PhysicsICSE Class 10 · Uniform Circular Motion

Uniform Circular Motion as Accelerated Motion: Speed, Velocity and Solved Questions (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

In uniform circular motion a body covers equal distances along a circle in equal times, so its speed is constant. But its velocity is along the tangent, and the tangent's direction changes at every point, so the velocity keeps changing. A changing velocity means acceleration, so the motion is accelerated, with the acceleration towards the centre.

The speed stays constant, but the velocity arrow turns at every point, so there is an acceleration pointing towards the centre.

Illustration drawn for QuestionBuddy.

Understand Uniform Circular Motion as Accelerated Motion

Uniform circular motion is the motion of a body along a circular path at constant speed: it covers equal lengths of the circle in equal intervals of time. Examples: the tip of a clock's hand, a point on a steadily turning fan blade, a satellite in a circular orbit, a stone whirled at a steady rate.

Speed. In one revolution the body covers the circumference $2 π r$ in the time period $T$ :

$v = \frac{2 π r}{T} .$

If the body makes $n$ revolutions per second, $T = \frac{1}{n}$ and $v = 2 π r n$ .

Velocity. Velocity is speed in a stated direction. At every point of the circle the body is moving along the tangent to the circle at that point. As the body goes round, the tangent turns, so the direction of the velocity changes continuously, even though its size (the speed) stays the same. On opposite ends of a diameter, the velocities are equal in size but opposite in direction.

Why it is accelerated. Acceleration is the rate of change of velocity, not of speed. Since the velocity changes (in direction) at every instant, the body has an acceleration all the time. This is why the syllabus calls uniform circular motion an example of constant speed though acceleration is present. By Newton's second law, an acceleration needs a net force; for circular motion that force is the centripetal force, directed towards the centre, and so is the acceleration.

What stays constant and what changes.

Constant: speed, kinetic energy ( $\frac{1}{2} m v^{2}$ ), the radius, the time for one revolution.
Changing: velocity (direction), momentum (direction), the direction of the acceleration (always towards the centre, so it turns as the body moves).

Distance and displacement. Distance is measured along the circle; displacement is the straight line from start to finish.

Quarter revolution: distance $\frac{π r}{2}$ , displacement $2 r$ .
Half revolution: distance $π r$ , displacement $2 r$ (the diameter).
One full revolution: distance $2 π r$ , displacement zero, so the average velocity over a full round is zero although the average speed is not.

A useful picture. Over half a revolution the velocity reverses: if the speed is $v$ , the change in velocity is $2 v$ , even though the speed never changed. That change is what the centripetal force produces.

Uniform Circular Motion as Accelerated Motion: formulas and key facts

Speed in uniform circular motion: $v = \frac{2 π r}{T} = 2 π r n$; $T$ = time for one revolution; $n$ = revolutions per second.
Distance and displacement in half a revolution: $s = π r, displacement = 2 r$; In a full revolution: $2 π r$ and zero.
Average velocity: $\overset{v}{ˉ} = \frac{displacement}{time}$; Not the same as average speed, which uses distance.
Change in velocity over half a revolution: $∣Δ v ∣ = 2 v$; The velocity reverses although the speed is unchanged.

How to solve Uniform Circular Motion as Accelerated Motion questions

Separate speed (size only) from velocity (size and direction) in every step.

1Find the circumference $2 π r$ , using the value of $π$ the question gives.
2Speed $= \frac{2 π r}{T}$ ; convert km/h to m/s by dividing by 3.6 if needed.
3For distance, use the fraction of a revolution times $2 π r$ .
4For displacement, draw the start and end points and join them: $2 r$ for half a turn, $2 r$ for a quarter or three-quarter turn, zero for a whole turn.
5For the direction of velocity at a point, draw the tangent at that point in the direction of motion.
6For 'is it accelerated?', argue from the changing direction of velocity, not from the speed.

Uniform Circular Motion as Accelerated Motion in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A steady speedometer is not zero acceleration

Suppose a car goes round a circular roundabout of radius 50 m with its speedometer steady at 36 km/h.

Find its speed in m/s and the time for one full round ( $π = 3.14$ ). Is the car accelerating?

