Where can I learn Centripetal acceleration for Cambridge 9702?

This free lesson on MarkScheme covers syllabus point 12.2 (Centripetal acceleration) for Cambridge A-Level Physics. It includes visual revision, key points, and links to real past-paper questions with mark-scheme marking at markscheme.app/courses/9702/12-2-centripetal-acceleration.

Is Centripetal acceleration (12.2) on 9702 free to study?

Yes. The Physics 9702 course on MarkScheme is 100% free — including premium visual lessons, exam tips, and past-paper marking. No paywall for Centripetal acceleration.

How do I revise Centripetal acceleration for 9702 exams?

Start with the visual steps on this page, note the key exam points, then attempt a 9702 past-paper question on Centripetal acceleration using MarkScheme's /mark tool. Mark strictly against the official scheme and re-read any marks you lost.

MarkScheme 9702

Opening…

12.2Physics20 min readPREMIUM

Centripetal acceleration

Welcome to the fascinating world of circular motion! You might intuitively think that if an object moves at a constant speed in a circle, it isn't accelerating. However, in physics, velocity is a vector, meaning it has both magnitude (speed) and direction. When the direction constantly changes, even if speed is constant, the object is accelerating. This specific acceleration is called centripetal acceleration, and it's key to understanding anything from satellites orbiting Earth to cars rounding a bend. This pilot aligns with Senpai Corner A Level notes (2025–2027) for topic 12.2.

BY THE END, YOU CAN…

exam checklist — tick these off

1Explain how an object moving at constant speed in a circle still accelerates.
2State and apply the formulas for centripetal acceleration in terms of linear and angular speed.
3Relate centripetal force to centripetal acceleration and solve problems involving both.
412.2.1: understand that a force of constant magnitude that is always perpendicular to the direction of motion causes centripetal acceleration
512.2.2: understand that centripetal acceleration causes circular motion with a constant angular speed
612.2.3: recall and use $a = rω^2$ and $a = v^2/r$
712.2.4: recall and use $F = mrω^2$ and $F = mv^2/r$

STUDY MODE

READING

TYPEFACE

Literata — a reading serif made for e-books; set like a textbook, sits well beside maths.

SIZE

AIRY SPACING

Wider letter and line spacing. The one change with solid evidence for readers who lose their place — worth a try whoever you are.

Pick whatever reads fastest for you — it differs from person to person, by up to a third.

EXPLAIN SIMPLER

The same notes in plain English — no jargon. Key points and exam tips stay.

DIAGRAM NOTES

A small sketch beside each block of notes.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Invisible Pull of Circular Motion

When an object moves in a circle, its direction constantly changes, even if its speed stays the same. This change in direction means it's accelerating, and that acceleration is always pulling it towards the centre of the circle.

THINK OF IT LIKE…

Imagine you're swinging a ball on a string in a horizontal circle above your head. You have to keep pulling the string towards your hand to make the ball go in a circle. If you let go, the ball flies off in a straight line. That pull is like the centripetal force, and it creates the centripetal acceleration.

1
Speed is constant, but velocity changes (direction!).
2
Change in velocity = acceleration (centripetal acceleration).
3
This acceleration points inwards, towards the circle's centre.
4
An inward force (centripetal force) causes this acceleration.

What this topic covers

The official Cambridge syllabus points this lesson works through.

12.2.1
Understand that a force of constant magnitude that is always perpendicular to the direction of motion causes centripetal acceleration
12.2.2
Understand that centripetal acceleration causes circular motion with a constant angular speed
12.2.3
Recall and use $a = rω^2$ and $a = v^2/r$
12.2.4
Recall and use $F = mrω^2$ and $F = mv^2/r$

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

Figure Live interactive1 / 4

Loading simulation…

Fetching from PhysicsHub (@mattqdev)…

Circular Motion

Acceleration points to the centre with magnitude v²/r; the velocity vector stays tangent.

If the embed does not load, use Open simulation.

