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Physics (IAL)

Circular Motion

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Edexcel IAL Physics — Unit 4

Circular Motion

The Big Idea Anything moving in a circle at constant speed is still accelerating — because its keeps changing — and something must always be pulling it toward the centre to keep it there.
Quick Summary
  • Radians measure angles using arc length ÷ radius — they're the "natural" unit for anything going in circles.
  • Angular displacement (Δθ) is how far round the circle something has swept, in radians.
  • Angular velocity (ω) is how fast that angle changes — the rotational equivalent of linear velocity.
  • Even at constant , an object going in a circle is always accelerating, because velocity is a vector and direction is constantly changing.
  • This acceleration is called centripetal acceleration — always pointing toward the centre of the circle.
  • Newton's Second Law says acceleration needs a force — that resultant force is the centripetal force, also always pointing toward the centre.
  • Centripetal force isn't a new type of force — it's whatever existing force (tension, friction, gravity...) happens to be doing the job of keeping the object curving.
  • In vertical circular motion, tension/normal force varies around the loop because gravity adds to it at the top and subtracts from it at the bottom (or vice versa depending on setup).
Topic 1 — Radians & Angular Displacement

What actually is a radian?

You're used to measuring angles in degrees, where a full circle is 360°. That number is completely arbitrary — it comes from ancient Babylonian counting systems, not from anything physical about circles. Radians, on the other hand, are defined , which is exactly why physicists love them for circular motion.

Here's the definition: one radian is the angle you get at the centre of a circle when the arc (the curved distance around the edge) is exactly as long as the radius. Imagine taking a piece of string the same length as the radius, laying it along the curved edge of the circle, and then drawing two lines from its ends back to the centre. The angle between those two lines is 1 radian — roughly 57.3°.

Because the full circumference of a circle is 2πr, and each "radius-length" of arc corresponds to 1 radian, a complete circle (360°) is exactly 2π radians (≈ 6.28 rad). That's not a coincidence you need to memorise separately — it falls straight out of the definition.

Angular Displacement
Δθ = Δs / r
  • Δθ = angular displacement (radians)
  • Δs = arc length / distance travelled around the circle (m)
  • r = radius of the circle (m)
In plain words: angular displacement is just "how much arc did you cover, relative to the size of the circle." A bigger circle needs a longer arc to sweep out the same angle.
Converting Between Degrees and Radians
Since 360° = 2π rad, to go from degrees to radians: θ° × (π/180) = θ rad. To go the other way: θ rad × (180/π) = θ°. Common ones worth knowing cold: 90° = π/2, 180° = π, 270° = 3π/2, 360° = 2π.
Calculator Trap
Your calculator has a Degree mode and a Radian mode (shown as "D" or "R" at the top of the screen). If you're finding sin, cos, or tan of an angle and you're in the wrong mode, you'll get a completely wrong — but plausible-looking — answer. Always check the mode before you touch trig functions in a circular motion question.
Practice Question

An angle is given as θ = π/3 radians. What is this in degrees?

Practice Question

A toy car drives around a circular track of radius 2.5 m. It travels an arc length of 4.0 m. What angle (in radians) has it swept through?

Topic 2 — Angular Velocity

How fast is it spinning?

Angular velocity, ω (the Greek letter omega), is the rotational cousin of linear velocity. Where linear velocity tells you metres covered per second, angular velocity tells you . It's defined as the rate of change of angular displacement.

Angular Velocity
ω = Δθ / Δt
  • ω = angular velocity (rad s⁻¹)
  • Δθ = change in angular displacement (rad)
  • Δt = time interval (s)

Angular velocity connects to the linear speed you're already comfortable with through a really useful equation. Think about it intuitively: if two people are on a spinning merry-go-round, one near the centre and one on the outer edge, they both sweep the in the same time (same ω) — but the person on the edge is physically moving much faster because they're covering more distance. That's why linear speed depends on how far out you are (the radius).

