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Free resources/Edexcel IAL Physics revision/Circular motion

Edexcel IAL Physics notes · A2 — Mechanics & fields

Circular motion

Circular motion requires an inward resultant force because velocity keeps changing direction. Centripetal describes the role of that resultant; it is not an extra force to add to a diagram.

By Dr Desouky Physics AcademyUpdated 25 September 202610 min reading + practice

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In this lesson
  1. 01The key idea
  2. 02Constant speed can still mean acceleration
  3. 03Name the interaction providing the inward resultant
  4. 04State what the circular model assumes
  5. 05Key definitions
  6. 06Equations & units
  7. 07Worked examples
  8. 08Practice & solutions
  9. 09Common mistakes
  10. 10Questions, explained
  11. 11Learning checklist
  12. 12Sources & citation

The key idea

The first A2 topic and the foundation for everything with 'fields' in the name. One idea does all the work: circular motion at constant speed is still acceleration, because the direction of the velocity is always changing.

Understand it · 01

Constant speed can still mean acceleration↗

Velocity includes direction as well as magnitude. An object moving around a circle at constant speed therefore accelerates because its velocity direction continually changes. The instantaneous velocity is tangent to the circle, while the acceleration points towards the centre. For a circle of radius r, the acceleration magnitude is v²/r. Moving twice as fast requires four times the inward acceleration at the same radius, which explains why speed matters so strongly on a bend.

Angular displacement describes the angle swept out at the centre, measured in radians. One complete revolution is 2π radians, so angular speed is 2π/T for period T and also 2πf for frequency f. Tangential speed is rω. Keep rotations per minute distinct from revolutions per second and from radians per second; many otherwise sound solutions lose their scale by substituting an unconverted rotation rate.

Understand it · 02

Name the interaction providing the inward resultant↗

The radial force equation is ΣFinward = mv²/r. The left side must contain actual forces: tension for a mass on a string, friction for a vehicle on a level curve, gravity for an orbit, or combinations of forces for a vertical circle. Do not draw another arrow labelled centripetal force in addition to those interactions. Doing so counts the same physical requirement twice and produces a wrong resultant.

At the top and bottom of a vertical circle, the weight has different signs relative to the inward direction. At the top, inward is downward, so tension and weight can both contribute inward. At the bottom, inward is upward, so weight opposes the inward tension. In non-uniform circular motion there may also be a tangential acceleration changing speed; the radial equation still uses the instantaneous speed at the position being considered.

A point on the right of a circle has tangential velocity upward and acceleration and resultant force towards the centre.
Figure 2. Velocity is tangential; the inward resultant changes its direction.Open full-size SVG diagram ↗

Pause & explain

What force acts outward on a stone after the string keeping it in a horizontal circle breaks?

Check your reasoning

No outward force is needed. With the inward tension removed, its immediate motion is along the tangent, before other forces change it.

Understand it · 03

State what the circular model assumes↗

A horizontal turn model may assume a level surface and sufficient static friction to maintain the path without slipping. If the required mv²/r exceeds the available friction, the vehicle cannot follow that radius at that speed. In a laboratory circular-motion investigation, measure radius to the centre of the rotating mass and time several revolutions. Dividing the total time by the number of turns reduces the relative effect of start–stop timing uncertainty.

Keep radius and the relevant force approximately constant while measuring speed, and account for any tilted string or non-horizontal motion. Tension may have a vertical component balancing weight as well as a horizontal component providing centripetal acceleration. A plot designed from the model, such as force against v² at fixed mass and radius, should be linear. Scatter alone does not prove the model false, but a systematic trend can reveal that one of the controlled quantities changed.

Pause & explain

If speed doubles at a fixed radius, how does required inward force change?

Check your reasoning

It becomes four times larger because F = mv²/r.

Key definitions

Radian
The angle subtended at the centre of a circle by an arc equal in length to the radius. 2π radians = 360°.
Angular velocity (ω)
The angle swept out per unit time: ω = 2π ÷ T. Units: rad/s.
Centripetal force
The resultant force acting towards the centre of the circle, required to keep a body moving in circular motion. It is not a new force — it is provided by tension, gravity, friction or a normal force.

Equations, units & conditions

Speed and angular velocity

v = r ω

Units: m/s

Centripetal acceleration

a = v² ÷ r = r ω²

Units: m/s²

Centripetal force

F = m v² ÷ r = m r ω²

Units: N

Open the Edexcel IAL Physics formula sheet →

Worked examples, step by step

Example 1

A 0.30 kg object slides on a smooth horizontal table in a circle of radius 0.80 m at 2.0 revolutions per second. A horizontal string supplies the inward force. Find its tension.

