Have you ever been on a merry-go-round and felt yourself being pushed outward? Or watched a satellite orbit the Earth? Or wondered why the Moon does not fly off into space? Or felt your car lean as you take a sharp corner? Every one of these is a story about circular motion.
Circular motion is everywhere — from the electrons orbiting atomic nuclei to planets orbiting stars, from a stone whirled on a string to a car navigating a roundabout. It is one of the most beautiful and counterintuitive topics in physics. Why does an object moving in a circle at constant speed still accelerate? Why does it feel like you are being pushed outward when the force actually points inward? In this chapter, we are going to explore these questions deeply. Let us begin.
I once asked a class, "If you are moving in a circle at constant speed, are you accelerating?" Most students said no. Then I asked, "Is your direction changing?" They said yes. "Then your velocity is changing, and if velocity changes, you are accelerating." That moment changed how they saw motion. Circular motion is where speed and velocity truly show their difference.By the end of this chapter, you should be able to:
Uniform circular motion is the motion of an object moving in a circle at a constant speed. The word "uniform" means the speed is constant. But even though the speed is constant, the velocity is not — because the direction is constantly changing.
Think of a car driving around a roundabout at a steady 30 km/h. Its speed is constant. But its velocity is changing every moment because it is constantly turning. This means it is accelerating.
Key examples of uniform circular motion:
| Quantity | Constant? | Why |
|---|---|---|
| Speed | Yes | Magnitude of velocity is constant |
| Velocity | No | Direction is constantly changing |
| Acceleration | Yes (in magnitude) | Directed toward the centre |
| Force | Yes (in magnitude) | Directed toward the centre |
When an object moves in a circle, we can describe its motion using angles instead of distances. This is often more convenient.
Angular displacement (θ): The angle through which an object has moved around a circle. It is measured in radians (rad).
Radian: The angle subtended at the centre of a circle by an arc equal in length to the radius. 1 radian = 57.3°.
Relationship between arc length and angle:
Arc length: s = r × θ
s = arc length (m), r = radius (m), θ = angular displacement (rad)
Conversion: 180° = π radians
Angular velocity (ω): The rate of change of angular displacement. It is measured in radians per second (rad/s).
Angular velocity: ω = θ / t = 2π / T = 2πf
ω = angular velocity (rad/s), θ = angular displacement (rad), t = time (s), T = period (s), f = frequency (Hz)
Period (T): The time taken for one complete revolution. Measured in seconds (s).
Frequency (f): The number of revolutions per second. Measured in hertz (Hz).
| Quantity | Symbol | SI Unit | Formula |
|---|---|---|---|
| Angular displacement | θ | radian (rad) | θ = s/r |
| Angular velocity | ω | rad/s | ω = 2π/T = 2πf |
| Period | T | second (s) | T = 1/f |
| Frequency | f | hertz (Hz) | f = 1/T |
Linear velocity (v) and angular velocity (ω) are related by the radius of the circle.
Linear velocity: v = r × ω
v = linear velocity (m/s), r = radius (m), ω = angular velocity (rad/s)
Linear acceleration (tangential): a_t = r × α
a_t = tangential acceleration (m/s²), α = angular acceleration (rad/s²)
This relationship tells us that for a given angular velocity, the linear speed is greater for points farther from the centre.
Think of a merry-go-round. The outer edge moves faster in a straight line than the inner part, even though both complete a revolution in the same time. The outer edge has a larger radius, so its linear speed is greater (v = rω).
| Quantity | Linear | Angular | Relationship |
|---|---|---|---|
| Displacement | s (m) | θ (rad) | s = rθ |
| Velocity | v (m/s) | ω (rad/s) | v = rω |
| Acceleration | a (m/s²) | α (rad/s²) | a = rα |
An object moving in a circle at constant speed is accelerating because its direction is changing. This acceleration is directed toward the centre of the circle and is called centripetal acceleration.
Centripetal acceleration: a = v² / r = ω² × r
a = centripetal acceleration (m/s²), v = linear velocity (m/s), r = radius (m), ω = angular velocity (rad/s)
Key points about centripetal acceleration:
Think of a car turning a corner. The tyres push the car toward the centre of the turn. This force causes the car to change direction. Without this force, the car would continue in a straight line.
| Change | Effect on Centripetal Acceleration |
|---|---|
| Double the speed | Quadruples acceleration (×4) |
| Double the radius | Halves acceleration (÷2) |
| Double the angular velocity | Quadruples acceleration (×4) |
According to Newton's Second Law (F = ma), if an object is accelerating, there must be a net force acting on it. In circular motion, this force is called centripetal force.
