Even though the speed of an object in uniform circular motion never changes, its velocity changes constantly — because the direction keeps shifting. That change in velocity is a real acceleration, always directed towards the centre of the circle. It’s called the centripetal acceleration.
📘 What you need to know
Centripetal acceleration always points towards the centre of the circle, perpendicular to the velocity.
In terms of linear speed: a = v² / r
In terms of angular speed: a = rω² = 4π²r / T²
The centripetal force is just F = ma applied to these expressions: F = mv² / r = mrω².
The three forms are equivalent — pick whichever matches what you’re given.
What centripetal acceleration is
At every instant, the object in a circle is changing direction. The rate at which the velocity changes is the acceleration, and because the velocity direction always swings towards the centre, the acceleration points inward — it is centripetal (centre-seeking).
The magnitude can be derived from the geometry of the velocity change, giving:
The 4π²r / T² version is handy whenever you know the period instead of the speed or angular speed.
Centripetal acceleration points towards the centre — always perpendicular to the velocity and parallel to the radius.
Connecting the three forms
All three expressions for centripetal acceleration come from the same geometry — they’re just written in terms of different variables. Use the form that matches what the question gives you:
a = v² / r given speed v
substitute v = rω
a = rω² given ang. speed ω
substitute ω = 2π/T
a = 4π²r / T² given period T
Quick reference:a = v²/r = rω² = 4π²r/T² • always directed towards the centre • multiply by m to get the centripetal force.
Worked examples
WE 1
Acceleration with doubled radius and doubled angular speed
A ball on a string of radius 1.5 m spins at 3.5 rad s⁻¹. Find the centripetal acceleration if the radius is doubled and the angular speed is also doubled.
Use a = rω²; new a = (2r)(2ω)²= 2r × 4ω² = 8rω²8 times the original accelerationOriginal a = 1.5 × 3.5² = 1.5 × 12.25 = 18.375 m s⁻²New a = 8 × 18.375a = 147 m s⁻²
WE 2
Using the period form
A satellite orbits at a radius of 7.0 × 10⁶ m with a period of 5800 s. Find its centripetal acceleration.
Use a = 4π²r / T²a = (4π² × 7.0×10⁶) ÷ 5800²= (4 × 9.87 × 7.0×10⁶) ÷ 33 640 000a ≈ 8.2 m s⁻²
WE 3
From acceleration to speed
A car travels around a circular bend of radius 80 m. Its centripetal acceleration is 3.6 m s⁻². Find its speed.
Rearrange a = v² / r → v = √(ar)v = √(3.6 × 80) = √288v ≈ 17 m s⁻¹
💡 Top tips
Choose the form that fits the given data — speed → v²/r; angular speed → rω²; period → 4π²r/T².
Centripetal acceleration is always inward, towards the centre — it’s never tangential.
Multiply by mass to get force — the formulas are on your data sheet, so know which to use and when.
Doubling ω quadruples a (ω is squared); doubling r only doubles a — proportional reasoning is faster than plugging in numbers.
⚠ Common mistakes
Confusing centripetal acceleration with the net acceleration. It is the net acceleration — not an extra one added to others.
Mixing T and f. The period T is in seconds per revolution; frequency f is revolutions per second. They’re reciprocals.
Using diameter instead of radius — always halve the diameter before substituting.
Thinking acceleration means speeding up. Centripetal acceleration changes direction, not magnitude of speed.
With centripetal acceleration in hand you have the full toolkit for uniform circular motion. The final page takes it one step further — what happens when the speed is not constant? Up next: Non-Uniform Circular Motion, where the tension in a vertical circle changes with position.
Need help with SL Forces & Momentum?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
