Circular Motion — Angular Quantities

Spec mapping: OCR H556 Module 5.3 — Circular motion (radian measure; angular displacement, angular velocity and angular frequency; period and frequency; the linear-angular bridge $v = r\omega$ ; sub-strand sits inside Module 5 alongside Oscillations and Gravitational fields). Refer to the official OCR H556 specification document for exact wording.

Circular motion is one of the most elegant topics in A-Level Physics. A body moving in a circle at constant speed is a perfect example of accelerated motion in which the speed is constant but the velocity is changing all the time. Before we can talk about the forces involved, we need a new language — the language of angular quantities: angle measured in radians, angular displacement, angular velocity, frequency and period.

This lesson is the foundation of OCR A-Level Physics A Module 5.3 (Circular motion). Everything else in the topic — centripetal force (lesson 2), simple harmonic motion (lessons 3-5), damping and resonance (lesson 6), and the orbital mechanics of Module 5.4 — depends on being able to move fluently between the linear world (metres and metres per second) and the angular world (radians and radians per second). Many students arrive at A-Level with a hazy notion of "the angle is just a number" — by the end of this lesson we want a sharper picture: an angle is a ratio of two lengths (arc to radius), and that ratio is what makes the calculus of trigonometric functions clean.

Why Not Just Use Degrees?

You have been using degrees since primary school. Why, at A-Level, do we suddenly switch to radians?

The honest answer is calculus. Differentiating the trigonometric functions gives the neatest possible formulae only when the angle is measured in radians. Specifically, $\frac{d}{d\theta}\sin\theta = \cos\theta$ holds only with $\theta$ in radians; in degrees the derivative picks up an awkward $\pi/180$ factor. Engineers and physicists need those neat formulae, so they use radians by default and leave degrees to navigators, surveyors and carpenters.

The practical answer is arc length. With radians, the relationship between the arc length $s$ , the radius $r$ and the angle $\theta$ is simply

$s = r\theta$

No awkward factor of $\pi/180$ . The radian was invented specifically so this equation would have no fudge factor in it. Compare with the degree formulation $s = r\theta \times \pi/180$ , which is correct but cluttered. Once you start chaining equations together (as we shortly will, to derive centripetal acceleration), the cleanness of the radian form pays off enormously.

Exam Tip: Whenever an OCR Physics question gives you an angle without explicitly saying "degrees", assume it is in radians. Make sure your calculator is in radian mode for any circular motion or SHM question. A single forgotten mode-switch destroys whole exam answers and is one of the most expensive silent errors at A-Level.

Defining the Radian

Imagine a circle of radius $r$ . Mark two radii so that the arc between them has length $s$ . The angle $\theta$ between the radii, measured in radians, is defined as

$\theta = \frac{s}{r}$

In words: one radian is the angle subtended at the centre of a circle by an arc equal in length to the radius. Because both $s$ and $r$ are lengths, the radian is dimensionless — it is really just a ratio. This is why you can drop the "rad" symbol when the radian is multiplied by a length to produce another length (as in $s = r\theta$ ): a metre times a (dimensionless) radian is still a metre. Strictly the radian is given its own SI symbol for clarity, but it has no SI base-unit dimensions.

If the arc length equals the radius ( $s = r$ ), the angle is exactly 1 rad. If the arc length is the full circumference ( $s = 2\pi r$ ), the angle is $2\pi$ rad, which corresponds to $360°$ .

Key Conversions

Degrees	Radians
$360°$	$2\pi$
$180°$	$\pi$
$90°$	$\pi/2$
$60°$	$\pi/3$
$45°$	$\pi/4$
$30°$	$\pi/6$
$1°$	$\pi/180 \approx 0.01745$ rad
$1$ rad	$180/\pi \approx 57.30°$

Conversion rules:

Degrees → radians: multiply by $\pi/180$

Radians → degrees: multiply by $180/\pi$

Memorise the "boundary" entries ( $30°, 45°, 60°, 90°, 180°, 360°$ and their radian images). The rest you can derive on the fly. Many OCR questions silently expect you to know that $\pi/3 = 60°$ when reading an angle off a diagram — getting the conversion wrong then propagates into every later calculation.