PhysicsHub (@mattqdev) · MIT

Step-synced diagram — highlights what to look for in the simulation above.

Step 1

Speed is constant, but velocity changes (direction!).

18 more simulations for this topic — run them in the Simulations section below

Simulations

Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.

18 simulations · 5 to start with

Start herein this order — each one shows a different piece of the topic

The Physics ClassroomStart here · 19702 12.1 · 9702 12.2 · IB A.2
Uniform Circular Motion
Change the mass, speed or radius and see the velocity, acceleration and net force on an object moving in a circle
Why this one: Change m, v and r one at a time and read the net force: F = mv²/r in numbers.
Try this
1. Double the speed and read the acceleration.
2. Double the radius and compare.
3. Change the mass and compare the net force.
Look for Centripetal acceleration equals v² divided by r.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
3JCN PhysicsStart here · 29702 12.1 · 9702 12.2 · IB A.2
Centripetal Force
Swing a mass in a horizontal circle; vary radius, speed and mass and read the centripetal force
Why this one: Doubling the speed quadruples the force needed; doubling the radius halves it.
Try this
1. Double the speed at fixed radius and mass; read the centripetal force.
2. Double the radius at fixed speed; read the force.
3. Double the mass; read the force.
Look for F = mv²/r: doubling the speed quadruples the force while doubling the radius halves it.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
oPhysicsStart here · 39702 12.1 · 9702 12.2 · IB A.2
The Conical Pendulum
Conical pendulum: adjust radius and view; see tension and weight resolving to give centripetal force
Why this one: Tension's vertical part balances the weight; its horizontal part is the centripetal force.
Try this
1. Increase the radius and watch the tension vector.
2. Change the view to look at the forces from the side.
3. Reduce the radius until the string hangs nearly vertical.
Look for The vertical component of tension balances the weight; the horizontal component supplies the centripetal force.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
The Physics ClassroomStart here · 49702 12.2 · IB A.2
Vertical Circle Simulation (Pendulum Motion)
A ball on a light string, a rigid rod or a loop wall moves in a vertical circle; change a variable and observe the tension and speed
Why this one: Tension is largest at the bottom and smallest at the top; slow down until the string goes slack.
Try this
1. Compare the tension at the top and the bottom.
2. Lower the speed until the string slackens.
3. Swap the string for the rod and repeat.
Look for Tension is largest at the bottom of the circle and smallest at the top.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
PhETStart here · 59702 12.1–12.2
Gravity and Orbits
A planet in a circular orbit — show the velocity and gravity vectors as it goes round.
Why this one: Gravity stays perpendicular to velocity, so the planet's speed is constant while its direction changes.
Try this
1. On “Model”, choose Sun–Earth; turn on “Velocity” and “Gravity Force” vectors.
2. Pause anywhere — the velocity is tangential and the force points to the centre.
3. Increase the Sun’s mass — the planet speeds up and the orbit tightens.
Look for Centripetal force is perpendicular to velocity, so speed stays constant while direction changes.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)

