Welcome to Circular Motion & Gravitation. The motion studied so far ran along straight lines, in Describing Motion and Uniform Acceleration & Free Fall, or along the open curve of a projectile. This topic follows bodies that go round and round: a point on a turning wheel, a stone whirled on a string, a satellite circling the Earth. Motion round a circle repeats itself, and that gives it quantities a straight-line motion does not have: the angle turned through, the time one turn takes and the number of turns made each second. This page defines them, links them to the ordinary speed and practises moving between them. The next page uses them to compare points on wheels, belts and gears; the third explains why a body on a circle needs a force at all; the last applies the whole topic to planets and satellites.
The topic has had two or three questions in every paper so far, 10 of the 192. Three of them are one-step conversions from this page. December 2025 Q12 and April 2026 Q13 are the same question with new numbers, an angular velocity from a speed and a radius, and January 2026 Q11 asks for the speed and the angular velocity from a radius and a period. All three are worked through below. Each takes under a minute, and the marks lost on them come from slips rather than from hard physics: a π written in where none belongs, a radius left out, two answers given in the wrong order. This page names each slip where it comes up.
What "uniform" means on a circle
A body in uniform circular motion moves round a circle and covers equal lengths of arc in equal times, however short the times are chosen. Its speed, the size of its velocity, therefore never changes. Its velocity does.
The Speed & Velocity page met this already. Velocity has a direction as well as a size, and on a curved path it points along the tangent, the way the body is heading at that moment. On a circle the tangent is at right angles to the radius, so as the body goes round, the direction of its velocity turns with it. Follow the body a quarter of the way round and its velocity has turned through a quarter turn as well. Two velocities with the same size but different directions are different velocities, so the velocity is changing all the time, even though the speed is steady.
A body going anticlockwise round a circle at a steady speed, shown once a second. Each second it covers the same length of arc, and its radius turns through the same angle, an eighth of a turn, so one turn takes 8 s. The velocity arrows all have the same length, but each points a different way: along the tangent, at right angles to the radius.
This has two consequences that the papers' wording leans on. A changing velocity means an acceleration, so uniform circular motion is not the uniform linear motion of Describing Motion, which needs a constant velocity. Nor is it uniformly accelerated motion, which needs a constant acceleration: the acceleration of a body on a circle turns as the body goes round, and the Centripetal Acceleration & Force page shows that it always points to the centre. "Uniform" here describes the speed and nothing else.
Uniform circular motion. Motion round a circle at a constant speed. The speed is constant; the velocity is not, because its direction, along the tangent and at right angles to the radius, turns as the body goes round. So the motion is accelerated, but not uniformly accelerated.
Try it! · Check-in question
A small ball moves round a horizontal circle at a constant speed of 2 m/s. Which statement about its motion is correct?
The speed is fixed, but the velocity points along the tangent, and the tangent turns as the ball goes round. A velocity whose direction changes is a changing velocity, and a changing velocity means an acceleration, so the answer is C. A treats speed and velocity as the same thing: the size stays the same, the direction does not. B describes the direction of the ball's acceleration, which the Centripetal Acceleration & Force page shows points to the centre. The velocity lies along the tangent, at right angles to the radius. D is half right, since the ball does accelerate. But its acceleration turns with it as it goes round, so it is not constant, and uniformly accelerated motion needs a constant acceleration.
Measuring angles in radians
How far a body has gone round a circle can be given as a length of arc or as the angle its radius has turned through. Physics measures that angle in radians, not degrees, because the radian ties the angle directly to the arc. The angle in radians is the arc length divided by the radius:
θ = s/r, or, rearranged, s = rθ.
An angle of 1 rad is the angle whose arc is exactly as long as the radius. A full turn has an arc equal to the whole circumference, 2πr, so a full turn is 2πr ÷ r = 2π rad. Half a turn is π rad, and a quarter turn is π/2 rad. Since 2π rad = 360°, 1 rad = 360° ÷ 2π, about 57.3°. This is where every π in circular motion comes from: a whole turn is 2π rad.
Because θ is one length divided by another, the radian has no dimension of its own. That is why an angular velocity in rad/s is sometimes written in s⁻¹.
