What is centripetal acceleration in the context of Motion in a Circle?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. It is essential for maintaining the object's circular motion and is calculated using the formula a_c = v^2/r, where v is the velocity of the object and r is the radius of the circle. This concept is crucial in the Cambridge International A Levels Physics curriculum under the topic of Motion in a Circle.

How does centripetal acceleration differ from linear acceleration?

Centripetal acceleration differs from linear acceleration in that it acts perpendicular to the velocity of an object, directing it towards the center of its circular path. In contrast, linear acceleration occurs in the direction of the object's velocity, changing its speed. Understanding this distinction is important for students studying Motion in a Circle in the Cambridge International A Levels Physics syllabus.

Why is centripetal acceleration always directed towards the center of the circle?

Centripetal acceleration is always directed towards the center of the circle because it is the result of the net force required to keep an object moving in a circular path. This inward force changes the direction of the object's velocity without altering its speed, ensuring the object remains in circular motion. This principle is a key aspect of Motion in a Circle in the Cambridge International A Levels Physics course.

What factors affect the magnitude of centripetal acceleration?

The magnitude of centripetal acceleration is affected by the object's speed and the radius of the circular path. Specifically, it increases with the square of the object's speed and decreases with a larger radius, as described by the formula a_c = v^2/r. This relationship is an important concept for students learning about Motion in a Circle in the Cambridge International A Levels Physics curriculum.

Can you provide an example of centripetal acceleration in real life?

A common example of centripetal acceleration is a car turning around a circular track. As the car moves, it experiences an acceleration towards the center of the track, allowing it to maintain its circular path. This real-world application helps illustrate the principles of Motion in a Circle, a topic covered in the Cambridge International A Levels Physics syllabus.

Why is "Saying centrifugal force pushes outward" a common mistake in Centripetal acceleration?

Why it happens: Apparent feeling in rotating frame. How to avoid it: In inertial frame, ONLY centripetal (inward) force acts. 'Centrifugal' is a fictitious force in rotating frame. Source: 9702 Examiner Reports 2022-2024

Motion in a CircleCambridge International A Levels Physics (9702)

Centripetal acceleration

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Centripetal Acceleration Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Acceleration toward centre. $a = v^2/r$ . Force needed to maintain circular motion.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

12.2.1 — Define centripetal acceleration.
12.2.2 — Apply $F = mv^2/r$ .
12.2.3 — Solve circular-motion problems.

Centripetal acceleration

Toward centre; changes direction.

Centripetal acceleration. Points TOWARDS the centre. Changes the DIRECTION of velocity, not its magnitude.

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

Centripetal force. $F = ma = mv^2/r$ . This is the NET force, supplied by:

Tension (string twirling ball).
Gravity (satellite orbit).
Friction (car cornering).
Normal reaction (banked curve).
Electrostatic attraction (electron in atom).

Vector picture. Velocity is tangent to circle. Acceleration is perpendicular to velocity (toward centre).

Example. 0.5 kg, 1.2 m string, $v = 4$ m/s. $a = 16/1.2 \approx 13.3$ m/s². Tension = 6.7 N.

Cambridge tip. Centripetal force is NOT a separate force — it's the net force.

Toward centre.
Changes direction only.
Supplied by various forces.

See the full worked example for centripetal acceleration →

Applications

Conical pendulum, banked curve.

Conical pendulum. String makes angle $\theta$ with vertical; mass moves in horizontal circle.

Vertical: $T\cos\theta = mg$ .
Horizontal: $T\sin\theta = mv^2/r$ where $r = L\sin\theta$ .

Banked curve (no friction). Normal reaction provides horizontal centripetal component.

$\tan\theta = v^2/(rg)$ .
$v = \sqrt{rg\tan\theta}$ is the 'designed' speed.

Satellite orbit. Gravity provides centripetal force.

$GMm/r^2 = mv^2/r \Rightarrow v = \sqrt{GM/r}$ .

Vertical loops. Tension/gravity together provide centripetal at top.

Cambridge tip. Always identify SOURCE of centripetal force.

Conical: vertical/horizontal split.
Banked: $\tan\theta = v^2/(rg)$ .
Satellite: $v = \sqrt{GM/r}$ .

