Circular Motion

Tie a conker to a string and whirl it round your head at a steady rate. The speedometer of the conker — if it had one — would read the same number all the way round: its speed never changes. And yet, second by second, the conker is doing the one thing we call accelerating. How can something moving at a constant speed be accelerating the whole time?

The answer is the whole reason this page exists. Velocity is a vector — a speed with a direction — and as the conker sweeps round the circle its direction is turning every instant. North, then north-east, then east, then south-east… The size of the velocity is pinned at one value, but the direction is in a constant state of change, and any change of velocity, in size or in direction, is an acceleration. An object in circular motion at constant speed is accelerating all the time, and this page works out exactly how much, in which direction, and what force is needed to produce it.

Measuring the turning: angular quantities

To describe going round, a distance in metres is awkward — far better to measure the angle swept out from the centre. We measure that angle in radians: one radian is the angle for which the arc length equals the radius, and a full turn is 2\pi radians (so 360^\circ = 2\pi\ \text{rad}, and 1\ \text{rad} \approx 57.3^\circ). The angle turned through is the angular displacement \theta.

How fast the angle grows is the angular velocity \omega (Greek "omega"), measured in radians per second. If the object sweeps out an angle \Delta\theta in a time \Delta t, then

\omega = \dfrac{\Delta\theta}{\Delta t}.

For steady circular motion it is often easier to think in whole laps. One complete revolution is 2\pi radians, and the time for one lap is the period T; the number of laps per second is the frequency f = 1/T. Putting one whole turn over one whole period,

\omega = \dfrac{2\pi}{T} = 2\pi f.

From angular speed to actual speed: v = \omega r

The angular velocity tells you how fast the angle turns, but two riders on a merry-go-round — one near the middle, one at the rim — share the same \omega while moving at very different speeds through the air. In one period each sweeps once round its own circle, covering a distance equal to that circle's circumference, 2\pi r. So the ordinary (tangential) speed is

v = \dfrac{2\pi r}{T} = \omega r.

The rider at the larger radius r moves faster, in exact proportion to how far out they sit. This little bridge, v = \omega r, lets us swap freely between the angular picture and the linear one — and we will need it in a moment.

The acceleration points to the centre

Now the key question: which way is the conker accelerating? Draw its velocity vector at two nearby instants and subtract them (acceleration is the rate of change of the velocity vector). Because the velocity always points along the tangent, and the tangent is swinging round, the little change \Delta\mathbf{v} from one instant to the next points squarely inwards, towards the centre of the circle. So the acceleration of anything moving in a circle is aimed straight at the middle. We call it the centripetal acceleration — "centripetal" is Latin for "centre-seeking".

Chasing the geometry through (comparing the triangle of velocity vectors with the triangle of positions) gives its size. For an object of speed v on a circle of radius r,

a = \dfrac{v^2}{r}.

Substituting v = \omega r gives the same acceleration written with the angular velocity, which is often more convenient:

a = \dfrac{v^2}{r} = \dfrac{(\omega r)^2}{r} = \omega^2 r.

Notice how sensitive it is to the speed: double v and the centripetal acceleration quadruples, because v is squared. Tighten the circle (smaller r) at the same speed and the acceleration grows too — a sharp bend is far more demanding than a gentle one.

The force that keeps it turning

An acceleration needs a force: by Newton's second law F = ma, a body accelerating towards the centre must be pushed or pulled towards the centre. That inward force is the centripetal force, and it has the size

F = ma = \dfrac{mv^2}{r} = m\omega^2 r.

An object of mass m moving at speed v (angular velocity \omega) around a circle of radius r has:

The single most important thing to grasp is that centripetal force is not a new kind of force. It is not an extra arrow you add to a free-body diagram. It is the name for the job that some perfectly ordinary force is already doing — the net inward force that any real push or pull happens to supply:

In every case you first ask "what real force is available?" — tension, friction, gravity, the normal force — and then set that force equal to \tfrac{mv^2}{r}. When the real force on offer is too small to supply the needed \tfrac{mv^2}{r}, the object simply cannot stay on the circle — it flies off along the tangent (which is exactly what a car does when it skids off an icy bend).