The angle θ in radians is the arc s divided by the radius r. The angle drawn is 1 rad, about 57°, where the arc is exactly as long as the radius. A full turn has an arc of 2πr, so it is 2π rad.
Try it! · Check-in question
A body moves 3 m along the arc of a circle of radius 2 m. Through what angle has the radius to the body turned?
The angle in radians is the arc divided by the radius: θ = s/r = 3 ÷ 2 = 1.5 rad, about 86°, just short of a quarter turn, so the answer is D. A multiplies the arc by the radius, which gives an area in m², not an angle. B adds a π. A π appears only when the angle is counted in turns, because a whole turn is 2π rad; here the angle comes straight from two lengths. C divides the wrong way round, r ÷ s. Check with s = rθ: 2 × 1.5 = 3 m, the arc given.
Linear speed and angular velocity
There are two natural ways to say how fast a body goes round. The linear speed v is the length of arc it covers each second, v = s/t, in m/s. It is the ordinary speed of Describing Motion, and the papers often call it the linear velocity. The angular velocity ω (the Greek letter omega) is the angle its radius turns through each second, ω = θ/t, in rad/s. In uniform circular motion both are constant.
The two are linked by the radius. In a time t the radius turns through θ, and the body covers the arc s = rθ. Divide both sides by t:
s/t = r × θ/t, so v = ωr.
The formula says something you can feel on a merry-go-round. Every point of a turning disc sweeps through the same angle each second, so they all have the same ω, but a point near the rim covers a longer arc in that time, so it has a larger v. The next page is built on that idea. On this page the formula is used to convert: ω = v/r, and v = ωr.
Linear speed and angular velocity. v = s/t is the arc covered per second, in m/s. ω = θ/t is the angle turned per second, in rad/s. They are linked by v = ωr, so ω = v/r, with the radius r in metres.
Worked example · the angular velocity from a speed and a radius (December 2025, Q12)
A particle moves in uniform circular motion with a speed of v = 6 m/s and a radius of R = 3 m. Its angular velocity is ( ).
A. 3 rad/s B. 3π rad/s C. 2 rad/s D. 2π rad/s
Step 1, the link: the speed and the radius are given, and v = ωR joins them to ω. Rearranged, ω = v/R.
Step 2: ω = 6 ÷ 3 = 2 rad/s. The answer is C.
Check: ωR = 2 × 3 = 6 m/s, the speed given. In each second the particle covers 6 m of arc, two radius-lengths, so its radius turns through 2 rad.
B and D are the trap. They put a π into the answer, 3π and 2π, as if every angular velocity carried one. A π enters only through a full turn, which is 2π rad, so it appears when the data include a period, a frequency or a number of revolutions. This question gives only a speed and a radius, and the answer is a plain number of radians per second. A, 3 rad/s, is the radius written down with the unit of angular velocity; it does not follow from any formula.
Try it! · Check-in question
A particle undergoes uniform circular motion with a linear velocity v = 4 m/s and a radius R = 2 m. Its angular velocity is ( ) (April 2026, Q13)
This is December's question with new numbers. "Linear velocity" is the paper's name for the speed v along the circle, so ω = v/R = 4 ÷ 2 = 2 rad/s, and the answer is A. Check: ωR = 2 × 2 = 4 m/s. B multiplies, v × R = 8, which turns the formula upside down; its unit would be m²/s, not rad/s. C copies the speed across as the angular velocity, as if v and ω were the same quantity. D divides by the diameter, 4 ÷ (2 × 2) = 1, but v = ωr uses the radius.
Period, frequency and revolutions per minute
Motion round a circle repeats, so it can also be described by how long one turn takes, or by how many turns are made in a given time.
The period T is the time for one full revolution, in seconds. The frequency f is the number of revolutions each second, in hertz (Hz, one per second), so f = 1/T: a wheel that takes 0.25 s per turn makes 4 turns a second. Machines are often rated by their rotational speed n in revolutions per minute (rev/min, or r/min). Dividing by 60 turns it into revolutions per second, which is the frequency.
Now apply the two definitions from the last section to one full turn, which takes one period. In that time the radius turns through 2π rad and the body covers one circumference, 2πr. So
ω = 2π/T = 2πf, and v = 2πr/T = 2πrf.
