recall and use a = rω2 and a = v2 / r

Centripetal Acceleration – Topic 12 (Motion in a Circle)

Learning Objectives (Syllabus 12.2)

  • Recall and use the three equivalent forms of the centripetal‑acceleration formula


    a = v² ⁄ r = r ω² = 4π² r ⁄ T².

  • Apply the formula to calculate the required centripetal force, its direction and magnitude in a range of contexts.

  • Link the concepts to angular kinematics (θ, ω, T) and Newton’s 2nd law.

1. Angular Kinematics Refresher

  • Radian: the angle subtended at the centre of a circle by an arc whose length equals the radius.


    Mathematically, θ (rad) = s ⁄ r, where s is the arc length.

  • Angular displacement, speed and period:

    • Angular speed ω = Δθ ⁄ Δt (rad s⁻¹)

    • For uniform circular motion, the period T (time for one revolution) is related to ω by


      ω = 2π ⁄ T and T = 2π ⁄ ω.

    • Linear speed v = r ω and v = 2π r ⁄ T.

2. Definition of Centripetal Acceleration

When an object moves in a circular path of radius r with constant speed, the direction of its velocity vector continuously changes. This change requires an inward (radially‑toward‑centre) acceleration called centripetal acceleration (a).

3. Vector Origin of the Acceleration

  1. Consider two successive velocity vectors v₁ and v₂ separated by a small angle Δθ (in radians). Both have magnitude v but different directions.

  2. The change in velocity is the vector difference


    Δv = v₂ − v₁.

    For a small Δθ, the magnitude of Δv is


    |Δv| ≈ v Δθ, and Δv points towards the centre of the circle.

  3. Acceleration is defined as a = Δv ⁄ Δt. Using Δθ = ω Δt (since ω = Δθ ⁄ Δt) gives


    a = v ω = v (v ⁄ r) = v² ⁄ r = r ω².

  4. Thus the acceleration vector is always radially inward.

4. Derivation of the Three Equivalent Forms

Starting relationSubstitutionResulting form of a
v = r ωInsert into a = v² ⁄ ra = r ω²
ω = 2π ⁄ TInsert into a = r ω²a = 4π² r ⁄ T²
v = 2π r ⁄ TInsert into a = v² ⁄ ra = 4π² r ⁄ T²

5. Summary Table of Kinematic Relationships

QuantitySymbolFormula
Radius of pathr–
Angular displacementθθ = s ⁄ r (rad)
PeriodTT = 2π ⁄ ω
Angular speedωω = 2π ⁄ T = v ⁄ r
Linear speedvv = r ω = 2π r ⁄ T
Centripetal accelerationaa = v² ⁄ r = r ω² = 4π² r ⁄ T²

6. Centripetal Force (Newton’s 2nd Law)

Applying F = ma to the inward acceleration gives the required centripetal force:

\$F_c = m a = m\frac{v^{2}}{r}= m r\omega^{2}= m\frac{4\pi^{2}r}{T^{2}}\$

  • Typical sources of the force:

    • Tension in a string (e.g. stone on a string).

    • Gravitational attraction for satellites.

    • Normal reaction or friction for a car rounding a curve.

  • The centripetal force does no work because it is always perpendicular to the instantaneous displacement.

7. Worked Example – Using the Period Form

Problem: A conical pendulum has a bob of mass 0.8 kg that makes 12 revolutions every 5 s. The length of the string is 1.5 m and the angle with the vertical is 30°. Find the magnitude of the centripetal acceleration of the bob.

  1. Calculate the period: T = 5 s ⁄ 12 = 0.417 s.

  2. Find the radius of the circular path: r = L sin 30° = 1.5 m × 0.5 = 0.75 m.

  3. Use the period form of the acceleration:

    \$a = \frac{4\pi^{2}r}{T^{2}} = \frac{4\pi^{2}\times0.75}{(0.417)^{2}} \approx 1.71\times10^{2}\ \text{m s}^{-2}.\$

  4. Result: a ≈ 1.7 × 10² m s⁻² directed towards the centre of the circular path (horizontal component of the tension).

8. Common Mistakes to Avoid

  • Confusing centripetal (real, inward) with centrifugal (apparent, outward) forces.

