Free Pearson Edexcel International A Level Physics lesson

Motion in a circle

A constant speed can hide a continuous change in velocity.

What you’ll learn

  • Relate angular and linear motion.
  • Derive inward acceleration from changing velocity.
  • Identify the actual force providing circular motion.

Follow an angle rather than a distance

A radian is defined by angle θ = s/r, where s is arc length. One complete circumference gives 2π radians, equivalent to 360°. The radian is dimensionless in an SI dimensional check, but writing rad distinguishes angular displacement from a numerical count. Convert degree values before using equations derived from arc length.

Angular velocity is the rate of change of angular displacement. For uniform rotation, ω = 2π/T = 2πf. A point at radius r travels arc length rθ, giving tangential speed v = rω. Two points fixed on a rigid disc complete a revolution together and therefore share ω, but the outer point travels further and has larger v.

Period can be found by timing many revolutions and dividing by the count. Mark one repeatable reference position and count complete cycles, not crossings without direction. If the disc accelerates, a many-turn average does not equal its angular velocity at every instant. Use short calibrated intervals or an encoder when the change during the observation matters.

θ = s/r; 2π rad = 360°; ω = 2π/T = 2πf; v = rω

Symbols and units

  • θ is angular displacement in rad
  • s is arc length in m and r is radius in m
  • ω is angular speed in rad s⁻¹
  • T is period in s
  • f is frequency in Hz
  • v is tangential speed in m s⁻¹.

How the equation explains the idea

  1. The radian definition is θ = s/r. In one revolution θ = 2π, so uniform rotation gives ω = 2π/T = 2πf.
  2. Divide s = rθ by time at fixed r to obtain v = rω. At fixed angular speed, a point twice as far from the centre has twice the tangential speed.

Worked example

Rotating scanner

  1. A scanner makes 5.0 turns per second.
  2. ω = 2π × 5.0 = 31.4 rad s⁻¹.
  3. At radius 0.080 m, v = 0.080 × 31.4 = 2.51 m s⁻¹.

An outer point has a greater speed than an inner point at the same angular velocity.

Keep this idea. Convert to radians, then use period for angular velocity and radius for linear speed.

Derive acceleration from a turning velocity vector

At two nearby positions in uniform circular motion, the velocity arrows have equal lengths but different directions. Translate them to a common tail without rotating them. Their difference points approximately towards the circle centre. As the interval shrinks, that direction becomes exactly radial and inward; velocity itself remains tangential.

The triangle of velocity changes is similar to the triangle formed by two radii for a small angular displacement. Hence Δv/v ≈ Δs/r. Divide by Δt and use Δs/Δt = v to obtain a = v²/r. Substituting v = rω gives a = rω². This derives the acceleration from geometry rather than assuming an inward force by name.

Be explicit about what stays fixed in a comparison. At fixed speed a larger radius means a gentler turn and smaller acceleration. At fixed angular velocity a larger radius means a larger speed, so acceleration increases. A speed change would add tangential acceleration; the radial expression alone then describes only one component of the total acceleration.

Δv/v ≈ Δs/r; a = lim(Δv/Δt) = v²/r = rω²

Symbols and units

  • v is speed in m s⁻¹
  • r is radius in m
  • Δv is the magnitude of the vector velocity change in m s⁻¹
  • a is inward acceleration in m s⁻²
  • ω is angular speed in rad s⁻¹.

How the equation explains the idea

  1. For a small angular step, similar triangles give Δv/v ≈ Δs/r. Divide by Δt and let the step shrink: a = (v/r)(ds/dt) = v²/r.
  2. Using v = rω gives a = rω². Doubling v at fixed radius quadruples a; changing radius at fixed ω is a different comparison from changing radius at fixed v.
Inward acceleration grows with speed squared

Conditions. Uniform circular-motion model a = v²/r with radii 1 m and 2 m. Each curve varies speed at a fixed radius; acceleration magnitude is shown.

Tangential speed / m s⁻¹; Inward acceleration / m s⁻².

At 2 m s⁻¹, the 1 m circle needs 4 m s⁻² while the 2 m circle needs 2 m s⁻². Doubling speed at fixed radius quadruples acceleration. This comparison does not hold angular speed fixed.

Open the interactive lesson to view the plotted curves and their coordinate tables.

Worked example

Two ways to calculate a turn

  1. A point moves at 3.0 m s⁻¹ on a circle of radius 0.50 m.
  2. a = 3.0²/0.50 = 18 m s⁻².
  3. ω = 3.0/0.50 = 6.0 rad s⁻¹; rω² = 0.50 × 36 = 18 m s⁻².

Both forms give the same inward acceleration.

Keep this idea. In a steady circle, velocity is tangential and acceleration points inward.

Find which real forces provide the inward resultant

Centripetal force is the name for the required inward resultant, not an additional interaction to add to a force diagram. Tension, friction, gravity or a combination of forces can supply it. First draw only real interactions, then resolve radially and set their inward resultant equal to mv²/r.

A mass on a string in a vertical circle has different radial equations at the top and bottom. At the bottom, tension is inward and weight outward, giving tension minus mg = mv²/r. At the top both point inward, giving tension plus mg = mv²/r. For a string to remain taut, its tension cannot be negative; a calculated negative value says the assumed circular constraint cannot be maintained.

In a circular-motion investigation, change one of mass, speed or radius while controlling the others. Measure the orbit radius to the object’s centre and time several turns. If a hanging mass supplies approximate tension through a tube, check that it remains stationary and account for friction. Use secure attachments, a clear swept area and modest speeds; an escaped object follows its instantaneous tangent.

ΣF inward = mv²/r = mrω²; bottom: T − mg = mv²/r; top: T + mg = mv²/r

Symbols and units

  • m is mass in kg
  • v is speed in m s⁻¹
  • r is radius in m
  • T is tension in N
  • g is gravitational acceleration in m s⁻².

How the equation explains the idea

  1. Choose inward positive at each position and sum actual radial forces. At the bottom of a vertical circle, T − mg = mv²/r; at the top, T + mg = mv²/r.
  2. Centripetal force is the required resultant, not an extra interaction. At the top of a taut-string limiting case, T = 0 gives v² = gr; a negative calculated tension means the taut-circle model fails.

Worked example

Bottom of a vertical circle

  1. A 0.20 kg mass moves at 4.0 m s⁻¹ on a 0.80 m radius circle.
  2. Required inward resultant = 0.20 × 4.0²/0.80 = 4.0 N.
  3. At the bottom T = 4.0 + 0.20 × 9.81 = 5.96 N.

Tension exceeds weight because the resultant must point upward towards the centre.

Keep this idea. Draw actual forces and use their signed inward resultant; never add a separate centripetal interaction.

Open the interactive lesson for free practice and reviewed three-question samples when available. New full question sets and timed unit checks are in preparation. Existing complete Physics revision packs retain paid access; existing free teaching PDFs remain free.

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