Circular Motion: Angular Velocity

When an object moves in a circle, we need new quantities to describe its motion. Linear velocity alone is not enough — we need angular velocity to describe how quickly the object sweeps through an angle. This lesson establishes the language and equations of circular motion that underpin everything that follows.

Radian Measure

Before we can define angular velocity, we need to be comfortable with radians. A radian is defined as the angle subtended at the centre of a circle by an arc equal in length to the radius.

For a circle of radius r, an arc of length s subtends an angle θ (in radians) given by:

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

Angle (degrees)	Angle (radians)	Exact Form	Arc Length (for r = 1 m)
0°	0	0	0 m
30°	0.524	π/6	0.524 m
45°	0.785	π/4	0.785 m
60°	1.047	π/3	1.047 m
90°	1.571	π/2	1.571 m
180°	3.142	π	3.142 m
270°	4.712	3π/2	4.712 m
360°	6.283	2π	6.283 m

To convert degrees to radians: multiply by π/180. To convert radians to degrees: multiply by 180/π.

Angular Displacement

Angular displacement is the angle through which an object has rotated, measured in radians. If a satellite completes one full orbit, its angular displacement is 2π rad. If a wheel rotates three complete turns, its angular displacement is 6π rad.

Angular Velocity

Angular velocity (ω, the Greek letter omega) is the rate of change of angular displacement:

$\omega = \frac{\Delta\theta}{\Delta t}$

It is measured in rad s⁻¹ (radians per second).

For an object moving in a complete circle of period T (the time for one complete revolution):

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

Since frequency f = 1/T, we can also write:

$\omega = 2\pi f$

flowchart LR
    A["Given rpm"] --> B["Divide by 60"]
    B --> C["Frequency f (Hz)"]
    C --> D["Multiply by 2π"]
    D --> E["ω (rad s⁻¹)"]
    F["Given period T (s)"] --> G["f = 1/T"]
    G --> C

Worked Example 1

A turntable rotates at 33⅓ rpm (revolutions per minute). Calculate its angular velocity.

Convert to revolutions per second: $f = \frac{33.33}{60} = 0.556 \text{ Hz}$

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

Worked Example 2

The Earth completes one rotation about its axis every 24 hours. What is its angular velocity?

$T = 24 \times 3600 = 86{,}400 \text{ s}$ $\omega = \frac{2\pi}{86{,}400} = 7.27 \times 10^{-5} \text{ rad s}^{-1}$

Relationship Between Linear and Angular Speed

Consider an object moving in a circle of radius r. In one complete revolution it covers a distance equal to the circumference, 2πr, in time T.

Linear speed: $v = \frac{2\pi r}{T} = r \times \frac{2\pi}{T} = r\omega$

This gives us the crucial relationship:

$v = r\omega$

This means that for a given angular velocity, points further from the centre move faster in terms of linear speed. On a spinning disc, the outer edge moves faster than a point near the centre, even though both have the same angular velocity.

Worked Example 3

A cyclist rides around a circular track of radius 25 m, completing one lap in 20 s.

Angular velocity: $\omega = \frac{2\pi}{20} = 0.314 \text{ rad s}^{-1}$

Linear speed: $v = r\omega = 25 \times 0.314 = 7.85 \text{ m s}^{-1}$

Worked Example 4

A centrifuge spins at 12,000 rpm with a tube radius of 0.10 m. Find the linear speed at the tip.

$f = \frac{12{,}000}{60} = 200 \text{ Hz}$ $\omega = 2\pi \times 200 = 1257 \text{ rad s}^{-1}$ $v = r\omega = 0.10 \times 1257 = 126 \text{ m s}^{-1}$

That is about 453 km/h — faster than most Formula 1 cars.

Period and Frequency

The period T is the time for one complete revolution, measured in seconds.

The frequency f is the number of complete revolutions per second, measured in hertz (Hz).

$f = \frac{1}{T} \qquad T = \frac{1}{f}$

Comparing Circular Motion Quantities

Object	ω (rad s⁻¹)	v (m s⁻¹)	Period	Notes
Hour hand of clock	1.45 × 10⁻⁴	~5 × 10⁻⁴ (for r = 3 cm)	12 h	Slowest common example
Minute hand of clock	1.75 × 10⁻³	~1.3 × 10⁻⁴ (for r = 7 cm)	60 min	12× faster than hour hand
Earth's rotation	7.27 × 10⁻⁵	465 (at equator)	24 h	r = 6.37 × 10⁶ m
Earth's orbit	1.99 × 10⁻⁷	29 800	365.25 days	r = 1.50 × 10¹¹ m
CD at 500 rpm	52.4	3.1 (at edge, r = 6 cm)	0.12 s	Inner tracks spin at same ω
Washing machine (1200 rpm)	126	31 (for r = 0.25 m)	0.050 s	Spin cycle
Ultracentrifuge (100 000 rpm)	10 470	1047 (for r = 0.10 m)	6 × 10⁻⁴ s	Supersonic tip speed

Uniform Circular Motion

An object moving in a circle at constant speed is said to undergo uniform circular motion. Its angular velocity is constant, its speed is constant, but its velocity is continuously changing because its direction is continuously changing.

