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A body in uniform circular motion travels at constant speed but its velocity is changing continuously, because velocity is a vector and the direction of motion is always tangent to the circle. A changing velocity means an acceleration, and an acceleration must be caused by a resultant force. This lesson derives the magnitude of that acceleration, identifies the direction in which it acts, and shows how a wide variety of real forces — tension, gravity, friction, normal contact, magnetic — can play the role of "centripetal" provider. By the end of the lesson, students should be able to write down a = v²/r = ω²r and F = mv²/r = mω²r without thinking and apply them confidently to ball-on-string, car-on-bend and conical-pendulum problems.
Spec mapping: This lesson sits under AQA 7408 section 3.6.1.1. It covers the magnitude and direction of centripetal acceleration, the centripetal force F = ma = mv²/r = mω²r, and the qualitative observation that the centripetal force is always perpendicular to the instantaneous velocity (so does no work). The defining principle that the centripetal force is the resultant of real forces (not a "new" force) is also introduced here. (Refer to the official AQA specification document for exact wording.)
Synoptic links:
- Section 3.6.1.2 (simple harmonic motion): the projection of uniform circular motion onto a diameter gives the kinematic content of SHM. The ω² that appears in the acceleration of an SHM oscillator (a = −ω²x) is the same ω² that appears here in a = ω²r.
- Section 3.7.2 (gravitational fields): for a satellite in a circular orbit, the gravitational force GMm/r² is the centripetal force mv²/r — equating these gives the orbital speed v² = GM/r and the period T² = 4π²r³/GM (Kepler's third law).
- Section 3.7.5 (magnetic fields): for a charged particle of charge Q and momentum p in a magnetic field B, the magnetic force BQv supplies the centripetal force mv²/r — giving the radius r = p/(BQ) used in mass spectrometers and the cyclotron.
Consider an object moving in a circle of radius r at constant speed v. At any instant its velocity vector v⃗ is tangent to the circle. A moment later — say, an angle Δθ further round — the velocity has the same magnitude but points in a slightly different direction. Because v⃗ has changed (in direction, not magnitude), there is a change Δv⃗, and the time-rate of this change is the acceleration.
The crucial observation is that Δv⃗ points towards the centre of the circle. To see why, sketch the two velocity vectors tail-to-tail: the resultant Δv⃗ closes the triangle, and for small Δθ this resultant is perpendicular to the velocity itself — that is, radially inward. A formal derivation follows.
There are two standard derivations. The first uses similar triangles; the second uses the polar-coordinate description from the previous lesson. Both end at the same result.
Let an object move from position P₁ to position P₂ in time Δt, sweeping out angle Δθ at the centre O. Its velocity at P₁ is v⃗₁ (tangent to the circle at P₁) and at P₂ is v⃗₂ (tangent at P₂). Both have magnitude v.
The position triangle OP₁P₂ is isoceles with sides r, r and chord P₁P₂. The velocity triangle, formed by placing v⃗₁ and v⃗₂ tail-to-tail and joining their tips, is also isoceles with sides v, v and base |Δv⃗|. The two triangles are similar (both have apex angle Δθ).
Similar triangles give the proportion:
|Δv⃗|/v = chord(P₁P₂)/r
For small Δθ, the chord P₁P₂ ≈ v·Δt (arc length equals chord at small angles). Therefore:
|Δv⃗|/v ≈ (v·Δt)/r, so |Δv⃗| ≈ v²·Δt/r.
The magnitude of the acceleration is
a = |Δv⃗|/Δt = v²/r.
The direction of Δv⃗, as Δθ → 0, becomes exactly perpendicular to v⃗ and points towards the centre. Hence:
The centripetal acceleration has magnitude a = v²/r and points radially inward.
Combining a = v²/r with the linear–angular link v = rω:
a = v²/r = (rω)²/r = ω²r.
So the centripetal acceleration can be written in two equivalent forms:
a = v²/r = ω²r (always directed towards the centre).
Choose the form that matches the variables the question gives you. If the problem specifies linear speed and radius, use a = v²/r. If it specifies angular speed and radius (e.g. centrifuge rpm, satellite period), use a = ω²r.
A 0.40 kg ball is whirled in a horizontal circle of radius 1.50 m at 2.0 revolutions per second. Calculate the centripetal acceleration.
f = 2.0 Hz → ω = 2πf = 4π = 12.57 rad s⁻¹.
a = ω²r = (12.57)² × 1.50 = 158 × 1.50 = 237 m s⁻² (about 24 g).
Equivalently, v = rω = 1.50 × 12.57 = 18.85 m s⁻¹, and a = v²/r = (18.85)²/1.50 = 237 m s⁻².
If a body of mass m has centripetal acceleration a, then by Newton's second law a resultant force F = ma must act on it, directed towards the centre. This is the centripetal force:
F = ma = mv²/r = mω²r
Critical conceptual point: the centripetal force is not a new kind of force. It is the name we give to the resultant of the real forces acting on the body, in the special case where that resultant happens to be directed radially inward. The real forces are tension, gravity, normal contact, friction, magnetic force, electrical force — these are the things you draw on a free-body diagram. The label "centripetal" merely describes the role those forces play in the geometry.
| Situation | What Provides the Centripetal Force |
|---|---|
| Satellite in orbit | Gravitational attraction (GMm/r²) |
| Car on a flat roundabout | Friction between tyres and the road |
| Ball on a string in a horizontal circle | Horizontal component of the tension |
| Conical pendulum | Horizontal component of the tension |
| Charged particle in a magnetic field | Magnetic Lorentz force (BQv) |
| Car on a banked road (frictionless case) | Horizontal component of the normal contact force |
| Electron in a Bohr-model atomic orbit | Electrostatic attraction to the nucleus |
| Roller-coaster loop (at top) | Combined weight + normal contact, both directed down toward centre |
This table is exam currency. When a centripetal-motion problem appears, the very first move is to draw the free-body diagram and write down "the centripetal force is provided by [name the real force or component]". Marks are routinely lost by candidates who write "the centripetal force is mv²/r" as if it were one of the diagram forces — it is the resultant, not an additional arrow.
