A car drives in a circle at constant speed. Its speedometer reads the same number throughout. Yet it is accelerating. Continuously, significantly, always toward the centre. Here is why, and what follows from it.

Circular Motion: Speed Without Acceleration Is Impossible Here

A stone tied to a string is swung in a horizontal circle at constant speed. The speed never changes — the stone covers equal arc lengths in equal times. Is it accelerating?

Yes. Continuously. The velocity vector is changing direction at every instant, even though its magnitude is constant. And from K01, acceleration is the rate of change of velocity — the full vector, not just its magnitude. A change in direction is a change in velocity. Therefore it is an acceleration.

This is the central insight of uniform circular motion: constant speed does not mean constant velocity, and non-constant velocity means acceleration. The acceleration here points not along the motion, but perpendicular to it — always inward, toward the centre of the circle.

The angular vocabulary

Before deriving the acceleration, one new set of quantities makes circular motion much easier to describe.

Angular displacement θ: the angle swept by the radius, measured in radians.

One full circle = 2π radians = 360°. The radian is defined so that arc length s = rθ, where r is the radius. This is why radians are natural for physics — they keep the geometry clean.

Angular velocity ω (omega): the rate at which angle is swept.

$\omega = \frac{d\theta}{dt}$

Unit: radians per second (rad/s). For uniform circular motion, ω is constant.

Period T: time for one complete revolution. Frequency f: number of revolutions per second. They are reciprocals:

$T = \frac{1}{f} \qquad \omega = \frac{2\pi}{T} = 2\pi f$

Connecting linear and angular quantities:

Arc length: s = rθ. Differentiate with respect to time:

$v = r\omega$

The linear speed v equals the radius times the angular velocity. For a fixed ω, points further from the centre move faster — the outer edge of a spinning disc covers more distance per revolution than the inner edge.

Deriving the centripetal acceleration

Consider a particle moving in a circle of radius r at constant speed v. At time t it is at position P with velocity v tangent to the circle. A short time Δt later it is at Q with velocity v' — same magnitude, but rotated by angle Δθ.

The change in velocity is Δv = v' − v.

For small Δθ, the vector Δv is approximately perpendicular to v (and therefore points toward the centre) and has magnitude:

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

This is because the velocity vector traces a circle of radius v in velocity space, and for a small angle Δθ, the chord length is v·Δθ.

The acceleration magnitude is:

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

Since v = rω:

a_c = \frac{v^2}{r} = r\omega^2 = v\omega

This acceleration points toward the centre — inward — at every instant. It is called centripetal acceleration (from the Latin: centre-seeking). It is not a new type of acceleration — it is ordinary acceleration (rate of change of velocity) applied to a vector that is constantly rotating.

The direction is always radially inward — perpendicular to the velocity, perpendicular to the motion. It changes the direction of velocity without changing its magnitude. This is exactly what an acceleration perpendicular to velocity does: it curves the path without speeding up or slowing down the object.

{/* DIAGRAM: Circle with particle at P and Q, velocity vectors v and v', triangle of velocities showing Δv pointing inward */}

Three equivalent forms

The centripetal acceleration has three equivalent expressions:

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

Which form to use depends on what is given:

Speed v and radius r given → use v²/r
Angular velocity ω and radius r given → use rω²
Speed v and angular velocity ω given → use vω (though this case is rare — usually r is known)

All three give the same number. Choose the one that avoids an extra step.

The time derivatives — angular analogy

Every linear kinematic quantity has an angular analogue. The structure is identical:

Linear	Angular
displacement x	angle θ
velocity v = dx/dt	angular velocity ω = dθ/dt
acceleration a = dv/dt	angular acceleration α = dω/dt
v = u + at	ω = ω₀ + αt
s = ut + ½at²	θ = ω₀t + ½αt²
v² = u² + 2as	ω² = ω₀² + 2αθ

For uniform circular motion, α = 0 (constant ω). For non-uniform circular motion — a car accelerating around a bend — α ≠ 0, and the linear kinematic equations above apply with angular variables substituted in.

This analogy is not cosmetic. The mathematics is identical. If you understand K02 deeply, you already understand angular kinematics — just replace x with θ, v with ω, a with α.

Two components of acceleration in non-uniform circular motion

When a particle moves in a circle with changing speed, the total acceleration has two components:

Centripetal (radial) component — always inward, magnitude v²/r. Changes direction of velocity.

Tangential component — along the circle (tangent), magnitude dv/dt = rα. Changes magnitude of velocity.

