A.2.26—Angular and linear speed

Syllabus
First assessment 2025
Objective
Level
SL

Link Angular and Linear Speed

Connect the descriptions

For uniform circular motion,

v=2πrT=ωrv=\frac{2\pi r}{T}=\omega r

Angular speed ω\omega is the same for all points on a rigid rotating body, while linear speed increases with distance from the axis.

Use the period

One revolution takes period TT, so ω=2π/T\omega=2\pi/T. Keep radians and seconds consistent.

Compare points on one disk

If one point is twice as far from the centre, its linear speed is twice as large at the same angular speed; its centripetal acceleration is also twice as large.

Common trap

Do not assume equal linear speeds for all points on a rotating disk. Equal angular speed does not mean equal tangential speed.

A.2.26 Exam Analysis

Assessment in practice

1–3 marks
How it is assessed

The evidence asks for ratios of linear speed and centripetal acceleration at two radii and asks for angular velocity from a 24-hour orbital period.

Command terms

Calculate / Identify

What earns marks

Use v=ωr and ω=2π/T. For a rigid disk, compare radii at the same angular speed; for an orbit, convert the period to seconds before calculating angular velocity.

Watch for

Using the same tangential speed at different radii on a rigid rotating disk or leaving a period in hours.

Representative question

Question 1

[Maximum number: 1]

A disk of radius R rotates about its axis with angular speed ω\omega. Point X is at a distance of R2\frac{R}{2} from the centre and point Y is on the circumference.

What are the ratios of the linear speeds and the centripetal acceleration of X to Y.
The linear speed of X is vXv_{X} and its acceleration is aXa_{X}; the linear speed of Y is vYv_{Y} and its acceleration is aYa_{Y}.

Linear speeds vXvY\frac{\boldsymbol{v}_{\mathbf{X}}}{\boldsymbol{v}_{\mathbf{Y}}}

Acceleration aXaY\frac{\mathbf{a}_{\mathbf{X}}}{\mathbf{a}_{\mathbf{Y}}}

12\frac{1}{2}

14\frac{1}{4}

12\frac{1}{2}

12\frac{1}{2}

1

14\frac{1}{4}

1

12\frac{1}{2}