This is a crucial point: constant speed does not mean constant velocity. Velocity is a vector. If the direction changes, the velocity changes, which means there must be an acceleration — even though the speed is unchanged. This acceleration is called centripetal acceleration, which we will explore in the next lesson.

Common Exam Mistakes

Mistake 1: Forgetting to convert rpm to rad s⁻¹. Always divide rpm by 60 to get Hz, then multiply by 2π to get ω.

Mistake 2: Confusing linear speed and angular velocity. Two points on a spinning disc have the same ω but different v (unless they are at the same radius).

Mistake 3: Using degrees in equations. All circular motion formulae require angles in radians. If θ = 90°, use π/2 rad.

Mistake 4: Stating that velocity is constant in uniform circular motion. Speed is constant; velocity is not — its direction continuously changes.

Summary

Angles in circular motion are measured in radians: θ = s/r
Angular velocity: ω = Δθ/Δt = 2π/T = 2πf (units: rad s⁻¹)
Linear speed and angular velocity: v = rω
Period T and frequency f are related by f = 1/T
In uniform circular motion, speed is constant but velocity changes continuously due to changing direction
rpm to rad s⁻¹: divide by 60, then multiply by 2π
All points on a rigid rotating body share the same ω, but outer points have greater linear speed v = rω

Worked Example 5: Comparing Points on a Rotating System

A bicycle wheel of radius 0.35 m rotates at 4.0 rev s⁻¹. Compare the linear speed at the tyre tread (outer edge) and at a point 0.10 m from the centre.

Angular velocity (same for all points on the wheel): $\omega = 2\pi f = 2\pi \times 4.0 = 25.1 \text{ rad s}^{-1}$

Outer edge (r = 0.35 m): $v = r\omega = 0.35 \times 25.1 = 8.8 \text{ m s}^{-1}$

Inner point (r = 0.10 m): $v = r\omega = 0.10 \times 25.1 = 2.5 \text{ m s}^{-1}$

The outer edge moves 3.5 times faster linearly, even though both points complete one revolution in the same time. This is because v ∝ r for a given ω.

Worked Example 6: Angular Velocity of Clock Hands

Calculate the angular velocity of the second hand, minute hand, and hour hand of a clock.

Second hand (T = 60 s): $\omega = \frac{2\pi}{60} = 0.105 \text{ rad s}^{-1}$

Minute hand (T = 3600 s): $\omega = \frac{2\pi}{3600} = 1.75 \times 10^{-3} \text{ rad s}^{-1}$

Hour hand (T = 43 200 s): $\omega = \frac{2\pi}{43{,}200} = 1.45 \times 10^{-4} \text{ rad s}^{-1}$

Each hand has an angular velocity 60 or 12 times smaller than the one before it. Despite these tiny angular velocities, the tips of the hands can have appreciable linear speeds on a large clock face.

Dimensional Analysis of Circular Motion Formulae

A useful exam check: verify the units of any equation you derive.

Formula	Units Check
ω = 2π/T	rad/s = 1/s = s⁻¹ ✓ (radians are dimensionless)
v = rω	m × rad s⁻¹ = m s⁻¹ ✓
θ = ωt	rad s⁻¹ × s = rad ✓
s = rθ	m × rad = m ✓

If your answer has the wrong units, re-check your working.

A-Level Deep Dive: Circular Motion: Angular Velocity

Spec mapping

Edexcel 9PH0 specification Topic 6 — Further Mechanics introduces angular displacement in radians, angular velocity $\omega = \Delta\theta/\Delta t$ , the relationships $\omega = 2\pi/T = 2\pi f$ and $v = r\omega$ , and the use of these quantities to describe uniform circular motion as a precursor to centripetal acceleration and force (refer to the official Pearson Edexcel specification document for exact wording). Angular velocity is examined directly in Paper 2 (Topics 5–8), where Topic 6 questions routinely begin "calculate the angular velocity of …" before progressing to centripetal-force analysis. Synoptic reuse appears in Paper 3 General and Practical Principles, which favours unfamiliar contexts (planetary orbits, ultracentrifuges, fairground rides). The Edexcel formula booklet supplies $\omega = v/r$ , $\omega = 2\pi f$ and $T = 1/f$ , but does not state the radian definition $\theta = s/r$ — that must be quoted from memory.

Worked example with full mark scheme

Question (8 marks):

A fairground "wave swinger" ride rotates passengers on horizontal swings around a vertical central tower. The swings hang from chains attached to a rotating canopy of effective radius 6.0 m. The ride completes one full revolution every 4.5 s.