A small bob of mass 0.20 kg hangs on a light string of length 0.80 m and is swung in a horizontal circle so that the string makes an angle of 30° with the vertical. Calculate (a) the tension in the string, (b) the radius of the circular path, (c) the angular speed of the bob.
Free-body diagram: the bob has two forces on it — tension T along the string (up and inward at 30° to the vertical), and weight mg = 0.20 × 9.81 = 1.962 N straight down.
(a) Resolving vertically (no vertical acceleration): T cos 30° = mg, so T = mg/cos 30° = 1.962/0.866 = 2.27 N.
(b) Geometry: r = L sin θ = 0.80 × sin 30° = 0.40 m.
(c) Resolving horizontally (towards centre): T sin 30° = mω²r, so
ω² = T sin 30° / (mr) = (2.27 × 0.5) / (0.20 × 0.40) = 1.134/0.080 = 14.17,
ω = √14.17 = 3.76 rad s⁻¹.
The period is T_period = 2π/ω = 2π/3.76 = 1.67 s.
A force does work when the body moves in the direction of the force. The centripetal force is always perpendicular to the velocity (radial vs tangential), so the angle between F and the displacement Δs is always 90°.
W = F · Δs · cos 90° = 0.
The centripetal force does no work on the body, and the kinetic energy ½mv² remains constant. This is consistent with the speed being constant — speed and kinetic energy are tied through v², so constant speed means constant KE.
This result is often tested in conceptual questions: "Explain why an object moving in uniform circular motion has zero net work done on it despite being accelerated." The answer is exactly the perpendicularity argument above. The acceleration changes the direction of v⃗ but not its magnitude; only forces with a tangential component change the magnitude (and hence the KE).
graph LR
A["Body at point P<br/>on circle radius r"] --> B["Velocity v⃗<br/>tangential"]
A --> C["Centripetal force F⃗<br/>radial (inward)"]
B -.-> D["F⃗ · v⃗ = 0<br/>(perpendicular)"]
C -.-> D
D --> E["No work done<br/>KE constant<br/>speed constant"]
style E fill:#27ae60,color:#fff
style D fill:#3498db,color:#fff
A 1,200 kg car negotiates a flat (unbanked) roundabout of radius 25 m at 12 m s⁻¹. Calculate (a) the centripetal force required, and (b) the minimum coefficient of static friction μ_s between the tyres and road for the car to round the corner without slipping. Take g = 9.81 m s⁻².
(a) F = mv²/r = 1200 × (12)²/25 = 1200 × 144/25 = 6,912 N (about 6.9 kN).
(b) The only horizontal force is the static friction f_s = μ_s N = μ_s mg. For the car to round the corner, friction must supply the centripetal force:
μ_s mg = mv²/r → μ_s = v²/(rg) = 144/(25 × 9.81) = 0.587.
A typical asphalt-rubber μ_s is about 0.7 (dry) or 0.4 (wet). Dry: the car corners fine. Wet: the car skids — at 12 m s⁻¹ the required μ_s of 0.59 exceeds the available wet value of 0.4. This is the physics of why speed limits on bends drop sharply in the wet.
An electron of mass m_e = 9.11 × 10⁻³¹ kg and charge e = 1.60 × 10⁻¹⁹ C travels at 4.0 × 10⁶ m s⁻¹ perpendicular to a uniform magnetic field of 8.0 mT. Calculate the radius of its circular path.
The magnetic force on the electron is F = evB = 1.60 × 10⁻¹⁹ × 4.0 × 10⁶ × 8.0 × 10⁻³ = 5.12 × 10⁻¹⁵ N. This force is perpendicular to v⃗ and so provides the centripetal force:
evB = m_e v²/r → r = m_e v/(eB) = (9.11 × 10⁻³¹ × 4.0 × 10⁶)/(1.60 × 10⁻¹⁹ × 8.0 × 10⁻³)
r = (3.644 × 10⁻²⁴)/(1.28 × 10⁻²¹) = 2.85 × 10⁻³ m = 2.85 mm.
This is the working principle of the cyclotron: charged particles trace circular arcs in a perpendicular B-field at the cyclotron frequency f = eB/(2πm), independent of speed (for non-relativistic speeds).
A car of mass 1,200 kg drives round a flat (unbanked) bend of radius 50 m at 15 m s⁻¹. The static-friction coefficient between tyre and road is μ_s = 0.7. (a) Calculate the centripetal force required. (b) Calculate the maximum frictional force available. (c) Decide whether the car can make the bend, and state what happens otherwise. (d) State the maximum safe speed for this bend under these conditions.
(a) Centripetal force required. The car follows a circular arc of radius 50 m, so the net horizontal force must be
F_c = mv²/r = 1,200 × (15)² / 50 = 1,200 × 225 / 50 = 5,400 N (directed toward the centre of the bend).
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