Centripetal and fictitious forces in Year 3

If you are in Year 3 at the liceo scientifico (the Italian science-focused high school), in these weeks of late September the teacher is probably finishing Galilean relativity and frames of reference, and this is the moment when centrifugal force appears — the one that seems clear in class and in a test becomes a black hole of lost marks. From here you move on to motion in the plane: projectile, circular, harmonic, pendulum. The test on this block falls in 4 to 7 weeks, so generally between the end of October and mid-November.

Physics in Year 3 goes from two to three hours a week and becomes quantitative: the jump compared with Year 2 is more abrupt than the one you experience in maths in the same year. Here I concentrate on one precise piece of that block — circular motion, centripetal force, fictitious forces — because it is the point where reasoning (not calculation) loses the most marks. I do not redo projectile motion: I already covered it in full, with a worked example, in the first physics test in Year 3.

Inertial frames: the starting point you need afterwards

Galileo’s transformations say something simple: if an observer is at rest (or moves with uniform straight-line motion) and a second observer moves relative to the first at constant velocity $v_{0}$ , velocities add vectorially,

v^{'} = v - v_{0}

and the laws of dynamics — in particular $F = ma$ — have the same form in both frames, provided both are inertial, that is, not accelerating. A frame that accelerates, brakes or rotates is not inertial, and in that frame $F = ma$ stops working as it stands: you have to add forces that do not exist for any outside observer. They are fictitious forces (or inertial forces), and centrifugal force is the commonest case you will meet in Year 3.

Keep this in mind as a guiding rule for the rest of the article: the question “which forces act?” does not have a single answer — it depends on where you place yourself to look.

Uniform circular motion: the quantities

In uniform circular motion the magnitude of the velocity stays constant, but its direction changes continuously: that is why, even though “uniform”, there is always an acceleration. The quantities:

T = period (time for one turn), f = \frac{1}{T} = frequency, ω = \frac{2 π}{T} = 2 π f

$ω$ is the angular velocity, measured in radians per second. The tangential velocity and the centripetal acceleration:

v = ω r, a_{c} = \frac{v ^{2}}{r} = ω^{2} r

The centripetal acceleration is always directed towards the centre of the path, never outwards: it is the first point on which all the rest of the chapter depends.

Centripetal force is not a new force

This is the most widespread conceptual mistake, and it is not a calculation error: it is an error of model. Centripetal force is not an additional type of force that adds to those already present in a drawing. It is the name we give to the role that an already existing force — real, applied by something — plays when it produces an acceleration directed towards the centre of a circular path.

In practice, when you draw the force diagram on a body in circular motion, you never add an arrow labelled “centripetal force”: you identify which real force (tension of a string, friction between tyre and tarmac, force of gravity, component of the normal reaction) is acting as centripetal, and you write $F_{real} = m a_{c}$ along the radial direction.

Car on a bend: the centripetal force is the static friction between tyre and road.
Ball tied to a string: it is the tension of the string.
Satellite in orbit: it is the force of gravity.
Passenger on a chair-swing ride: it is the horizontal component of the tension in the chains.

If you cannot name which force acts as the centripetal one, you have not understood the system — and no calculation will save you.

Centrifugal force exists, but only if you change your point of view

Here is the second half of the mistake, the one the test almost always exploits with a theory question. Take a person on a rotating roundabout.

Seen from the ground (inertial frame): a real force towards the centre acts on the person — the one that produces the centripetal acceleration. Full stop. There is no outward force: if at that instant the centripetal force disappeared, the person would carry on in a straight line by inertia (tangent to the circle), not “outwards”.

Seen by the person sitting on the roundabout (non-inertial frame, because it rotates): they feel at rest relative to their own seat, and yet a real force (the belt, the backrest) pushes them towards the centre. Why do they not accelerate, if they feel a net force? Because in their frame, which is non-inertial, you have to introduce an equal and opposite fictitious force — centrifugal force — which balances the real one and accounts for the sensation of being “pushed outwards”.