Angular Displacement

Angular displacement $\theta$ is the angle, measured in radians, through which a rotating body has turned from some reference direction. It is the angular equivalent of linear displacement $s$ .

Just as linear displacement is a vector, angular displacement is also a vector — by convention, its direction lies along the axis of rotation and its sense is given by the right-hand rule (curl the fingers in the direction of rotation; the thumb gives the angular vector direction). For OCR A-Level purposes you only need to treat angular displacement as a signed scalar: positive for anticlockwise, negative for clockwise (or vice versa, depending on the choice of axis). Vector treatments are reserved for undergraduate mechanics and the "Going further" section below.

Worked Example 1 — Arc Length

A car tyre of radius $0.33$ m rolls through $12$ complete revolutions without slipping. Calculate the distance travelled.

One revolution $= 2\pi$ rad, so $12$ revolutions $= 24\pi$ rad.

$s = r\theta = 0.33 \times 24\pi = 24.88 \text{ m}$

Notice how painless the calculation is in radians. In degrees you would have needed $12 \times 360 = 4320°$ , then converted to radians anyway. The no-slip rolling condition $s = r\theta$ is the same relationship that unwinds string from a pulley, or threads paper through a printer roller: every turn through angle $\theta$ delivers an arc length $r\theta$ of perimeter. This is the same equation as for a fixed-axis rotation, but applied along the tyre's own rim rather than at a fixed point in space.

Worked Example 2 — Pendulum Arc

A simple pendulum of length $L = 1.20$ m is pulled aside through an angle of $4.5°$ from the vertical. How far has the bob moved along its arc?

Convert: $\theta = 4.5° \times \pi/180 = 0.0785$ rad.

$s = L\theta = 1.20 \times 0.0785 = 0.0942 \text{ m} = 9.4 \text{ cm}$

For small angles, the bob's arc length and its straight-line displacement from equilibrium differ by less than $0.1\%$ . This is exactly the "small-angle approximation" that turns a pendulum into an SHM oscillator — a result we will exploit heavily in lesson 3.

Angular Velocity

Just as linear velocity is the rate of change of linear displacement, angular velocity $\omega$ (the Greek letter omega) is the rate of change of angular displacement:

$\omega = \frac{\theta}{t} \qquad \text{(uniform circular motion)} \qquad \omega = \frac{d\theta}{dt} \qquad \text{(instantaneous)}$

The SI unit is radians per second (rad s $^{-1}$ ). Numerically $\omega$ is a scalar for our purposes, but strictly it is also a vector along the axis of rotation (again, right-hand rule). When OCR writes "the wheel has angular velocity $20$ rad s $^{-1}$ " without specifying axis, the magnitude is what they want.

Angular Velocity and Period

A body that travels once around a circle ( $\theta = 2\pi$ ) in a time $T$ — the period — has angular velocity

$\omega = \frac{2\pi}{T}$

Rearranging,

$T = \frac{2\pi}{\omega}$

The period tells you how many seconds each revolution takes. It is by far the most physically intuitive of the three angular measures: ask anyone how fast a flywheel spins, and they will instinctively reach for "revolutions per minute" — a period-style quantity — rather than $\omega$ in rad s $^{-1}$ .

Angular Velocity and Frequency

Frequency $f$ is the number of complete revolutions per second, measured in hertz (Hz). Since one revolution takes $T$ seconds, there are $1/T$ revolutions per second, so

$f = \frac{1}{T} \qquad \omega = 2\pi f$

You will use $\omega = 2\pi f$ constantly, both in circular motion and later in simple harmonic motion. The factor $2\pi$ converts "revolutions per second" into "radians per second" — exactly $2\pi$ radians per revolution.

