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A Level Physics 9702 · A Level · Topic 12 · ~5 min read

Motion in a circle

Radians and angular speed, why uniform circular motion needs a resultant force towards the centre, centripetal acceleration a = rω² = v²/r, and applying F = mv²/r to cars on bends, vertical circles and orbits.

RadiansAngular displacementAngular speed ωv = rωCentripetal accelerationCentripetal forceVertical circlesBanked & flat bends
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In 30 seconds

  • Angles in radians: a full turn is 2π rad. Angular speed ω = Δθ/Δt = 2π/T = 2πf.
  • Speed v = rω. Even at constant speed, the direction changes, so the object accelerates towards the centre.
  • Centripetal acceleration a = rω² = v²/r; resultant force F = mv²/r = mrω², always towards the centre.
  • Centripetal force is not a new force. It is the resultant of real forces: tension, friction, gravity, a normal reaction.

By the end you can

  • Define the radian and express angular displacement in radians.
  • Understand and use angular speed ω, and v = rω.
  • Explain that a resultant force of constant size perpendicular to the velocity produces uniform circular motion.
  • Recall and use a = rω² = v²/r and F = mrω² = mv²/r.

1Describing circular motion

Radians and angular speed

Radian — the angle subtended at the centre of a circle by an arc equal in length to the radius.

Angular speed ω — the rate of change of angular displacement: ω = Δθ/Δt.

ω = 2π/T = 2πf ; v = rω
T = period, f = frequency of rotation
Worked example

Find the angular speed of the Earth's rotation, and the centripetal acceleration of a point on the equator (radius 6.37 × 10⁶ m).

  1. ω = 2π/(24 × 3600) = 7.27 × 10⁻⁵ rad s⁻¹ C1 A1
  2. a = rω² = 6.37 × 10⁶ × (7.27 × 10⁻⁵)² = 0.0337 m s⁻² A1
Common mistakeUsing degrees, or rpm, in ω. Convert to rad s⁻¹ first: 300 rpm = 300 × 2π/60 = 31.4 rad s⁻¹.

2Centripetal force

Why it accelerates

rv (tangent)F, atowards the centrev = 2πr/T = ωra = v²/r = ω²r = 4π²r/T²F = mv²/rThe force is at 90° to v,so it changes direction,not speed. No work done.Something must supply it:tension, friction, gravity…
Velocity is along the tangent; acceleration and resultant force point to the centre.
a = rω² = v²/r ; F = mrω² = mv²/r
directed towards the centre of the circle
  • Why no work is done: the force is perpendicular to the velocity, so the speed (and kinetic energy) stays constant.
  • If the force is removed, the object moves off along the tangent (Newton's first law), not outwards.
Worked example

A car of mass 1200 kg travels round a flat bend of radius 50 m at 15 m s⁻¹. Find the friction force needed.

  1. F = mv²/r = 1200 × 15²/50 C1
  2. F = 5400 N, provided by friction towards the centre A1

Vertical circles

  • At the top: T + mg = mv²/r (both point down, to the centre). At the bottom: T − mg = mv²/r. So the tension is greatest at the bottom.
  • Least speed at the top (string just slack, T = 0): mg = mv²/r, so v = √(gr).
Worked example

A 0.20 kg mass whirls in a vertical circle on a string of length 0.80 m. (a) Find the least speed at the top for the string to stay taut. (b) At the bottom its speed is 6.0 m s⁻¹; find the tension.

  1. (a) v = √(gr) = √(9.81 × 0.80) = 2.80 m s⁻¹ C1 A1
  2. (b) T − mg = mv²/r ⇒ T = 0.20 × 9.81 + 0.20 × 36/0.80 C1
  3. T = 11.0 N A1
Common mistakeDrawing "centripetal force" as an extra arrow on a free-body diagram. Draw only real forces; their resultant is the centripetal force. This was a common 2024 Paper 4 error.
Linking questionGravity as the centripetal force for orbits is in Gravitational fields; charged particles moving in circles are in Magnetic fields; and SHM is the projection of circular motion: Oscillations.

Test yourself — cover the answers first

Quick check

  1. Convert 90° to radians.
  2. A wheel turns at 5 revolutions per second. Find ω.
  3. What is the direction of centripetal acceleration?
  4. Why does the kinetic energy of an object in uniform circular motion stay constant?
  5. Find v for r = 0.5 m and ω = 8 rad s⁻¹.

Exam-style

  1. A 50 g stone on a 0.40 m string moves in a horizontal circle at 3.0 revolutions per second. Find the tension (ignore gravity).
  2. Explain why passengers feel lighter at the top of a hump-backed bridge.
  3. A satellite orbits at 7.5 km s⁻¹ at radius 7.0 × 10⁶ m. Find its centripetal acceleration.
Show answers
  1. π/2
  2. 10π = 31.4 rad s⁻¹
  3. towards the centre of the circle
  4. the force is perpendicular to the velocity, so it does no work
  5. 4 m s⁻¹
  6. ω = 6π = 18.85 rad s⁻¹; T = mrω² = 0.050 × 0.40 × 355 = 7.1 N
  7. mg − R = mv²/r, so R < mg; the reaction (felt weight) is reduced
  8. a = v²/r = (7500)²/7.0 × 10⁶ = 8.0 m s⁻²
Practise this chapter
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Independent notes written by Riyaz Sir. Not produced, endorsed or approved by Cambridge University Press & Assessment. Syllabus codes 9709 and 9702 refer to the Cambridge International AS & A Level syllabuses.