Centripetal acceleration and radial force models
Key idea: H2 Physics lessons on angular displacement, angular velocity, tangential speed, centripetal acceleration and radial force models.
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The core idea
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Big question: What force keeps velocity turning toward the centre?
Uniform circular motion has inward acceleration a = v²/r = rω² because velocity changes direction. 'Centripetal force' is not a new interaction: it is the inward resultant supplied by tension, gravity, friction, normal reaction or a combination. Draw the real forces, resolve inward, then set their radial resultant equal to mv²/r.
Acceleration can change direction without changing speed
In uniform circular motion, velocity is tangent to the path and continually changes direction. Comparing velocity vectors at nearby points shows that their change points towards the centre, so the acceleration is centripetal.
Its magnitude is a = v²/r = rω². At fixed radius, doubling speed makes the required acceleration four times larger; at fixed speed, a tighter curve needs greater acceleration.
Check your understanding: A body's circular speed doubles at the same radius. What happens to centripetal acceleration?
It becomes four times larger because a = v²/r.
Find the real inward resultant
Centripetal force is not an extra interaction. Draw only real forces such as tension, friction, gravity or normal contact, resolve them radially and set their inward resultant equal to mv²/r or mrω².
If the available inward force is too small, the body follows a less curved path. It does not need a new outward force; if the inward force vanishes, its instantaneous motion is tangential.
Check your understanding: What force supplies centripetal acceleration for a car on a flat bend?
Static friction between tyres and road supplies the inward resultant.
Key ideas to keep
- Centripetal acceleration points inward while velocity is tangent.
- Do not add a separate centripetal force to the free-body diagram.
- Constant speed does not mean zero acceleration.
See the reasoning
Worked example
Use angular data to find the inward force
Question: A 0.40 kg mass moves in a horizontal circle of radius 0.80 m with angular velocity 15 rad s⁻¹. Find its centripetal acceleration and the tension if tension is the inward resultant.
Step 1: Choose the angular form
Why: Radius and angular velocity are given directly.
Working: a = rω² = 0.80(15²) = 180 m s⁻².
Step 2: Name the real radial force
Why: Tension, not a separate 'centripetal force', supplies the inward resultant.
Working: T = ma = 0.40(180) = 72 N.
Step 3: Cross-check through tangential speed
Why: The equivalent form must agree.
Working: v = rω = 12 m s⁻¹ and mv²/r = 0.40(12²)/0.80 = 72 N.
Answer: Centripetal acceleration is 180 m s⁻² and tension is 72 N inward.
Check: Both radial forms give the same result and the force arrow points towards the centre.
Another worked model
Question
A 0.30 kg mass on a string moves in a vertical circle of radius 0.80 m. Find the string tension at the top when its speed is 5.0 m s⁻¹ and at the bottom when its speed is 7.0 m s⁻¹.
Check the worked solution
At the top, inward is down: T + mg = mv²/r, so T = 0.30(5.0²)/0.80 − 0.30(9.81) = 6.43 N. At the bottom, inward is up: T − mg = mv²/r, so T = 0.30(7.0²)/0.80 + 0.30(9.81) = 21.3 N.
Use a hint if needed
Practise with support
Try this
A 1200 kg car takes a level bend of radius 60 m at 20 m s⁻¹. Find its acceleration and the frictional force required.
Hint: Static friction is the real force providing the horizontal inward resultant.
Check your answer
a = v²/r = 20²/60 = 6.67 m s⁻². The required inward friction is F = ma = 1200(6.67) = 8.00 × 10³ N.
Now work without the hint
Practise independently
Your turn
A frictionless road is banked at 14° for a bend of radius 65 m. Derive the design-speed relation and calculate the speed for which no sideways friction is needed.
Check your answer
Resolve the normal contact force: N cos θ = mg and N sin θ = mv²/r. Dividing gives tan θ = v²/(rg), so v = √(rg tan θ) = √[65(9.81)tan 14°] = 12.6 m s⁻¹.
Avoid these traps
Common mistakes
Common mistake
The velocity and acceleration of a body in uniform circular motion both point towards the centre.
What is wrong with this reasoning?
Show better thinking
Velocity is tangent to the path. Centripetal acceleration is perpendicular to that velocity and points towards the centre.
Common mistake
Centripetal force is an extra force that must be added to weight, tension or friction.
What is wrong with this reasoning?
Show better thinking
Centripetal force names the inward resultant. Resolve the real forces radially and set their inward resultant equal to mv²/r or mrω².
Common mistake
An inward force must increase a body's speed because it causes acceleration.
What is wrong with this reasoning?
Show better thinking
In uniform circular motion the inward force is perpendicular to velocity, so it changes direction without doing work or changing speed.
Write for the examiner
Exam guidance
Label the centre and write ΣFinward = mv²/r before choosing which real forces contribute.
Exam-style practice [7 marks]
A 950 kg car rounds a flat curve of radius 55 m at 14 m s⁻¹. Find its centripetal acceleration and the friction force required. If maximum friction is 4200 N, calculate the greatest safe speed and explain what happens above it.
Plan before you answer
- Treat friction as the real inward force.
- At the limit set mv²/r equal to maximum friction.
- Describe the motion without inventing an outward force.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
a = 14²/55 = 3.56 m s⁻², so friction must be 950(3.56) = 3.39×10³ N inward. At the limiting friction, v = √(Fr/m) = √[4200(55)/950] = 15.6 m s⁻¹. Above this speed the required inward force exceeds the available friction, so the car follows a less curved path, initially tending towards the tangent; no additional outward force acts.
Come back in three days
Check what stayed with you
Recall question 1
Where do velocity and acceleration point in uniform circular motion?
Check the answer
Velocity is tangential and acceleration is towards the centre.
Recall question 2
State both centripetal-acceleration forms.
Check the answer
a = v²/r = rω².
Recall question 3
What is meant by centripetal force?
Check the answer
The inward resultant of the real forces acting on the body.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. The official topic states no explicit exclusions. Radial equations in this chain are always built from the real forces acting on the body; centripetal force is not added as a separate force.
- GCE A-Level H2 PhysicsTopic 7(d) / Topic 7(e) / Topic 7(f) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027