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Circular-Motion Force Models

Key idea: Build circular-motion force models from real interactions, a radial direction and Newton's second law.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Express angular displacement in radians and use s = rθ.
  • Relate angular velocity, period, frequency and tangential speed using v = rω.
  • Explain and apply centripetal acceleration and resultant-force relationships.

Circular-motion dynamics is Newton’s second law applied along a changing radial direction. The reliable method is to model the interactions first and introduce mv²/r only as the radial ma term.

1. The Five-Step Method

Step 1: Isolate the object

Choose the object whose motion you are analysing. Draw only forces exerted on that object by other bodies: weight, tension, normal reaction, friction, or another stated interaction.

Never add a centripetal-force arrow

“Centripetal” describes the direction of the resultant force. It is not another interaction to place beside tension, friction, weight or normal reaction.

Step 2: Mark the centre and radial direction

At the instant shown, draw an axis pointing towards the centre and define it as positive. The inward direction can be horizontal, upward, downward or inclined depending on the object’s position.

Step 3: Resolve the real forces

Take the component of each real force along the radial axis. Forces perpendicular to that axis do not enter the radial equation. If the speed changes, use a separate tangential equation for the tangential component.

Step 4: Write Newton’s second law symbolically

With inward positive,

∑ F_inward = maᵣ = mv²/r = mrω².

The left side contains named forces or components. The right side contains the required radial acceleration.

Step 5: Solve and check the model

Substitute only after the symbolic equation is correct. Then check:

  • units are consistent;
  • the radius belongs to the path of the chosen object;
  • the answer respects physical constraints such as tension ≥ 0;
  • increasing speed at fixed radius demands a larger inward resultant proportional to v².
Real forces in two circular-motion modelsA top-view car on a level bend has static friction directed towards the centre. A ball at the top of a vertical circle has both tension and weight directed towards the centre. Neither diagram includes an additional centripetal-force arrow.Level bend: top viewfrictionradial: friction = mv²/rVertical circle: at toptensionweightradial: tension + weight = mv²/r
Scroll diagram horizontally to read all labels.
Draw only interactions on the free-body diagram. Their inward resultant equals mv²/r; centripetal force is not an extra interaction.

2. Choosing the Radial Equation

SituationReal radial forcesInward equation
Car on a level bendhorizontal frictionf = mv²/r
Mass on a horizontal table, tied to the centretensionT = mv²/r
Ball at the top of a vertical circletension and weight both inwardT + mg = mv²/r
Ball at the bottom of a vertical circletension inward, weight outwardT-mg = mv²/r
Frictionless banked bendhorizontal component of normal reactionN sin θ = mv²/r

These are not separate formulae to memorise. Each row follows from the same free-body-diagram method.

4. Common Mistakes

  • Drawing mv²/r as a force arrow instead of putting it on the ma side.
  • Assuming tension, friction or weight always equals mv²/r without checking other radial forces.
  • Using the same signed equation at the top and bottom of a vertical circle.
  • Resolving forces relative to the page rather than relative to the radial axis.
  • Using the diameter instead of the radius of the object’s path.

5. Exam Tips

  • Draw and label only real interactions before writing a radial equation.
  • Redraw the inward direction at every position in a vertical circle.
  • Keep the equation symbolic until the force signs and components are settled.

6. Worked Examples

Modelled example 1

Level bend with a given friction limit

Core

Problem

A 900 kg car takes a level bend of radius 45 m. Maximum horizontal friction is 6.5 kN. Find the greatest speed without skidding.
Study the worked solution

  1. Model real interactions

    Method

    Balance weight and normal reaction vertically and use friction as the horizontal inward resultant.

    Reason

    Only horizontal friction points towards the centre.

    Working

    f = mv²/r

  2. Apply the limiting condition

    Method

    Set f = 6500 N and solve for speed.

    Reason

    The greatest speed occurs when all available friction is used.