Linear Speed Around a Circle

If a body moves in a circle of radius $r$ with angular velocity $\omega$ , in time $t$ it sweeps through an angle $\omega t$ and covers an arc length

$s = r\theta = r(\omega t)$

Its linear (tangential) speed is therefore

$v = \frac{s}{t} = r\omega$

This is arguably the single most important equation in circular motion. It is the bridge between the angular world and the linear world. Memorise it.

Angular quantity	Linear equivalent	Relationship
$\theta$ (rad)	$s$ (m)	$s = r\theta$
$\omega$ (rad s $^{-1}$ )	$v$ (m s $^{-1}$ )	$v = r\omega$
$\alpha$ (rad s $^{-2}$ )	$a_{\text{tangential}}$ (m s $^{-2}$ )	$a_t = r\alpha$

The last row — angular acceleration $\alpha$ — only matters if the rotation is speeding up or slowing down; for uniform circular motion $\alpha = 0$ and the entire row can be ignored. For OCR you will almost always work in uniform circular motion (constant $\omega$ ), so the first two rows are the workhorses.

Worked Example 3 — Spinning Vinyl Disc

A vinyl record spins at $33\frac{1}{3}$ rpm (revolutions per minute). A speck of dust sits $0.12$ m from the centre.

(a) Angular velocity:

$33\frac{1}{3}$ rpm $= 33.33 / 60 = 0.556$ rev s $^{-1}$

$\omega = 2\pi f = 2\pi \times 0.556 = 3.49 \text{ rad s}^{-1}$

(b) Period:

$T = \frac{2\pi}{\omega} = \frac{2\pi}{3.49} = 1.80 \text{ s}$

Sanity check: $60$ s $/ 33.33$ rev $\approx 1.80$ s per revolution. ✓

(c) Linear speed of the dust:

$v = r\omega = 0.12 \times 3.49 = 0.419 \text{ m s}^{-1}$

Notice the dust is moving at about $0.42$ m s $^{-1}$ , but a speck at $0.06$ m from the centre would move at only $0.21$ m s $^{-1}$ . Different radii, same angular velocity: $v \propto r$ at fixed $\omega$ . This is why a car's tyre rim moves faster than its hub even though both share the same wheel.

Worked Example 4 — The Earth's Rotation

Estimate the linear speed of a point on the Earth's equator. Take the Earth's radius as $6.37 \times 10^{6}$ m and its period as $24$ hours.

$\begin{aligned} T &= 24 \times 3600 = 86\,400 \text{ s} \\ \omega &= \frac{2\pi}{T} = \frac{2\pi}{86\,400} = 7.27 \times 10^{-5} \text{ rad s}^{-1} \\ v &= r\omega = 6.37 \times 10^{6} \times 7.27 \times 10^{-5} = 463 \text{ m s}^{-1} \end{aligned}$

So you and your chair are moving eastward at roughly $463$ m s $^{-1}$ (about $1670$ km/h) without feeling a thing, because the ground beneath you is moving at exactly the same speed (Galilean relativity, see Newton 1 in the Newton-momentum lesson). At higher latitudes the radius of the circle of latitude is smaller — at $50°$ N, the effective radius is $R\cos 50° \approx 4.10 \times 10^{6}$ m, and the surface speed drops to $\approx 298$ m s $^{-1}$ . This latitude-dependence has practical engineering importance: rockets are launched eastward and from low latitudes wherever possible, to harvest as much of Earth's rotation as a free initial speed.

Exam Tip: OCR frequently asks candidates to compute the linear speed of a satellite, spinning wheel or ride at a funfair using $v = r\omega$ . Make sure your angular velocity is in rad s $^{-1}$ , not revolutions per minute. Marks are routinely lost when candidates substitute rpm directly.

Worked Example 5 — Bicycle Wheel

A bicycle wheel has diameter $0.66$ m. The cyclist pedals at a steady $25$ km h $^{-1}$ on level ground. Find the angular velocity of the wheel in rad s $^{-1}$ and the period of one rotation.

