# Angular displacement and speed

> Physics · CIE A-Level
> Source: https://www.owlsprep.com/study/cie-9702-u16-angular-displacement-and-speed/

This sub-topic introduces the core kinematic quantities used to describe uniform circular motion, covering angular measurement in radians, relationships between angular speed, period, frequency, and conversion to linear tangential speed.

**Prerequisites:** Basic trigonometry and radians as angle units; Linear kinematics (speed, distance, time relationships)

## Learning objectives

- Define angular displacement and angular speed using radians as the unit of measurement
- Relate angular speed to the period and frequency of rotation
- Convert between angular speed and tangential linear speed for points on a rotating body

## Angular Displacement in Radians

**Angular Displacement** — The angle (in radians) that a rotating body has turned from its initial position around a fixed axis, with positive values for counter-clockwise rotation by convention

*Notation:* \theta

*Example:* A full rotation gives \theta = 2\pi radians

Radians are the standard unit for circular motion because they create a simple proportional relationship between arc length (the distance a point travels along the circular path), radius, and angular displacement:

$$s = r\theta$$

**Worked example:** A point on a bicycle wheel of radius 30 cm rotates by 2.5 radians. Calculate the arc length the point travels.

1. Identify known values, convert radius to SI units:
2. $$r = 30 \text{ cm} = 0.30 \text{ m}, \quad \theta = 2.5 \text{ rad}$$
3. Substitute into the arc length formula:
4. $$s = r\theta = 0.30 \times 2.5 = 0.75 \text{ m}$$
5. The point travels 0.75 m (75 cm) along its path.

*Calculator:* allowed

## Angular Speed, Period and Frequency

**Angular Speed** — The rate of change of angular displacement over time, measured in radians per second (rad s⁻¹)

*Notation:* \omega

*Example:* A ceiling fan rotating 4π radians per second has \omega = 4\pi rad s⁻¹

**Derivation:** Derive the relationship between angular speed, period and frequency

*Starting from:* Angular speed is defined as \omega = \frac{\Delta \theta}{\Delta t}

1. For one full rotation, the total angular displacement is \(2\pi\) radians, and the time taken is the period \(T\). Substitute into the definition:
2. $$\omega = \frac{2\pi}{T}$$
3. Frequency \(f\) is defined as the number of rotations per second, so \(f = \frac{1}{T}\). Substitute to get:
4. $$\omega = 2\pi f$$

*Conclusion:* Angular speed is directly proportional to rotation frequency and inversely proportional to the period of rotation.

**Worked example:** A car engine crankshaft rotates at 3000 revolutions per minute (rpm). Calculate its angular speed in rad s⁻¹.

1. Convert rotation rate from rpm to frequency in Hz (revolutions per second):
2. $$f = \frac{3000}{60} = 50 \text{ Hz}$$
3. Substitute into the angular speed formula:
4. $$\omega = 2\pi f = 2\pi (50) = 100\pi \approx 314 \text{ rad s}^{-1}$$

*Calculator:* allowed

## Relationship Between Angular and Linear Speed

For a rigid rotating body, all points rotate with the same angular speed, but the linear (tangential) speed of a point depends on its distance from the axis of rotation. Starting from the arc length relationship \(s = r\theta\), differentiate both sides with respect to time to get:

$$v = r\omega$$

> **tip**
>
> This \(v\) is the tangential speed along the circular path, not angular speed. Even if angular speed is constant, the direction of velocity changes continuously, which causes centripetal acceleration.

**Worked example:** A merry-go-round rotates with constant angular speed of 0.8 rad s⁻¹. Calculate the tangential speed of a rider sitting 3.0 m from the center of rotation.

1. Identify known values: \(r = 3.0\) m, \(\omega = 0.8\) rad s⁻¹
2. Substitute into the relationship between angular and linear speed:
3. $$v = r\omega = 3.0 \times 0.8 = 2.4 \text{ m s}^{-1}$$

**Check your understanding**

Test your understanding:

1. A wheel of diameter 0.8 m rotates with angular speed 10 rad s⁻¹. What is the tangential speed of a point on its edge?

   - 4 m s⁻¹
   - 8 m s⁻¹
   - 16 m s⁻¹
   - 10π m s⁻¹

   *Answer:* 4 m s⁻¹

   *Why:* Correct: First find radius \(r = 0.8/2 = 0.4\) m, then \(v = 0.4 \times 10 = 4\) m s⁻¹.

*Calculator:* allowed

## Common pitfalls

- **Wrong:** Using degrees instead of radians in the formulas \(s = r\theta\) or \(v = r\omega\)
  - Why it fails: These proportional relationships are only valid when angular displacement is measured in radians, not degrees
  - Correct: Always convert angles from degrees to radians by multiplying by \(\pi/180\) before substituting into formulas
- **Wrong:** Assuming all points on a rigid rotating body have the same linear speed
  - Why it fails: Only angular speed is constant across all points on a rigid rotating body; linear speed increases with distance from the axis
  - Correct: Remember that for any point on a rigid rotating body, \(v = r\omega\), so linear speed scales with radius
- **Wrong:** Forgetting to convert frequency from rpm to Hz before calculating angular speed
  - Why it fails: SI units for frequency are revolutions per second (Hz); using rpm directly gives an incorrect result 60 times too small
  - Correct: Always divide rpm by 60 to get frequency in Hz before substituting into \(\omega = 2\pi f\)
- **Wrong:** Using diameter instead of radius in \(v = r\omega\)
  - Why it fails: Exam questions commonly give the diameter of wheels/circles to test attention to detail
  - Correct: Always check if the given length is radius or diameter, and divide diameter by 2 to get radius before substitution

## Cheatsheet

| Quantity | Symbol | Formula | SI Unit |
| --- | --- | --- | --- |
| Angular Displacement | $\theta$ | $\theta = s/r$ | radians (rad) |
| Angular Speed | $\omega$ | $\omega = \Delta\theta/\Delta t = 2\pi/T = 2\pi f$ | rad s⁻¹ |
| Tangential Speed | $v$ | $v = r\omega$ | m s⁻¹ |
| Period of Rotation | $T$ | $T = 1/f = 2\pi/\omega$ | seconds (s) |
| Frequency of Rotation | $f$ | $f = 1/T = \omega/2\pi$ | hertz (Hz) |

## What's next

The concepts of angular displacement and speed covered here form the foundation for all further work on circular motion, the next core topic being centripetal acceleration and force, which explain why circular motion requires a net force even when speed is constant. Mastery of these angular-linear relationships is essential for all subsequent circular motion problems, and they extend to rotational dynamics, torque, and moment of inertia topics that appear in later sections of the CIE 9702 syllabus. Exam questions regularly combine these basic kinematic relationships with force calculations, so consistent recall is key.

- [Centripetal Acceleration](https://www.owlsprep.com/study/cie-9702-u16-centripetal-acceleration/)
- [Centripetal Force](https://www.owlsprep.com/study/cie-9702-u16-centripetal-force/)
- [Gravitational fields](https://www.owlsprep.com/study/cie-9702-u17-overview/)

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