This equation links the rotational description ($\omega$) to the translational description (v)
Circular Motion - OCR A-Level Physics
$$v = r\omega$$
- This equation links the rotational description ($\omega$) to the translational description (v).
- An object further from the centre travels a greater distance per revolution, so it has a greater linear speed even $though \omega$is the same.
- At the equator, the Earth's surface has a greater linear speed than at higher latitudes, even though the angular velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹.The rate of change of angular displacementThe distance moved in a particular direction from a starting point. A vector quantity. Measured in metres (m).. The angle swept per unit time for an object moving in a circle. Measured in rad s⁻¹. is the same everywhere.
- The direction of the linear velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹. is always tangentialIn the direction of the tangent to the circular path — the direction of the instantaneous velocity. to the circular path at any instant.
Worked Example
A disc rotates at 3000 rpm. Calculate the linear speed of a point 0.12 m from the centre.
Show Solution
1
$Convert rpm to rad s^{-1}
\omega = 3000 \times \frac{2\pi}{60} = 100\pi rad s^{-1} = 314 rad s^{-1}.$
2
$Apply v = r\omega: v = 0.12 \times 314 = 37.7 m s^{-1}.$
3
$$v \approx 38$ $m s^{-1}$.$
Answer
$$v \approx 38$ $m s^{-1}$$