Centripetal force: F = m × v² / r = m × ω² × r
F = centripetal force (N), m = mass (kg), v = linear velocity (m/s), r = radius (m), ω = angular velocity (rad/s)
Key points about centripetal force:
| Situation | Source of Centripetal Force |
|---|---|
| Stone whirled on a string | Tension in the string |
| Car turning a corner | Friction between tyres and road |
| Satellite orbiting Earth | Gravitational force |
| Electron orbiting nucleus | Electrostatic force |
| Rider on a merry-go-round | Friction and normal force |
| Ball on a string | Tension |
Many people talk about "centrifugal force" — the apparent force that seems to push you outward when you are in circular motion. But in physics, centrifugal force is not a real force. It is a fictitious force that appears only in a rotating frame of reference.
| Feature | Centripetal Force | Centrifugal Force |
|---|---|---|
| Nature | Real force | Fictitious force |
| Direction | Toward the centre | Away from the centre |
| Frame of reference | Inertial (non-rotating) | Rotating |
| Example | Tension in a string | The feeling of being pushed outward |
Think of a car turning a corner. You feel like you are being pushed outward against the door. But what is really happening is that the car is turning inward, and your body is trying to continue in a straight line (Newton's First Law). The door pushes you inward, providing the centripetal force. The sensation of being pushed outward is your body's inertia resisting the change in direction.
The confusion between centripetal and centrifugal force is one of the most common misconceptions in physics. Centripetal force is real and points inward. Centrifugal force is a feeling, not a force — it is the sensation of your inertia when you are in a rotating frame. Once you understand this, you will never be confused again.When an object moves in a vertical circle (like a roller coaster loop or a bucket of water swung in a circle), the forces are different at different points.
At the top of the circle:
At the bottom of the circle:
Minimum speed at top of vertical circle: v_min = √(g × r)
v_min = minimum speed (m/s), g = gravitational field strength (N/kg), r = radius (m)
Think of a roller coaster loop. At the top, you feel lighter because gravity is helping to provide the centripetal force. At the bottom, you feel heavier because the track must push you up against gravity to keep you in the circle.
| Position | Forces | Centripetal Force | Sensation |
|---|---|---|---|
| Top | Gravity + tension/normal (both down) | Sum of both | Feel lighter |
| Bottom | Tension/normal (up) − gravity (down) | Difference | Feel heavier |
| Sides | Tension/normal (horizontal) | Tension/normal | Normal |
| Application | How It Works |
|---|---|
| Satellites in orbit | Gravity provides centripetal force |
| Car on a roundabout | Friction provides centripetal force |
| Washing machine spin cycle | Centripetal force separates water from clothes |
| Centrifuge | Centripetal force separates substances of different densities |
| Racing cars on banked tracks | Normal force and friction provide centripetal force |
| Hammer throw in athletics | Tension provides centripetal force before release |
| Orbiting space stations | Gravity provides centripetal force |
| Electrons in magnetic fields | Magnetic force provides centripetal force |
A satellite orbiting the Earth is in uniform circular motion. The centripetal force is provided by the gravitational force between the satellite and the Earth.
Orbital speed: v = √(G × M / r)
v = orbital speed (m/s), G = gravitational constant (6.67 × 10⁻¹¹ N m²/kg²), M = mass of Earth (5.97 × 10²⁴ kg), r = orbital radius (m)
Key points about satellite motion:
Think of a satellite as a ball thrown so fast that it never hits the ground — it falls around the Earth. The curved surface of the Earth falls away at the same rate as the satellite falls toward it.
| Orbit Type | Altitude | Period | Use |
|---|---|---|---|
| Low Earth orbit | 200–2 000 km | 90 min – 2 hours | ISS, imaging |
| Medium Earth orbit | 2 000–35 000 km | 2–24 hours | GPS, navigation |
| Geostationary orbit | 35 786 km | 24 hours | Communication, weather |
A centrifuge is a device that uses circular motion to separate substances of different densities. It spins a sample at high speed, and the denser components move outward while the less dense components stay near the centre.
How a centrifuge works:
Applications of centrifuges:
When a road or track is banked (tilted), the normal force from the surface can provide part or all of the centripetal force needed for a vehicle to turn. This reduces the reliance on friction and allows vehicles to turn at higher speeds safely.