See the full worked example for centripetal acceleration →

How it’s examined

Circular motion central to A2. Most-tested: centripetal force (5-7 marks), conical pendulum (8 marks), banked curve (6-8 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/42 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Centripetal acceleration

Step-by-step solutions to past-paper-style questions on centripetal acceleration, written exactly the way a tutor would explain them at the board.

1Centripetal acceleration (5 marks)
Extended• Adapted from 9702/42 May/Jun 2024• centripetal
Question
A 0.50 kg ball on a 1.2 m string is whirled at 4.0 m/s in a horizontal circle. Find (a) centripetal acceleration, (b) tension. (5 marks)
Step-by-step solution
1. Step 1
  $a = v^2/r$ .
  $a = \dfrac{4.0^2}{1.2} = 13.3 \text{ m/s}^2$
2. Step 2
  $F = ma$ .
  $T = 0.50 \times 13.3 = 6.67 \text{ N}$
Answer
$a = 13.3$ m/s²; $T \approx 6.7$ N.
2Conical pendulum (8 marks)
Extended• conical pendulum
Question
A 0.20 kg mass on a 0.50 m string moves in a horizontal circle. The string makes $30°$ with the vertical. Find the speed of the mass. Take $g = 9.81$ . (8 marks)
Step-by-step solution
1. Step 1
  Radius. $r = L\sin\theta = 0.50 \times 0.5 = 0.25$ m.
2. Step 2
  Vertical: $T\cos\theta = mg$ .
  $T = mg/\cos\theta = 0.20 \times 9.81/\cos 30° \approx 2.27 \text{ N}$
3. Step 3
  Horizontal: $T\sin\theta = mv^2/r$ .
4. Step 4
  Solve for $v$ .
  $v = \sqrt{\dfrac{T\sin\theta \cdot r}{m}} = \sqrt{\dfrac{2.27 \times 0.5 \times 0.25}{0.20}} \approx 1.19 \text{ m/s}$
Answer
$v \approx 1.19$ m/s.
3Banked curve (6 marks)
Extended• banked
Question
A car travels in a horizontal circle of radius 50 m on a road banked at $20°$ . For NO friction needed, find the speed. Take $g = 9.81$ . (6 marks)
Step-by-step solution
1. Step 1
  Bank angle gives horizontal component of normal reaction as centripetal force. No friction means $\tan\theta = v^2/(rg)$ .
2. Step 2
  Solve.
  $v = \sqrt{rg\tan\theta} = \sqrt{50 \times 9.81 \times \tan 20°} \approx 13.4 \text{ m/s}$
Answer
$v \approx 13.4$ m/s.

Key Formulae — Centripetal acceleration

The formulae you need to memorise for centripetal acceleration on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Centripetal acceleration
$a = \dfrac{v^2}{r} = r\omega^2$
When to use
Acceleration toward centre of circle.
Centripetal force
$F = ma = \dfrac{mv^2}{r} = mr\omega^2$
When to use
Net force toward centre. NOT a separate force — supplied by tension, gravity, friction, normal, etc.

Key Definitions and Keywords — Centripetal acceleration

Definitions to memorise and the exact keywords mark schemes credit for centripetal acceleration answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Centripetal acceleration
Examiner keyword
Acceleration directed toward centre of circular path. Causes change in direction of velocity (not magnitude).
Centripetal force
Examiner keyword
Net force directed toward centre, providing centripetal acceleration.

Common Mistakes and Misconceptions — Centripetal acceleration

The traps other students keep falling into on centripetal acceleration questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Saying centrifugal force pushes outward
9702 Examiner Reports 2022-2024
Why it happens
Apparent feeling in rotating frame.
How to avoid it
In inertial frame, ONLY centripetal (inward) force acts. 'Centrifugal' is a fictitious force in rotating frame.

Practice questions

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Centripetal acceleration — frequently asked questions

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Centripetal acceleration

Short Study Notes in the form of Slides

Detailed Study Notes

What you’ll learn

How it’s examined

1Centripetal acceleration (5 marks)

2Conical pendulum (8 marks)

3Banked curve (6 marks)

Centripetal acceleration

Centripetal force

Centripetal acceleration

Centripetal force

✕Saying centrifugal force pushes outward

Practice questions

Past paper style quiz

Assess your understanding

Centripetal acceleration — frequently asked questions