These are the traps that catch almost everyone — check yourself against all three:

Worked examples

Example 1 — a conker on a string. A conker of mass m = 0.20\ \text{kg} is whirled on a string of radius r = 0.80\ \text{m} at a speed of v = 4.0\ \text{m/s}. Find the centripetal acceleration and the tension in the string.

a = \dfrac{v^2}{r} = \dfrac{4.0^2}{0.80} = \dfrac{16}{0.80} = 20\ \text{m/s}^2. F = \dfrac{mv^2}{r} = m a = 0.20 \times 20 = 4.0\ \text{N}.

So the string must pull inwards with 4.0\ \text{N} of tension. Pull any harder than the string can bear and it snaps — and the conker leaves along the tangent.

Example 2 — a car on a bend. A car of mass m = 1000\ \text{kg} takes a bend of radius r = 30\ \text{m} at v = 12\ \text{m/s}. What centripetal force must friction provide?

F = \dfrac{mv^2}{r} = \dfrac{1000 \times 12^2}{30} = \dfrac{144\,000}{30} = 4800\ \text{N}.

The tyres' grip must supply 4800\ \text{N} pointing towards the centre of the bend. On a wet or icy road the available friction can drop below this, and the car understeers straight on — off the tangent of the curve.

Example 3 — a spinning disc (angular version). A vinyl record turns once every T = 1.8\ \text{s}. Find its angular velocity, and the speed and centripetal acceleration of a speck of dust r = 0.12\ \text{m} from the centre.

\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{1.8} = 3.5\ \text{rad/s}. v = \omega r = 3.5 \times 0.12 = 0.42\ \text{m/s}, \qquad a = \omega^2 r = 3.5^2 \times 0.12 = 1.5\ \text{m/s}^2.

Example 4 — reading off a frequency. A drum in a washing machine spins at f = 20\ \text{Hz} (20 turns per second). Its angular velocity is \omega = 2\pi f = 2\pi \times 20 \approx 126\ \text{rad/s}; a shirt at r = 0.25\ \text{m} then feels a centripetal acceleration a = \omega^2 r = 126^2 \times 0.25 \approx 3900\ \text{m/s}^2 — nearly 400 times gravity. That enormous inward pull, supplied by the drum wall, is what wrings the water out.

Try it: the two arrows of circular motion

The object below circles at a fixed point on its path. Its velocity v always points along the tangent (the way it is heading), while the centripetal force F always points inwards, to the centre — the two are permanently at right angles. Drag the radius and speed sliders and watch the live readouts of a = \tfrac{v^2}{r} and F = \tfrac{mv^2}{r} (taking a mass of m = 2\ \text{kg}). Notice how raising the speed grows the inward arrow far faster than it grows the velocity arrow — because the acceleration depends on v^2 — and how tightening the radius makes the same speed far more demanding.

Where the force comes from — and where it fails

Because the needed force is \tfrac{mv^2}{r}, circular motion is always a contest between the force a situation can supply and the force it demands. Speed up, or tighten the turn, and the demand climbs steeply. When the real force can no longer keep up, the circle breaks.

On a dry road the friction between four tyres and the tarmac can reach roughly the car's own weight — plenty to bend its path round an ordinary corner. Ice changes everything: the grip can fall to a tenth of that. Take our 1000\ \text{kg} car from Example 2, needing 4800\ \text{N} to round the 30\ \text{m} bend at 12\ \text{m/s}. If the icy road can offer only, say, 2000\ \text{N} of friction, the demand outstrips the supply. With too little centripetal force, the car cannot follow the curve — it carries straight on along the tangent, sliding off the outside of the bend. This is also why signs tell you to slow down for a curve: halving v quarters the required force \tfrac{mv^2}{r}, bringing it back within what the tyres can grip. And a banked track (a tilted road, or the walls of a velodrome) tips the normal force inwards so that it helps provide the centripetal force — letting vehicles corner far faster than friction alone would allow.