Centripetal Acceleration and Force

Spec mapping: OCR H556 Module 5.3 — Circular motion (centripetal acceleration $a = v^2/r = r\omega^2$ ; centripetal force $F = mv^2/r = mr\omega^2$ as the resultant inward force on a body in uniform circular motion; applications to horizontal and vertical circles, banked tracks, conical pendulums and orbital motion). Refer to the official OCR H556 specification document for exact wording.

In the previous lesson we discovered that an object moving in a circle at constant speed has a changing velocity, because the direction of motion is continuously turning. A changing velocity means an acceleration, and by Newton's second law (Module 3.2) an acceleration requires a net force. This lesson derives the famous formula for centripetal acceleration and applies it to real situations: cars on bends, buckets swung overhead, conical pendulums, planets in orbit, and the spinning samples in a medical centrifuge.

This is OCR A-Level Physics A Module 5.3 — the heart of circular motion, and the conceptual bridge to gravitational orbits (Module 5.4) and magnetic circular motion (Module 6.4). The single most important idea is this: the centripetal force is not a new force. It is whatever resultant inward force the existing real forces happen to add up to. The mistake of treating it as an extra force is by far the most common conceptual error at A-Level.

What is "Centripetal"?

The word centripetal comes from the Latin centrum (centre) + petere (to seek). A centripetal force is any force that seeks the centre — it points from the moving object towards the centre of the circle it is travelling on.

Centripetal force is not a new kind of force. It is the net (resultant) inward force on the body, made up of the ordinary forces — gravity, tension, friction, normal contact, electrostatic, magnetic — that you already know about from Module 3.2. The job of a circular-motion problem is to identify which real forces have an inward component, and to set their inward resultant equal to $mv^2/r$ .

Situation	Centripetal force provided by
Ball on a string swung in a horizontal circle	Tension in the string
Car rounding a flat bend	Friction between tyre and road
Earth orbiting the Sun	Gravitational attraction $GMm/r^2$
Electron in a (classical) Bohr orbit	Electrostatic Coulomb attraction $ke^2/r^2$
Rider on a fairground "wall of death"	Normal contact force from the wall
Bob of a conical pendulum	Horizontal component of the string tension
Charged particle in a uniform magnetic field	Magnetic Lorentz force $qvB$ (Module 6.4)
Bicycle on a banked velodrome	Inward component of normal contact and friction

Exam Tip: When an OCR question asks "What provides the centripetal force?" the correct answer is the physical force — tension, friction, gravity, normal reaction, lift, magnetic. Writing "centripetal force" is never the answer and will score zero. Always name the origin of the inward pull.

Deriving Centripetal Acceleration

Consider an object moving at constant speed $v$ in a circle of radius $r$ . In a short time $\Delta t$ it sweeps through a small angle $\Delta\theta = \omega\Delta t$ . Its velocity vector changes from $\vec{v}_1$ to $\vec{v}_2$ , both of magnitude $v$ , but pointing in slightly different directions (each is tangential to the circle at its own point).

The change in velocity, $\Delta\vec{v} = \vec{v}_2 - \vec{v}_1$ , is what we need. Two equal-magnitude vectors with a small angle $\Delta\theta$ between them subtract to give a third vector of magnitude $v\Delta\theta$ (chord length on a circle of radius $v$ in velocity space) pointing approximately radially inward — perpendicular to either velocity arrow, in the limit $\Delta\theta \to 0$ . This is exactly the small-angle isoceles-triangle geometry: two sides of length $v$ , apex angle $\Delta\theta$ , base of length $v\Delta\theta$ for small $\Delta\theta$ .

$|\Delta\vec{v}| \approx v\,\Delta\theta$

Dividing by $\Delta t$ and taking the limit gives the magnitude of the acceleration:

$a = \lim_{\Delta t \to 0}\frac{|\Delta\vec{v}|}{\Delta t} = v\,\frac{d\theta}{dt} = v\omega$

Substituting $v = r\omega$ (from lesson 1) gives the two canonical forms used throughout the topic:

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

Both formulae give the same number; you choose whichever fits the data you are given. If the question hands you $v$ and $r$ , use $a = v^2/r$ ; if it hands you $\omega$ and $r$ (or rpm), use $a = r\omega^2$ .

Direction of the Acceleration

Because $\Delta\vec{v}$ points towards the centre of the circle (in the limit), so does the acceleration. The centripetal acceleration is always radially inward, perpendicular to the instantaneous velocity. It has no tangential component — and that is why the speed remains constant even though the velocity is changing. Tangential acceleration would speed the body up or slow it down; perpendicular acceleration only rotates the velocity vector.

Centripetal Force

Applying Newton's second law $\vec{F} = m\vec{a}$ to the centripetal acceleration gives the net inward force required to maintain the circular motion:

$F_\text{net, inward} = \frac{mv^2}{r} = mr\omega^2$

This is the resultant force required to keep a body of mass $m$ moving at speed $v$ (or angular velocity $\omega$ ) on a circle of radius $r$ . It acts towards the centre of the circle. It is not a separate force added to gravity, tension, friction etc. — it is what the inward components of those real forces must sum to. Get this conceptual point right and 90% of A-Level circular-motion mistakes evaporate.