More simulations13 more on this topic — core ones first

PhETCoreJava · best on a laptop9702 12.1–12.2 · 1.4 · IB A.1
Motion in 2D
Drag an object, or pick circular or elliptical motion, and watch the velocity and acceleration vectors.
Try this
1. Choose circular motion — velocity is tangential and acceleration points inward.
2. Choose elliptical — the acceleration no longer points at the centre.
3. Drag freely — speed up and watch the acceleration arrow swing forward.
Look for Uniform circular motion has constant speed but an acceleration toward the centre: v²/r.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
PhETCoreJava · best on a laptop9702 12.1–12.2 · IB A.4 · A.1
Ladybug Revolution
Spin a turntable with a ladybug on it and read angular position, velocity and acceleration alongside the linear values.
Try this
1. Set a steady angular velocity — the ladybug’s speed stays constant.
2. Move the ladybug outward — same ω, larger linear speed.
3. Turn on the acceleration vector — it points to the centre.
Look for v = ωr and a = ω²r: further out means faster, with acceleration always toward the axis.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
oPhysicsCore9702 12.2 · IB A.2
Conical Pendulum: 3D
3D conical pendulum: adjust string length, velocity and view angle; relate speed to cone angle
Try this
1. Increase the velocity and watch the cone angle.
2. Keep the velocity and lengthen the string.
3. Rotate the view angle to see the circle from above.
Look for A faster bob needs more centripetal force, so the string makes a wider cone angle.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
The Physics ClassroomCore9702 12.2 · IB A.2
Horizontal Circle Simulation
A ball on a string, a car on a banked turn and a plane in a horizontal circle; change a variable and observe its effect
Try this
1. Change the string length and watch the ball.
2. Change the bank angle of the turn.
3. Change the speed and compare.
Look for The inward force needed grows with the square of the speed.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
The Physics ClassroomCore9702 12.2 · 9702 5.2 · IB A.2
Roller Coaster Design
Change the design parameters of a coaster and investigate its safety and thrill: hill height, loop radius and speed
Try this
1. Lower the loop radius and watch the top of the loop.
2. Raise the hill height and compare the speed.
3. Compare the speeds at the top and bottom of the loop.
Look for At the top of a loop the car stays on the track when v² is at least gr.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
SimuPhysicsCore9702 12.1 · 9702 12.2 · IB A.2
Circular Motion Simulation
The velocity and acceleration vectors are drawn as an object circles at steady speed
Try this
1. Watch the velocity vector round the circle.
2. Watch the acceleration vector.
3. Change the speed and compare the acceleration.
Look for Velocity is tangent to the circle and acceleration points towards the centre.
Open on SimuPhysicsRuns on their site
SimuPhysics by Mohamed Abdelsalam · Licensed to MarkScheme (site publishes no licence; served with frame-ancestors self)

Key formulas

Tap any symbol to reveal exactly what it means and its units.

KEY RELATION

$a_c = \frac{v^2}{r}$

Tap a symbol — great for exam definitions

KEY RELATION

$a_c = \omega^2 r$

Tap a symbol — great for exam definitions

KEY RELATION

$F_c = m a_c$

Tap a symbol — great for exam definitions

KEY RELATION

$F_c = \frac{mv^2}{r}$

Tap a symbol — great for exam definitions

KEY RELATION

$F_c = m\omega^2 r$

Tap a symbol — great for exam definitions

Full topic notes

Formal explanation with the rigour you need for the exam.

The Surprise of Acceleration

Imagine an object tracing a perfect circle. While its speedometer might read a steady value, its motion isn't uniform in the purest sense. Velocity, being a vector, is defined by both how fast an object is moving and in which direction. In circular motion, the direction of travel is continuously shifting. This continuous alteration of the velocity vector means there's a constant rate of change of velocity, which is, by definition, an acceleration.

Not clear?

Centripetal Acceleration: Always Inwards

This unique acceleration experienced by objects in circular motion is termed centripetal acceleration ( $a_c$ ). Crucially, its direction is always perpendicular to the instantaneous linear velocity of the object and points directly towards the absolute centre of the circular path. Without this inward acceleration, the object would simply fly off tangentially in a straight line, as dictated by Newton's First Law of Motion.

Not clear?

Calculating Centripetal Acceleration (Linear Speed)

To quantify centripetal acceleration, we can use the object's linear speed. Linear speed ( $v$ ) is how fast the object is moving along the circumference, measured in metres per second (m s⁻¹). The radius ( $r$ ) is the distance from the object to the centre of its circular path, measured in metres (m).

$a_c = \frac{v^2}{r}$

Not clear?

Calculating Centripetal Acceleration (Angular Speed)

Alternatively, centripetal acceleration can be expressed using angular speed. Angular speed ( $\omega$ ) is the rate at which the object's angular position changes, measured in radians per second (rad s⁻¹). This formula is particularly useful when dealing with rotational systems.