Convert speed: $v = 25 \times 1000/3600 = 6.94$ m s $^{-1}$ . Radius $r = 0.33$ m.

$\omega = \frac{v}{r} = \frac{6.94}{0.33} = 21.0 \text{ rad s}^{-1}$

$T = \frac{2\pi}{\omega} = \frac{2\pi}{21.0} = 0.299 \text{ s}$

So the wheel completes a full rotation in just under one third of a second, about $3.3$ revolutions per second. This is the kind of feel-it estimate that good A-Level physicists develop — when a question gives a speed and a wheel, the period is "fraction of a second", not "many seconds". Anything else suggests an arithmetic slip.

Uniform Circular Motion — Constant Speed, Changing Velocity

A body moving at constant speed in a circle has a constant $|v|$ but a continuously changing direction of $v$ . Because velocity is a vector, any change in direction is a change in velocity, and any change in velocity is — by Newton's first law — an acceleration. This is one of the most important conceptual insights in A-Level mechanics, and one that GCSE rarely emphasises.

The velocity vector is always tangential to the circle. At the top of the circle it points one way; a quarter-turn later it points perpendicular to the first. The speed is unchanged, but the velocity is not.

This observation is the gateway to the next lesson, where we will quantify this "changing velocity" as the centripetal acceleration and use Newton's second law to deduce the force required to keep the object in its circular path.

Synoptic Links

Newton 1 and 2 (Module 3.2): Uniform circular motion is not a counterexample to Newton 1. Constant speed in a circle is not constant velocity; the velocity vector rotates, so a non-zero resultant force is required by Newton 2 — supplied by gravity, tension, normal contact or friction depending on context. This is treated quantitatively in lesson 2.
Simple harmonic motion (Module 5.3, lessons 3-5): The constant $\omega$ of uniform circular motion is the same $\omega$ that appears in SHM as angular frequency. The projection of uniform circular motion onto a diameter is exactly an SHM oscillator — a deep mathematical identity that gives SHM its sinusoidal solutions.
Gravitational fields and orbits (Module 5.4): Planetary and satellite motion is uniform circular motion (to a good approximation) with the centripetal force supplied by gravity, giving the relationship $v = \sqrt{GM/r}$ and Kepler's third law $T^2 \propto r^3$ . The angular framework set up here is the launching pad for all orbital calculations.
Magnetic forces on charged particles (Module 6.4): A charged particle in a uniform magnetic field moves in a circle with $r = mv/(qB)$ and $\omega = qB/m$ . The angular framework transfers directly across.

Specimen question modelled on the OCR H556 paper format

Question (8 marks): A child's roundabout in a playground has radius $1.6$ m. It is rotating uniformly, completing one full revolution every $2.5$ s. A child sits on a horse fixed to the rim.

(a) Calculate (i) the period $T$ , (ii) the frequency $f$ , and (iii) the angular velocity $\omega$ of the roundabout, stating SI units. [3]

(b) Find the linear speed $v$ of the child at the rim. [2]

(c) Without further calculation, describe and explain what happens to (i) the child's angular velocity and (ii) the child's linear speed if she stands up and walks along a radial line from the rim to half-way to the centre. [3]

AO breakdown

Mark	AO	Awarded for
1	AO2	$T = 2.5$ s with unit (or stated directly)
2	AO2	$f = 1/T = 0.40$ Hz
3	AO2	$\omega = 2\pi/T = 2.51$ rad s $^{-1}$
4	AO2	$v = r\omega$ used (or $v = 2\pi r/T$ )
5	AO2	$v = 4.0$ m s $^{-1}$ to $2$ s.f.
6	AO1	Angular velocity unchanged (rigid-body rotation)
7	AO2	Linear speed at new radius $v' = r'\omega = 0.80 \times 2.51 = 2.0$ m s $^{-1}$
8	AO3	Reasoning: same $\omega$ for all points on rigid roundabout; $v \propto r$ at fixed $\omega$

AO split: AO1 = 1, AO2 = 6, AO3 = 1.