Follow the method

  1. ω = 2π × 2.0 = 12.6 rad/s.
  2. F = mrω² = 0.30 × 0.80 × (12.6)² = 0.30 × 0.80 × 158.

T ≈ 38 N

Practice questions with worked solutions

Try each question on paper before opening the solution. These are original academy questions; the suggested marks are a self-checking guide, not an official exam-board mark scheme. Difficulty labels describe this practice set, not predicted grades.

  1. Build · 3 suggested marks

    1. Rotation rate to force

    A 0.20 kg object travels in a horizontal circle of radius 0.50 m at 120 revolutions per minute. Find its speed and inward resultant force.

    Show solution and self-check: question 1
    1. 120 revolutions per minute = 2.0 Hz.
    2. v = 2πrf = 2π × 0.50 × 2.0 = 6.283 m s⁻¹.
    3. F = mv²/r = 0.20 × 6.283²/0.50.

    6.3 m s⁻¹ and 16 N to two significant figures.

    Self-check · one mark per point

    • Convert the rotation rate to 2.0 Hz.
    • Calculate tangential speed.
    • Calculate approximately 15.8 N inward.
  2. Build · 3 suggested marks

    2. Bottom of a vertical circle

    A 0.30 kg mass moves at 4.0 m s⁻¹ at the bottom of a vertical circle of radius 0.80 m. Find the string tension, taking g = 9.81 m s⁻².

    Show solution and self-check: question 2
    1. At the bottom, inward is upward: T − mg = mv²/r.
    2. T = 0.30 × 9.81 + 0.30 × 4.0²/0.80.

    8.94 N, approximately 8.9 N.

    Self-check · one mark per point

    • Write T − mg as the inward resultant.
    • Calculate the required inward force of 6.0 N.
    • Add the weight to obtain 8.94 N.

Common mistakes & how to avoid them

  • Adding a 'centripetal force' to the free-body diagram. Centripetal force is the resultant of the real forces — drawing it as an extra force double-counts and wrecks the algebra.
  • Saying the speed changes in uniform circular motion. Speed is constant; velocity changes because direction changes. The acceleration is perpendicular to the velocity, so it never speeds the object up.
  • At the top of a vertical circle, forgetting both weight and tension (or normal force) point towards the centre: mg + T = mv²/r. The minimum-speed condition is T = 0, giving v² = gr.
  • Degrees left in the calculator. Angular equations assume radians — one mode error can void a whole question.

Understand the why

Questions, explained

Start with the short answer, then follow the reasoning. Each explanation has a permanent link you can share with a classmate or return to when revising.

  1. What provides the centripetal force?
  2. Why is there acceleration if the speed is constant?
  3. What is the minimum speed at the top of a vertical circle?
  4. Why is string tension greater at the bottom of a vertical circle?
  5. How do I convert rpm into angular speed and tangential speed?
  6. Does increasing the radius increase or decrease centripetal force?
Browse all physics questions →

What provides the centripetal force?

Centripetal force is the inward resultant required for circular motion, supplied by real interactions such as gravity, tension, friction or contact forces. It is a role those forces play, not an extra force added to them.

Start with a free-body diagram containing the actual interactions. Then identify their components towards the centre and equate the radial resultant to mv²/r. For a satellite, gravity supplies it; for a car on a level bend, static friction can supply it. In a vertical circle, weight may assist or oppose the inward tension depending on position. If you draw both tension and a separate centripetal arrow as additional physical forces, you count the inward effect twice. Non-uniform circular motion can also have a tangential resultant that changes speed.

  • Read the supporting explanation ↑

Why is there acceleration if the speed is constant?

Velocity is a vector, so changing its direction produces acceleration even when its magnitude stays constant. In uniform circular motion the acceleration points towards the centre and has magnitude v²/r.

Compare velocity arrows at two nearby points on a circle. Their magnitudes are equal, but their difference points approximately inward; taking the limit of a short interval gives the radial acceleration. This acceleration is perpendicular to instantaneous velocity, so it changes direction without increasing kinetic energy. If the inward force disappears, the object initially moves along a tangent rather than radially outward. Other forces may then curve that path, as gravity does for a released object near Earth. Constant speed is therefore not the same condition as constant velocity.