$a_c = \omega^2 r$

Not clear?

The Centripetal Force

According to Newton's Second Law ( $F=ma$ ), if there's an acceleration, there must be a resultant force causing it. This resultant force, which is responsible for maintaining circular motion, is known as the centripetal force ( $F_c$ ). Just like centripetal acceleration, the centripetal force always acts towards the centre of the circular path. It’s important to remember that centripetal force isn't a new fundamental force; rather, it’s a role played by existing forces like tension, gravity, or friction.

$F_c = m a_c$

By substituting the expressions for $a_c$ , we get two primary formulas for centripetal force:

$F_c = \frac{mv^2}{r}$

$F_c = m\omega^2 r$

12.2 Centripetal acceleration.
During a uniform circular motion, an object is continuously changing direction .
Since velocity is a vector , the change in direction would imply that there is an acceleration on the object.
This acceleration is called centripetal acceleration .
The centripetal acceleration is caused by centripetal force .
Centripetal force means centre seeking force as it always acts towards the centre.

Not clear?

Worked examples

See the formulas applied — reveal one step at a time, like the exam.

reveal slowly — mark each step

EXAMPLE 1

A 0.50 kg mass is swung in a horizontal circle of radius 0.80 m at a constant linear speed of 4.0 m s⁻¹. Calculate: a) its centripetal acceleration, and b) the centripetal force acting on it.

1
Identify knowns: $m = 0.50 \text{ kg}$ , $r = 0.80 \text{ m}$ , $v = 4.0 \text{ m s}^{-1}$ .

EXAMPLE 2

A car rounds a bend of radius 50 m at $12\,\text{m s}^{-1}$ . Calculate centripetal acceleration using $a = v^2/r$ and state what provides the centripetal force.

1
$a = v^2/r = 12^2/50 = 2.9\,\text{m s}^{-2}$ (2 s.f.).

How it all connects

The big idea sits in the middle — tap a linked idea to explore the link.

MAIN IDEACentripetal acceleration

Tap a linked idea to see how it connects back to the main topic — that connection is what examiners reward.

Glossary

Key terms for this topic — skim now; the Check step will test them.

centripetal acceleration: This unique acceleration experienced by objects in circular motion is termed centripetal acceleration ( $a_c$ ).
centripetal force: This resultant force, which is responsible for maintaining circular motion, is known as the centripetal force ( $F_c$ ).
In which direction does: Towards the centre of the circular path.

10Apply

Quick check

Write your answer first, then compare it with the model one — the gap is what you would have lost.

0 of 6 answered

11Understand

Teach it back

If you can explain it simply, you own it — gaps here are marks you’d lose.

Teach it back

Explain this topic as if teaching a friend. We name the gaps an examiner would still dock.

Your explanation — no notes

12Remember

Revision flashcards

Guess first, then flip — retrieval beats re-reading.

Your guess — write it before you look

1 / 9 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

close the tab only after these stick

12.2 Centripetal acceleration.
During a uniform circular motion, an object is continuously changing direction .
Since velocity is a vector , the change in direction would imply that there is an acceleration on the object.
This acceleration is called centripetal acceleration .
The centripetal acceleration is caused by centripetal force .
Centripetal force means centre seeking force as it always acts towards the centre.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Question 1 of 29702/42 · Q1(a)(i)

PhysicsOctober/November 2024

(a) (i)

For point X, determine: (i) the speed

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/O/N/24

Question 2 of 29702/42 · Q1(c)(ii)

PhysicsMay/June 2024

Calculate the acceleration a of the centre of gravity of the modelling clay.

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/M/J/24

Extra simulations & links

PhET, GeoGebra and other curated tools — open in a new tab.

Frequently asked

Why does an object moving at constant speed in a circle still accelerate?

What is the key difference between centripetal acceleration and linear acceleration?

Can an object moving in a circle have zero centripetal force?