    Working

    v = square root of (6500(45)/900) = 18.0 m s⁻¹

Guided practice 2

Rebuild the radial equation at each position

About 8 min

Problem

A 0.30 kg ball moves in a vertical circle of radius 0.80 m at 5.0 m s⁻¹ at the top and 7.0 m s⁻¹ at the bottom. Find both tensions; g = 9.81 m s⁻².

Try this before viewing the solution

Hints

Hint 1: top: inward is downward
At the top, tension and weight are both inward.
Hint 2: bottom: inward is upward
At the bottom, tension is inward but weight is outward.
View solution step by step

  1. Top equation

    Method

    Write Tₜₒₚ + mg = mvₜₒₚ²/r.

    Reason

    Both forces point towards the centre.

    Working

    Tₜₒₚ = 6.43 N

  2. Bottom equation

    Method

    Write T_bottom-mg = mv_bottom²/r.

    Reason

    Weight is opposite the upward inward direction.

    Working

    T_bottom = 21.3 N

Common misconception 3

An extra “centripetal force” arrow

Find and correct the mistake

Learner free-body diagram

For a mass tied to the centre of a horizontal table, a learner draws inward tension and a second inward arrow labelled mv²/r. Locate the first modelling error and write the radial equation.

Try this before viewing the solution

First error

View solution step by step

  1. List real interactions

    Method

    Keep tension, weight and normal reaction only.

    Reason

    Forces must arise from physical interactions.

    Working

    Vertically N = mg; radially only T remains.

  2. Apply Newton's second law

    Method

    Put mv²/r on the radial ma side.

    Reason

    It is the product of mass and centripetal acceleration.

    Working

    T = ∑ F_radial = mv²/r

Examiner practice 4

Banked bend from force components

4 marks

Examination derivation

On a frictionless banked bend, the normal reaction is at angle θ to the vertical. Derive the design-speed relation. [4 marks]

Try this before viewing the solution

View solution step by step

  1. Vertical balance

    1 mark

    Method

    Resolve N cos θ = mg.

    Reason

    There is no vertical acceleration.

    Working

    N cos θ = mg.

  2. Radial component

    1 mark

    Method

    Resolve N sin θ = mv²/r.

    Reason

    The horizontal component supplies the inward resultant.

    Working

    N sin θ = mv²/r.

  3. Eliminate shared factors

    1 mark

    Method

    Divide the radial equation by the vertical equation.

    Reason

    This removes both N and m.

    Working

    tan θ = v²/(rg).

  4. State design relation

    1 mark

    Method

    Give v = square root of (rg tan θ) if speed is required.

    Reason

    It follows from the force model rather than a memorised extra law.

    Working

    v = square root of (rg tan θ)

Challenge 5

Conical-pendulum force model

Minimal support

Component-model transfer

A mass on a string moves as a conical pendulum, with the string at angle θ to the vertical and circular radius r. Derive a relation between v, r, g and θ.

Try this before viewing the solution

Hints

Hint 1: reuse the component structure
Tension has a vertical component balancing weight and a horizontal component supplying radial acceleration.
View solution step by step

  1. Resolve real forces

    Method

    Write T cos θ = mg vertically and T sin θ = mv²/r radially.

    Reason

    The mass has no vertical acceleration but does have horizontal inward acceleration.

    Working

    T cos θ = mg, T sin θ = mv²/r

  2. Eliminate tension

    Method

    Divide the radial equation by the vertical equation.

    Reason

    The shared tension and mass cancel.

    Working

    tan θ = v²/rg, v = square root of (rg tan θ)

7. Mind Stretchers

Mind stretcher 1: Diagnose an impossible modelExtension

At the top of a vertical circle, a student’s calculation gives a string tension of -2.0 N. What does the result mean physically?

Show Answer

A string cannot push, so negative tension is impossible. The assumed taut circular path cannot be maintained at that speed. The string goes slack and the subsequent motion is no longer the same circular-motion model.

After this method lesson, use the quick fluency warm-up for retrieval, the model-evidence form for representation and model discrimination, and the structured set for the independent production evidence required before a strong topic interpretation.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027