Introduction to Circular Motion Revision Notes | A Level AQA Physics

1. The Puzzle of Constant Speed with Changing Direction

Picture a car travelling around a perfectly circular roundabout at a steady 10 m s⁻¹. The speedometer never moves. Yet something is definitely changing — the car is constantly turning. That change is the heart of circular motion, and it leads to one of the most important ideas in A Level mechanics.

Velocity is a vector: it has both magnitude (speed) and direction. Even if the speed stays fixed, a change in direction means a change in velocity. A change in velocity means there is an acceleration. So an object moving in a circle at constant speed is always accelerating, even though its speed never changes.

Because there is an acceleration, Newton's second law ($F = ma$) tells us there must be a resultant force in the same direction as that acceleration — towards the centre. This is the centripetal force. Without it, the object would travel in a straight line (Newton's first law). The centripetal force is not a mysterious extra force; it is whatever real force happens to point towards the centre in that situation — tension in a string, friction from a road surface, or gravitational attraction.

This spec point — that constant speed in a circle implies acceleration and requires a centripetal force — is tested frequently. Examiners expect you to state explicitly that it is the direction of velocity that changes, not the magnitude, and that the resultant force is directed towards the centre of the circle.

2. Measuring Angles in Radians

Before calculating angular speed, you need a reliable way to measure angles. Degrees work fine for everyday life, but circular motion equations require a more mathematically natural unit: the radian.

The key conversions to memorise are: $360° = 2\pi \text{ rad}$, $180° = \pi \text{ rad}$, and $90° = \frac{\pi}{2} \text{ rad}$. These appear constantly when you link period, frequency, and angular speed.

One practical point: the AQA specification states that the direction of angular velocity will not be assessed. In calculations, $\omega$ represents the magnitude of angular velocity — a scalar value. You do not need to consider the vector direction of angular velocity at A-level; this keeps the maths clean and focused on the physics.

3. Angular Speed and Its Equations

Now that angles are measured in radians, angular speed has a precise definition. Think of it as the rotational equivalent of linear speed: instead of metres per second, it measures radians per second.

The link $\omega = \frac{v}{r}$ comes directly from the definition of arc length: in one second, an object moving at speed $v$ around a circle of radius $r$ sweeps through an angle of $\frac{v}{r}$ radians. The link $\omega = 2\pi f$ comes from the fact that one complete revolution covers $2\pi$ radians, and $f$ revolutions happen each second.

Notice that $\omega$ is the same for every point on a rigid rotating object, but $v$ is not — points further from the centre have a larger $v$ because $v = \omega r$ and $r$ is bigger. This distinction matters in questions about rotating discs or fairground rides.

4. Worked Calculations with Angular Speed

Theory is only useful when you can apply it under exam conditions. The two forms of the angular speed equation are tested in different ways — sometimes you are given the period, sometimes the linear speed. Here is how to handle both confidently.

Worked example: Calculating Angular Speed from Period and Radius

A stone is tied to a string and swung in a horizontal circle of radius 0.80 m. It completes one full revolution in 0.50 s. Calculate (a) the angular speed and (b) the linear speed of the stone.

1. $f = \frac{1}{T} = \frac{1}{0.50} = 2.0 \text{ Hz}$ — Period $T = 0.50$ s means 2 complete revolutions per second. [M1]
2. $\omega = 2\pi f = 2\pi \times 2.0 = 4\pi \approx 12.6 \text{ rad s}^{-1}$ — Apply $\omega = 2\pi f$ directly. Leave as $4\pi$ for exactness or give to 3 significant figures. [M1, A1]
3. $v = \omega r = 4\pi \times 0.80 = 3.2\pi \approx 10.1 \text{ m s}^{-1}$ — Rearrange $\omega = \frac{v}{r}$ to get $v = \omega r$. Substitute $\omega = 4\pi$ rad s⁻¹ and $r = 0.80$ m. [M1, A1]
4. $[\omega r] = \text{rad s}^{-1} \times \text{m} = \text{m s}^{-1}$ ✓ — Radians are dimensionless, so the unit of $v$ is correctly m s⁻¹.

