Centripetal Acceleration and Force

An object moving in a circle at constant speed is not in equilibrium. Its velocity is constantly changing direction, which means it is accelerating. This acceleration is always directed towards the centre of the circle and is called centripetal acceleration. Understanding what provides the centripetal force in each scenario is the single most important skill in circular motion questions.

Why There Must Be an Acceleration

Consider a satellite orbiting the Earth. At one instant it is moving east. A quarter of an orbit later, it is moving north. Its speed has not changed, but its velocity vector has rotated by 90°. A change in velocity — whether in magnitude or direction — means acceleration.

The direction of this acceleration can be found by considering the change in velocity vector, Δv. For a small time interval, the change in velocity points towards the centre of the circle. In the limit (infinitesimally small time interval), the acceleration is exactly directed towards the centre.

This is centripetal acceleration: always directed towards the centre of the circular path, perpendicular to the velocity at every point.

Centripetal Acceleration Formulae

For an object moving in a circle of radius r at speed v with angular velocity ω:

$a = \frac{v^2}{r} = \omega^2 r = v\omega$

These are all equivalent (since v = rω). Use whichever form is most convenient for the data given.

Worked Example 1

A car travels around a roundabout of radius 20 m at 10 m s⁻¹. What is its centripetal acceleration?

$a = \frac{v^2}{r} = \frac{10^2}{20} = \frac{100}{20} = 5.0 \text{ m s}^{-2}$

This is about half of g — a noticeable acceleration that you feel as a sideways pull when driving around a roundabout.

Worked Example 2

A washing machine drum spins at 1200 rpm with radius 0.25 m. What centripetal acceleration do the clothes experience?

$\omega = \frac{2\pi \times 1200}{60} = 40\pi \text{ rad s}^{-1} = 125.7 \text{ rad s}^{-1}$

$a = \omega^2 r = (125.7)^2 \times 0.25 = 3950 \text{ m s}^{-2}$

That is about 403g — the enormous acceleration forces water out of the clothes through the drum holes.

Centripetal Force

By Newton's second law, if there is a centripetal acceleration, there must be a net force causing it. This is the centripetal force:

$F = ma = \frac{mv^2}{r} = m\omega^2 r$

Crucially, centripetal force is not a new type of force. It is the name we give to the resultant force (or the component of the resultant force) directed towards the centre of the circle. The actual force providing the centripetal force is always one of the fundamental forces you already know.

flowchart TD
    A["Object moving in a circle"] --> B{"What scenario?"}
    B -->|"Car on flat road"| C["Friction between\ntyres and road"]
    B -->|"Ball on string\n(horizontal)"| D["Tension in\nthe string"]
    B -->|"Satellite\norbiting planet"| E["Gravitational\nattraction"]
    B -->|"Electron orbiting\nnucleus"| F["Electrostatic\n(Coulomb) force"]
    B -->|"Car on banked\ncurve"| G["Component of\nnormal reaction"]
    B -->|"Clothes in\nspinning drum"| H["Normal contact\nforce from drum wall"]
    B -->|"Planet orbiting\nstar"| E
    B -->|"Charged particle\nin magnetic field"| I["Magnetic\n(Lorentz) force"]
    C --> J["F = mv²/r = mω²r"]
    D --> J
    E --> J
    F --> J
    G --> J
    H --> J
    I --> J

Centripetal Force Reference Table

Situation	Force Providing Centripetal Force	Key Equation
Car on flat road going around a bend	Friction: f = μmg	mv²/r ≤ μmg
Ball on a string swung horizontally	Tension in the string	T = mv²/r
Satellite orbiting a planet	Gravitational attraction	GMm/r² = mv²/r
Electron orbiting a nucleus	Electrostatic force	kQq/r² = mv²/r
Car on a banked curve (no friction)	Component of normal reaction	N sin θ = mv²/r
Clothes in a spinning drum	Normal contact force from drum wall	N = mω²r
Conical pendulum	Horizontal component of tension	T sin θ = mω²r

Worked Example 3

A 1200 kg car goes around a flat curve of radius 50 m. The maximum friction force between the tyres and the road is 8000 N. What is the maximum safe speed?

$F = \frac{mv^2}{r}$ $8000 = \frac{1200 \times v^2}{50}$ $v^2 = \frac{8000 \times 50}{1200} = 333$ $v = 18.3 \text{ m s}^{-1} \approx 66 \text{ km/h}$

If the car exceeds this speed, the required centripetal force exceeds the maximum friction available, and the car skids outward.

Worked Example 4: Centripetal Acceleration on Earth's Surface

What centripetal acceleration does a person standing on the equator experience due to Earth's rotation? (R_Earth = 6.37 × 10⁶ m, ω = 7.27 × 10⁻⁵ rad s⁻¹)

$a = \omega^2 r = (7.27 \times 10^{-5})^2 \times 6.37 \times 10^6 = 5.29 \times 10^{-9} \times 6.37 \times 10^6 = 0.034 \text{ m s}^{-2}$

This is only 0.35% of g. This small centripetal acceleration is why objects on the equator weigh very slightly less than at the poles — a tiny fraction of apparent weight goes towards providing the centripetal force for rotation.