Try it first, then show the solutionSolution

$v = \frac{36}{3.6} = 10$ m/s.
$T = \frac{2 π r}{v} = \frac{2 \times 3.14 \times 50}{10} = 31.4$ s.
Its direction of motion changes all the time, so its velocity changes: it is accelerating towards the centre.

Answer: 10 m/s; 31.4 s; yes, it is accelerating.

Memory hook: The speedometer shows speed, not velocity.

Speed of a clock hand

Imagine the minute hand of a wall clock is 7 cm long.

Find the speed of its tip in cm/s ( $π = \frac{22}{7}$ ). Is the tip's velocity constant?

Try it first, then show the solutionSolution

One revolution takes 60 min = 3600 s; circumference $= 2 \times \frac{22}{7} \times 7 = 44$ cm.
$v = \frac{44}{3600} \approx 0.0122$ cm/s.
The direction keeps changing, so the velocity is not constant.

Answer: About 0.012 cm/s; the velocity changes in direction.

Memory hook: Even the slowest hand is accelerating.

Tricks to remember Uniform Circular Motion as Accelerated Motion

Short memory aids. Say each one aloud once, then use it on a question the same day.

Speed same, path bends
Bending the path changes the velocity; changing velocity is acceleration.
Use it when: Any 'why is it accelerated?' question.
Velocity hugs the tangent
Velocity is always along the tangent; force and acceleration point to the centre.
Use it when: Drawing arrows on a circle.
Half turn, double v
After half a turn the velocity has reversed: change = 2v, displacement = 2r.
Use it when: Displacement and change-of-velocity questions.
Full circle, zero shift
One complete round: distance 2πr, displacement zero, average velocity zero.
Use it when: Average speed versus average velocity.

Uniform Circular Motion as Accelerated Motion: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Direction of velocity · 1 mark · MCQBoard level

For a body in uniform circular motion, the direction of its velocity at any point is:
(a) towards the centre
(b) away from the centre
(c) along the tangent at that point
(d) along the radius

Try it first, then show the solutionSolution

At each instant the body is moving along the circle, whose direction at that point is the tangent.
Force and acceleration are along the radius, towards the centre; velocity is not.

Answer: (c)

Example 2: Speed of a fan blade tip · 2 marks · Short answerBoard level

The tip of a ceiling-fan blade is 0.7 m from the axis. The fan makes 5 revolutions per second. Find the speed of the tip. ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Circumference $= 2 \times \frac{22}{7} \times 0.7 = 4.4$ m. (1)
$v = 2 π r n = 4.4 \times 5 = 22$ m/s. (1)

Answer: 22 m/s

Example 3: Speed versus average velocity · 3 marks · Short answerHarder board level

A toy train runs round a circular track of radius 7 m at constant speed, taking 88 s per round. ( $π = \frac{22}{7}$ )
(a) Find its speed.
(b) For half a round, find the distance, the displacement and the average velocity.

Try it first, then show the solutionSolution

(a) $v = \frac{2 \times \frac{22}{7} \times 7}{88} = \frac{44}{88} = 0.5$ m/s. (1)
(b) Half a round takes 44 s; distance $= π r = 22$ m; displacement $= 2 r = 14$ m. (1)
Average velocity $= \frac{14}{44} \approx 0.32$ m/s, along the diameter; less than the speed, because the displacement is shorter than the path. (1)

Answer: (a) 0.5 m/s (b) 22 m, 14 m, about 0.32 m/s

Example 4: Same speed, different velocity · 3 marks · Short answerHarder board level

A stone whirled at a steady 5 m/s in a horizontal circle moves due east at point $A$ and due west at point $B$ .
(a) Where are $A$ and $B$ on the circle?
(b) Find the change in velocity from $A$ to $B$ .
(c) Has the speed changed?

