Physics 9702 · syllabus 12
Cambridge A Level Physics 9702 — Motion in a circle topical past papers
Uniform circular motion measured in radians: angular speed, centripetal acceleration and the resultant force that keeps an object on a circular path.
- Questions
- 48
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 23
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Motion in a circle covers
An object moving in a circle at constant speed is still accelerating, because the direction of its velocity keeps changing. Angles are measured in radians, arc length divided by radius, and the angular speed ω is the angle swept out per second, so ω = 2π/T and the speed along the path is v = rω.
The acceleration points towards the centre and has magnitude a = rω² = v²/r. It needs a resultant force towards the centre, F = mv²/r, which is always perpendicular to the velocity and so changes the direction of motion without changing the speed. Centripetal force is not an extra force: it is the resultant that the real forces, such as tension, friction, gravity or a normal contact force, provide.
Motion in a circle opens the A Level course and is examined on Paper 4. It relies on dynamics from AS and leads straight into gravitational fields, where gravity supplies the centripetal force on a satellite, and magnetic fields, where the force on a moving charge does the same job and bends its path into a circle.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 12 Motion in a circle
- 12.1 Kinematics of uniform circular motion
- 12.2 Centripetal acceleration
What you should be able to do
- Convert between degrees, revolutions and radians
- Use ω = 2π/T and v = rω
- Use a = rω² = v²/r and F = mrω² = mv²/r
- Identify which real force or combination of forces provides the centripetal force
- Analyse motion at the top and bottom of a vertical circle
- Explain why motion in a circle at constant speed involves an acceleration
Key ideas
- Angle in radians θ = s/r; 2π rad = 360°
- ω = 2π/T = 2πf (rad s⁻¹); v = rω
- Centripetal acceleration a = rω² = v²/r, directed towards the centre
- Resultant force F = mrω² = mv²/r, perpendicular to the velocity
How Motion in a circle has been examined
From 2022 to 2026, Motion in a circle was on Paper 4 · A Level Structured Questions in 9 of 11 exam series, worth about 8% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · A Level Structured Questions | 23 | 8% | 9 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 11.
Filed in Momentum under: Motion in a Circle.
A worked example
A 0.20 kg ball on a string moves in a vertical circle of radius 0.80 m. At the top of the circle its speed is 4.0 m s⁻¹. Find its angular speed there and the tension in the string.
- Angular speed: ω = v/r = 4.0 / 0.80 = 5.0 rad s⁻¹
- Resultant force needed: mv²/r = 0.20 × 4.0² / 0.80 = 4.0 N towards the centre, which is downwards at the top
- At the top, weight and tension both act downwards: T + mg = mv²/r
- T = 4.0 − 0.20 × 9.81 = 4.0 − 1.96 = 2.0 N
Answer: Angular speed 5.0 rad s⁻¹; tension 2.0 N
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Drawing centripetal force as an extra arrow alongside tension and weight — the centripetal force is the resultant of the forces already shown
- Saying an object in uniform circular motion is in equilibrium because its speed is constant — its velocity changes direction, so there must be a resultant force
- Using the diameter for r, or a period in minutes, in v = 2πr/T — both give answers wrong by a fixed factor
- At the top of a vertical circle, subtracting the weight from the tension — both act downwards, towards the centre, so T + mg = mv²/r
- Saying the object is flung outwards when the string breaks — with no resultant force it continues in a straight line along the tangent
Exam technique
- Draw the forces at the exact point asked about and write 'resultant force towards the centre = mv²/r' before substituting
- Convert revolutions per minute to rad s⁻¹ by multiplying by 2π and dividing by 60 before using any circular-motion formula
- In 'explain' questions, use the phrases 'velocity changes direction', 'acceleration towards the centre' and 'force perpendicular to the velocity'
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.