Understand uniform circular motion the easy way, with plain English intuition, an animated diagram, centripetal acceleration and force, banked tracks, worked examples and an auto marked practice test. VCE Physics Units 3 and 4.
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Swing a ball on a string in a circle and let go, and it does not curve away. It shoots off in a dead straight line. That tells you something deep: to keep anything moving in a circle, something must constantly pull it toward the centre. The speed can stay perfectly steady, yet the ball is accelerating the whole time, because its direction never stops changing.
A steady speed that is still accelerating
Here is the part that trips everyone up. In uniform circular motion the speed is constant, but the velocity is not, because velocity has a direction and that direction is forever turning.
The velocity always points along the circle, tangent to the path, in the direction of travel.
The acceleration always points straight in toward the centre. We call it the centripetal acceleration.
Because the velocity keeps getting bent toward the centre, there must be a net force pointing the same way. That inward force is the centripetal force, and it is supplied by something real: tension in a string, friction under a tyre, gravity on a satellite, or a normal force on a banked track.
The blue arrow is the velocity, always tangent to the circle, pointing where the ball is heading. The red arrow is the centripetal force, always aimed at the centre. As the ball travels around, both arrows swing with it, but the blue one stays tangent and the red one keeps pointing inward.
Acceleration toward the centre
The size of that inward acceleration grows fast with speed and shrinks as the circle gets bigger. Double the speed and the acceleration goes up by four times, because the speed is squared.
You can write the centripetal acceleration two ways, and they say the same thing. One uses the speed, the other uses how long one lap takes, the periodT. The frequencyf=T1 counts laps per second, and the speed around the circle is the lap distance over the period, v=T2πr.
See it for yourself
Watch the ball travel around the circle in the diagram above. The blue velocity arrow stays tangent to the path the whole way round, while the red force arrow keeps pointing straight at the centre. Notice that the two arrows are always at right angles to each other: the force never speeds the ball up or slows it down, it only changes the direction.
A planet orbiting the Sun is circular motion in action. Gravity is the centripetal force, always pulling toward the centre and bending the path into a circle. Speed the planet up or change the masses and watch the orbit respond.
Circular motion questions follow a short, repeatable recipe.
Find the speed if you are not given it, using v=T2πr for something going round once every period T.
Work out the centripetal acceleration with ac=rv2.
The net inward force is Fc=rmv2. Ask which real force provides it: tension, friction, gravity or a normal force.
For a frictionless banked track, the angle that lets the normal force do the job satisfies tanθ=rgv2.
Always point the acceleration and the net force toward the centre, never along the direction of motion. The force changes the direction of travel, not the speed.
See the recipe in action in the Worked Examples tab, then test yourself in Try It.
Lock it in with active recall
Cover the answer and say each one out loud before you flip. Rate yourself honestly — the cards you find hard come back sooner, the ones you know are spaced further out.
Active recall
Answer from memory first, then flip. Rate yourself and each card returns on a spaced schedule (1 → 3 → 7 → 16 days).
In uniform circular motion, in what direction does the acceleration point?
Straight toward the centre of the circle (centripetal), even though the speed never changes.
Write the formula for centripetal acceleration.
ac=rv2 (equivalently T24π2r), directed toward the centre.
If the speed is constant, how can there be an acceleration?
Velocity includes direction, and the direction is always turning. A changing velocity means an acceleration, even with constant speed.
What real forces can supply the centripetal force?
It is not a new force — it is the net inward force, supplied by tension, friction, gravity, or a normal force. Take it away and the object flies off along a tangent.
What is the speed of an object going once around a circle of radius r in period T?
v=T2πr — the lap distance 2πr divided by the time for one lap.
What angle makes a frictionless banked track work at speed v and radius r?
tanθ=rgv2, so the normal force alone supplies the centripetal force.
Recall · Newton's Laws and Forces
State Newton’s second law, the rule behind centripetal force.
Fnet=ma. The centripetal force is just Fnet=mac=rmv2 applied toward the centre.
Recall · Projectile Motion
What is the only quantity the horizontal and vertical parts of a projectile share?
The time of flight — found from the vertical motion, then used in the horizontal motion.
Worked examples
Worked Example 1A ball on a string
A 0.50 kg ball on a string moves in a horizontal circle of radius 0.80 m at a constant 4.0 m/s. Find the centripetal acceleration and the tension in the string.
