Skip to main content
ExamExplained
SA · Physics
Physics study scene
§-Syllabus dot point
SAPhysicsSyllabus dot point

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.

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section
  1. What this dot point is asking
  2. Why constant speed still means accelerating
  3. Period, frequency and speed
  4. Identifying the centripetal force
  5. Banked tracks
  6. The "centrifugal" misconception
  7. How SACE assesses this

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 TT is the time for one revolution and the frequency f=1/Tf = 1/T. The speed is the circumference divided by the period:

v=2πrT=2πrf.v = \frac{2\pi r}{T} = 2\pi r f.

Substituting into a=v2/ra = v^2/r gives an equivalent form in terms of period:

a=4π2rT2.a = \frac{4\pi^2 r}{T^2}.

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 in a straight line.

Banked tracks

On a banked curve the road is tilted so the normal force gains a horizontal component pointing toward the centre. This horizontal component supplies part or all of the centripetal force, reducing the reliance on friction. At the design speed, the horizontal component of the normal force alone supplies the entire mv2/rmv^2/r, so the curve can be taken safely even on a frictionless (icy) surface.

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.

How SACE assesses this

SACE Stage 2 circular-motion questions usually give two of speed, radius and period and ask for the centripetal acceleration or force, often in a real context such as a turbine blade, a fairground ride or a car on a bend. The dependable approach is to identify which real force provides the centripetal force (friction, tension, gravity or the normal force), then equate it to mv2r\dfrac{mv^2}{r}. Where the period is involved, convert between vv and TT with v=2πrTv = \dfrac{2\pi r}{T} first. Several "show that" parts ask you to verify a period or acceleration, so quote the formula, substitute with units, and arrive visibly at the printed value rather than just stating it.

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.

SACE 20254 marksA wind turbine has three blades that move in a circular path. Each blade is 32.0 m32.0\ \text{m} long and the tips move at a constant speed of 49.3 m s149.3\ \text{m s}^{-1}. (a) Show that the period of rotation of the blades is 4.08 s4.08\ \text{s}. (b) Calculate the magnitude of the acceleration of the tip of a blade.
Show worked answer →

The blade tip moves in a circle of radius r=32.0 mr = 32.0\ \text{m} at constant speed v=49.3 m s1v = 49.3\ \text{m s}^{-1}.

(a) For circular motion v=2πrTv = \dfrac{2\pi r}{T}, so

T=2πrv=2π(32.0)49.3=201.149.3=4.08 s.T = \dfrac{2\pi r}{v} = \dfrac{2\pi(32.0)}{49.3} = \dfrac{201.1}{49.3} = 4.08\ \text{s}.
(2 marks)

(b) The centripetal acceleration is

a=v2r=(49.3)232.0=2430.532.0=75.9 m s2.a = \dfrac{v^2}{r} = \dfrac{(49.3)^2}{32.0} = \dfrac{2430.5}{32.0} = 75.9\ \text{m s}^{-2}.
(2 marks)
This acceleration points toward the centre (the hub) even though the speed is constant, because the velocity direction changes continuously.

SACE 20233 marksA racing car travels along a circular section of track of radius 85.0 m85.0\ \text{m} at a constant speed of 38.0 m s138.0\ \text{m s}^{-1}. The friction force between the tyres and road, which provides the circular motion, has magnitude 1.35×104 N1.35\times10^4\ \text{N}. Determine the mass of the car.
Show worked answer →

Friction supplies the centripetal force, so F=mv2rF = \dfrac{mv^2}{r}.

Rearrange for mass: m=Frv2m = \dfrac{Fr}{v^2}.

m=(1.35×104)(85.0)(38.0)2=1.1475×1061444=794 kg.m = \dfrac{(1.35\times10^4)(85.0)}{(38.0)^2} = \dfrac{1.1475\times10^6}{1444} = 794\ \text{kg}.

1 mark for identifying friction as the centripetal force, 1 mark for the rearrangement, 1 mark for the answer of about 794 kg794\ \text{kg} (or 7.9×102 kg7.9\times10^2\ \text{kg}).

ExamExplained