Mid-band response (4/8):

(a) (i) $T = 2.5$ s. (ii) $f = 1/T = 0.4$ Hz. (iii) $\omega = 2\pi/T = 2.51$ rad/s.

(b) $v = r \times \omega = 1.6 \times 2.51 = 4.0$ m s $^{-1}$ .

Examiner commentary: The next-band move is to state the principle that all points on a rigid rotating body share the same $\omega$ , and only then deduce that $v$ scales with $r$ . Marks 1-5 awarded for the calculation. Marks 6-8 lost — the candidate gives a one-line assertion in (c) but does not separate $\omega$ from $v$ , does not name "rigid rotation", and does not state the half-radius numerical answer ( $\approx 2.0$ m s $^{-1}$ ).

Stronger response (7/8):

(a) (i) $T = 2.5$ s. (ii) $f = 1/T = 0.40$ Hz. (iii) $\omega = 2\pi/T = 2\pi/2.5 = 2.51$ rad s $^{-1}$ .

(b) $v = r\omega = 1.6 \times 2.51 = 4.0$ m s $^{-1}$ (to 2 s.f.).

(c) (i) Her angular velocity is unchanged at $2.51$ rad s $^{-1}$ because the roundabout is a rigid body — every fixed point on it rotates at the same $\omega$ . (ii) Her linear speed drops because $v = r\omega$ and $r$ has halved while $\omega$ is constant; at $r = 0.80$ m, $v = 0.80 \times 2.51 \approx 2.0$ m s $^{-1}$ .

Examiner commentary: To lift to top-band, the answer would make explicit the vector interpretation — that $\omega$ is a property of the rigid body, not of an individual particle — and would note that the child is not in fact a point on the rigid body once she is moving radially, so during her walk her angular velocity does briefly differ from the roundabout's. Marks 1-7 awarded. Mark 8 partial: the candidate has the right answer with reasoning, but the implicit assumption that the child stays locked to the roundabout's $\omega$ during the walk is not justified.

Top-band response (8/8):

(a) (i) $T = 2.5$ s. (ii) $f = 1/T = 1/2.5 = 0.40$ Hz. (iii) $\omega = 2\pi f = 2\pi \times 0.40 = 2.513$ rad s $^{-1}$ .

(b) The child is fixed to the rim of a rigid rotating body, so $v = r\omega$ . With $r = 1.6$ m and $\omega = 2.513$ rad s $^{-1}$ :

$v = 1.6 \times 2.513 = 4.02 \text{ m s}^{-1} \approx 4.0 \text{ m s}^{-1}$

(c) (i) Angular velocity of the roundabout itself is unchanged: a rigid body has a single $\omega$ that all of its fixed material points share. If the child is standing on the roundabout (not propelling it independently), then at any instant her angular velocity matches the roundabout's at $2.51$ rad s $^{-1}$ — though strictly her path is not a circle of constant radius during the walk, since $r(t)$ is decreasing. After she has come to rest at the new radius, her steady-state $\omega = 2.51$ rad s $^{-1}$ again.

(ii) Linear speed is governed by $v = r\omega$ . With $\omega$ fixed and $r$ halved from $1.6$ m to $0.80$ m, her linear speed at the new radius is $v' = 0.80 \times 2.513 = 2.01$ m s $^{-1} \approx 2.0$ m s $^{-1}$ .

Examiner commentary: Full marks. The discriminator moves are (1) the principle "rigid body shares a single $\omega$ " stated up front; (2) the subtlety that during the walk itself the child's instantaneous angular velocity is not strictly constant — only the steady-state endpoint is. Top-band answers do not just compute; they reason about which idealisation applies and surface its limits. Note also the careful use of $2.513$ rad s $^{-1}$ in working with rounding only at the final answer — this avoids accumulated rounding errors.

Common errors and mark-loss patterns

Pedagogical observations from teaching A-Level circular motion, with no fabricated examiner-report percentages.