1
The speed is steady but the direction keeps turning, so there is an acceleration pointing to the centre. Use the centripetal acceleration formula.
ac=rv2=0.80(4.0)2=20 m/s2
2
The net force toward the centre is mass times that acceleration. Here the string tension provides it.
F=mac=0.50×20=10 N
Answer
ac=20 m/s2,F=10 N
Worked Example 2A car rounding a bend
A car rounds a circular bend of radius 50 m at a constant 15 m/s. Find its centripetal acceleration.
1
The car moves at a steady speed but its direction changes, so it still accelerates toward the centre of the bend.
ac=rv2=50(15)2
2
Work out the numbers.
ac=50225=4.5 m/s2
Answer
ac=4.5 m/s2
Practice questions
Practice test
Try it yourself
6 questions, 12 marks
Choose your answers, then submit to see your score and the full worked solutions.
Multiple choice is marked for you, just like Exam 2 Section A.
Q1.A ball moves in a horizontal circle at a constant speed. The direction of its acceleration is:
1mark
Need a hint?
A steady speed still means a changing velocity direction. Acceleration points the way the velocity is being pulled.
Show worked solution
Even though the speed never changes, the direction of the velocity is always turning, so there is an acceleration. It points straight toward the centre of the circle and is called the centripetal acceleration. Thinking the acceleration is zero because the speed is constant is the single most common circular motion error.
Q2.A ball moves once around a circle of radius 2.0 m every 4.0 s. Using v=T2πr, its speed is closest to:
1mark
Need a hint?
Speed is the distance once around, 2πr, divided by the period T.
Show worked solution
The distance once around is 2πr=2π(2.0)≈12.6 m, and this takes T=4.0 s, so v=T2πr=4.012.6≈3.1 m/s. Forgetting the factor of 2π, or dividing by 2T, gives the smaller options.
Q3.A 2.0 kg object moves in a circle of radius 1.5 m at 3.0 m/s. The net force toward the centre is closest to:
1mark
Need a hint?
Use Fc=rmv2.
Show worked solution
Fc=rmv2=1.52.0×(3.0)2=1.518=12 N. Leaving out the square on the speed, or dividing by the wrong quantity, produces the other options.
Q4.A car travels at 10 m/s around a circular track of radius 25 m. Its centripetal acceleration is closest to:
1mark
Need a hint?
Use ac=rv2.
Show worked solution
ac=rv2=25(10)2=25100=4.0 m/s2. Forgetting to square the speed gives 0.4 m/s squared, and swapping the speed and radius gives the other wrong answers.
Q5.A 0.20 kg ball on a string moves in a horizontal circle of radius 0.40 m at a constant 6.0 m/s. Taking the string as the only horizontal force, find the centripetal acceleration and the tension in the string. Show your working.
3marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
The centripetal acceleration points to the centre, so
ac=rv2=0.40(6.0)2=0.4036=90 m/s2.
The tension provides the net centripetal force, so
F=mac=0.20×90=18 N.
Q6.A racing car of total mass 800 kg rounds a circular turn of radius 240 m on a flat track at a constant 30 m/s. (a) Calculate the magnitude of the total sideways force the road exerts on the tyres. (b) The track is then banked so that no sideways friction is needed at the same speed and radius. Determine the banking angle θ (g=9.8 m/s2).
5marks
Work this on paper. The worked solution appears once you submit.
Show worked solution
(a) The sideways force supplies the centripetal force: F=rmv2=240800×302=3000 N.
(b) For a banked turn needing no friction, tanθ=rgv2=240×9.8302=0.383, so θ=20.9∘.
VCAA 2025 Physics Exam, Section B Q4
Frequently asked questions
What is centripetal force, really?
It is not a new force you add on. It is just the name for the net force that already points toward the centre, supplied by whatever is doing the pulling, such as tension, friction, gravity or a normal force. Take that force away and the object flies off in a straight line.
If the speed is constant, how can there be an acceleration?
Acceleration means any change in velocity, and velocity includes direction. In a circle the direction is always turning, so the velocity is always changing even when the speed stays the same. That change is the centripetal acceleration, directed toward the centre.
Why do race tracks and bends get banked?
Tilting the surface lets part of the normal force point toward the centre of the turn, which helps supply the centripetal force without relying only on friction. On a frictionless banked track the correct angle satisfies tan of the angle equals v squared over r times g.
