Why does an object moving at constant speed in a circle still accelerate, and what force keeps it on the circular path?
Describe uniform circular motion using centripetal acceleration and force, relating them to speed, radius and period.
Why circular motion at constant speed is accelerated motion, how centripetal acceleration and force point toward the centre, and how to relate them to speed, radius and period.
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What this dot point is asking
You need to explain why circular motion is accelerated even at constant speed, identify the centripetal force in a given situation, and calculate centripetal acceleration and force using speed, radius and period.
Why constant speed still means accelerating
Velocity is a vector. In circular motion the speed (the size of the velocity) is constant, but the direction is continuously changing, always tangent to the circle. A changing velocity is an acceleration, so the object is accelerating even though it is not speeding up or slowing down.
This acceleration points toward the centre of the circle. It is called centripetal ("centre-seeking") acceleration.
Period, frequency and speed
For an object completing circles, the period is the time for one revolution and the frequency . The speed is the circumference divided by the period:
Substituting into gives an equivalent form in terms of period:
Identifying the centripetal force
The centripetal force is whatever real force (or combination) provides the inward net force:
- A ball on a string: the tension points inward.
- A car turning on a flat road: friction between tyres and road.
- A satellite orbiting Earth: gravity.
- A passenger on a fairground ride: the normal force from the seat or wall.
If the required inward force is not available (e.g. not enough friction), the object cannot follow the circle and moves off tangentially.
The "centrifugal" misconception
There is no outward force on the object. In the rotating frame of a passenger it feels as if they are pushed outward, but in an inertial frame the only forces are real and the net force is inward. What the passenger feels is their own inertia (tendency to go straight) resisting the inward push of the seat.
Exam-style practice questions
Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.
2025 SACE Stage 24 marksA wind turbine has three blades that move in a circular path. Each blade is 32.0 m long and the tips move at a constant speed of 49.3 m s-1. (a) Show that the period of rotation of the blades is 4.08 s. (b) Calculate the magnitude of the acceleration of the tip of the blades.Show worked answer →
The blade tip moves in a circle of radius r = 32.0 m at constant speed v = 49.3 m s-1.
(a) For circular motion v = 2 pi r / T, so T = 2 pi r / v = 2 pi (32.0) / 49.3 = 201.06 / 49.3 = 4.08 s. (2 marks)
(b) The centripetal acceleration is a = v squared / r = (49.3 squared) / 32.0 = 2430.5 / 32.0 = 75.9 m s-2. (2 marks)
This acceleration is directed toward the centre of the circle (the hub), even though the speed is constant, because the direction of the velocity is continuously changing.
2023 SACE Stage 23 marksA racing car travels along a section of a circular path of radius 85.0 m at a constant speed of 38.0 m s-1. The friction force between the tyres and road, which causes the circular motion, has a magnitude of 1.35 x 10^4 N. Determine the mass of the car.Show worked answer →
The friction force supplies the centripetal force, so F = m v squared / r.
Rearrange for mass: m = F r / v squared.
Substitute: m = (1.35 x 10^4)(85.0) / (38.0 squared) = (1.1475 x 10^6) / 1444 = 794 kg.
1 mark for identifying friction as the centripetal force, 1 mark for the rearrangement, 1 mark for the answer of about 794 kg (or 7.9 x 10^2 kg).