These agree with v = ωr: multiply ω = 2π/T by r and you get 2πr/T. That is a useful check, and it means v = 2πr/T needs no separate learning. Find ω from the period, then multiply by r.
The four ways to say how fast a body goes round, and the links between them. Along the row, read the label above an arrow when going right and the label below it when going left: f = 1/T, ω = 2πf, v = ωr, and back again. The curved arrows are the two shortcuts from the period, ω = 2π/T and v = 2πr/T.
Period, frequency and rotational speed. T is the time for one revolution, in s; f = 1/T is the number of revolutions per second, in Hz. A rotational speed of n rev/min is n/60 revolutions per second. One revolution is 2π rad and a distance of 2πr, so ω = 2π/T = 2πf and v = 2πr/T = ωr.
Try it! · Check-in question
What is the angular velocity of the minute hand of a clock?
The minute hand goes once round the dial in an hour, so its period is T = 3600 s, and ω = 2π/T = 2π/3600 = π/1800 rad/s, so the answer is D. A uses T = 60 s, the period of the second hand. The minute hand moves on by one mark each minute, but it needs 60 minutes for a full turn. B uses π rad for a full turn, π/T, but π rad is only half a turn. C is the hour hand, which takes 12 h = 43 200 s per turn: 2π/43 200 = π/21 600 rad/s.
Try it! · Check-in question
The blades of an electric fan turn at a steady 1200 rev/min. What is their angular velocity?
First the frequency: 1200 revolutions a minute is 1200 ÷ 60 = 20 revolutions a second, f = 20 Hz. Each revolution is 2π rad, so ω = 2πf = 2π × 20 = 40π rad/s, about 126 rad/s, and the answer is B. A stops at the frequency and writes 20 revolutions a second as 20 rad/s; a revolution is 2π rad, not 1 rad. C forgets to change minutes to seconds: 2π × 1200 = 2400π is the angle turned in a minute. D counts a revolution as π rad instead of 2π rad.
Two answers at once: "respectively"
January 2026 asked for two quantities in one question, and a question like that has two extra hazards. The options are pairs of numbers, so the two answers must come in the order the stem names them, which is what "respectively" means. And in this paper the options were printed without units, so nothing in an option shows which number is meant to be the speed and which the angular velocity.
1
Angular velocity from the period. ω = 2π/T. The radius plays no part in ω.
2
Speed from ω. Convert the radius to metres, then v = ωr.
3
Match the order. Write "v = …, ω = …" (or whatever the stem names first) and only then look at the options.
Worked example · v and ω from a radius and a period (January 2026, Q11)
An object moves in a uniform circular motion with a radius of 0.1m and a period of 5s. The linear velocity and angular velocity of the object during motion are, respectively,
A. 100π, 0.4π B. 0.4π, 0.4π
C. 0.5π, 100π D. 0.04π, 0.4π
Step 2, the speed: the radius is already in metres, so v = ωr = 0.4π × 0.1 = 0.04π m/s. The shortcut gives the same: v = 2πr/T = (2π × 0.1) ÷ 5 = 0.04π m/s.
Step 3, the order: the stem asks for the linear velocity first, then the angular velocity, so the pair is 0.04π m/s, 0.4π rad/s. The answer is D.
Both answers keep their π, because the data include a period and one turn is 2π rad. Leave π as a symbol, as the options do; 0.04π m/s is about 0.13 m/s.
B gives both quantities the value of ω, 0.4π. It leaves out the factor r: v and ω have the same number only on a circle of radius 1 m. A puts the period on top and the radius underneath, 2πT/r = 2π × 5 ÷ 0.1 = 100π. A sense check rules it out: an object going round a 0.1 m circle once every 5 s cannot be moving at over 300 m/s. C pairs 0.5π with that same upside-down 100π; neither of its numbers follows from the right formulas.
Four slips in a "respectively" question. Leaving out r: v = ωr, so v and ω share a number only when r = 1 m. Turning a formula upside down: ω = 2π/T, not 2πT; a longer period means slower turning, so T belongs underneath. Leaving a radius in centimetres: ω does not change, but v comes out 100 times too large. Swapping the pair: when no units are printed, the right two numbers in the wrong order look like a correct option.