  • Read the supporting explanation ↑

What is the minimum speed at the top of a vertical circle?

For a mass on a flexible string at the top of a vertical circle, the limiting contact condition is zero tension. Gravity then supplies the inward force, giving vmin = √(gr). This condition depends on the stated constraint.

A string can pull but cannot push. At a speed below the limiting value, maintaining the circular path would require a negative tension, so the taut-string model fails. A rigid rod can exert a force in either direction, so the same minimum-speed condition does not automatically apply. Likewise, a vehicle’s contact with the inside or outside of a track changes which normal-force direction is possible. Always draw the forces at the actual position. The speed at the bottom needed to reach this top condition also depends on the energy transferred during the ascent.

The top and bottom of a vertical circle show inward directions reversing, with weight downward at both positions and tension inward.
At the top, T + mg supplies the inward force; at the bottom, T − mg does.Open full-size SVG diagram ↗

Top-of-loop limit

A mass on a flexible string moves in a vertical circle of radius 0.60 m. Find the limiting speed at the top, using g = 9.81 m s⁻².

  1. At the limit T = 0, so mg = mv²/r.
  2. v = √(gr) = √(9.81 × 0.60).

2.43 m s⁻¹.

  • Read the supporting explanation ↑

Why is string tension greater at the bottom of a vertical circle?

At the bottom of a vertical circle, tension must both oppose weight and provide the inward resultant. At the top, weight assists the inward tension. For equal speeds, the bottom tension is therefore greater by 2mg.

Choose inward as positive separately at each point. At the bottom, Tbottom − mg = mv²/r; at the top, Ttop + mg = mv²/r. In a freely moving system the speeds at the two positions are usually not equal, because gravitational potential energy changes. Then substitute the appropriate local speed into each equation rather than using the equal-speed difference blindly. This is a good example of a sign convention that changes direction around the path while the actual weight arrow remains downward everywhere.

Bottom tension

A 0.25 kg mass moves at 5.0 m s⁻¹ at the bottom of a 1.0 m radius circle. Find tension using g = 9.81 m s⁻².

  1. T − mg = mv²/r.
  2. T = 0.25 × 9.81 + 0.25 × 5.0²/1.0.

8.70 N.

  • Read the supporting explanation ↑

How do I convert rpm into angular speed and tangential speed?

Convert revolutions per minute to revolutions per second by dividing by 60, then multiply by 2π to obtain angular speed in radians per second. Tangential speed follows from v = rω.

One revolution is a complete turn of 2π radians, while a radian is defined by an arc length equal to the radius. The formulas a = rω² and v = rω assume radian measure because that definition links arc length directly to angle. Using the numerical rpm value as ω can produce an enormous squared error in force. Frequency in hertz and angular frequency are related but are not numerically interchangeable. Keep units beside each conversion, and check whether a quoted diameter needs halving before it is used as r.

  • Read the supporting explanation ↑

Does increasing the radius increase or decrease centripetal force?

At fixed speed, increasing the circular radius reduces the required inward acceleration and force. At fixed angular speed, increasing radius increases them instead, because tangential speed then rises with radius. State what is held constant before comparing.

The two expressions a = v²/r and a = rω² are consistent because v = rω. They answer different controlled comparisons. A car taking a wider bend at the same road speed needs less lateral force. Points farther from the centre of a rigid rotating disc share angular speed, but have greater tangential speed and acceleration. In an experiment, changing radius may unintentionally change rotation rate or tension, so identify the measured and controlled variables before describing a proportional relationship. A formula without its fixed quantities is not a complete prediction.

  • Read the supporting explanation ↑

Your learning checklist

Use these goals to check your understanding. Requirements vary by specification, tier and exam year; the official references below define your full course.

  • Define the radian; convert between degrees and radians
  • Define angular velocity ω; relate it to period and speed: v = rω
  • State that centripetal acceleration a = v²/r = rω² is directed towards the centre
  • Identify the physical force providing the centripetal force in each situation
  • Analyse motion in horizontal and vertical circles
Check which exam board you study

Sources & how to cite these notes

Original explanations, illustrations and practice by Dr Desouky Physics Academy. Official specifications guide the course scope; questions and suggested marks on this page are our own revision material.

  • Pearson Edexcel International AS/A Level Physics — specification

Reference this page

Dr Desouky Physics Academy. Edexcel IAL Physics: Circular motion. Updated 2026-09-25. https://www.drdesouky.com/revision/a-level-physics/circular-motion

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