Calculator in degree mode. A silent killer. Set radian mode for the entire topic and check it is still in radian mode at the start of every question; trig in degree mode produces wildly wrong answers without any obvious warning.
rpm not converted to rad s $^{-1}$ . " $1500$ rpm" needs $\times 2\pi/60$ to give $\omega = 157$ rad s $^{-1}$ . Substituting $1500$ directly into $v = r\omega$ inflates the speed by a factor of $\approx 9.5$ .
Period and frequency confused. $T$ and $f$ are reciprocals, not equal. If $f = 50$ Hz then $T = 0.02$ s, not $50$ s.
Using $s = r\theta$ with $\theta$ in degrees. The formula is radian-only. The degree-mode version is $s = r\theta\pi/180$ .
"Speed is changing" for uniform circular motion. Examiners explicitly look for velocity (vector) versus speed (scalar). Saying "the speed is changing because the direction is changing" loses the mark.
Quoting $v = r\omega$ but forgetting to convert $r$ to metres. "The arm is $40$ cm long, $\omega = 5$ rad s $^{-1}$ " gives $v = 0.40 \times 5 = 2.0$ m s $^{-1}$ , not $200$ .
Misreading a "frequency" given as cycles per minute. $300$ cpm $= 5$ Hz, not $300$ Hz. Always convert per-minute quantities to per-second before plugging into $\omega = 2\pi f$ .
Forgetting that all points on a rigid rotor share $\omega$ but not $v$ . Useful in centrifuge, hard-drive, and wind-turbine questions where different radii give different linear speeds at the same $\omega$ .

Going further

The radian as a ratio of arc to radius is one of the cleanest examples in school physics of why dimensionless angles are the natural choice — and the rabbit hole goes deep.

At undergraduate level, angular velocity is treated as a vector $\vec{\omega}$ along the rotation axis, and the velocity of any material point in a rigid body is given by the elegant cross-product $\vec{v} = \vec{\omega} \times \vec{r}$ . This single formula contains $v = r\omega$ as its magnitude (for $\vec{r}$ perpendicular to $\vec{\omega}$ ), but also recovers the direction of $\vec{v}$ automatically: tangential, perpendicular to both $\vec{\omega}$ and $\vec{r}$ , sense given by the right-hand rule. This is the standard formulation in Classical Mechanics by John Taylor (chapter 9) and in Mechanics by Kleppner and Kolenkow (a favourite of Cambridge entrance).

A further deep idea: in Lagrangian mechanics, the choice of "generalised coordinate" for a rotating system is naturally the angle $\theta$ (rather than the linear position $x$ ), and the conjugate "generalised momentum" turns out to be the angular momentum $L = I\omega$ , with $I$ the moment of inertia. The conservation of angular momentum then follows from the rotational symmetry of the Lagrangian — a particular case of Noether's theorem, one of the most beautiful results in theoretical physics. (Symmetry under rotation $\Rightarrow$ angular momentum conservation; symmetry under spatial translation $\Rightarrow$ linear momentum conservation; symmetry under time translation $\Rightarrow$ energy conservation.)

Two Oxbridge interview prompts that probe this material:

"A coin of radius $r$ rolls without slipping around the inside of a circular track of radius $R$ . How far does the coin's centre travel for one complete circuit, and how many times does the coin rotate about its own axis?" (Answer: the centre travels $2\pi(R - r)$ ; the coin rotates $(R-r)/r$ times about its own axis relative to the lab frame, but $R/r$ times if you count the orbital rotation as well — the "coin rotation paradox" of the 1982 SAT.)
"A bug walks at constant ground speed $v$ along a record turntable rotating at $\omega$ . Describe the bug's path in the lab frame." (Answer: an Archimedean spiral if walking radially; a complicated trochoid in general. The point is that the bug's motion couples its own walk with the carrier rotation, and decomposes into radial and tangential components.)
"Why does Earth's day on average get longer over geological time?" (Tidal friction between the Earth, Moon and oceans converts rotational kinetic energy of Earth into orbital kinetic energy of the Moon, slowing Earth's $\omega$ and recessing the Moon. The current rate is about $1.7$ ms per century.)

