Circular Motion - Physics for Pilot Assessments

Something moving in a circle at a perfectly steady speed is accelerating, and the force causing that acceleration points inwards, towards the centre. Both of those statements fight intuition, and between them they account for one of the most reliably wrong-answered topics in mechanics. This lesson is mostly about dismantling the intuition.

Why a steady circle is an acceleration

Acceleration was defined as the rate of change of velocity, and velocity includes direction. An object going round a circle is changing direction continuously, so its velocity is changing continuously even when the speedometer reading never moves — and a changing velocity is, by definition, an acceleration. If you found this claim suspicious when it appeared in the velocity lesson, this is where it earns its keep.

Which way the force points

Since there is an acceleration there must be a net force, and it points the same way the acceleration does: towards the centre of the circle. It is called the centripetal force, from Latin for centre-seeking. Whirl a ball on a string and the string's tension supplies it; drive round a bend and friction between tyres and road supplies it; the Moon is held in its path by gravity. There is no single new kind of force here — centripetal simply describes the job an ordinary force is doing.

Centripetal acceleration

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

$a$: acceleration towards the centre, in metres per second squared (m/s²)
$v$: speed along the circular path, in metres per second (m/s)
$r$: radius of the circle, in metres (m)

The speed is squared, so this grows sharply: double the speed round the same bend and the required acceleration — and therefore the required force — becomes four times as large, not twice. That is why cornering fast is so much more demanding than cornering slowly, and it is the same squared relationship that made stopping distances grow so alarmingly in the earlier lesson. A larger radius reduces the requirement, which is why a gentle, sweeping bend is easier than a tight one at the same speed.

There is no outward force

The feeling of being flung outwards in a turning car is real, but the explanation usually attached to it is not. Nothing is pushing you outwards. Your body is trying to continue in a straight line, exactly as Newton's first law says it should, while the car turns beneath you — and what you feel is the door pushing you inwards, changing your direction. The word centrifugal describes this apparent outward effect, and it is not a force acting on you.

The test for this is what happens when the inward force is removed. Cut the string on a whirling ball and it does not fly outwards along the radius; it travels off in a straight line along the tangent, carrying on with the velocity it had at the instant of release. If an outward force existed, the ball would curve away from the centre — it does not.

The velocity points along the tangent and the net force points to the centre — always at right angles to each other. Release the object and it continues along the tangent, not outwards.

Putting it to work

Example 3 — can the tyres supply it?

A 1,200 kg car takes a bend of radius 100 m at 20 m/s. How much centripetal force does the bend demand — and can the tyres deliver it with a μ of 0.8 on the dry road, or 0.1 on ice? Take g = 10.

Acceleration demanded: $a = \frac{v^2}{r} = \frac{400}{100} = 4\ \text{m/s}^2$
Force demanded: $F = ma = 1{,}200 \times 4 = 4{,}800$ N, pointing towards the centre of the bend.
What friction can supply: on a level road the normal force is the weight, 1,200 × 10 = 12,000 N, so the ceiling is μN = 0.8 × 12,000 = 9,600 N. The dry road delivers the 4,800 N with room to spare.
On ice: $0.1 \times 12{,}000 = 1{,}200\ \text{N}$ That is a quarter of what the bend demands. The car cannot be turned that tightly at that speed, so it runs wide. It is not flung outwards; it simply fails to be turned.

What actually supplies the force

Because centripetal is a job description rather than a kind of force, it is worth running through what fills the role in each familiar case. For a ball on a string it is the tension in the string. For a car on a bend it is friction between the tyres and the road, acting sideways. For a passenger in that car it is the seat and the door pushing them round. For a satellite or the Moon it is gravity. For water in a spun bucket it is the bucket's base pushing inwards. In every case, if you are asked which force is centripetal, the answer is whichever real force happens to point towards the centre.

That framing also explains the limits. A car can only corner as hard as the available friction allows, and the friction lesson gave the ceiling: mu multiplied by the normal force. Ask for more centripetal force than the tyres can supply — by entering too fast or meeting a wet surface where mu has dropped — and the requirement simply cannot be met, so the car cannot follow the curve and continues on a straighter path than the bend. It is not thrown outwards; it fails to be turned inwards, which looks similar from inside and is a different thing.

Speed, radius, and what to change

Since the required acceleration is the speed squared divided by the radius, there are two ways to reduce the demand on a bend: go slower, or take a wider line. Halving the speed cuts the requirement to a quarter, because of the square; doubling the radius merely halves it. Speed is by far the stronger lever, which is why a small reduction in cornering speed makes such a large difference to how much grip is being asked of the tyres.

A common source of confusion is the difference between the centripetal force and the total force. On a bend the centripetal force is the part pointing towards the centre — the part responsible for turning. There may well be other forces acting at the same time: a car cornering while also accelerating has a forward force as well, and their combination is what the tyres must supply. When a question asks for the centripetal force specifically, it wants only the inward component, not everything acting on the object. And when it asks what would happen if that inward force vanished, the answer is always the tangent: straight on, at the speed it had, exactly as Newton's first law requires.

Why it matters in aviation

An aircraft turns by tilting its lift. Rolling into a bank swings the lift vector away from the vertical so that part of it points horizontally, towards the inside of the turn, and that horizontal part is the centripetal force doing the turning. The consequence follows straight from this lesson. If only a portion of the lift is now holding the aircraft up, the total lift must be increased so that its vertical part still matches the weight, and the steeper the bank the more total lift is needed. That is why the load factor climbs as bank angle increases, and why the speed at which the wing stalls climbs with it: the wing is being asked for more, so it reaches its limit at a higher speed than it would in level flight.

The same relationship explains why instrument turns are specified as a rate — so many degrees of heading change per second — rather than as a bank angle. A faster aircraft holding a given rate sweeps a larger circle, and a larger radius at a higher speed demands more centripetal force, so the bank required to maintain that rate increases with airspeed rather than being a single number.

What the test asks

Expect the released-object question, with the outward radial path offered as the answer — the correct one is the tangent. Expect an item claiming that constant speed round a bend means zero acceleration. And expect the squared relationship to be tested by doubling a speed and offering a doubled force, where the true answer is four times.