  • Using the radius of curvature of a non‑circular path; the formula requires the actual radius of the circular motion.

  • Mixing units – ensure v in m s⁻¹, r in m, ω in rad s⁻¹, T in s, and m in kg.

  • When ω is given, remember v = rω before using the v²⁄r form; when T is given, use ω = 2π⁄T or the 4π²r⁄T² form.

  • Neglecting the direction of the acceleration/force – it is always radially inward.

9. Practice Questions (AO2 & AO3)

  1. Angular‑speed problem:

    A stone is tied to a string and whirled in a horizontal circle of radius 0.75 m at an angular speed of 12 rad s⁻¹.

    • (a) Find the centripetal acceleration.

    • (b) Find the tension in the string if the stone’s mass is 0.50 kg.

    • (c) State the direction of the tension.

  2. Satellite problem:

    A satellite orbits Earth in a circular orbit of radius 7.0 × 10⁶ m with a speed of 7.5 × 10³ m s⁻¹. Determine the required centripetal acceleration and the magnitude of the gravitational (centripetal) force if the satellite’s mass is 1.2 × 10³ kg.

  3. Period‑form derivation:

    Show that the period T of uniform circular motion can be expressed as

    \$T = 2\pi\sqrt{\frac{r}{a}}\$

    using the relation a = rω² and the definition ω = 2π⁄T.

  4. Direction & sign convention:

    A car of mass 800 kg rounds a flat curve of radius 30 m at 15 m s⁻¹.

    • (a) Calculate the required centripetal force.

    • (b) Indicate the direction of this force with respect to the car’s motion (draw a labelled diagram).

10. Suggested Laboratory Activity (AO3)

Measuring centripetal acceleration with a rotating platform

  1. Equipment: low‑friction turntable, small mass m, string, ruler, stopwatch (or photogate), optional force sensor.

  2. Procedure

    • Attach the mass to the string and place it on the turntable so that it moves in a horizontal circle of known radius r.

    • Increase the turntable speed in steps, recording the revolutions per minute (rpm) each time.

    • Convert rpm to angular speed: ω = 2π (rpm) ⁄ 60 (rad s⁻¹).

    • Calculate the theoretical centripetal acceleration a = r ω² and the corresponding force F_c = m a.

    • If a force sensor is available, measure the tension in the string and compare with the calculated F_c.

  3. Analysis

    • Plot a (or F_c) against ω². The graph should be a straight line through the origin with gradient r (or mr).

    • Discuss sources of error (friction, non‑uniform radius, timing uncertainty) and their effect on the gradient.

11. Cross‑References to Other Syllabus Topics

  • Topic 3 – Dynamics: Newton’s second law and the concept of force (centripetal force = ma).

  • Topic 5 – Work, Energy & Power: The centripetal force does no work because it is perpendicular to the displacement.

  • Topic 9 – Momentum: In uniform circular motion the momentum vector changes direction, giving rise to the centripetal acceleration.

12. Quick Summary

  • Centripetal acceleration: a = v² ⁄ r = r ω² = 4π² r ⁄ T² (always towards the centre).

  • Centripetal force: F_c = m a = m v² ⁄ r = m r ω² = m 4π² r ⁄ T².

  • Key relationships: v = r ω = 2π r ⁄ T, ω = 2π ⁄ T, θ = s ⁄ r (rad).

  • Always state the direction (radially inward) and keep units consistent.

Suggested diagram: top‑view of a particle moving in a circle of radius r, showing the tangent velocity v, the inward centripetal acceleration a, and the corresponding centripetal force F_c.