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do 9702/42 · Q1(a)(i) on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS12.1 Kinematics of uniform circular motion NEXT →13.1 Gravitational field

Discuss Centripetal acceleration

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Revise Centripetal acceleration for Cambridge 9702 (12.2)

Looking for free Physics 9702 notes on Centripetal acceleration? This A-Level lesson covers official syllabus point 12.2 on Paper 4: A Level Structured Questions. Use the visual steps, key terms, and quick checks below — then practise a real 9702 past-paper question with mark-scheme marking. Built for students searching "Centripetal acceleration 9702 revision", "9702 12.2 notes", or a free alternative to paid note sites.

Full Physics 9702 course Mark a past paper →9702 subject hub

12.2Physics20 min readPREMIUM

Centripetal acceleration

BY THE END, YOU CAN…

exam checklist — tick these off

1Explain how an object moving at constant speed in a circle still accelerates.
2State and apply the formulas for centripetal acceleration in terms of linear and angular speed.
3Relate centripetal force to centripetal acceleration and solve problems involving both.
412.2.1: understand that a force of constant magnitude that is always perpendicular to the direction of motion causes centripetal acceleration
512.2.2: understand that centripetal acceleration causes circular motion with a constant angular speed
612.2.3: recall and use $a = rω^2$ and $a = v^2/r$
712.2.4: recall and use $F = mrω^2$ and $F = mv^2/r$

STUDY MODE

READING

TYPEFACE

Literata — a reading serif made for e-books; set like a textbook, sits well beside maths.

SIZE

AIRY SPACING

Wider letter and line spacing. The one change with solid evidence for readers who lose their place — worth a try whoever you are.

Pick whatever reads fastest for you — it differs from person to person, by up to a third.

EXPLAIN SIMPLER

The same notes in plain English — no jargon. Key points and exam tips stay.

DIAGRAM NOTES

A small sketch beside each block of notes.

In simple terms

A friendly intro before the formal notes — no formulas yet.

The Invisible Pull of Circular Motion

THINK OF IT LIKE…

1
Speed is constant, but velocity changes (direction!).
2
Change in velocity = acceleration (centripetal acceleration).
3
This acceleration points inwards, towards the circle's centre.
4
An inward force (centripetal force) causes this acceleration.

What this topic covers

The official Cambridge syllabus points this lesson works through.

12.2.1
Understand that a force of constant magnitude that is always perpendicular to the direction of motion causes centripetal acceleration
12.2.2
Understand that centripetal acceleration causes circular motion with a constant angular speed
12.2.3
Recall and use $a = rω^2$ and $a = v^2/r$
12.2.4
Recall and use $F = mrω^2$ and $F = mv^2/r$

Explore the concept

Use the live diagram, PhET or GeoGebra sim, and synced steps — play it, drag controls, or tap a step.

Figure Live interactive1 / 4

Loading simulation…

Fetching from PhysicsHub (@mattqdev)…

Circular Motion

Acceleration points to the centre with magnitude v²/r; the velocity vector stays tangent.

If the embed does not load, use Open simulation.

PhysicsHub (@mattqdev) · MIT

Step-synced diagram — highlights what to look for in the simulation above.

Step 1

Speed is constant, but velocity changes (direction!).

18 more simulations for this topic — run them in the Simulations section below

Simulations

Every simulation here runs the real model — try the steps on a card, then check what you see against the notes.

