Uniform Circular Motion & Centripetal Acceleration

You’ll be able to

Explain why an object in uniform circular motion is accelerating despite constant speed
Relate period, frequency, linear speed and angular speed for circular motion
Compute centripetal acceleration and identify its direction

Constant speed is not constant velocity

Velocity is a vector, so it changes if either its magnitude or its direction changes. An object moving in a circle at unchanging speed is changing direction continuously, therefore its velocity is changing, therefore it is accelerating — even though a speedometer would read a fixed number. This acceleration points toward the center of the circle and is called centripetal, from the Latin for "center-seeking". It is a description of the direction of the acceleration, not the name of a new force.

Centripetal acceleration and the kinematics of the circle

a_c = v²/r = ω²r v = ωr ω = 2π/T = 2πf

a_c always points toward the center, perpendicular to the velocity. In uniform circular motion the speed is constant, so there is no tangential component of acceleration.

Where v²/r comes from

Over a small time Δt the velocity vector rotates through the same angle Δθ that the position vector sweeps. The change in velocity Δv is therefore an arc of a circle of radius v, giving |Δv| ≈ vΔθ. Dividing by Δt and taking the limit gives a = v(dθ/dt) = vω, and substituting ω = v/r gives a = v²/r. The direction of Δv in that limit points from the object toward the center — which is why the result is centripetal rather than tangential.

When motion is non-uniform

If the speed is also changing — a car accelerating around a bend, a pendulum bob away from the lowest point — the acceleration has two perpendicular components: a centripetal component v²/r toward the center that changes direction, and a tangential component dv/dt along the path that changes speed. The total magnitude is √(a_c² + a_t²). Recognizing that these are independent and perpendicular is what makes non-uniform circular motion tractable.

Worked example

A stone on a string of radius 0.50 m completes one revolution every 0.40 s. Find its speed and its centripetal acceleration, and express the acceleration as a multiple of g.

1.One revolution covers a circumference 2πr in one period T, so v = 2πr/T = 2π(0.50)/0.40.
2.v = π/0.40 ≈ 7.85 m/s.
3.a_c = v²/r = (7.85)²/0.50 ≈ 61.7/0.50 ≈ 123 m/s².
4.Compare with g: 123/9.8 ≈ 12.6, so the acceleration is about 13g.

Answer: v ≈ 7.9 m/s and a_c ≈ 1.2 × 10² m/s², roughly 13g — which is why the string tension is far larger than the stone's weight

Checkpoint

If the speed of an object in uniform circular motion is doubled while the radius is unchanged, the centripetal acceleration —

Watch out

There is no such thing as a "centripetal force" in a free-body diagram. Centripetal is a direction. The force producing the acceleration is always an identifiable real force — tension, friction, gravity, the normal force — and your diagram must name it. Writing "F_c" beside a real force double-counts it.

Checkpoint

A car speeds up as it rounds a curve. Its acceleration vector points —

Answer the 2 checkpoints as you read.