The operating rule: choose a frame of reference and stay there for the whole exercise. Either you work from outside (only real forces, the centripetal is the role of one of them, the body accelerates) or you work from inside the rotating frame (you add the centrifugal as a fictitious force, and the body is in equilibrium in its frame). Mixing the two — putting both in the same drawing seen from the ground — is the most frequent mistake, and it makes no physical sense: it is counting the same thing twice from two different perspectives.

Other fictitious forces: the lift and the braking bus

The rotating frame is not the only case of a non-inertial frame. You find a second, more intuitive one in any linear acceleration.

You are on a bus stopped at a traffic light, the light turns green and the bus accelerates forward: in your frame (the bus’s, non-inertial) you feel a fictitious force pushing you backwards, even though nobody is really pushing you — it is your body that, by inertia, “stays behind” relative to the floor that accelerates under you. When the bus brakes the opposite holds: the fictitious force pushes you forward.

The lift works the same way along the vertical, and is the basis of exercise 3: when it accelerates upwards there appears, in its frame, a fictitious force downwards that adds to the real weight and makes the passenger feel heavier; in accelerated descent the opposite holds.

In all these cases the fictitious force appears only in the accelerated frame, has a direction opposite to the acceleration of the frame, and has no reaction according to the third law: it is not a real force, it is a corrective to make $F = ma$ work in a frame that is accelerating.

Four worked examples

1. Car on a bend: the maximum speed

A car goes round a flat bend of radius $r = 50 m$ . The coefficient of static friction between tyres and tarmac is $μ_{s} = 0.80$ . What is the maximum speed at which the car can take the bend without skidding? ( $g \approx 9.8 m/ s^{2}$ )

The only horizontal force on the car is static friction, and it is what acts as the centripetal. The maximum value static friction can provide is $μ_{s} N = μ_{s} m g$ (on a flat road the normal force balances the weight). The maximum speed is the one at which friction is already at its limit value:

μ_{s} m g = \frac{m v _{m a x}^{2}}{r} ⟹ v_{m a x} = μ_{s} g r

Note that the mass cancels: it does not matter how much the car weighs, only the available friction and the radius count.

v_{m a x} = 0.80 \cdot 9.8 \cdot 50 \approx 19.8 m/s \approx 71 km/h

If the car enters the bend faster than this, static friction is no longer enough to provide all the centripetal force required, and the car starts to slide towards the outside of the bend — not because a force “pushes it out”, but because the force towards the centre is no longer sufficient to make it follow the circular path.

2. Ball on a vertical string: the minimum speed at the top

A ball is tied to a string of length $r = 0.80 m$ and makes a complete turn in a vertical plane (the classic loop-the-loop). What is the minimum speed it must have at the highest point of the path for the string to stay taut? ( $g \approx 9.8 m/ s^{2}$ )

At the highest point, two forces act on the ball, both directed downwards, that is, towards the centre of the circle: the weight $m g$ and the tension in the string $T$ . Together they act as the centripetal force:

m g + T = \frac{m v ^{2}}{r}

The string stays taut as long as $T \geq 0$ . The limit condition, that of the minimum speed, is $T = 0$ : in that case the weight alone provides the whole centripetal force needed.

m g = \frac{m v _{m i n}^{2}}{r} ⟹ v_{m i n} = g r = 9.8 \cdot 0.80 \approx 2.8 m/s

Below this speed the string would go slack before even reaching the top: it is not that “the centrifugal force is missing to hold it up”, it is that without enough speed the weight alone would provide more acceleration towards the centre than is needed, and the ball would fall inside the circular path instead of following it.