Notice that centripetal force depends on the square of the speed. Doubling the speed quadruples the required force. This is why sharp bends become dangerous so quickly: a driver entering a corner at $60$ mph instead of $30$ mph needs four times as much friction — and friction has an absolute ceiling set by $\mu_s N$ , beyond which the tyres slide.

The Inward-Force Audit — A Universal Method

For any circular-motion problem at A-Level, run the same procedure:

Draw a free-body diagram of the object at one instant.
Identify every real force on it: gravity, normal contact, tension, friction, drag, electrostatic, magnetic.
Resolve forces into radial (inward/outward) and tangential components.
Set the resultant radial component equal to $mv^2/r$ (inward positive).
Solve for the unknown (often the maximum speed, minimum tension, or required radius).

This is a Newton-2 calculation — there is no "centripetal force" to add to the FBD; only $mv^2/r$ on the right-hand side of $F_\text{net} = ma$ . If your diagram has an " $F_c$ " arrow in addition to gravity and tension, you have already gone wrong.

flowchart TD
  A[Object in uniform circular motion] --> B[Draw FBD: list real forces only]
  B --> C{Resolve each into radial + tangential}
  C --> D[Sum inward components: F_net,in]
  D --> E[Set F_net,in = mv²/r]
  E --> F[Solve for unknown]
  F --> G[Sanity-check: is the required force achievable? friction ≤ μN, tension ≤ breaking strength]

Worked Example 1 — Ball on a String

A $0.25$ kg ball is swung in a horizontal circle of radius $0.80$ m at a constant speed of $4.0$ m s $^{-1}$ . Calculate (a) the centripetal acceleration and (b) the tension in the string.

(a) Using $a = v^2/r$ :

$a = \frac{v^2}{r} = \frac{(4.0)^2}{0.80} = \frac{16}{0.80} = 20 \text{ m s}^{-2}$

The acceleration is $20$ m s $^{-2}$ ( $\approx 2g$ ) directed toward the centre of the circle at every instant.

(b) If the ball is moving in a horizontal circle on a frictionless smooth surface, the only horizontal force on the ball is the string tension, pulling inward toward the centre.

$T = ma = 0.25 \times 20 = 5.0 \text{ N}$

So the string tension equals the required centripetal force: $T = mv^2/r = 5.0$ N. (For a ball swung in a conical pendulum the string is tilted upward and only the horizontal component of tension supplies the centripetal force — see Example 4 below.)

Worked Example 2 — Car Rounding a Flat Bend

A car of mass $1200$ kg drives round a flat (unbanked) bend of radius $45$ m. The coefficient of static friction between the tyres and the road is $\mu_s = 0.70$ . Calculate the maximum speed at which the car can round the bend without skidding. Take $g = 9.81$ m s $^{-2}$ .

The centripetal force is supplied entirely by friction between the tyres and the road — there is nothing else with an inward horizontal component. The maximum static friction available is

$F_\text{max} = \mu_s N = \mu_s mg = 0.70 \times 1200 \times 9.81 = 8\,240 \text{ N}$

Setting this equal to the required centripetal force at the limit:

$\begin{aligned} \frac{mv^2}{r} &= \mu_s mg \\ v^2 &= \mu_s g r = 0.70 \times 9.81 \times 45 = 309 \text{ m}^2\text{ s}^{-2} \\ v_\text{max} &= \sqrt{309} = 17.6 \text{ m s}^{-1} \approx 63 \text{ km h}^{-1} \end{aligned}$

Notice that the mass cancels — the maximum speed on a flat bend does not depend on how heavy the car is. Examiners like this result because it is counter-intuitive. An empty Mini and a fully laden lorry share the same skid-speed at a given $\mu$ and $r$ . The same principle is why an HGV and a sports car spin out at similar speeds on the same icy corner.

Wet road sanity check. If rain drops $\mu_s$ from $0.7$ to $0.3$ , $v_\text{max}$ falls by a factor of $\sqrt{0.3/0.7} \approx 0.65$ — a 35% reduction in safe cornering speed. The squared dependence ( $v^2 \propto \mu$ ) is the physical basis for "slow down in the wet" advice.

Worked Example 3 — Bucket of Water Swung Overhead

A bucket of water of total mass $m$ is swung in a vertical circle of radius $r = 1.0$ m. What is the minimum speed at the top of the circle for the water to stay in the bucket? Take $g = 9.81$ m s $^{-2}$ .

At the top of the circle, both gravity ( $mg$ downward, toward the centre) and the normal contact force from the bottom of the bucket on the water ( $N$ , downward, also toward the centre) act radially inward. Newton 2 (radial direction, inward positive):

$N + mg = \frac{mv^2}{r}$

For the water to just stay in contact with the bucket — the borderline case — the contact force $N = 0$ . Then gravity alone supplies the centripetal force:

$\begin{aligned} mg &= \frac{mv^2}{r} \\ v^2 &= gr = 9.81 \times 1.0 = 9.81 \text{ m}^2\text{ s}^{-2} \\ v_\text{min} &= 3.13 \text{ m s}^{-1} \end{aligned}$

Swing any slower and gravity is more than the required centripetal force; the surplus accelerates the water out of the bucket and you get wet. Swing faster and the bucket-bottom has to push the water inward (positive $N$ ) to make up the additional centripetal requirement.