How banking works:
Ideal speed on a banked track: v = √(r × g × tan θ)
v = ideal speed (m/s), r = radius of turn (m), g = gravitational field strength (N/kg), θ = banking angle
Think of a velodrome where cyclists race. The track is steeply banked so that cyclists can maintain high speeds around the curves without slipping. The banking provides the centripetal force.
| Banking Angle | Effect |
|---|---|
| Small angle | Low ideal speed |
| Large angle | High ideal speed |
| Zero angle (flat) | Friction must provide all centripetal force |
Angular displacement: θ = s / r
Angular velocity: ω = θ / t = 2π / T = 2πf
Linear velocity: v = r × ω
Centripetal acceleration: a = v² / r = ω² × r
Centripetal force: F = m × v² / r = m × ω² × r
Period and frequency: T = 1 / f
Minimum speed at top of vertical circle: v_min = √(g × r)
Orbital speed: v = √(G × M / r)
Ideal speed on banked track: v = √(r × g × tan θ)
Conversion: 180° = π radians
Question: A wheel completes 60 revolutions in 12 seconds. Calculate its angular velocity.
Answer:
Frequency: f = 60 / 12 = 5 Hz
Formula: ω = 2πf
Substitution: ω = 2π × 5 = 10π rad/s ≈ 31.4 rad/s
Units check: the answer is in rad/s. Notice that we first found the frequency in Hz, then used ω = 2πf. This is a common approach in circular motion problems.Question: A point on the edge of a wheel of radius 0.5 m has an angular velocity of 10 rad/s. Calculate its linear velocity.
Answer:
Formula: v = r × ω
Substitution: v = 0.5 m × 10 rad/s
Answer: v = 5 m/s
Units check: m × rad/s = m/s. The answer is in m/s, correct for linear velocity. Does 5 m/s make sense? Yes, it is a reasonable speed for a point on a spinning wheel.Question: A car of mass 1 200 kg moves around a circular track of radius 50 m at a speed of 20 m/s. Calculate the centripetal acceleration and the centripetal force.
Answer:
Centripetal acceleration: a = v² / r = (20)² / 50 = 400 / 50 = 8 m/s²
Centripetal force: F = m × a = 1 200 × 8 = 9 600 N
Answer: a = 8 m/s², F = 9 600 N
Units check: m²/s² / m = m/s² for acceleration; kg × m/s² = N for force. Notice that the force is quite large — this is why cars need good tyres and why high-speed turns are dangerous.Question: A stone of mass 0.5 kg is whirled in a horizontal circle of radius 1.5 m at a speed of 6 m/s. Calculate the tension in the string.
Answer:
Formula: F = m × v² / r
Substitution: F = 0.5 × (6)² / 1.5 = 0.5 × 36 / 1.5 = 18 / 1.5 = 12 N
Answer: The tension in the string is 12 N.
Units check: the answer is in newtons. Notice that the tension provides the centripetal force. If the string breaks, the stone will fly off in a straight line — this is Newton's First Law in action.Question: A satellite orbits the Earth at a radius of 7 × 10⁶ m. Calculate its orbital speed. Use G = 6.67 × 10⁻¹¹ N m²/kg² and M = 5.97 × 10²⁴ kg.
Answer:
Formula: v = √(G × M / r)
Substitution: v = √((6.67 × 10⁻¹¹) × (5.97 × 10²⁴) / (7 × 10⁶))
v = √(3.98 × 10¹⁴ / 7 × 10⁶) = √(5.69 × 10⁷)
Answer: v ≈ 7.54 × 10³ m/s = 7.54 km/s
Units check: the answer is in m/s. Does 7.54 km/s make sense? Yes, this is a typical orbital speed for a satellite in low Earth orbit. The International Space Station orbits at about 7.66 km/s.Graphs help us understand the relationships in circular motion.
Centripetal force vs speed: If you plot centripetal force on the vertical axis and speed on the horizontal axis, you get a curve. Force is proportional to the square of the speed, so the graph is a parabola.
Centripetal force vs radius: If you plot centripetal force on the vertical axis and radius on the horizontal axis, you get a curve. Force is inversely proportional to radius.
Centripetal force vs 1/r: If you plot force on the vertical axis and 1/r on the horizontal axis, you get a straight line through the origin. The gradient is m × v².
Angular velocity vs period: If you plot angular velocity on the vertical axis and 1/T on the horizontal axis, you get a straight line through the origin. The gradient is 2π.
These graphs help us visualise the relationships in circular motion. They turn equations into pictures.
Think of a graph as a map of a relationship. The shape tells you how one quantity changes with another.
Apparatus: A rubber bung, a string, a glass tube, a set of weights, a metre rule, and a stopwatch.
Method:
Calculation: If the bung has mass 0.05 kg, the radius is 0.5 m, and the period is 0.4 s:
v = 2π × 0.5 / 0.4 = 7.85 m/s
F = 0.05 × (7.85)² / 0.5 = 0.05 × 61.6 / 0.5 = 6.16 N
This should equal the weight of the hanging mass (if it is 0.63 kg, then weight = 6.3 N).