3. Lift with a scale: the apparent weight

A person of mass $m = 70 kg$ steps on a scale inside a lift. What does the scale read (a) when the lift accelerates upwards with $a = 2.0 m/ s^{2}$ , and (b) when it accelerates downwards with the same acceleration? ( $g \approx 9.8 m/ s^{2}$ )

It is best solved from outside, that is, from the inertial frame of someone watching the lift from the ground: on the person act the weight $m g$ downwards and the normal reaction of the scale $N$ upwards, and $N$ is exactly what the scale reads (by the third law, it equals in magnitude the force the person exerts on the scale).

(a) Upward acceleration. Taking the positive direction upwards, $N - m g = ma$ :

N = m (g + a) = 70 \cdot (9.8 + 2.0) = 826 N

(b) Downward acceleration. Same balance, but now the acceleration has the opposite direction to the chosen positive one, $N - m g = - ma$ :

N = m (g - a) = 70 \cdot (9.8 - 2.0) = 546 N

Unloaded, that is, at rest or at constant speed, the scale would read $m g = 686 N$ . In the lift’s frame (non-inertial) the same result reads as “apparent weight”: in accelerated ascent a fictitious force downwards appears that adds to the weight, in accelerated descent a fictitious force upwards that reduces it. They are two ways of telling the same number, not two different physics.

4. A roundabout: from period to angular velocity

A roundabout rotates with radius $r = 4.0 m$ and makes a complete turn every $T = 8.0 s$ . Calculate the angular velocity, the tangential speed of a passenger and their centripetal acceleration.

ω = \frac{2 π}{T} = \frac{2 π}{8.0} \approx 0.785 rad/s

v = ω r \approx 0.785 \cdot 4.0 \approx 3.1 m/s

a_{c} = \frac{v ^{2}}{r} \approx \frac{3. 1 ^{2}}{4.0} \approx 2.5 m/ s^{2}

An acceleration of about a quarter of $g$ : enough to be felt, not enough to worry — it is roughly the order of magnitude of a bend taken decisively in a car.

If you want to check for yourself whether the method has become automatic, at this point in the syllabus — on motion in the plane — you will find the mock test on motion in the plane in Year 3: the correct answer to every question is free, the guided step-by-step solution unlocks with the first lesson.

The mistakes that come back every year

Drawing a “centrifugal force” arrow in the frame seen from the ground. From outside it does not exist: only the real force acting as centripetal exists.
Mixing the two frames in the same exercise — putting both the real centripetal and the fictitious centrifugal in the same drawing “from the ground”, counting them as if both were true together.
Forgetting that “uniform” does not mean “without acceleration”. The magnitude of the velocity is constant, the direction is not: there is always a centripetal acceleration.
Confusing the highest and the lowest point in vertical circular motion: at the lowest point weight and tension (or normal force) are in opposite directions, not in the same direction as at the highest point — the limit condition changes sign.
Leaving degrees in place of radians in $ω$ , which must always be expressed in rad/s in the formulas with $v = ω r$ .

How to tell whether you are ready

The most reliable test is not rereading the theory: it is trying to explain out loud, without looking at your notebook, why centrifugal force “does not exist from the ground”. If the explanation comes out fluently, the concept is yours. If you get stuck on the words, that is where marks are lost in tests — not in the calculations.

Remember also what awaits you straight afterwards, in December: work, energy and collisions, the most feared test of the year according to most of the students I teach — I talk about it in the dedicated guide to work, energy and collisions in Year 3. The motion in the plane you are studying now is also the training ground in which the method (frame of reference, resolution of forces, final check) that you will use there is consolidated.

If you want a hand

On this chapter the most useful work is almost always the same: two sessions to fix the difference between inertial and non-inertial frames firmly, then mixed circular-motion exercises until recognising “who acts as the centripetal” becomes automatic. I work online, one-to-one. You will find the full path on the page about physics tutoring, and to start you can book a free first chat via contact: send me the syllabus covered and the date of the test, and I will tell you honestly where it is best to begin.

Centripetal force and fictitious forces in Year 3