. The droplet trajectories, the dryness of the clothes, and every measurable number agree between the two descriptions — only the naming of causes differs. Fictitious forces have real consequences because they encode real inertia.

Which way would I actually fly off a merry-go-round — outward along the radius?

No — and this is a beautiful test of whether you have the inertial picture. The instant you let go, no real horizontal force acts on you, so you travel in a straight line along your instantaneous velocity, which is tangent to the circle, not along the radius. In the rotating frame that straight line appears to curve outward and backward — an apparent motion assembled from the centrifugal and Coriolis forces acting together. "Flung radially outward" is the one thing that happens in neither frame.

Does the centrifugal force do work? I thought fictitious forces couldn't have energies.

The centrifugal force genuinely does work in the rotating frame — that is exactly why it has a potential, $U_{\mathrm{cf}}=-\tfrac12 m\omega^2\rho^2$ , and why rotating-frame energy conservation includes it. A bead sliding outward along a frictionless spoke gains kinetic energy at the expense of $U_{\mathrm{cf}}$ , in perfect analogy to falling in gravity. It is the Coriolis force that never does work (always perpendicular to velocity). Of course, energies differ between frames — the inertial observer attributes the bead's speed-up to the spoke's sideways push — but each frame's ledger is internally consistent.

Why does a centrifuge separate by density rather than just pushing everything to the rim?

Everything is pushed outward, in proportion to its mass — but the sample is a fluid, and fluids push back with buoyancy. In the effective gravity $\omega^2\rho$ , the surrounding liquid exerts an inward buoyant force on any parcel, equal to the centrifugal force the displaced liquid would feel. A parcel denser than its surroundings wins the tug and sinks outward; a lighter one is displaced inward toward the axis. It is Archimedes's principle verbatim, with $g$ replaced by $\omega^2\rho$ — turned up to thousands of $g$ so that separations which would take gravity months happen in minutes.

At the equator the centrifugal force pushes me up — could a fast-spinning planet actually throw things off?

Yes, in principle: the condition is $\omega^2 R > g$ , when the required centripetal acceleration exceeds what gravity can supply. For Earth, $\omega^2 R\approx0.034\ \mathrm{m/s^2}$ , about 290 times too small — you'd need a day of roughly 84 minutes for loose equatorial objects to become orbital. Astronomy meets this limit for real: millisecond pulsars spin near the break-up rate where centrifugal support balances their ferocious gravity, and asteroids spun up past the limit by sunlight (the YORP effect) shed rubble from their equators. Note that a "day" of 84 minutes is precisely the period of a satellite skimming Earth's surface — break-up and orbit are the same condition.

Why do textbooks say the centrifugal force acts on stationary objects when I only notice it while moving in a car turn?

"Stationary" means stationary in the rotating frame. Sitting still on a spinning carousel, you are stationary in its frame, and the centrifugal force acts on you the whole time — it is what makes you lean inward and grip the rail. In the car, the turning cab is momentarily a rotating frame, and you sitting "still" in your seat feel pressed outward against the door: same situation. What needs motion within the frame to switch on is the Coriolis force. The clean statement: centrifugal reads your position, Coriolis reads your velocity.

Is the outward force I feel in a loop-the-loop at the top of the loop centrifugal or something else?

In the rotating (car) frame it is exactly the centrifugal force, magnitude $mv^2/r$ pointing away from the loop's centre — upward at the top. Whether you stay on the track depends on whether gravity ( $mg$ downward, toward the centre at the top) can supply the needed centripetal pull: the famous minimum-speed condition $v^2 \geq gr$ is just "centrifugal must not exceed gravity" in that frame, or "gravity must not exceed the required centripetal force" in the inertial one. Same inequality, two dialects — a good habit is to solve such problems both ways until the translation is automatic.

The centrifugal potential decreases outward without bound — doesn't that mean infinite energy is available?

Within the idealization of a frame rotating rigidly forever, yes, $U_{\mathrm{cf}}\to-\infty$ as $\rho\to\infty$ — but the idealization is doing the smuggling. Any real rotating system has finite angular momentum and finite kinetic energy: a bead flung down the centrifugal hill gains energy that is really being withdrawn from the rotation of the platform, which slows unless a motor keeps feeding it torque. The "bottomless" potential is the signature of pretending the rotation source is infinitely stiff. The same caution applies to the rotating-frame treatment of orbits: the Jacobi integral is conserved, but it mixes energy and angular momentum precisely to keep this bookkeeping honest.

Centrifugal force

Where this lives today

The physical picture

The mathematics of the outward push

Limiting cases

Worked examples

1 · Spinning up artificial gravity (confidence builder)

2 · The bench-top ultracentrifuge (exam level)

Going further

Before this

Where this leads