understand and use the concept of angular speed

Kinematics of Uniform Circular Motion

When an object moves in a circle at constant speed, its motion is called uniform circular motion (UCM). Although the speed is constant, the direction of the velocity continuously changes, giving the object a centripetal (centre‑seeking) acceleration.

Angular Quantities

In circular motion it is convenient to describe the motion using angular quantities rather than linear ones.

• Angular displacement (\$\theta\$) – the angle swept out by the radius vector, measured in radians (rad).
• Angular speed (\$\omega\$) – the rate of change of angular displacement,

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

  with SI unit rad s\(^{-1}\). For uniform circular motion \$\omega\$ is constant.

• Angular acceleration (\$\alpha\$) – the rate of change of angular speed,

  \$\alpha = \frac{\Delta\omega}{\Delta t}\;,\$

  measured in rad s\(^{-2}\). In uniform circular motion \$\alpha = 0\$.

Relation Between Angular and Linear Quantities

The linear (tangential) speed \$v\$ of a point at a distance \$r\$ from the centre of the circle is directly proportional to the angular speed:

\$v = \omega r\$

Similarly, the linear (centripetal) acceleration \$a_c\$ can be expressed in two equivalent forms:

\$a_c = \frac{v^{2}}{r} = \omega^{2} r\$

Key Equations Summary

Using Angular Speed in Problems

Typical steps when solving A‑Level questions involving angular speed:

1. Identify the radius \$r\$ of the circular path and any given linear quantities (e.g., speed \$v\$ or period \$T\$).
2. Convert any linear data to angular form using \$v = \omega r\$ or \$T = 2\pi/\omega\$.
3. Apply the appropriate formula from the table above to find the required quantity (e.g., \$\omega\$, \$v\$, \$a_c\$).
4. Check units and, if necessary, convert between revolutions per minute (rpm), hertz (Hz), or rad s\(^{-1}\).
5. State the final answer with the correct number of significant figures.

Example Problem

Question: A car travels round a circular track of radius \$30\,\$m at a constant speed of \$15\,\$m s\(^{-1}\). Determine the angular speed \$\omega\$ and the centripetal acceleration \$a_c\$ of the car.

Solution:

1. Calculate angular speed using \$v = \omega r\$:

   \$\omega = \frac{v}{r} = \frac{15\;\text{m s}^{-1}}{30\;\text{m}} = 0.50\;\text{rad s}^{-1}.\$

2. Find centripetal acceleration:

   \$a_c = \omega^{2} r = (0.50\;\text{rad s}^{-1})^{2}\times 30\;\text{m}=7.5\;\text{m s}^{-2}.\$

3. Answer: \$\omega = 0.50\;\text{rad s}^{-1}\$, \$a_c = 7.5\;\text{m s}^{-2}\$.

Common Misconceptions

• Confusing angular speed \$\omega\$ with angular displacement \$\theta\$. \$\omega\$ is a rate (rad s\(^{-1}\)), whereas \$\theta\$ is an angle (rad).
• Assuming that a constant linear speed implies zero acceleration. In circular motion the direction changes, giving a non‑zero centripetal acceleration.
• Neglecting the factor \$2\pi\$ when converting between period \$T\$, frequency \$f\$, and angular speed \$\omega\$.

Summary

Understanding angular speed is essential for analysing uniform circular motion. The core relationships \$v = \omega r\$ and \$a_c = \omega^{2} r\$ link angular and linear descriptions, allowing you to move seamlessly between them in A‑Level physics problems.