Exam wording. The English papers call v the "linear velocity" (January 2026 Q11, April 2026 Q13) or simply the "speed" (December 2025 Q12); both mean the speed along the circle, in m/s. In the Chinese papers uniform circular motion is 匀速圆周运动, linear velocity is 线速度, angular velocity is 角速度 and the period is 周期. A textbook also uses 频率 for frequency, 转速 for rotational speed and 弧度 for the radian.
Try it! · Check-in question
A toy car runs round a circular track of radius 20 cm at a steady speed, taking 4 s for each lap. Its linear speed and angular velocity are, respectively,
ω = 2π/T = 2π ÷ 4 = 0.5π rad/s. Convert the radius, 20 cm = 0.2 m, and then v = ωr = 0.5π × 0.2 = 0.1π m/s. The speed comes first, as asked: 0.1π m/s, 0.5π rad/s, so the answer is C. A leaves out the factor r and gives v the same number as ω. B leaves the radius in centimetres, 0.5π × 20 = 10π, which would be over 30 m/s for a toy car. D has the right two numbers in the wrong order. Here the units give the swap away; on a paper that prints no units, only the order of the stem does.
Working backwards, and when π appears
Each formula can be run the other way. From an angular velocity, the period is the time to turn through 2π rad, T = 2π/ω. From a speed and a radius, it is the time to cover one circumference, T = 2πr/v. And from the period, f = 1/T as before.
Looking back over the examples, one pattern decides whether π belongs in an answer. π enters circular motion in one place only: a full revolution is 2π rad and 2πr of arc. So π appears when the working passes through a whole turn, going from a period, a frequency, a rate in rev/min or a number of revolutions to an angle or a speed, or the other way round. It does not appear when the data and the answer are linked by v = ωr alone; December's 6 m/s on 3 m gives 2 rad/s, not 2π. A π in the data can also cancel: a period of π s gives ω = 2π ÷ π = 2 rad/s.
When to expect π. Passing through a full turn (a period, a frequency, rev/min or a count of revolutions) brings in 2π: ω = 2π/T, v = 2πr/T, T = 2π/ω = 2πr/v. Going between v and ω with v = ωr brings in no π. An option with a π that your working did not produce is a trap, not a better answer.
Try it! · Check-in question
A wheel turns at a steady angular velocity of 4 rad/s. How long does it take to make one revolution?
One revolution is 2π rad, turned at 4 rad each second, so T = 2π/ω = 2π ÷ 4 = 0.5π s, about 1.6 s, and the answer is A. Check: 2π/T = 2π ÷ 0.5π = 4 rad/s. B is 1/ω, which treats ω as a frequency; but 4 rad/s is 4 radians a second, not 4 revolutions. C multiplies, 2π × 4, so a faster wheel would take longer per turn. D counts one revolution as π rad, which is only half a turn.
See also. Three kinds of circle question in the papers belong to other topics. The distance and the displacement for one full turn (January 2026 Q7, the Ferris wheel; March 2026 Q18) are worked on the Particles, Frames & Displacement page: one turn is a distance of 2πR and a displacement of 0. The average velocity along two semicircles (December 2025 Q13, April 2026 Q14) is on the Speed & Velocity page. And the directions of the velocity and the net force on a curved path (January 2026 Q36, March 2026 Q23, April 2026 Q48) belong to the Projectile Motion topic.
Your turn! · Practice questions
Practice · Question 1
A ball on a string moves round a horizontal circle of radius 0.6 m at a steady speed of 3 m/s. What is the period of its motion?
One revolution is one circumference, 2πr = 2π × 0.6 = 1.2π m, covered at 3 m/s, so T = 2πr/v = 1.2π ÷ 3 = 0.4π s, about 1.3 s, and the answer is B. A takes the circumference as πr, leaving out the 2. C turns the formula upside down, 2πv/r = 2π × 3 ÷ 0.6 = 10π; a faster ball would then take longer per lap. D finds ω = v/r = 5 rad/s correctly but then takes T = 1/ω, as if ω counted revolutions. It counts radians, and one revolution is 2π rad: T = 2π/5 = 0.4π s.
