recall and use ω = 2π / T and v = rω

Kinematics of Uniform Circular Motion

Learning Objective

Recall and use the relationships

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

and

$$v = r\omega$$

where:

• $\omega$ – angular speed (rad s⁻¹)
• $T$ – period of one revolution (s)
• $r$ – radius of the circular path (m)
• $v$ – linear (tangential) speed (m s⁻¹)

1. Angular Speed and Period

The angular speed $\omega$ describes how quickly an object sweeps out angle in radians per second. For uniform circular motion the angular displacement after one complete revolution is $2\pi$ radians, so the definition of period $T$ gives

$$\omega = \frac{\Delta\theta}{\Delta t} = \frac{2\pi}{T}$$

Key points:

1. Period $T$ is the time for one full circle.
2. Frequency $f = 1/T$ (revolutions per second).
3. Angular speed can also be written $\omega = 2\pi f$.

2. Linear Speed from Angular Speed

The linear (tangential) speed $v$ is the distance travelled along the circumference per unit time. The circumference of a circle is $2\pi r$, so in one period the object travels that distance:

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

Thus, for a given radius, the linear speed is directly proportional to the angular speed.

3. Worked Example

Problem: A car moves around a circular track of radius $r = 50\ \text{m}$ and completes a lap in $T = 20\ \text{s}$. Find the angular speed $\omega$ and the linear speed $v$.

1. Calculate $\omega$: $$\omega = \frac{2\pi}{T} = \frac{2\pi}{20\ \text{s}} = 0.314\ \text{rad s}^{-1}$$
2. Calculate $v$ using $v = r\omega$: $$v = (50\ \text{m})(0.314\ \text{rad s}^{-1}) = 15.7\ \text{m s}^{-1}$$

4. Common Mistakes to Avoid

• Confusing angular speed $\omega$ (rad s⁻¹) with frequency $f$ (Hz). Remember $\omega = 2\pi f$.
• Using the diameter instead of the radius in $v = r\omega$.
• Omitting the factor $2\pi$ when converting between period and angular speed.

5. Summary Table of Key Relationships

Quantity	Symbol	Formula	Units
Period	$T$	Time for one revolution	s
Frequency	$f$	$f = \dfrac{1}{T}$	Hz (s⁻¹)
Angular speed	$\omega$	$\omega = \dfrac{2\pi}{T} = 2\pi f$	rad s⁻¹
Linear (tangential) speed	$v$	$v = r\omega = \dfrac{2\pi r}{T}$	m s⁻¹

6. Suggested Diagram

7. Quick Revision Questions

1. If a satellite orbits Earth at a radius of $7.0\times10^6\ \text{m}$ with an angular speed of $1.1\times10^{-3}\ \text{rad s}^{-1}$, what is its orbital period?
2. A turntable rotates at $120\ \text{rpm}$. Convert this to angular speed in rad s⁻¹ and find the linear speed of a point $0.15\ \text{m}$ from the centre.