Circular motion and radial forces
Connect angular speed and linear speed, then resolve forces toward the centre in horizontal and vertical circles.
Exam scope and assessment notes
Mapped to Cambridge International AS & A Level Further Mathematics 9231. These original explanations and self-checks are independent study support, not Cambridge questions, marks or grades. The syllabus assumes prior knowledge of the relevant 9709 Mathematics content; check each component's prerequisites with your teacher. Full-paper assessment and human subject-specialist review remain outstanding.
Before you begin
- Cambridge expects learners to know the whole Cambridge International A Level Mathematics (9709) content. Paper 1 further pure mathematics draws on 9709 Papers 1 and 3. Confirm your exam year and component route with your school.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What makes a Further Mathematics solution convincing?
What you will learn
- Use v=rω and the radial acceleration v²/r=ω²r
- Model constant-speed motion in horizontal circles
- Solve vertical-circle questions about tension, normal reaction and complete motion
Acceleration points toward the centre
For a particle moving at constant speed in a circle of radius r, v=rω and the acceleration magnitude is v²/r=ω²r, directed radially inward. The velocity direction changes even when speed does not.
The resultant radial force is mv²/r. Identify the real forces first; the phrase ‘centripetal force’ names their inward resultant rather than an extra force.
Vertical circles combine radial force and energy
At the top of a vertical circle, inward is downward; at the bottom, inward is upward. Tension or normal reaction is not automatically directed toward the centre at every point, so draw the force diagram at the specified position.
For a string to remain taut at the top, T≥0, so v²/r≥g. At the limiting case T=0 and v=√(gr); use conservation of energy to relate speeds at different heights when energy is conserved.
Pause and explain
Which habit best checks your work on circular motion and radial forces?
Worked example
A particle moves in a vertical circle of radius 2 m on a light string. Find the least speed at the top for the string to remain taut; take g=10 m s⁻².
Show the worked solution
- At the top, the inward direction is downward.
- At limiting tautness, tension is zero and weight alone supplies mv²/r=mg.
- Therefore v²=gr=20.
Answer Least speed √20=2√5 m s⁻¹.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A 0.5 kg particle moves at 4 m s⁻¹ in a horizontal circle of radius 2 m. Find the required inward resultant force.
Give me a hint
Use F=mv²/r.
Compare my reasoning
- The radial acceleration is 4²/2=8 m s⁻².
- Multiply by mass 0.5 kg.
- State the direction toward the centre.
4 N toward the centre.
Look for these in your work
- Radial acceleration calculated
- Mass included
- Direction stated
A particle moves in a vertical circle of radius 5 m. At the top its speed is 8 m s⁻¹. With g=10 m s⁻², find the string tension there for mass 2 kg.
Give me a hint
At the top both tension and weight point inward.
Compare my reasoning
- The inward resultant is mv²/r=2(64)/5=25.6 N.
- At the top, T+mg=25.6.
- Subtract the weight 20 N.
T=5.6 N; the string remains taut.
Look for these in your work
- Top direction identified
- Radial resultant used
- Tension remains non-negative
A bead on a smooth circular track has zero normal reaction at the top. Explain what this says about its speed there and why it matters.
Give me a hint
With no normal reaction, gravity alone supplies the inward force.
Compare my reasoning
- At the top, mg=mv²/r.
- Thus v²=gr and v=√(gr).
- This is the limiting speed for maintaining contact; a lower speed would require the track to pull, which it cannot do.
v=√(gr), the limiting contact speed at the top.
Look for these in your work
- Force balance correct
- Speed relation derived
- Contact limitation explained
Common mistakes
- Adding a fictitious centripetal force to the real force diagram
- Using v²/r with the wrong inward direction
- Assuming a string can push when the calculated tension is negative
What can you explain now?
What provides the centripetal force in a circular-motion model?
Compare with the explanation
The inward resultant of the real forces
It is a name for the net radial force, not an additional interaction.
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Make it stick
How do direction and position change the radial force equation in a vertical circle?
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