Turning is an acceleration, and constant speed does not help

A stone whirled on a string at a steady rate is doing something that ordinary language has no comfortable word for. Its speed never changes. Its velocity changes constantly, at a ferocious rate, and the difference between those two sentences is not a quibble — it is the reason the string is under tension and the reason it hurts when the string is let go.

What is actually changing #

Velocity is a vector: a magnitude and a direction. Acceleration is the rate at which that vector changes. Nothing in the definition says the magnitude has to be the part that changes.

Going round a circle at a steady rate, the magnitude is fixed and the direction is turning at a constant rate, so the velocity vector is changing at a constant rate, so there is a constant acceleration. That is the whole argument, and it is complete. Everything below is about how large the acceleration is and which way it points.

The reason the conclusion feels wrong is that “acceleration” in ordinary speech means getting faster, and a car going round a roundabout at a constant thirty is not getting faster by any measure a passenger would name. The passenger nevertheless feels something, and what they feel is exactly this.

That resolution is the useful one here, and it is a different choice from the one that makes a projectile easy or a slope easy. Horizontal and vertical are convenient when the force has a fixed direction. When the constraint is a circle, the natural axes turn with the object, and the rule is the same one as always: choose the axes that make one equation trivial.

The construction that gives the size #

The magnitude of the acceleration comes out of a picture with two arrows in it and no calculus at all.

Take the velocity at two moments separated by a small turn $\Delta\theta$. Both have the same length $v$. Redrawn tail to tail, they form an isosceles triangle whose apex angle is $\Delta\theta$ — the same angle the object turned through, because the velocity turns with the radius.

Now the key step, which is a statement about similar triangles rather than about physics. The triangle of the two velocities and the triangle of the two radii have the same shape: both isosceles, both with apex angle $\Delta\theta$. So the ratios of corresponding sides are equal:

$$\frac{|\Delta \mathbf{v}|}{v} = \frac{|\Delta \mathbf{r}|}{r}.$$

Divide both sides by the time taken. On the left, $|\Delta\mathbf{v}|/\Delta t$ is the acceleration. On the right, $|\Delta\mathbf{r}|/\Delta t$ is the speed. So

$$a = \frac{v^2}{r},$$

and the direction of $\Delta\mathbf{v}$, as the figure shows, is toward the centre.

The square is what makes the result consequential. Doubling the speed around the same bend does not double the acceleration; it quadruples it. A car taking a 200-metre bend at 30 m/s needs 4.5 m/s², which is about 0.46 of gravity and comfortably within what tyres supply. At 60 m/s the same bend demands 18 m/s², which is 1.8 g, which no tyre on a public road has ever delivered. Nothing about the corner changed.

The force is a requirement, not a new kind of force #

The phrase centripetal force invites a misreading worth heading off, because it is the single most common error about this subject.

There is no centripetal force in the sense that there is a gravitational force or an electric force. “Centripetal” describes a role, not an origin. Any force that happens to point at the centre of the curve is playing that role, and the object curves because something is pushing it inward — the tension in a string, the friction under a tyre, the normal force from a banked road, gravity on a satellite, the electromagnetic attraction on an electron in a magnetic field.

Read the equation in the right direction and the confusion evaporates. It does not say “circular motion generates a force”. It says: to move on a circle of radius $r$ at speed $v$, something must supply a force of $mv^2/r$ toward the centre, and if nothing does, the motion is not circular. The stone whose string breaks does not fly outward; it flies off along the tangent, in a straight line, because the only force acting on it has stopped and Newton’s first law takes over.

The outward feeling in a cornering car is real and it is a feeling about the car, not about the passenger. The passenger continues straight; the car turns; the door arrives. Describing this from inside the turning car requires an outward “centrifugal force” that has no source, acts on everything in proportion to mass, and vanishes the moment the description is made from outside. It is a bookkeeping term introduced by insisting on a rotating frame, and its family includes the Coriolis force — the same kind of term, in the same kind of frame, and the reason a Foucault pendulum appears to turn — a term of exactly the kind the free-body diagram has no way to represent.

What the turn costs, and who pays it #

The requirement has to be met by real hardware, and the limits of that hardware are what the equation is used for in practice.

On a flat road the only inward force available is friction, capped at $\mu N = \mu m g$. Setting that equal to $mv^2/r$ makes the mass cancel — the same cancellation that makes the sliding angle on a slope a property of the materials — and leaves

$$v_{\max} = \sqrt{\mu g r}.$$

With a good dry coefficient of 0.9 and a 200-metre bend, that is 42 m/s, or about 150 km/h. In the wet, with $\mu$ nearer 0.4, it falls to 28 m/s. The bend has not moved; the available grip has halved and the safe speed has dropped by a third, because of the square root.

Banking removes the dependence on friction entirely. Tilting the road tilts the normal force, and the normal force is supplied by the ground pushing, which it will do as hard as required. Balancing vertically gives $N\cos\theta = mg$ and horizontally $N\sin\theta = mv^2/r$; dividing one by the other cancels both $N$ and $m$ and leaves

$$\tan\theta = \frac{v^2}{rg}.$$

A bank of 22° on a 200-metre curve is the design speed of 28 m/s, or about 100 km/h — and at exactly that speed a vehicle would hold the curve on ice. Faster, and friction must make up the difference outward; slower, and it must hold the vehicle from sliding down the bank. Every motorway interchange and every railway curve is built to this equation, and the choice of design speed is the choice of which traffic gets the frictionless ride.

