understand and use the concept of angular speed

Kinematics of Uniform Circular Motion (UCM)

1. Introduction

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

2. Fundamental Definitions

  • Radian (rad): the angle subtended at the centre of a circle when the arc length equals the radius; \(s = r\theta\).
  • Angular displacement \(\theta\): the angle swept out by the radius vector, measured in radians (rad).
  • Angular speed \(\omega\): rate of change of angular displacement, \[\omega = \frac{\Delta\theta}{\Delta t}\qquad\text{(rad s}^{-1}\text{)}.\]
  • Angular acceleration \(\alpha\): rate of change of angular speed, \[\alpha = \frac{\Delta\omega}{\Delta t}\qquad\text{(rad s}^{-2}\text{)}.\] For uniform circular motion \(\alpha = 0\) (i.e. \(\omega\) is constant).

3. Key Relationships Between Angular and Linear Quantities

  • Linear (tangential) speed \(v = \omega r\)
  • Centripetal (radial) acceleration \(a_c = \dfrac{v^{2}}{r} = \omega^{2}r\)
  • Centripetal force (Newton’s 2nd law) \(F_c = m a_c = m\omega^{2}r = \dfrac{m v^{2}}{r}\)

Derivation of \(\displaystyle\omega = \frac{2\pi}{T}\)

One complete revolution corresponds to an angular displacement of \(2\pi\) rad. If \(T\) is the period (time for one revolution), then

\[ \omega = \frac{\Delta\theta}{\Delta t}= \frac{2\pi\;\text{rad}}{T}\;, \] so \(\displaystyle\omega = \frac{2\pi}{T}\). Consequently \[ f = \frac{1}{T}= \frac{\omega}{2\pi},\qquad\omega = 2\pi f . \]

4. Summary of Formulae

Quantity Symbol Formula Units
Angular displacement \(\theta\) \(\theta = s/r\) rad
Angular speed \(\omega\) \(\omega = \Delta\theta/\Delta t = 2\pi/T = 2\pi f\) rad s\(^{-1}\)
Angular acceleration \(\alpha\) \(\alpha = \Delta\omega/\Delta t\) rad s\(^{-2}\)
Linear (tangential) speed \(v\) \(v = \omega r\) m s\(^{-1}\)
Centripetal acceleration \(a_c\) \(a_c = \omega^{2}r = v^{2}/r\) m s\(^{-2}\)
Centripetal force \(F_c\) \(F_c = m\omega^{2}r = mv^{2}/r\) N
Period of revolution \(T\) \(T = 2\pi/\omega = 1/f\) s
Frequency of revolution \(f\) \(f = 1/T = \omega/2\pi\) Hz

5. Solving Problems Involving Angular Speed

  1. Read the question carefully and list the given data (radius \(r\), linear speed \(v\), period \(T\), mass \(m\), etc.).
  2. Convert any linear information to angular form using \(v = \omega r\) or \(T = 2\pi/\omega\) (or \(f = \omega/2\pi\)).
  3. Select the appropriate formula from the table – e.g. \(\omega = v/r\), \(a_c = \omega^{2}r\), \(F_c = m\omega^{2}r\).
  4. Carry out the calculation, keeping track of units. Remember to convert between rev min\(^{-1}\), Hz and rad s\(^{-1}\) when required.
  5. Round the final answer to the correct number of significant figures and attach the proper unit.

6. Worked Example

Question: A car travels round a circular track of radius \(30\;\text{m}\) at a constant speed of \(15\;\text{m s}^{-1}\). Determine (a) the angular speed \(\omega\), (b) the centripetal acceleration \(a_c\), and (c) the centripetal force if the car’s mass is \(1200\;\text{kg}\).

Solution:

  1. Angular speed \[ \omega = \frac{v}{r}= \frac{15\;\text{m s}^{-1}}{30\;\text{m}} = 0.50\;\text{rad s}^{-1}. \]
  2. Centripetal acceleration \[ a_c = \omega^{2}r = (0.50\;\text{rad s}^{-1})^{2}\times30\;\text{m}=7.5\;\text{m s}^{-2}. \]
  3. Centripetal force \[ F_c = m a_c = 1200\;\text{kg}\times7.5\;\text{m s}^{-2}=9.0\times10^{3}\;\text{N}. \]

Answers: \(\omega = 0.50\;\text{rad s}^{-1}\), \(a_c = 7.5\;\text{m s}^{-2}\), \(F_c = 9.0\times10^{3}\;\text{N}\).

Top‑view diagram (not to scale): radius \(r\), car moving with speed \(v\), angular displacement \(\theta\), angular speed \(\omega\), and centripetal force \(F_c\) directed toward the centre.

7. Common Misconceptions & How to Avoid Them

  • Angular speed vs. angular displacement: \(\omega\) is a rate (rad s\(^{-1}\)), whereas \(\theta\) is the angle itself (rad).
  • Zero acceleration? A constant linear speed in a circle does not mean zero acceleration; the continuously changing direction gives a centripetal acceleration \(a_c\).
  • Missing \(2\pi\) factor: When converting between period \(T\), frequency \(f\) and angular speed \(\omega\), always use \(\omega = 2\pi f = 2\pi/T\).
  • Units of angular speed: The radian is dimensionless, but it should be retained in the notation to remind you that the quantity is an angular rate.
  • Angular acceleration in UCM: For uniform circular motion \(\alpha = 0\); \(\omega\) remains constant throughout the motion.

8. Quick Revision Checklist

  • Radian definition and \(s = r\theta\).
  • Angular displacement \(\theta\), speed \(\omega\), acceleration \(\alpha\) (zero for UCM).
  • Key relations: \(\omega = 2\pi/T = 2\pi f\), \(v = \omega r\), \(a_c = \omega^{2}r\), \(F_c = m\omega^{2}r\).
  • Be fluent in converting between linear and angular forms and between \(T\), \(f\) and \(\omega\).
  • Check units and significant figures in every answer.

Mastering these points will enable you to tackle any Cambridge 9702 exam question on uniform circular motion, from straightforward speed calculations to more complex force problems.

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