Module 3: Kinematics · Theory · Prerequisites: K14, Foundations 6 and 8
A new set of coordinates for motion that turns — and one result that applies to every curved path, not just circles.
Mahavakya
Acceleration has two jobs: changing how fast you are going, and changing where you are going. One dimension only ever showed you the first.
On this page
1. Angular coordinates
A particle moving on a circle of fixed radius has only one degree of freedom — K14’s language for it. Its position is fully specified by one number: the angle.
So describe the motion in that number instead of in x and y:
Angular position θ — where on the circle, measured from a chosen zero.
Angular velocity ω = dθ/dt — how fast the angle changes.
Angular acceleration α = dω/dt — how fast that changes.
The same chain as K4, applied to a different coordinate. Nothing new is being claimed about motion; a different variable is being differentiated.
The sign convention is a choice, exactly as in K1. Anticlockwise positive is conventional and not obligatory — but having chosen, hold it.
ω is properly a vector, directed along the axis of rotation and fixed by the right-hand rule. Foundation 6 argued that the right-hand rule is an agreement rather than a discovery, and introduced axial vectors — quantities that behave differently under reflection. Angular velocity is exactly such a quantity. For motion in a plane the axis never moves, so a signed scalar carries all the information and that is what we use here.
2. Why radians
Measure the angle in radians and the arc length is simply
s = Rθ
That is the definition of the radian, not a result — the angle in radians is the arc divided by the radius. In degrees the same relation carries a factor of π/180, which is why every formula in this part would acquire one.
A useful consequence: since a radian is a length divided by a length, it is dimensionless. So:
[θ] = 1 [ω] = T−1 [α] = T−2
check: v = Rω gives L T−1 = L × T−1 ✓
Every angular relation in this part can be checked in a few seconds this way. Foundation 1 installed the habit; here is a topic where it pays repeatedly, because the formulas are similar enough to confuse.
Period and frequency follow directly. One revolution is 2π radians, so
ω = 2π/T = 2πf and v = 2πR/T
3. v = Rω, and what it means
Differentiate s = Rθ with respect to time, holding R fixed:
v = Rω
and again: at = Rα
where at is the tangential acceleration — the part that changes the speed. Section 6 will name the other part.
The relation looks like a conversion factor and is worth more than that.
One angular velocity, two different speeds.
Every point on a rigid rotating body shares one ω — they all turn through the same angle in the same time. But their speeds differ, in proportion to how far out they sit.
ω is a property of the body; v is a property of the point. That single distinction is the basis of every gear, pulley, belt-drive and rotating-disc problem. Two gears meshing share a rim speed; two points on one gear share an angular velocity. Getting those the right way round is most of the work.
4. The dictionary, and its limits
The angular quantities map onto the linear ones so neatly that it is worth tabulating — and worth saying what the table does not mean.
| Linear | Angular | Link |
|---|---|---|
| s | θ | s = Rθ |
| v | ω | v = Rω |
| a | α | at = Rα |
| v = u + at | ω = ω0 + αt | constant α only |
The last row deserves the caution it carries. Those equations are K8’s, and they hold under exactly the same condition — constant α and nothing else. The analogy is a convenience of notation, not a new physical law, and it inherits every restriction the originals had.
5. Centripetal acceleration, geometrically
K11 obtained this by parametrising the position and differentiating twice. Here is the geometric route, which is shorter and shows where the v² comes from.
Take two nearby instants. The speed is unchanged, so the two velocity vectors have the same length and differ only in direction — by a small angle Δθ. Draw them with their tails together and they form a thin isosceles triangle whose third side is Δv. For a small angle:
|Δv| ≈ v Δθ
Meanwhile the particle has travelled an arc Δs = RΔθ in time Δt, and Δs = vΔt. So Δθ = vΔt/R, giving
|Δv|/Δt ≈ v²/R
ac = v²/R = Rω²
The v appears twice for two different reasons — once because a faster particle turns its velocity vector through the same angle in less time, and once because the vector being turned is itself longer. That is why the dependence is quadratic, and the geometry shows it in a way the algebra does not.
Two routes, one result. This geometric argument and K11’s parametric differentiation give the same answer by entirely different means — and Foundation 8 obtained it a third way, using rotating basis vectors. Three independent derivations is strong evidence that ac = v²/R is a consequence of circular motion rather than a fact about it.
Which to reach for depends on the problem: the geometry when you want to see why, the parametrisation when you need components.
Centripetal acceleration is not a force. Circular motion requires an inward acceleration, and something physical must supply the corresponding force — tension for a stone on a string, friction for a car turning, gravity for a satellite. “Centripetal” names the direction and the job, not the cause. There is no such thing as a centripetal force in the sense of a new kind of interaction.