Linear Speed from Angular Speed
v = ωr
  • v = linear speed (m s⁻¹)
  • ω = angular speed (rad s⁻¹)
  • r = radius of the circular path (m)

There's also a neat relationship between ω and how long a full revolution takes. If one entire lap is 2π radians, and it takes time T (the period), then:

Angular Velocity from Period / Frequency
ω = 2π / T = 2πf
  • T = time period — time for one full revolution (s)
  • f = frequency — revolutions per second (Hz)
This is why ω is sometimes called "angular frequency" — it's directly built from f, just scaled by 2π to convert revolutions into radians.
Velocity vs Speed — Don't Mix Them Up
Just like in normal kinematics, angular velocity is a vector (magnitude + direction of rotation), while angular speed is just the magnitude. In practice, exam questions use v = ωr and a = rω² treating v, ω as their magnitudes — but always remember direction is what's changing that causes the acceleration in the first place.
Worked-Style Practice

A bird flies in a horizontal circle of radius 650 m with angular speed 5.25 rad s⁻¹. (a) Find its linear speed. (b) Find its frequency of rotation.

Topic 3 — Centripetal Acceleration

Why "constant speed" doesn't mean "no acceleration"

This is the idea that trips almost everyone up the first time they meet it. In everyday language, "accelerating" means "speeding up." But in physics, acceleration is the rate of change of velocity — and velocity is a vector. A vector has both size (speed) and direction. If the direction changes, even while the speed stays exactly the same, the velocity has still changed — which means there's still an acceleration.

Picture a ball on a string being swung in a horizontal circle at a steady speed. At every instant, its velocity arrow points in a new direction — tangent to the circle. Since that arrow is constantly rotating, there's a constant acceleration happening, even though the length of the arrow (the speed) never changes. This acceleration is called centripetal acceleration, and — as the derivation in your notes shows using vector diagrams and the small-angle approximation (sin θ ≈ θ for tiny θ) — it always points of the circle.

"Centripetal" literally means "centre-seeking." It's a slightly misleading name in one sense: it doesn't make the object move toward the centre and crash into it — it just continuously bends the object's path so that it curves around the centre rather than flying off in a straight line.

Centripetal Acceleration — Two Equivalent Forms
a = v² / r  =  rω²
  • a = centripetal acceleration (m s⁻²)
  • v = linear speed (m s⁻¹)
  • r = radius of the circular path (m)
  • ω = angular speed (rad s⁻¹)
Use v²/r when you know the linear speed; use rω² when you know the angular speed. They give identical answers because v = ωr — pick whichever the question gives you data for.
Common Mistake
Students often think a bigger radius means bigger acceleration because "it's a bigger circle." But look at a = v²/r carefully — r is on the bottom. For a , a larger radius actually gives centripetal acceleration (a gentler curve). It's only when angular speed ω is held fixed that a bigger radius gives more acceleration (a = rω² has r on top).
Practice Question

A ball on a string moves in a horizontal circle, radius 1.5 m, angular speed 3.5 rad s⁻¹. If both the radius and the angular speed are doubled, by what factor does the centripetal acceleration change? Then calculate the new value.

Topic 4 — Maintaining Circular Motion (Centripetal Force)

What actually keeps something moving in a circle?

Newton's First Law says an object keeps moving in a straight line unless a resultant force acts on it. An object going around in a circle is clearly moving in a straight line — its direction is constantly being bent inward. That bending requires a resultant force, and by Newton's Second Law (F = ma), that force must point in the exact same direction as the acceleration it causes — toward the centre. This resultant force is the centripetal force.

The Single Most Important Idea in This Chapter
Centripetal force is NOT a new, separate kind of force. It is simply the name we give to whichever resultant force happens to be pointing toward the centre and doing the job of curving the object's path. It's a role, not a force type — like how "the driver" could be different people depending on the situation, but it's always someone doing the driving.
SituationWhat provides the centripetal force?
Car going round a roundaboutFriction between tyres and road
Ball on a string swung in a circleTension in the string
Earth orbiting the SunGravitational force
Charged particle in a magnetic fieldMagnetic force (always perpendicular to velocity)
Don't Draw "Centrifugal Force"
You'll often hear people talk about a "centrifugal force" flinging you outward in a car going round a bend. That's not a real force acting on you — it's your body's inertia (Newton's First Law) trying to keep going in a straight line while the car curves underneath/around you. On a free-body diagram in an exam, never draw an outward "centrifugal" arrow — only draw the real, physical forces present (tension, friction, gravity, normal force, etc.), and their resultant is what points inward.
Topic 5 — Centripetal Force & Vertical Circular Motion

Calculating the force, and what happens when gravity gets involved

Since F = ma, and we already have three equivalent expressions for centripetal acceleration, we get three equivalent expressions for centripetal force just by multiplying by mass:

Centripetal Force
F = mv² / r  =  mrω²  =  mvω
  • F = centripetal force (N)
  • m = mass of the object (kg)
  • v = linear speed (m s⁻¹)
  • ω = angular speed (rad s⁻¹)
  • r = radius of the circular path (m)
All three forms are the same equation rearranged using v = ωr — use whichever matches the variables the question gives you.