Try it first, then show the solutionSolution

(a) The velocities are opposite, so $A$ and $B$ are at opposite ends of a diameter. (1)
(b) Taking east as positive: $Δ v = (- 5) - (+ 5) = - 10$ m/s, that is 10 m/s towards the west. (1)
(c) No, the speed is 5 m/s at both points; only the direction has changed. (1)

Answer: (a) Ends of a diameter (b) 10 m/s, westward (c) No

Example 5: Where does the change in velocity point? · 4 marks · Long answerHOTS / competency

A body moves round a circle at a steady 4 m/s, taking 8 s per revolution. In a quarter of a revolution its velocity turns from due north to due west.
(a) Find the size of the change in velocity.
(b) Find the average acceleration over this quarter turn.
(c) The change in velocity points south-west. Show that this direction points from the middle of the arc towards the centre of the circle, and say what this suggests.

Try it first, then show the solutionSolution

(a) $Δ v = v_{west} - v_{north}$ : 4 m/s west and 4 m/s south, size $4^{2} + 4^{2} = 42 \approx 5.66$ m/s. (1)
(b) Quarter turn takes $\frac{8}{4} = 2$ s: average acceleration $= \frac{5.66}{2} \approx 2.83$ m/s². (1)
(c) Moving north then west means the body goes anticlockwise, from the east point of the circle to the north point; the middle of this arc is to the north-east of the centre, so south-west points from there towards the centre. (1)
So the acceleration of uniform circular motion points towards the centre: it is centripetal. (1)

Answer: (a) About 5.66 m/s (b) About 2.83 m/s² (c) It points to the centre: the acceleration is centripetal.

Example 6: Case study: a Ferris wheel · 4 marks · Case-basedHOTS / competency

A Ferris wheel of radius 14 m turns at a steady rate, taking 88 s per revolution. A rider of mass 50 kg sits in a car at its rim. ( $π = \frac{22}{7}$ )
(a) Find the rider's speed.
(b) Is the rider accelerated? Give a reason.
(c) Find the distance and the size of the displacement in 22 s.
(d) Find the rider's kinetic energy and say whether it changes.

Try it first, then show the solutionSolution

(a) $v = \frac{2 \times \frac{22}{7} \times 14}{88} = \frac{88}{88} = 1$ m/s. (1)
(b) Yes. The direction of motion changes continuously, so the velocity changes. (1)
(c) 22 s is a quarter turn: distance $= \frac{88}{4} = 22$ m; displacement $= 2 \times 14 \approx 19.8$ m. (1)
(d) $K = \frac{1}{2} \times 50 \times 1^{2} = 25$ J; it stays constant because the speed is constant. (1)

Answer: (a) 1 m/s (b) Yes, the velocity's direction changes (c) 22 m; about 19.8 m (d) 25 J, constant

Where students lose marks in Uniform Circular Motion as Accelerated Motion

The slips examiners see most often, and what to write instead.

Mistake: Saying the velocity is constant
Only the speed is constant; the direction, and therefore the velocity, changes all the time.
Wrong working
Uniform circular motion has uniform velocity.
Right working
Uniform circular motion has uniform speed but variable velocity.
Do this: Write 'speed' when you mean size only and 'velocity' when direction matters.
Usually costs: Definition and reasoning marks in 1-2 mark questions.
Mistake: Drawing the velocity towards the centre
The velocity is along the tangent; the force and acceleration are towards the centre.
Wrong working
Velocity arrow pointing along the string to the hand.
Right working
Velocity arrow along the tangent; force arrow along the string to the centre.
Do this: Draw two arrows at right angles at the same point.
Usually costs: The diagram mark.
Mistake: Using distance for average velocity
Average velocity uses displacement; average speed uses distance.
Wrong working
Half a round of a 7 m track in 44 s: average velocity $= \frac{22}{44} = 0.5$ m/s.
Right working
Average velocity $= \frac{14}{44} \approx 0.32$ m/s.
Do this: Draw the straight line from start to end before dividing.
Usually costs: The answer mark.
Mistake: Thinking the displacement makes the work zero
Work by the centripetal force is zero because the force is perpendicular to the motion at every instant, not because the displacement after a round is zero.
Wrong working
Work is zero because the stone comes back to the start.
Right working
Work is zero because the force (towards the centre) is at $90°$ to the displacement (along the tangent) at every instant.
Do this: Give the angle reason, $W = F s cos 90° = 0$ .
Usually costs: Reported for 2024 Q2(ii)(b), where candidates gave zero for the wrong reason.