18 simulations · 5 to start with

Start herein this order — each one shows a different piece of the topic

The Physics ClassroomStart here · 19702 12.1 · 9702 12.2 · IB A.2
Uniform Circular Motion
Change the mass, speed or radius and see the velocity, acceleration and net force on an object moving in a circle
Why this one: Change m, v and r one at a time and read the net force: F = mv²/r in numbers.
Try this
1. Double the speed and read the acceleration.
2. Double the radius and compare.
3. Change the mass and compare the net force.
Look for Centripetal acceleration equals v² divided by r.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
3JCN PhysicsStart here · 29702 12.1 · 9702 12.2 · IB A.2
Centripetal Force
Swing a mass in a horizontal circle; vary radius, speed and mass and read the centripetal force
Why this one: Doubling the speed quadruples the force needed; doubling the radius halves it.
Try this
1. Double the speed at fixed radius and mass; read the centripetal force.
2. Double the radius at fixed speed; read the force.
3. Double the mass; read the force.
Look for F = mv²/r: doubling the speed quadruples the force while doubling the radius halves it.
Open on 3JCN Physics
3JCN Physics Simulation by Thomas Nguyen · CC BY 4.0
oPhysicsStart here · 39702 12.1 · 9702 12.2 · IB A.2
The Conical Pendulum
Conical pendulum: adjust radius and view; see tension and weight resolving to give centripetal force
Why this one: Tension's vertical part balances the weight; its horizontal part is the centripetal force.
Try this
1. Increase the radius and watch the tension vector.
2. Change the view to look at the forces from the side.
3. Reduce the radius until the string hangs nearly vertical.
Look for The vertical component of tension balances the weight; the horizontal component supplies the centripetal force.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
The Physics ClassroomStart here · 49702 12.2 · IB A.2
Vertical Circle Simulation (Pendulum Motion)
A ball on a light string, a rigid rod or a loop wall moves in a vertical circle; change a variable and observe the tension and speed
Why this one: Tension is largest at the bottom and smallest at the top; slow down until the string goes slack.
Try this
1. Compare the tension at the top and the bottom.
2. Lower the speed until the string slackens.
3. Swap the string for the rod and repeat.
Look for Tension is largest at the bottom of the circle and smallest at the top.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
PhETStart here · 59702 12.1–12.2
Gravity and Orbits
A planet in a circular orbit — show the velocity and gravity vectors as it goes round.
Why this one: Gravity stays perpendicular to velocity, so the planet's speed is constant while its direction changes.
Try this
1. On “Model”, choose Sun–Earth; turn on “Velocity” and “Gravity Force” vectors.
2. Pause anywhere — the velocity is tangential and the force points to the centre.
3. Increase the Sun’s mass — the planet speeds up and the orbit tightens.
Look for Centripetal force is perpendicular to velocity, so speed stays constant while direction changes.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)