$a_{total} = \sqrt{a_c^2 + a_t^2} = \sqrt{\left(\frac{v^2}{r}\right)^2 + (r\alpha)^2}$

For uniform circular motion: a_t = 0, so a_total = a_c = v²/r.

For non-uniform: both components are present. The total acceleration points neither purely inward nor purely tangentially — it points somewhere between, depending on the magnitudes.

{/* DIAGRAM: Particle on circle with both centripetal (inward) and tangential (along circle) acceleration components shown */}

Note: 1. Identify r, v, ω, T or f — usually at least two of these are given. 2. Convert between them as needed: v = rω, ω = 2π/T = 2πf. 3. Centripetal acceleration: a_c = v²/r = rω². Direction: always toward the centre. 4. For non-uniform circular motion: find the tangential acceleration separately (a_t = dv/dt or rα), then combine with a_c for total acceleration. 5. Angular kinematics: if ω is changing, use the rotational analogues of the K02 equations — same structure, different symbols. 6. Never say "centrifugal force" in an inertial frame. The centripetal acceleration requires a centripetal force (from Newton's second law, Chapter 3) — that force is real. The outward "centrifugal force" is a fictitious force that appears only in a rotating (non-inertial) reference frame. In the ground frame, it does not exist.

Warning: "Constant speed means no acceleration." In circular motion, the speed is constant but the velocity vector is continuously changing direction. That rate of change of direction is the centripetal acceleration. Speed and acceleration are independent quantities.

> > **Centripetal acceleration pointing outward.** It always points inward — toward the centre. The velocity points tangentially; the acceleration points radially inward. They are perpendicular. > > **Confusing centripetal acceleration with centripetal force.** Centripetal acceleration is a kinematic quantity — it describes the geometry of the motion. Centripetal force is what Newton's second law (F = ma) tells you must be present to produce that acceleration. The force is provided by something real: tension, gravity, friction, normal force. "Centripetal force" is not a new type of force — it is whatever real force is directed inward. > > **"At the top of a vertical circle, centripetal acceleration is zero."** The particle is still moving in a circle at the top — centripetal acceleration is still v²/r, still pointing downward (toward the centre, which is below the top). Its direction is downward at the top of the loop, not upward. > > **Using degrees instead of radians in ω.** Angular velocity must be in radians per second for v = rω and a = rω² to be dimensionally consistent. Convert degrees to radians before computing.

A worked example

A particle moves in a circle of radius 2 m. Its speed at a certain instant is 4 m/s and is increasing at 3 m/s². Find: (a) the centripetal acceleration, (b) the tangential acceleration, (c) the total acceleration and its angle with the radius.

Part (a) — centripetal acceleration:

$a_c = \frac{v^2}{r} = \frac{16}{2} = 8 \text{ m/s}^2 \quad \text{(directed inward)}$

Part (b) — tangential acceleration:

$a_t = \frac{dv}{dt} = 3 \text{ m/s}^2 \quad \text{(directed along the circle, forward)}$

Part (c) — total acceleration:

$a_{total} = \sqrt{8^2 + 3^2} = \sqrt{64 + 9} = \sqrt{73} \approx 8.54 \text{ m/s}^2$

Angle with the radius (centripetal direction):

$\phi = \tan^{-1}\left(\frac{a_t}{a_c}\right) = \tan^{-1}\left(\frac{3}{8}\right) \approx 20.6°$

The total acceleration points 20.6° away from the inward radial direction, toward the forward tangential direction.

Twist variant

The same particle, but now the question: at what point does the centripetal acceleration equal the tangential acceleration?

a_c = a_t requires v²/r = 3. With r = 2: v² = 6 → v = √6 m/s.

At the instant the speed reaches √6 ≈ 2.45 m/s, the two components are equal — the total acceleration points at 45° to the radius. Before that speed, a_t > a_c (tangential dominates); after, a_c > a_t (centripetal dominates). The crossover happens at a specific speed, not at a specific time — which is why expressing it in terms of v is cleaner.

What to remember

Speed constant does not mean velocity constant. Changing direction is changing velocity. Changing velocity is acceleration.

Centripetal acceleration = v²/r, directed inward. Always. For any speed, any radius.

The angular quantities ω, α, θ mirror the linear quantities v, a, x exactly. The K02 equations apply with angular substitution.

For non-uniform circular motion, centripetal and tangential accelerations coexist — perpendicular to each other, combined by Pythagoras.

This completes the Kinematics chapter. Every concept here — position, velocity, acceleration, projectile motion, relative motion, circular motion — is a description of motion. No forces yet. Chapter 3 begins with Newton's laws and asks: why does motion change? The answer is force. The language is everything built in these five articles.