6. Any curved path at all
Now the result that makes this part more than a special case.
For a particle on any curved path, the acceleration splits into two perpendicular components:
a = att̂ + ann̂
at = dv/dt an = v²/R
with R the instantaneous radius of curvature — the radius of the circle that best fits the path at that point.
Any curved path: acceleration splits into two jobs.
The tangential component changes how fast the particle moves.
The normal component changes where it is going.
A circle is now visible as the special case where R never changes. Uniform circular motion is the further case where at = 0 as well, leaving only the inward component.
And straight-line motion — everything from K1 to K10 — is the case where the direction never changes, so an = 0 and a = dv/dt accounts for the whole acceleration.
That is why one-dimensional intuition fails here. It was never wrong; it was restricted, and the restriction was invisible because nothing in one dimension can turn. Every rule you learned about signs and speeding up was the tangential component in disguise.
7. What exactly is changing?
One question handles this entire topic.
What exactly is changing?
- If the magnitude of the velocity changes — there is tangential acceleration.
- If the direction of the velocity changes — there is normal acceleration.
- If both change — there are both components.
- If neither changes — the acceleration is zero, and only then.
The four cases give a table worth carrying:
| Path | Speed | Acceleration |
|---|---|---|
| straight | constant | zero |
| straight | changing | tangential only |
| curved | constant | normal only |
| curved | changing | both |
Only one row gives zero. A car rounding a bend at a steady 40 km/h is accelerating, and the speedometer will not tell you so.
Worked example
A particle moves on a circle of radius 2 m. Its speed increases uniformly from rest at 3 m/s². Find the magnitude of its total acceleration when it has completed a quarter revolution.
Arc travelled: s = Rθ = 2 × (π/2) = π m.
Speed there, from v² = u² + 2ats: v² = 0 + 2(3)(π) = 6π, so v = 4.34 m/s.
at = 3 m/s² an = v²/R = 6π/2 = 9.42 m/s²
|a| = √(3² + 9.42²) = 9.89 m/s²
The components are perpendicular, so they combine by Pythagoras — never by addition. Note also that an grows as the particle speeds up, while at stays fixed: the turning becomes the dominant part of the acceleration.
Contemplation
A scalar tells you how much. A vector tells you how much and where. For ten parts that difference has looked like a technicality — a matter of remembering to write arrows and take components.
Circular motion is where it stops being one. An object can travel at exactly the same speed, second after second, while its state of motion changes continuously. Ordinary language has no way to say that; “it is going at a steady forty” sounds like a complete description and is not.
So the vector definition of velocity is not a more complicated version of speed. It carries information that speed cannot carry, and this is the topic where you can see what that information was for.
Common misconceptions
1. Constant speed means zero acceleration
Only if the direction is also constant. Uniform circular motion has constant speed and non-zero acceleration throughout.
2. Centripetal acceleration is a force
It is an acceleration that circular motion requires. Tension, friction or gravity supplies the force; “centripetal” describes the direction, not the cause.
3. All points on a rotating body move at the same speed
They share ω, not v. Speed is Rω and grows with distance from the axis.
4. The angular equations always apply
ω = ω0 + αt requires constant α, exactly as its linear counterpart requires constant a. The analogy inherits the conditions.
5. Tangential and normal accelerations add arithmetically
They are perpendicular, so |a| = √(at² + an²). Adding them gives a number that is always too large.
Reflection
- A projectile’s path is curved. What is its radius of curvature at the apex, in terms of the launch speed and angle?
- Section 5 said the v appears twice for two different reasons. State each one in a sentence without using algebra.
- Can a particle have an = 0 at one instant on a curved path? What would that mean about the path there?
Key takeaways
- θ, ω = dθ/dt, α = dω/dt — the same chain as K4 in a different coordinate.
- Radians make s = Rθ exact and are dimensionless, so [ω] = T−1.
- v = Rω: ω belongs to the body, v belongs to the point.
- ac = v²/R = Rω², derived here geometrically and in K11 by parametrisation.
- Centripetal acceleration is not a force. Something physical must supply one.
- For any curved path: at = dv/dt changes speed; an = v²/R changes direction, with R the instantaneous radius of curvature.
- Perpendicular components combine by Pythagoras.
What comes next
That is the physics of Kinematics complete. K16 gathers it — every equation with the condition it depends on, the traps collected in one place, and a checklist for deciding which tool a problem needs before reaching for any of them.
Prerequisites: K14, Foundations 6 and 8 · Reading: 18 min · Practice: 40 min · Difficulty: Intermediate to Advanced