(a) Calculate the angular velocity of the canopy. (2)

(b) Calculate the linear speed of a swing at radius 6.0 m. (2)

(c) A child sits on a swing at radius 4.0 m (closer to the central tower). State, with calculation, the linear speed of this swing and the angular velocity it experiences. (3)

(d) The operator increases the period to 6.0 s. Determine the new linear speed at radius 6.0 m. (1)

Solution with mark scheme:

(a) Step 1 — apply $\omega = 2\pi/T$ .

$\omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{4.5}$

M1 — correct substitution into $\omega = 2\pi/T$ with $T$ in seconds. Common error: candidates write $\omega = 1/T$ or $\omega = 2\pi T$ , confusing the relationship.

A1 — $\omega = 1.40$ rad s $^{-1}$ (3 s.f.), with units quoted explicitly.

(b) Step 2 — apply $v = r\omega$ .

$v = r\omega = 6.0 \times 1.40 = 8.4 \text{ m s}^{-1}$

M1 — substitution into $v = r\omega$ using the angular velocity from (a). Examiners reward "carry forward" credit if (a) was wrong but (b)'s method is correct.

A1 — $v = 8.4$ m s $^{-1}$ , units stated.

B1 — angular velocity at the inner swing is the same as at the outer swing: $\omega = 1.40$ rad s $^{-1}$ . Both swings complete one revolution in the same period, so they must share $\omega$ .

M1 — apply $v = r\omega$ at the new radius: $v = 4.0 \times 1.40$ .

A1 — $v = 5.6$ m s $^{-1}$ . Crucially, the inner swing moves slower linearly even though it shares $\omega$ with the outer swing — this is the signature of rigid rotation.

(d) Step 4 — recompute $\omega$ for the new period and apply $v = r\omega$ .

$\omega' = 2\pi/6.0 = 1.047$ rad s $^{-1}$ , so $v' = 6.0 \times 1.047 = 6.28$ m s $^{-1}$ .

A1 — $v' \approx 6.3$ m s $^{-1}$ (one mark, awarded for correct method and answer with units).

Total: 8 marks (M3 A4 B1, distributed as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 2 format

Question (6 marks): A car tyre of outer radius 0.32 m has a small stone wedged in the tread. The car travels at a steady 22 m s $^{-1}$ along a straight road.

(a) Calculate the angular velocity of the tyre. (2)

(b) Determine the period of rotation of the tyre. (2)

(c) The stone breaks loose and is, momentarily, at the top of the tyre. State and justify the velocity of the stone (relative to the road) at the instant it leaves the tread. (2)

Mark scheme decomposition by AO:

(a)

M1 (AO2 — application of $v = r\omega$ rearranged): $\omega = v/r = 22/0.32$ .
A1 (AO1.1 — numerical answer with units): $\omega = 69$ rad s $^{-1}$ (2 s.f.).

(b)

M1 (AO2): apply $T = 2\pi/\omega$ or $T = 2\pi r/v$ .
A1 (AO1.1): $T = 2\pi/68.75 = 0.091$ s, or equivalently $T = 2\pi(0.32)/22 = 0.091$ s.

(c)

B1 (AO3.1 — recognising the velocity of a point on a rolling wheel): the centre of the wheel moves forward at 22 m s $^{-1}$ ; the top of the tyre moves forward at 22 m s $^{-1}$ relative to the wheel's centre (since $v = r\omega$ at the rim), giving 44 m s $^{-1}$ relative to the road.
B1 (AO3.2 — interpretation): the stone leaves the tread horizontally at twice the car's speed; this is why grit thrown by the front wheels can damage following vehicles.

Total: 6 marks split AO1 = 2, AO2 = 2, AO3 = 2. This is a balanced AO question typical of Paper 2 Topic 6: AO1 for the calculation, AO2 for selecting and rearranging the correct relationship, AO3 for reasoning about the kinematic combination of translation and rotation.

Synoptic links

Connects to:

Linear kinematics analogy (Topic 1): every linear-kinematics relationship has a rotational counterpart. Displacement $s$ ↔ $\theta$ ; velocity $v$ ↔ $\omega$ ; acceleration $a$ ↔ $\alpha = d\omega/dt$ . The SUVAT equations have direct analogues ( $\omega_f = \omega_i + \alpha t$ , $\theta = \omega_i t + \tfrac{1}{2}\alpha t^2$ ) used in extended A-Level treatments and in undergraduate mechanics.
Centripetal acceleration (Topic 6, next lesson): the centripetal acceleration of a point on a rotating body is $a = r\omega^2 = v^2/r$ . Both forms follow directly from $v = r\omega$ . Without fluency with $v$ and $\omega$ , the centripetal-force calculations dominating Topic 6 are inaccessible.
Simple harmonic motion as projection of circular motion (Topic 7): SHM can be derived as the shadow (projection onto a diameter) of a point in uniform circular motion. The angular frequency $\omega$ of the SHM equals the angular velocity of the underlying circular motion, and the expressions $x = A\cos(\omega t)$ , $v = -A\omega\sin(\omega t)$ , $a = -\omega^2 x$ inherit $\omega$ directly from this lesson.