Answer: (a) $\omega = 4\pi \approx 12.6$ rad s⁻¹; (b) $v = 3.2\pi \approx 10.1$ m s⁻¹

Worked example: Finding Angular Speed from Linear Speed

A car travels around a circular track of radius 120 m at a constant speed of 24 m s⁻¹. Calculate the angular speed of the car and the time taken for one complete lap.

1. $\omega = \frac{v}{r} = \frac{24}{120} = 0.20 \text{ rad s}^{-1}$ — Divide linear speed by radius. Units: m s⁻¹ ÷ m = rad s⁻¹. [M1, A1]
2. $f = \frac{\omega}{2\pi} = \frac{0.20}{2\pi} \approx 0.03183 \text{ Hz}$ — Rearrange $\omega = 2\pi f$ to give $f = \frac{\omega}{2\pi}$. [M1]
3. $T = \frac{1}{f} = \frac{1}{0.03183} \approx 31.4 \text{ s}$ — Alternatively: $T = \frac{2\pi}{\omega} = \frac{2\pi}{0.20} = 10\pi \approx 31.4$ s. Both routes give the same answer. [A1]
4. $\text{circumference} = 2\pi r = 2\pi \times 120 = 240\pi \approx 754 \text{ m}$; $T = \frac{754}{24} \approx 31.4 \text{ s}$ ✓ — Cross-check: distance divided by speed gives the same period. Consistency confirmed.

Answer: $\omega = 0.20$ rad s⁻¹; $T = 10\pi \approx 31.4$ s

5. The Direction of Centripetal Acceleration and Force

Earlier we established that circular motion at constant speed requires an acceleration. Now consider its direction more carefully. At any instant, the velocity of the object is tangential to the circle — it points along the edge of the path. The acceleration must be perpendicular to this velocity, pointing directly towards the centre of the circle.

This is why the term centripetal is used — it comes from the Latin for 'centre-seeking'. The acceleration (and therefore the net force) always points inward, towards the centre, regardless of where on the circle the object currently is. As the object moves, the direction of the centripetal acceleration rotates continuously to keep pointing at the centre.

A critical consequence: centripetal force does no work on the object. Work requires a force component in the direction of displacement. Since centripetal force is always perpendicular to the velocity (and therefore to the instantaneous displacement), it cannot transfer energy. This is why speed — and hence kinetic energy — remains constant in uniform circular motion.

6. Identifying the Centripetal Force in Real Situations

One of the most common exam tasks is identifying which physical force acts as the centripetal force in a given scenario. The centripetal force is always a real, named force — never an additional invented one. The table below summarises the most frequently tested cases.

Notice that in the loop-the-loop case, both the normal reaction and weight point towards the centre at the top of the loop. The centripetal force is their resultant. This is a favourite exam scenario because students often forget to include both forces in their analysis.

7. Exam Technique and Common Mistakes

Circular motion questions reward precision in language and careful handling of units. The following guidance addresses the errors that most frequently cost marks in AQA examinations.

For a 6-mark extended response on circular motion — for example, 'Explain why a car travelling at constant speed around a circular bend is accelerating and describe the force responsible' — structure your answer as: (1) velocity is a vector so direction change means acceleration; (2) acceleration is directed towards the centre; (3) by Newton's second law, a resultant force must act towards the centre; (4) this is friction between tyres and road; (5) if friction is insufficient, the car cannot maintain circular motion and moves in a straight line. That chain of reasoning, with correct terminology at each step, secures full marks.

Every spec requirement in this topic connects back to one central idea: changing direction at constant speed is still acceleration, and acceleration requires a force. Master that principle, apply the angular speed equations fluently in both forms, and work confidently in radians — and circular motion becomes one of the most straightforward topics in A Level mechanics.

The core AQA equations for circular motion are a = v^2/r = ω^2r and F = mv^2/r = mω^2r. They show that centripetal acceleration and centripetal force both increase if speed or angular velocity increases, and both decrease if the radius is made larger while everything else stays the same.