Centripetal Acceleration and Force Revision Notes | A Level AQA Physics

1. Why Circular Motion Needs a Special Kind of Force

Picture a ball on a string being swung in a horizontal circle. The ball moves at constant speed, yet something is clearly changing — its direction. And a change in velocity (which is a vector) means there must be an acceleration, even when the speed stays the same.

That acceleration always points inward, towards the centre of the circle. The force responsible for it — whatever physical force is doing the job — is called the centripetal force. Understanding both the acceleration and the force is the core of this topic.

The direction of centripetal acceleration is crucial: it is always perpendicular to the velocity of the object. This is why the speed does not change — the force does no work on the object in the direction of motion.

2. The Centripetal Acceleration Formulae

There are two equivalent expressions for centripetal acceleration, and you need to be comfortable choosing between them depending on which quantities the question gives you.

Both forms are equivalent because $v = \omega r$. Substituting into $a = \frac{v^2}{r}$ gives $a = \frac{(\omega r)^2}{r} = \omega^2 r$. The AQA specification states that the derivation of these formulae will not be examined — you need to be able to use them, not prove them.

Notice that for a fixed radius, doubling the speed quadruples the centripetal acceleration. The $v^2$ dependence is a favourite source of exam questions, so keep it in mind when checking whether your answer is physically reasonable.

3. From Acceleration to Centripetal Force

Once you have the centripetal acceleration, Newton's second law ($F = ma$) connects it directly to the centripetal force. Multiplying both acceleration expressions by mass $m$ gives the two centripetal force formulae.

A critical point: centripetal force is not a new type of force. It is the label given to whatever resultant force happens to be pointing towards the centre. In different physical situations, different forces play this role.

• Ball on a string swung in a circle: tension in the string provides the centripetal force
• Planet orbiting the Sun: gravitational attraction provides the centripetal force
• Car cornering on a flat road: friction between tyres and road provides the centripetal force
• Electron in a magnetic field: the magnetic force provides the centripetal force

4. Worked Calculations

Theory is only useful when you can apply it under exam conditions. The two worked examples below cover the most common question types: one using the linear speed form and one using the angular velocity form.

Worked example: Centripetal Force on a Satellite

A satellite of mass 800 kg orbits Earth in a circular path of radius $7.2 \times 10^6$ m at a linear speed of $7.4 \times 10^3$ m s$^{-1}$. Calculate the centripetal force acting on the satellite. [3 marks]

1. $F = \frac{mv^2}{r}$ — [M1] Correct formula selected. Use the linear speed form because $v$ and $r$ are given directly.
2. $m = 800$ kg, $v = 7.4 \times 10^3$ m s$^{-1}$, $r = 7.2 \times 10^6$ m — All values are already in SI units, so no conversion is needed.
3. $v^2 = (7.4 \times 10^3)^2 = 7.4 \times 7.4 \times 10^6 = 54.76 \times 10^6 = 5.476 \times 10^7$ m$^2$ s$^{-2}$ — Expand the square carefully: $7.4 \times 7.4 = 54.76$, and $(10^3)^2 = 10^6$. Never skip this step — squaring errors are very common.
4. $F = \frac{800 \times 5.476 \times 10^7}{7.2 \times 10^6}$ — [M1] Correct substitution with units.
5. $800 \times 5.476 \times 10^7 = 4380.8 \times 10^7 = 4.3808 \times 10^{10}$ — $800 \times 5.476 = 4380.8$, then carry the power of ten.
6. $F = \frac{4.3808 \times 10^{10}}{7.2 \times 10^6} = \frac{4.3808}{7.2} \times 10^{4} \approx 0.6084 \times 10^{4} = 6080$ N — [A1] $4.3808 \div 7.2 \approx 0.6084$, and $10^{10} \div 10^6 = 10^4$. Final answer to 3 significant figures: $6080$ N (or $6.08 \times 10^3$ N).

Answer: $F \approx 6080$ N (or $6.08 \times 10^3$ N)

Worked example: Centripetal Acceleration Using Angular Velocity

A fairground ride spins riders in a horizontal circle of radius 4.0 m. The ride completes one full revolution every 3.0 s. Calculate the centripetal acceleration of a rider. [3 marks]

1. $\omega = \frac{2\pi}{T} = \frac{2\pi}{3.0}$ rad s$^{-1}$ — The period $T = 3.0$ s. Angular velocity is $2\pi$ radians divided by the period. This gives $\omega \approx 2.094$ rad s$^{-1}$.
2. $a = \omega^2 r$ — [M1] Use the angular velocity form because $\omega$ and $r$ are now both known.
3. $\omega^2 = \left(\frac{2\pi}{3.0}\right)^2 = \frac{4\pi^2}{9.0} \approx \frac{39.478}{9.0} \approx 4.386$ rad$^2$ s$^{-2}$ — $4\pi^2 = 4 \times 9.8696 \approx 39.478$. Dividing by 9.0 gives approximately 4.386.
4. $a = 4.386 \times 4.0 \approx 17.5$ m s$^{-2}$ — [M1][A1] $4.386 \times 4.0 = 17.544$, rounded to 3 significant figures gives $17.5$ m s$^{-2}$.

Answer: $a \approx 17.5$ m s$^{-2}$

5. Applying the Formulae to Real Contexts

Exam questions rarely ask you to calculate centripetal force in isolation. More often, they embed it in a physical scenario and ask you to identify which force is acting as the centripetal force, then apply the formula to find an unknown quantity such as speed, radius, or mass.

Consider a car travelling over the top of a hill modelled as a circular arc. At the top, both weight ($mg$) and the normal contact force ($N$) act vertically. Weight acts downward (towards the centre of the arc), while the normal force acts upward (away from the centre). The resultant of these two forces provides the centripetal force: $mg - N = \frac{mv^2}{r}$.

This kind of force-balance reasoning — identifying which forces contribute to the centripetal resultant and in which direction — is exactly what AQA mark schemes reward. Getting the direction of each force correct before substituting is the key step.

6. Exam Technique and Common Mistakes

Centripetal acceleration and force questions can appear in AQA A Level Physics Papers 1 and 2. The marks are often straightforward to collect — but only if you avoid the traps that catch most students.

Across this lesson, every spec requirement maps directly to exam marks: knowing both forms of $a$ and $F$, understanding that the derivation is not assessed, and being able to identify the physical source of the centripetal force in context. Practise applying the formulae in unfamiliar scenarios — that is where AQA questions most often push beyond straightforward recall.

In analysis questions, always identify the real force that is providing the centripetal force. In a satellite it is gravity, in a car on a bend it is friction, and in a mass on a string it is tension. Saying that centripetal force is a separate extra force is wrong: it is the name given to the resultant inward force required for circular motion.