At the bottom of the same loop, gravity acts away from the centre (downward, but the centre is now above), while normal force acts toward the centre. Newton 2 gives $N - mg = mv^2/r$ , so $N = mg + mv^2/r > mg$ . The water experiences "apparent weight" greater than $mg$ — a familiar sensation at the bottom of a roller-coaster loop, and an OCR favourite for follow-up questions.

Worked Example 4 — The Conical Pendulum

A bob of mass $m = 0.20$ kg hangs from a light string of length $L = 0.50$ m and traces out a horizontal circle of radius $r$ at constant speed. The string makes a constant angle $\theta = 25°$ with the vertical. Find (a) the period of the motion and (b) the tension in the string.

Geometry: $r = L\sin\theta = 0.50 \times \sin 25° = 0.211$ m.

The bob has two forces on it: tension $T$ along the string (inward and upward) and weight $mg$ vertically downward. Resolve into vertical (balanced) and horizontal (provides centripetal):

Vertical: $T\cos\theta = mg \quad \Rightarrow \quad T = mg/\cos\theta = 0.20 \times 9.81 / \cos 25° = 2.17$ N.
Horizontal: $T\sin\theta = mv^2/r = mr\omega^2$ .

From the horizontal equation, $\omega^2 = T\sin\theta/(mr) = 2.17 \times \sin 25°/(0.20 \times 0.211) = 21.7$ rad $^2$ s $^{-2}$ , so $\omega = 4.66$ rad s $^{-1}$ and $T_\text{period} = 2\pi/\omega = 1.35$ s. Notice the string angle and the period are linked by a single condition — they cannot be set independently.

Important conceptual point: the tension itself is not "the centripetal force". Only its horizontal component $T\sin\theta$ is. The vertical component $T\cos\theta$ does no circular-motion work; it balances gravity. This is the inward-force audit in action.

Worked Example 5 — Centrifuge

A laboratory centrifuge rotates at $4500$ rpm. A sample tube is $8.0$ cm from the rotation axis. Find the centripetal acceleration experienced by the sample in units of $g$ .

First convert rpm to rad s $^{-1}$ :

$\omega = 4500 \times \frac{2\pi}{60} = 471 \text{ rad s}^{-1}$

Then

$a = r\omega^2 = 0.080 \times (471)^2 = 0.080 \times 2.22 \times 10^5 = 1.78 \times 10^4 \text{ m s}^{-2}$

In units of $g$ :

$\frac{a}{g} = \frac{1.78 \times 10^4}{9.81} \approx 1810\,g$

A typical medical centrifuge can generate thousands of $g$ , which is why centrifugation separates blood cells from plasma so effectively. Plasma proteins, red blood cells, white blood cells and platelets have slightly different densities, and at $\sim 2000\,g$ they stratify into clean layers within minutes rather than the days it would take under Earth's $1g$ .

Ultracentrifuges used in biochemistry research routinely run at $50\,000$ – $100\,000$ rpm, generating up to $10^6\,g$ — enough to separate molecules by molecular weight.

Worked Example 6 — Banked Track (no friction)

A car of mass $m$ rounds a curve of radius $r = 80$ m on a road banked at angle $\theta$ to the horizontal. Find the angle $\theta$ at which the car can negotiate the bend at $v = 22$ m s $^{-1}$ ( $\approx 50$ mph) with no friction required.

Forces on the car: weight $mg$ down, normal contact $N$ perpendicular to the road surface. Resolve into vertical and horizontal:

Vertical: $N\cos\theta = mg$
Horizontal (inward): $N\sin\theta = mv^2/r$

Dividing equation 2 by equation 1:

$\tan\theta = \frac{v^2}{gr} = \frac{(22)^2}{9.81 \times 80} = \frac{484}{785} = 0.617 \quad \Rightarrow \quad \theta = 31.7°$

So at $32°$ banking, $v = 22$ m s $^{-1}$ is the natural speed: a frictionless car holds the curve without any sideways pull. Below that speed, friction must point up-slope to keep the car from sliding down; above it, friction must point down-slope. This is exactly the physics behind highway banking, racetrack design and velodrome geometry.

Centripetal Acceleration and Force

Centripetal Acceleration and Force

What is "Centripetal"?

Deriving Centripetal Acceleration

Direction of the Acceleration

Centripetal Force

The Inward-Force Audit — A Universal Method

Worked Example 1 — Ball on a String

Worked Example 2 — Car Rounding a Flat Bend

Worked Example 3 — Bucket of Water Swung Overhead

Worked Example 4 — The Conical Pendulum

Worked Example 5 — Centrifuge

Worked Example 6 — Banked Track (no friction)

Does Centripetal Force Do Work?

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