Direction of Centripetal Force and Acceleration

Both the centripetal acceleration and centripetal force point towards the centre of the circle at all times. They are perpendicular to the velocity. Because the force is always perpendicular to the displacement, the centripetal force does no work — it changes the direction of motion but not the speed. This is why an object in uniform circular motion has constant kinetic energy.

Common Misconceptions

"Centrifugal force": Students often speak of a force pushing outward — the feeling of being "thrown outward" in a car going around a bend. This is not a real force. It is the sensation of your body trying to continue in a straight line (Newton's first law) while the car is accelerating inward. In the non-inertial (rotating) reference frame of the car, you need to invent a fictitious outward force to explain why objects seem to accelerate outward, but in the inertial (ground) frame, the only real force is inward.

"The centripetal force is an additional force": Centripetal force is not something extra. It is the resultant of the real forces acting. When drawing free-body diagrams for circular motion, you should draw the actual forces (weight, tension, normal force, friction) and not draw an extra arrow labelled "centripetal force."

"Doubling the speed doubles the centripetal force": Since F = mv²/r, doubling v quadruples F. The relationship is with v², not v.

Proportionality Problems

These are common exam questions requiring careful reasoning:

Change	Effect on Centripetal Force	Reasoning
Speed doubles, r constant	F × 4	F ∝ v²
Radius doubles, v constant	F × ½	F ∝ 1/r
Radius doubles, ω constant	F × 2	F = mω²r, so F ∝ r
Mass doubles	F × 2	F ∝ m
Speed and radius both double	F × 2	F = mv²/r → m(2v)²/(2r) = 2mv²/r

Worked Example 5: Choosing the Right Formula

A satellite orbits a planet at 5000 m s⁻¹ in a circle of radius 8.0 × 10⁶ m. What is its centripetal acceleration?

Use a = v²/r (given v and r): $a = \frac{5000^2}{8.0 \times 10^6} = \frac{2.5 \times 10^7}{8.0 \times 10^6} = 3.13 \text{ m s}^{-2}$

If instead you were given ω = 6.25 × 10⁻⁴ rad s⁻¹ and r = 8.0 × 10⁶ m, use a = ω²r: $a = (6.25 \times 10^{-4})^2 \times 8.0 \times 10^6 = 3.91 \times 10^{-7} \times 8.0 \times 10^6 = 3.13 \text{ m s}^{-2}$

Both give the same answer. Choose the formula that matches the data you are given.

Centripetal Force in Everyday Life

Situation	Mass	Speed or ω	Radius	Centripetal Force
Car on roundabout	1200 kg	10 m s⁻¹	20 m	6 000 N (friction)
Electron in atom	9.1 × 10⁻³¹ kg	2.2 × 10⁶ m s⁻¹	5.3 × 10⁻¹¹ m	8.2 × 10⁻⁸ N (electrostatic)
ISS in orbit	420 000 kg	7 670 m s⁻¹	6.77 × 10⁶ m	3.65 × 10⁶ N (gravity)
Person on equator	70 kg	465 m s⁻¹	6.37 × 10⁶ m	2.4 N (gravity component)
Ball on string	0.50 kg	4.0 m s⁻¹	0.80 m	10 N (tension)
Washing machine clothes	3.0 kg	126 rad s⁻¹	0.25 m	11 900 N (normal force)

Summary

Centripetal acceleration: a = v²/r = ω²r, directed towards the centre
Centripetal force: F = mv²/r = mω²r, directed towards the centre
Centripetal force is always provided by a real force (gravity, tension, friction, etc.)
The centripetal force does no work because it is perpendicular to the velocity
"Centrifugal force" is not a real force — it is a fictitious force in a rotating reference frame
F ∝ v² (not v), so doubling the speed quadruples the required force
Choose between a = v²/r and a = ω²r based on the data given in the question

A-Level Deep Dive: Centripetal Acceleration and Force

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics introduces centripetal acceleration $a = v^2/r = \omega^2 r$ directed towards the centre of a circular path, and the resultant centripetal force $F = mv^2/r = m\omega^2 r$ acting in the same direction; candidates must identify the real force (gravity, tension, friction, normal contact, electromagnetic) that supplies this resultant in each context (refer to the official Pearson Edexcel specification document for exact wording). Centripetal-force analysis is examined directly in Paper 2 (Topics 5–8) and reused synoptically in Paper 3 General and Practical Principles, where unfamiliar contexts (looped rollercoasters, banked motorways, ultracentrifuges, particles in magnetic fields) demand confident free-body analysis. The Edexcel formula booklet supplies $a = v^2/r$ , $a = r\omega^2$ , $F = mv^2/r$ and $F = mr\omega^2$ , but does not define which real force is centripetal in any given scenario — that recognition is the AO2 / AO3 focus of Topic 6.

Worked example with full mark scheme

Question (8 marks):

A car of total mass $1\,200$ kg drives over a humped-back bridge whose upper surface is approximately a circular arc of radius $r = 25$ m in the vertical plane.