Comparing the two figures makes a point about method that outlasts the example. They are the same object at the same angle, and the resolutions differ because the questions differ. On the static slope the constraint forbids motion perpendicular to the surface, so those are the useful axes. On the banked turn the acceleration is horizontal and known, so horizontal and vertical are the useful axes and the surface direction is a nuisance. Choosing the axes is the whole of the technique, and the geometry is indifferent to which choice is made.

Where the model stops #

Uniform circular motion is a special case that has been given a great deal of attention because it is solvable, and four of its assumptions fail routinely.

The speed is constant. Almost no real circular motion has this. A pendulum, a ball on a string swung in a vertical circle, a car accelerating out of a bend — all have a tangential acceleration as well as a radial one, and the total acceleration is the vector sum. The radial part is still $v^2/r$, with $v$ the instantaneous speed, so the picture generalises; it simply stops being a picture of arrows that never change length. A pendulum is the everyday instance: its bob is on a circle of fixed radius, and its speed varies through every swing.

The radius is constant. A path that curves by a varying amount has a different $r$ at each point — the radius of the circle that best fits the curve there — and $v^2/r$ still gives the radial acceleration with that local value. This is why the equation is used for road design at all, since real roads use easement curves whose curvature increases gradually rather than jumping from zero.

The object is a point. A rotating rigid body has parts at different radii moving at different speeds, and the stress that holds it together must supply $v^2/r$ at every radius. That requirement sets the bursting speed of flywheels, turbine discs and centrifuges, and it is why the limit on such machines is a tip speed rather than a rotation rate. The same requirement sets the maximum useful speed of a centrifuge separating gases by molecular mass. The same requirement, applied to a planet, is why fast-rotating bodies bulge at the equator.

The frame is inertial. Every statement above is made from outside. Inside the turning frame, the description needs the fictitious terms discussed earlier, and their status is genuinely subtle rather than merely a convention: an observer sealed in a rotating room can detect the rotation, which is not true of uniform motion. Rotation is absolute in a way that velocity is not, and the reason why is one of the older open questions in the subject.

The orbit, which is the same picture #

The most striking application is the one where nothing is touching anything.

A satellite in a circular orbit is doing exactly what the first figure draws. Its speed is constant, its velocity turns, and the acceleration required is $v^2/r$ directed at the centre of the Earth. The force supplying it is gravity, and gravity is not doing anything unusual — it is pulling with the same strength it would pull on a stationary object at that height.

The numbers are worth checking, because the agreement is not approximate. The International Space Station orbits at about 7.66 km/s at a radius of 6,771 km from the Earth’s centre. That demands an acceleration of $v^2/r = 8.7$ m/s². Gravity at that radius, scaled from its surface value by the inverse square, is 8.7 m/s². The two match, which is what “in orbit” means: the speed is exactly the one whose required acceleration equals the gravity available.

That also disposes of the idea that astronauts float because gravity is weak up there. Gravity at station altitude is 89 per cent of its value at the ground. They float because they are in free fall along with everything around them, and free fall is the condition in which the only force present is exactly the one being used to turn. An orbit is a projectile trajectory that misses.

Newton drew this in the Principia as a cannon on a mountain, firing progressively harder — the same parabola-and-its-limits argument, pushed until the ground’s curvature stops being negligible. The shots fall further away, then much further, then the ground curves away as fast as the ball falls and the ball never lands. The figure did what a good figure does — it made a discontinuity in the reader’s mind, between “falling” and “orbiting”, visibly not a discontinuity in the physics.

The same requirement, with no string and no gravity #

The third instance is the one that turned the construction into an instrument.

A charged particle moving through a magnetic field feels a force perpendicular to its velocity and perpendicular to the field, of size $qvB$. Perpendicular to the velocity is precisely the role this essay has been describing, so the force cannot change the speed and can only turn the particle — which means the path is a circle, without anything having been assumed about circles.

Setting $qvB = mv^2/r$ and cancelling one factor of $v$ gives

$$r = \frac{mv}{qB},$$

the radius in proportion to the momentum. That is the working principle of a mass spectrometer, of the magnetic analysers on every accelerator, and of the bubble-chamber photographs in which the curvature of each track is a direct reading of the momentum of whatever made it. A track that curves tightly is a slow or light particle; one that curves the other way carries the opposite charge; a track that barely curves at all is carrying most of the energy. The pictures were read that way for forty years, and the reading is the equation above.

There is a further consequence that looks like a coincidence and is not. Cancelling $v$ from both sides removes it from the time as well: the period of the orbit, $2\pi m/qB$, contains no speed at all. Faster particles travel bigger circles at exactly the rate that keeps the time per lap the same. That is why the cyclotron works — an accelerating voltage alternating at one fixed frequency stays in step with particles of every energy, and Lawrence built one on that observation in 1932.

It stops working at high energy, and the reason is the one thing this essay’s mechanics cannot supply. As the speed approaches that of light the momentum grows faster than the velocity does, so the period is no longer independent of speed, and the particles fall out of step with the fixed frequency. Fixing it means ramping the frequency as the particles speed up, which is the synchrocyclotron, and every large machine since has been built around that correction. The failure of a constant to stay constant is how relativity announced itself in engineering.

The ladder from here #

Later rungs on this anchor: non-uniform circular motion, and the tangential and radial components of acceleration treated together. The vertical circle, and the minimum speed at the top, where the required acceleration is supplied by gravity alone. Angular velocity and angular acceleration as the natural variables. Torque and moment of inertia, which are the rotational versions of force and mass. Angular momentum, and why a spinning skater speeds up on pulling in. The conical pendulum, which is a circular orbit held by a string. Rotating frames done properly, with the centrifugal and Coriolis terms derived rather than described. And the orbit as an energy problem, where the effective potential folds the turning into the landscape and the circle becomes the bottom of a well.