Recommended reading: Six Easy Pieces (Feynman) Chapter 1 for conceptual clarity; Kleppner and Kolenkow Mechanics Chapter 8 for the rigid-body formulation; Penrose The Road to Reality Chapter 11 for the abstract algebra of rotations and the introduction to SO(3) and quaternions.

A-Level misconceptions

"Uniform circular motion has zero acceleration because speed is constant." Wrong. Acceleration is the time-derivative of velocity, which is a vector. A change in direction of $\vec{v}$ at constant $|\vec{v}|$ is still a change in $\vec{v}$ , hence a non-zero acceleration. (This is the conceptual seed of the next lesson.)
"Angular velocity is just speed in rad s $^{-1}$ ." Loose. $\omega$ has units of rad s $^{-1}$ (which are formally s $^{-1}$ since rad is dimensionless), whereas linear speed $v$ has units of m s $^{-1}$ . They are related by $v = r\omega$ but are not interchangeable.
"The radian is a unit like the metre." No — the radian is dimensionless (a pure ratio). It is given a symbol for clarity but carries no SI dimensions. This is why $s = r\theta$ produces a length (metres) and not a length-times-an-angle.
"A rigid body has one angular velocity but many speeds." Correct! Every fixed material point on a rigid rotor shares the same $\omega$ , but their linear speeds $v = r\omega$ depend on their distance from the rotation axis. This is the engineering principle behind the centrifuge, the wind turbine and the spinning record.
"You can use degrees in $\omega = 2\pi f$ ." No — that formula assumes radian measure. In degree measure you'd write " $\omega_\text{deg} = 360 f$ " but no one does, because the calculus is then ugly.
"Angular displacement is just the angle." Loose — angular displacement is a signed (or vector) angle measured from a reference direction. It is the angular analogue of linear displacement, not of the "angle subtended at a point" in everyday geometry.

Summary

An angle in radians is the ratio of arc length to radius: $\theta = s/r$ , dimensionless, and the only choice that makes the calculus of trig functions clean.
$360° = 2\pi$ rad; conversion is multiply by $\pi/180$ or $180/\pi$ .
Angular displacement $\theta$ (rad), angular velocity $\omega = d\theta/dt$ (rad s $^{-1}$ ), period $T = 2\pi/\omega$ (s), frequency $f = 1/T$ (Hz), with $\omega = 2\pi f$ throughout.
The linear-angular bridge: $s = r\theta$ , $v = r\omega$ , $a_t = r\alpha$ . For uniform circular motion the third row is zero.
Uniform circular motion: constant speed $|\vec{v}|$ , but the direction of $\vec{v}$ rotates at $\omega$ — so velocity is changing and there must be an acceleration. This sets up the next lesson.
A rigid rotor has a single $\omega$ shared by all its fixed material points; their linear speeds differ as $v = r\omega$ .

Master these five relationships and you will never be stuck on a circular-motion calculation. The next lesson builds directly on them to derive centripetal acceleration — the acceleration required to keep an object moving in a circle.

Reference: OCR A-Level Physics A (H556) specification 5.3 — Circular motion (refer to the official OCR H556 specification document for exact wording).

Circular Motion — Angular Quantities

Circular Motion — Angular Quantities

Why Not Just Use Degrees?

Defining the Radian

Key Conversions

Angular Displacement

Worked Example 1 — Arc Length

Worked Example 2 — Pendulum Arc

Angular Velocity

Angular Velocity and Period

Angular Velocity and Frequency

Linear Speed Around a Circle

Worked Example 3 — Spinning Vinyl Disc

Worked Example 4 — The Earth's Rotation

Worked Example 5 — Bicycle Wheel

Uniform Circular Motion — Constant Speed, Changing Velocity

Synoptic Links

Specimen question modelled on the OCR H556 paper format

AO breakdown

Common errors and mark-loss patterns

Going further

A-Level misconceptions

Summary

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