(a) Draw a free-body diagram for the car at the highest point of the hump and write Newton's second law in the radial direction. (2)

(b) Hence determine the maximum speed $v_{\max}$ at which the car can pass over the highest point while remaining in contact with the road. Take $g = 9.81$ m s $^{-2}$ . (4)

(c) The driver crests the hump at $15$ m s $^{-1}$ . Calculate the magnitude of the normal contact force on the car at that instant. (2)

Solution with mark scheme:

(a) Step 1 — identify the real forces and their directions.

At the top of the hump the centre of the circle lies directly below the car. Two real forces act on the car: weight $W = mg$ acting vertically downward (toward the centre), and the normal contact force $N$ from the road surface acting vertically upward (away from the centre).

B1 — clear free-body diagram showing $mg$ down and $N$ up, with the centre of the circle marked beneath the car.

Step 2 — Newton's second law toward the centre.

Taking "toward the centre" (i.e. downward here) as positive:

$mg - N = \dfrac{mv^2}{r}$

B1 — correct radial equation with $mg$ and $N$ on the LHS in the right signs and $mv^2/r$ on the RHS. Common error: writing $N - mg = mv^2/r$ (treating up as the centripetal direction).

(b) Step 3 — apply the contact condition.

The car loses contact at the instant the road can no longer pull the car downward — i.e. when $N = 0$ . Setting $N = 0$ :

M1 — recognising the contact condition $N \to 0$ as the threshold for loss of contact.

$mg = \dfrac{mv_{\max}^2}{r}$

M1 — substituting $N = 0$ into the radial equation and cancelling $m$ .

Step 4 — solve for $v_{\max}$ .

$v_{\max} = \sqrt{gr} = \sqrt{9.81 \times 25}$

A1 — correct algebraic rearrangement.

$v_{\max} = \sqrt{245.25} = 15.7 \text{ m s}^{-1} \;\; (3 \text{ s.f.})$

A1 — final value with units, three significant figures.

$N = mg - \dfrac{mv^2}{r} = 1\,200 \times 9.81 - \dfrac{1\,200 \times 15^2}{25}$

M1 — correct substitution into the rearranged radial equation $N = mg - mv^2/r$ .

$N = 11\,772 - 10\,800 = 972 \text{ N} \approx 970 \text{ N}$

A1 — final value with units. Note $N < mg$ , so the driver feels lighter than usual at the crest — the qualitative check that the answer is sensible.

Total: 8 marks (B2 M3 A3, distributed as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A small bob of mass $0.20$ kg hangs from a light inextensible string of length $L = 0.80$ m attached to a fixed point. The bob swings in a horizontal circle (a conical pendulum) such that the string makes a constant angle $\theta = 30°$ with the vertical.

(a) Show that the period of motion is $T = 2\pi\sqrt{L\cos\theta / g}$ . (3)

(b) Calculate the period for the values given. (1)

Mark scheme decomposition by AO:

(a)

M1 (AO2 — resolve forces): identify two forces on the bob (tension $T_s$ along the string, weight $mg$ down). Write vertical equilibrium $T_s\cos\theta = mg$ and horizontal Newton's second law $T_s\sin\theta = m\omega^2 r$ where $r = L\sin\theta$ .
M1 (AO2 — eliminate $T_s$ ): divide the horizontal equation by the vertical to obtain $\tan\theta = \omega^2 L\sin\theta / g$ , cancel $\sin\theta$ to give $\omega^2 = g/(L\cos\theta)$ .
A1 (AO1.1): apply $T = 2\pi/\omega$ to obtain the printed result $T = 2\pi\sqrt{L\cos\theta / g}$ .

(b)

A1 (AO1.1): $T = 2\pi\sqrt{0.80 \times \cos 30° / 9.81} = 2\pi\sqrt{0.0706} = 1.67$ s (3 s.f.).

(c)

B1 (AO3.1): the derived expression $T = 2\pi\sqrt{L\cos\theta / g}$ contains no mass term — $m\$ cancels because both the centripetal force and the weight scale with mass.
B1 (AO3.2): therefore the period is unchanged by doubling $m$ , provided the geometry (angle $\theta$ and length $L$ ) is held fixed.

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. This is a balanced AO question typical of Paper 2 Topic 6: AO1 for the algebraic finishing line, AO2 for the simultaneous resolution, AO3 for the mass-independence reasoning.

Synoptic links

Connects to:

Angular velocity $\omega$ (Topic 6, previous lesson): every centripetal-force expression $F = mr\omega^2$ inherits $\omega$ directly from the kinematic definitions of the previous lesson. The two interchangeable forms $F = mv^2/r$ and $F = mr\omega^2$ are linked by $v = r\omega$ . Selecting the form that matches the data given is an AO2 skill examiners reward.
Gravitational fields and orbits (Topic 8): for a satellite in circular orbit, gravity is the centripetal force: $GMm/r^2 = mv^2/r$ . Cancelling and rearranging gives $v = \sqrt{GM/r}$ , and combining with $T = 2\pi r/v$ recovers Kepler's third law $T^2 \propto r^3$ . Without the centripetal-force balance, all orbital mechanics is inaccessible.