More simulations13 more on this topic — core ones first

PhETCoreJava · best on a laptop9702 12.1–12.2 · 1.4 · IB A.1
Motion in 2D
Drag an object, or pick circular or elliptical motion, and watch the velocity and acceleration vectors.
Try this
1. Choose circular motion — velocity is tangential and acceleration points inward.
2. Choose elliptical — the acceleration no longer points at the centre.
3. Drag freely — speed up and watch the acceleration arrow swing forward.
Look for Uniform circular motion has constant speed but an acceleration toward the centre: v²/r.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
PhETCoreJava · best on a laptop9702 12.1–12.2 · IB A.4 · A.1
Ladybug Revolution
Spin a turntable with a ladybug on it and read angular position, velocity and acceleration alongside the linear values.
Try this
1. Set a steady angular velocity — the ladybug’s speed stays constant.
2. Move the ladybug outward — same ω, larger linear speed.
3. Turn on the acceleration vector — it points to the centre.
Look for v = ωr and a = ω²r: further out means faster, with acceleration always toward the axis.
Open on PhET
Simulation by PhET Interactive Simulations, University of Colorado Boulder · Licensed to MarkScheme (public licence CC BY-NC 4.0 since 2026-03-30)
oPhysicsCore9702 12.2 · IB A.2
Conical Pendulum: 3D
3D conical pendulum: adjust string length, velocity and view angle; relate speed to cone angle
Try this
1. Increase the velocity and watch the cone angle.
2. Keep the velocity and lengthen the string.
3. Rotate the view angle to see the circle from above.
Look for A faster bob needs more centripetal force, so the string makes a wider cone angle.
Open on oPhysics
Simulation by Tom Walsh, oPhysics.com — made with GeoGebra · Licensed to MarkScheme (site: free for non-profit educational use; applets made with GeoGebra)
The Physics ClassroomCore9702 12.2 · IB A.2
Horizontal Circle Simulation
A ball on a string, a car on a banked turn and a plane in a horizontal circle; change a variable and observe its effect
Try this
1. Change the string length and watch the ball.
2. Change the bank angle of the turn.
3. Change the speed and compare.
Look for The inward force needed grows with the square of the speed.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
The Physics ClassroomCore9702 12.2 · 9702 5.2 · IB A.2
Roller Coaster Design
Change the design parameters of a coaster and investigate its safety and thrill: hill height, loop radius and speed
Try this
1. Lower the loop radius and watch the top of the loop.
2. Raise the hill height and compare the speed.
3. Compare the speeds at the top and bottom of the loop.
Look for At the top of a loop the car stays on the track when v² is at least gr.
Open on The Physics Classroom
Physics Interactives by The Physics Classroom · Licensed to MarkScheme (site terms otherwise permit linking only)
SimuPhysicsCore9702 12.1 · 9702 12.2 · IB A.2
Circular Motion Simulation
The velocity and acceleration vectors are drawn as an object circles at steady speed
Try this
1. Watch the velocity vector round the circle.
2. Watch the acceleration vector.
3. Change the speed and compare the acceleration.
Look for Velocity is tangent to the circle and acceleration points towards the centre.
Open on SimuPhysicsRuns on their site
SimuPhysics by Mohamed Abdelsalam · Licensed to MarkScheme (site publishes no licence; served with frame-ancestors self)

Key formulas

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KEY RELATION

$a_c = \frac{v^2}{r}$

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KEY RELATION

$a_c = \omega^2 r$

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KEY RELATION

$F_c = m a_c$

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KEY RELATION

$F_c = \frac{mv^2}{r}$

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KEY RELATION

$F_c = m\omega^2 r$

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Full topic notes

Formal explanation with the rigour you need for the exam.

The Surprise of Acceleration

Not clear?

Centripetal Acceleration: Always Inwards

Not clear?

Calculating Centripetal Acceleration (Linear Speed)

$a_c = \frac{v^2}{r}$

Not clear?

Calculating Centripetal Acceleration (Angular Speed)

$a_c = \omega^2 r$

Not clear?

The Centripetal Force

$F_c = m a_c$

By substituting the expressions for $a_c$ , we get two primary formulas for centripetal force:

$F_c = \frac{mv^2}{r}$

$F_c = m\omega^2 r$

12.2 Centripetal acceleration.
During a uniform circular motion, an object is continuously changing direction .
Since velocity is a vector , the change in direction would imply that there is an acceleration on the object.
This acceleration is called centripetal acceleration .
The centripetal acceleration is caused by centripetal force .
Centripetal force means centre seeking force as it always acts towards the centre.

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Worked examples

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reveal slowly — mark each step

EXAMPLE 1

A 0.50 kg mass is swung in a horizontal circle of radius 0.80 m at a constant linear speed of 4.0 m s⁻¹. Calculate: a) its centripetal acceleration, and b) the centripetal force acting on it.

1
Identify knowns: $m = 0.50 \text{ kg}$ , $r = 0.80 \text{ m}$ , $v = 4.0 \text{ m s}^{-1}$ .

EXAMPLE 2

A car rounds a bend of radius 50 m at $12\,\text{m s}^{-1}$ . Calculate centripetal acceleration using $a = v^2/r$ and state what provides the centripetal force.

1
$a = v^2/r = 12^2/50 = 2.9\,\text{m s}^{-2}$ (2 s.f.).

How it all connects

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MAIN IDEACentripetal acceleration

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Glossary

Key terms for this topic — skim now; the Check step will test them.

centripetal acceleration: This unique acceleration experienced by objects in circular motion is termed centripetal acceleration ( $a_c$ ).
centripetal force: This resultant force, which is responsible for maintaining circular motion, is known as the centripetal force ( $F_c$ ).
In which direction does: Towards the centre of the circular path.