Precautions:
Likely source of error: Friction in the glass tube and air resistance can affect the results. Keep the tube smooth and the speed moderate.
This experiment is a classic way to verify the centripetal force formula. It connects the tension in the string to the weight of the hanging mass and the circular motion of the bung. Once you see the numbers match, the formula becomes real.Uniform circular motion is the motion of an object moving in a circle at a constant speed.
Because its direction is constantly changing, so its velocity (a vector) is changing. Acceleration is the rate of change of velocity.
Angular velocity is the rate of change of angular displacement. Its SI unit is rad/s.
a = v²/r = ω²r.
F = mv²/r = mω²r.
Friction between the tyres and the road.
Gravitational force between the satellite and the Earth.
F = mv²/r = 2 × (4)² / 2 = 2 × 16 / 2 = 16 N.
(a) Define uniform circular motion and explain why an object in circular motion is accelerating. [4]
(b) A car of mass 1 500 kg moves around a circular bend of radius 60 m at a speed of 15 m/s. Calculate the centripetal acceleration and the centripetal force. [5]
(c) State what provides the centripetal force for the car. [2]
(a) Uniform circular motion is motion in a circle at constant speed. It is accelerating because the direction of velocity is constantly changing. (4 marks.)
(b) a = v²/r = (15)² / 60 = 225 / 60 = 3.75 m/s². F = m × a = 1 500 × 3.75 = 5 625 N. (5 marks.)
(c) Friction between the tyres and the road. (2 marks.)
(a) Define angular velocity and state its relationship with linear velocity. [3]
(b) A wheel of radius 0.4 m rotates at 300 revolutions per minute. Calculate its angular velocity and the linear speed of a point on its rim. [5]
(c) Explain why the outer edge of a rotating wheel moves faster than a point closer to the centre. [3]
(a) Angular velocity is the rate of change of angular displacement (ω = θ/t). It is related to linear velocity by v = rω. (3 marks.)
(b) f = 300 / 60 = 5 Hz. ω = 2πf = 2π × 5 = 10π ≈ 31.4 rad/s. v = rω = 0.4 × 31.4 = 12.56 m/s. (5 marks.)
(c) All points on a rotating wheel have the same angular velocity, but linear speed depends on radius (v = rω). The outer edge has a larger radius, so it moves faster. (3 marks.)
(a) Explain the difference between centripetal and centrifugal force. [4]
(b) A stone of mass 0.2 kg is whirled in a horizontal circle of radius 0.8 m at a speed of 5 m/s. Calculate the tension in the string. [4]
(c) What happens to the stone if the string breaks? Explain using Newton's First Law. [3]
(a) Centripetal force is a real force directed toward the centre of the circle. Centrifugal force is a fictitious force that appears only in a rotating frame of reference and is directed away from the centre. (4 marks.)
(b) F = mv²/r = 0.2 × (5)² / 0.8 = 0.2 × 25 / 0.8 = 5 / 0.8 = 6.25 N. (4 marks.)
(c) If the string breaks, the stone moves in a straight line tangent to the circle. This is because there is no longer a centripetal force to change its direction, so it continues with the velocity it had at the moment the string broke (Newton's First Law). (3 marks.)
Explain why a satellite in orbit is said to be in free fall, and why astronauts inside it appear weightless.
A satellite is in free fall because the only force acting on it is gravity, which pulls it toward the Earth. However, it also has a horizontal velocity that keeps it moving around the Earth rather than falling straight down. Astronauts inside the satellite are also falling at the same rate, so they float relative to the spacecraft — they appear weightless because they are not being pushed against any surface.
Explain why a centrifuge is effective at separating substances of different densities.
In a centrifuge, particles are spun in a circle. The centripetal force required to keep each particle in the circle depends on its mass (F = mv²/r). Denser particles have more mass, so they require more force. If the force is not enough, they move outward. Less dense particles require less force, so they stay closer to the centre. This causes the components to separate into layers based on density.
A student says, "Centripetal force does no work on an object in circular motion." Is this correct? Explain.
Yes, this is correct. Work done is defined as force × distance moved in the direction of the force. In circular motion, the centripetal force is always perpendicular to the velocity (and therefore to the motion). Since the force has no component in the direction of motion, no work is done. This is why the speed of the object remains constant — the centripetal force only changes the direction of motion, not its magnitude.
Circular motion is the dance of the universe — from electrons to galaxies, everything moves in circles, held by invisible forces toward the centre.