recall and use ω = 2π / T and v = rω

Published by Patrick Mutisya · 8 days ago

Cambridge A-Level Physics 9702 – Kinematics of Uniform Circular Motion

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

QuantitySymbolFormulaUnits
Period\$T\$Time for one revolutions
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

Suggested diagram: A circle of radius \$r\$ with an arrow indicating angular displacement \$\Delta\theta\$ and a tangent arrow showing linear speed \$v\$ at a point on the circumference.

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.