Exam tips

Use the value of $π$ given in the question; choose radii like 7, 14, 21 with $\frac{22}{7}$ .
For 'why is uniform circular motion accelerated?', write: velocity changes in direction at every point, acceleration is rate of change of velocity, so it is accelerated though the speed is constant.
Distinguish speed, velocity, distance and displacement in each answer; examiners mark the word you use.
The syllabus describes it as an example of constant speed with acceleration present: learn that phrase.

Practice questions on Uniform Circular Motion as Accelerated Motion

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A body in uniform circular motion has:
(a) zero acceleration
(b) constant velocity
(c) constant speed but changing velocity
(d) changing speed but constant velocity
Hint
Velocity includes direction.
Answer
(c)
Q2 · 2 marks · Short answerBoard level
The second hand of a clock is 2.1 cm long. Find the speed of its tip. ( $π = \frac{22}{7}$ )
Hint
One revolution takes 60 s.
Answer
$\frac{2 \times \frac{22}{7} \times 2.1}{60} = \frac{13.2}{60} = 0.22$ cm/s.
Q3 · 2 marks · Short answerHarder board level
A student says: 'The car went round the roundabout at a steady 20 km/h, so it had no acceleration.' Correct the statement and give the reason.
Hint
What does acceleration measure?
Answer
The car was accelerating. Acceleration is the rate of change of velocity; going round, the direction of motion changes continuously, so the velocity changes even though the speed is steady.
Q4 · 3 marks · Short answerHarder board level
A cyclist rides round a circular track of radius 21 m at a constant 6.6 m/s. ( $π = \frac{22}{7}$ ) Find (a) the time for one round, (b) the number of rounds in 5 minutes.
Hint
Circumference first.
Answer
(a) $2 \times \frac{22}{7} \times 21 = 132$ m; $T = \frac{132}{6.6} = 20$ s. (b) $\frac{300}{20} = 15$ rounds.
Q5 · 2 marks · Short answerHarder board level
A body moves three-quarters of the way round a circle of radius 10 m. Find the distance travelled and the size of the displacement. ( $π = 3.14$ )
Hint
Start and end points are a quarter-circle apart the other way.
Answer
Distance $= \frac{3}{4} \times 2 \times 3.14 \times 10 = 47.1$ m; displacement $= 2 \times 10 \approx 14.1$ m.
Q6 · 3 marks · Short answerHarder board level
State, with a reason, whether each is uniform circular motion: (a) the tip of a clock's hour hand, (b) a car speeding up on a circular track, (c) a stone whirled steadily on a string in a horizontal circle.
Hint
Is the speed constant? Is the path a circle?
Answer
(a) Yes: circular path, constant speed. (b) No: the speed changes. (c) Yes: circular path, constant speed.
Q7 · 3 marks · Short answerHOTS / competency
The tip of a fan blade moves at a steady 10 m/s. Half a revolution takes 0.05 s. Find (a) the change in the tip's speed, (b) the size of the change in its velocity, (c) its average acceleration over the half revolution.
Hint
After half a turn the velocity has reversed.
Answer
(a) Zero. (b) $10 - (- 10) = 20$ m/s. (c) $\frac{20}{0.05} = 400$ m/s².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Which is constant in uniform circular motion: speed or velocity?Show answer
Speed.
2.Displacement after one full revolution of radius r?Show answer
Zero.
3.Displacement after half a revolution of radius 5 m?Show answer
10 m.
4.Direction of acceleration in uniform circular motion?Show answer
Towards the centre.
5.Speed for r = 7 m and T = 22 s ( $π = \frac{22}{7}$ )?Show answer
2 m/s.

Uniform Circular Motion as Accelerated Motion: frequently asked questions

How can a body accelerate if its speed is constant?

Acceleration is the rate of change of velocity. In circular motion the direction of the velocity changes at every point, so the velocity changes and the body accelerates even though its speed stays the same.

What is the direction of velocity in uniform circular motion?

Along the tangent to the circle at the body's position, in the direction of motion.

Is the kinetic energy constant in uniform circular motion?

Yes. Kinetic energy depends on speed, which is constant, so it does not change.