10Apply

Quick check

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0 of 6 answered

11Understand

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Teach it back

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Your explanation — no notes

12Remember

Revision flashcards

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Your guess — write it before you look

1 / 9 · ← → space

Key takeaways

Review these before you close the topic — retrieval beats re-reading.

close the tab only after these stick

12.2 Centripetal acceleration.
During a uniform circular motion, an object is continuously changing direction .
Since velocity is a vector , the change in direction would imply that there is an acceleration on the object.
This acceleration is called centripetal acceleration .
The centripetal acceleration is caused by centripetal force .
Centripetal force means centre seeking force as it always acts towards the centre.

Practice — then mark it

The whole point: a real Cambridge question, marked mark-by-mark.

Question 1 of 29702/42 · Q1(a)(i)

PhysicsOctober/November 2024

(a) (i)

For point X, determine: (i) the speed

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/O/N/24

Question 2 of 29702/42 · Q1(c)(ii)

PhysicsMay/June 2024

Calculate the acceleration a of the centre of gravity of the modelling clay.

Do it on paper → mark itMARKED MARK-BY-MARK · B1 / M1 / A1 · OFFICIAL SCHEME

9702/42/M/J/24

Extra simulations & links

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Frequently asked

Why does an object moving at constant speed in a circle still accelerate?

What is the key difference between centripetal acceleration and linear acceleration?

Can an object moving in a circle have zero centripetal force?

Checkpoint

One marked question is worth ten re-reads — close the loop before you move on.

Checkpoint

Reading it isn’t knowing it — prove it.

Before you move on: do 9702/42 · Q1(a)(i) on paper, snap a photo, and get examiner-style feedback on exactly where you win and lose marks.

Attempt & get marked →Takes about a minute · you’ll land back on this lesson

← PREVIOUS12.1 Kinematics of uniform circular motion NEXT →13.1 Gravitational field

Discuss Centripetal acceleration

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Revise Centripetal acceleration for Cambridge 9702 (12.2)

In simple terms

What this topic covers

Explore the concept

Simulations

Uniform Circular Motion

Centripetal Force

The Conical Pendulum

Vertical Circle Simulation (Pendulum Motion)

Gravity and Orbits

Motion in 2D

Ladybug Revolution

Conical Pendulum: 3D

Horizontal Circle Simulation

Roller Coaster Design

Circular Motion Simulation

Key formulas

Full topic notes

The Surprise of Acceleration

Centripetal Acceleration: Always Inwards

Calculating Centripetal Acceleration (Linear Speed)

Calculating Centripetal Acceleration (Angular Speed)

The Centripetal Force

Worked examples

How it all connects

Glossary

Quick check

Teach it back

Revision flashcards

Key takeaways

Practice — then mark it

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Centripetal acceleration

Revise Centripetal acceleration for Cambridge 9702 (12.2)

In simple terms

What this topic covers

Explore the concept

Simulations

Uniform Circular Motion

Centripetal Force

The Conical Pendulum

Vertical Circle Simulation (Pendulum Motion)

Gravity and Orbits

Motion in 2D

Ladybug Revolution

Conical Pendulum: 3D

Horizontal Circle Simulation

Roller Coaster Design

Circular Motion Simulation

Key formulas

Full topic notes

The Surprise of Acceleration

Centripetal Acceleration: Always Inwards

Calculating Centripetal Acceleration (Linear Speed)

Calculating Centripetal Acceleration (Angular Speed)

The Centripetal Force

Worked examples

How it all connects

Glossary

Quick check

Teach it back

Revision flashcards

Key takeaways

Practice — then mark it

Extra simulations & links

Frequently asked

Checkpoint

Reading it isn’t knowing it — prove it.

Discuss Centripetal acceleration