Things get more interesting when the circle is vertical instead of horizontal — think of a ball on a string being swung in a vertical loop, or a rollercoaster doing a loop-the-loop. Now there are two forces acting on the object at all times: the tension (or normal force) pointing toward the centre, and the object's own weight (mg), which always points straight down no matter where the object is in the loop.

Because weight never changes direction but the required centripetal force always points toward the centre, the relationship between tension and weight — and therefore the value of the tension itself — changes continuously as the object goes around.

At the Bottom of the Loop
Tmax = mv² / r + mg
Here, tension has to supply the centripetal force support the weight (which is pulling the opposite way, downward/outward from the loop's centre at this point) — so tension is at its maximum.
At the Top of the Loop
Tmin = mv² / r − mg
Here, gravity is already pointing toward the centre, doing some of the work for free — so tension only has to make up the rest, meaning tension is at its minimum.
The "Just Barely Making It" Condition
A classic exam question type: "what's the minimum speed at the top of the loop so the string doesn't go slack / the water doesn't spill?" The trick is realising that at the critical minimum speed, the string or surface is providing zero force (T = 0), and gravity alone is providing 100% of the required centripetal force. Set T = 0 in the top-of-loop equation and solve for v.
Worked-Style Practice

A bucket of mass 8.0 kg filled with water is attached to a string of length 0.5 m and swung in a vertical circle. What is the minimum speed the bucket must have at the top of the circle so that no water spills out?

Practice Question

A 0.20 kg ball on a 0.80 m string is swung in a vertical circle, passing through the bottom point at 4.0 m s⁻¹. Calculate the tension in the string at that instant. (g = 9.81 m s⁻²)

What to Memorise
Radian (definition)
The angle at the centre of a circle subtended by an arc equal in length to the radius. Full circle = 2π rad = 360°.
Δθ = Δs / r
Angular displacement = arc length ÷ radius.
ω = Δθ / Δt
Angular velocity = rate of change of angular displacement (rad s⁻¹).
v = ωr
Linear speed = angular speed × radius.
ω = 2π/T = 2πf
Links angular speed to the period and frequency of rotation.
a = v²/r = rω²
Centripetal acceleration, directed toward the centre of the circle.
F = mv²/r = mrω² = mvω
Centripetal force — the resultant force toward the centre, not a new force type.
Vertical loop: top & bottom
Tmin = mv²/r − mg (top); Tmax = mv²/r + mg (bottom).
Small angle approximation
For very small θ (in radians), sin θ ≈ θ — key step in deriving a = v²/r.
Key fact
Centripetal force and centripetal acceleration always point in the same direction — toward the centre — by Newton's Second Law.
Concepts Checklist
Exam Tips & Common Mistakes
Calculator mode
Always check Deg/Rad mode before using trig functions — this single slip loses more marks on circular motion papers than almost anything else.
Don't draw a "centrifugal force"
Examiners specifically look for correct free-body diagrams. Only real forces belong on the diagram (tension, weight, friction, normal force, etc.) — never an outward "centrifugal" arrow.
"Resultant force" wording
When asked to state what provides the centripetal force, always name the actual physical force (e.g. "tension in the string," "friction between tyres and road") — not the word "centripetal force" itself, since that's just the name for the resultant, not a source.
Vertical circle questions
Always draw the forces at the specific point being asked about (top, bottom, or side) — the direction of weight relative to tension changes around the loop, so a diagram at the correct point avoids sign errors when combining mv²/r with mg.
Choosing the right formula
If a question gives you angular speed (ω) directly, use the rω² or mrω² forms — don't waste time converting to linear speed first unless the question specifically asks for v.
Radius errors
Watch for questions that give you a diameter instead of a radius, or a "distance from the axis" that isn't simply the length of a string (e.g. conical pendulum problems) — always double check what r actually refers to in the geometry of the problem.
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