Practice · Question 2
A disc spins at a steady 300 rev/min. What is the speed of a point on the disc 6 cm from its centre?
The frequency is 300 ÷ 60 = 5 revolutions a second, so ω = 2πf = 10π rad/s. With r = 6 cm = 0.06 m, v = ωr = 10π × 0.06 = 0.6π m/s, about 1.9 m/s, so the answer is C. A counts a revolution as π rad instead of 2π: π × 5 × 0.06 = 0.3π. B forgets to change minutes to seconds: 2π × 300 × 0.06 = 36π is the length of arc covered in a minute, not in a second. D leaves the radius in centimetres: 10π × 6 = 60π.
Practice · Question 3
A cyclist rides round a circular track of radius 50 m at a steady 5 m/s. Her period and her angular velocity are, respectively,
The period passes through a full lap: T = 2πr/v = 2π × 50 ÷ 5 = 20π s, about 63 s. The angular velocity comes from v = ωr alone: ω = v/r = 5 ÷ 50 = 0.1 rad/s, with no π. Period first, as asked, so the answer is A. Check: 2π/T = 2π ÷ 20π = 0.1 rad/s. B adds a π to the angular velocity because the period has one; ω = v/r gives a plain number. C takes a lap as πr, half the circumference. D has the two numbers in the wrong order: a period of 0.1 s would mean ten laps every second.
Practice · Question 4
A wheel turns at a steady angular velocity of 0.5π rad/s. How many revolutions does it make in one minute?
In 60 s the wheel turns through θ = ωt = 0.5π × 60 = 30π rad. Each revolution is 2π rad, so it makes 30π ÷ 2π = 15 revolutions, and the answer is D. The frequency gives the same: f = ω/2π = 0.25 revolutions a second, and 0.25 × 60 = 15. A divides by π instead of 2π, as if a revolution were π rad. B is the angle, 30π rad, not the number of turns. C is the number of revolutions in one second, not yet multiplied by the 60 s in a minute.
Practice · Question 5
A toy train runs round a circular track at a steady speed. The track is replaced by a circle with twice the radius, and the train keeps the same speed. What happens to its angular velocity ω and its period T?
With v fixed, ω = v/r, so doubling r halves ω. The period is T = 2πr/v, so doubling r doubles T: the lap is twice as long, covered at the same speed. The answer is B. The two results agree, because T = 2π/ω: half the angular velocity means twice the time per turn. A has both changes the wrong way round, as if the bigger circle were turned faster. C ties ω to the speed, but ω is the angle swept per second, and on a larger circle the same speed sweeps a smaller angle. D cannot happen at all: since T = 2π/ω, when ω falls T must rise.
Things to remember
Concept
What to know
Uniform circular motion
Constant speed round a circle. The velocity lies along the tangent, at right angles to the radius, and keeps turning, so the motion is accelerated but not uniformly accelerated. "Uniform" describes the speed only.
The radian
θ = s/r, so s = rθ. A full turn is 2π rad = 360°; 1 rad ≈ 57.3°. Every π in circular motion comes from the 2π of a full turn.
Linear speed and angular velocity
v = s/t in m/s (the papers' "linear velocity"); ω = θ/t in rad/s. v = ωr, so ω = v/r. December 2025 Q12 and April 2026 Q13: both 2 rad/s.
Period and frequency
T is the time for one revolution; f = 1/T, in Hz. n rev/min is n/60 revolutions per second.
Through a full turn
ω = 2π/T = 2πf and v = 2πr/T = ωr; back again, T = 2π/ω = 2πr/v. January 2026 Q11: 0.04π m/s and 0.4π rad/s.
When π appears
Only when the working passes through a full turn: a period, a frequency, rev/min or a count of revolutions. From v and r alone, ω has no π.
"Respectively"
Give the answers in the order the stem names them. When the options carry no units, write "v = …, ω = …" before matching.
The common wrong options
A π added to ω = v/r; v × r or v ÷ diameter for ω; r left out so that v = ω; T upside down (2πT); a revolution taken as π rad; minutes or centimetres not converted.
