The big idea: Anything moving in a circle is always changing direction, so it is always accelerating — even at a steady speed.
That acceleration points toward the centre of the circle. We call it the centripetal acceleration ('centre-seeking').
It needs a real net force, also pointing to the centre — the centripetal force.
Spot it: Centripetal = 'toward the centre'. The force and the acceleration both point inward, along the radius — never along the direction of motion.
The data booklet gives the centripetal acceleration and the speed. Combine them with F = ma to get the force that must point to the centre.
- centripetal acceleration — points to the centre (m s⁻²)
- speed around the circle (m s⁻¹)
- radius of the circle (m)
- angular speed — radians turned per second (rad s⁻¹)
- period — time for one full lap (s)
- speed around the circle (m s⁻¹)
- radius of the circle (m)
- period — time for one full lap (s)
- angular speed (rad s⁻¹)
Put them together: Newton's second law (F = ma, also given) with a = v²/r gives the centripetal force:
F_c = mv²/r — the net force pointing to the centre.
- centripetal force — the NET force toward the centre (N)
- mass of the object (kg)
- speed around the circle (m s⁻¹)
- radius of the circle (m)
[Diagram: phys-formula-triangle] - Available in full study mode
Worked example — force on a car in a bend
A 1200 kg car drives round a flat bend of radius 50 m at 15 m s⁻¹. Find the centripetal force needed.
Solution
- Start with the centripetal-force result:
- Put in the numbers (m = 1200, v = 15, r = 50):
- Work it out — keep the unit:
Final answer
Fc = 5400 N, directed toward the centre of the bend (here it is supplied by friction).
[Diagram: phys-free-body] - Available in full study mode
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How this is tested: Circular motion is examined in two main ways.
- Paper 1A: a multiple-choice question asking you to pick the correct free-body diagram — e.g. a car on a banked (tilted) road, showing weight, the normal force and friction. - Paper 2: a structured vertical circle calculation — find the string tension at the lowest point, where tension and weight act in opposite directions.
Classic trap: Fc is the net force toward the centre, not an extra force you add to the diagram. At the bottom of a vertical circle: tension − weight = mv²/r, so the tension is bigger than the weight.
Vertical circle, lowest point: At the bottom, the tension pulls up (toward the centre) and the weight pulls down (away from it).
The net upward force is the centripetal force:
T − mg = mv²/r, so T = mg + mv²/r.
| Force at the lowest point | Direction | Toward centre? |
|---|---|---|
| Tension T (string) | Up — toward the centre | + (helps) |
| Weight mg | Down — away from the centre | − (opposes) |
| Net = T − mg | Up | = mv²/r |
IB-style question — tension at the bottom of a vertical circle
A 0.40 kg ball on a string is swung in a vertical circle of radius 0.80 m. At the lowest point its speed is 6.0 m s⁻¹. Find the tension in the string there. (Take g = 9.8 m s⁻².)
Solution
- At the lowest point the net upward force is the centripetal force:
- Rearrange for the tension:
- Put in the numbers (m = 0.40, v = 6.0, r = 0.80):
- Work it out — weight 3.92 N plus centripetal 18 N:
Final answer
T ≈ 22 N — bigger than the weight (3.9 N), because the tension must both support the ball AND supply the centripetal force.