Physics › Kinematics ›

Angular Velocity and Acceleration

Angular velocity and acceleration, including centripetal and coriolis acceleration.

View other versions (4)

Contents

Overview
Angular Velocity
Angular Acceleration
Page Comments

Overview

Key facts

Angular velocity is the rate of change of the position-specific angle $\theta$ with respect to time:

$\omega = \frac{d\theta}{dt}$

Angular acceleration is the rate of change of angular velocity with respect to time:

$\alpha = \frac{d\omega}{dt}=\frac{d^2\theta}{dt^2}$

The centripetal acceleration can be defined as:

$a_c = r\omega^2$

The Coriolis component of acceleration, or compound supplementary acceleration can be defined as:

$a_C = 2v\omega$

Angular velocity and acceleration are vector quantities that describe an object in circular motion. When an object, like a ball attached to a length of string, is rotated at a constant angular velocity, it is said to be in uniform circular motion. However, if the object is rotated at increasing or decreasing speeds, it can be said to be in a state of angular acceleration. The acceleration can have a centripetal component (acting inwards toward the axis of rotation) or a Coriolis component(acting perpendicular to the direction of velocity and the axis of rotation).

Angular Velocity

In order to define angular velocity, consider a particle $P$ moving in a $XOY$ reference plane, as diagramed in Figure 1. Let $x$ be the projection of $P$ on the $OX$ axis, $y$ the projection of $P$ on the $OY$ axis, $r$ the length of $OP$ , and $\theta$ the $\angle{POX}$ angle.

We can then write that:

$x = r\cos\theta$

$y = r\sin\theta$

By differentiating these equations with respect to time, we have:

$\dot x = \frac{dx}{dt} = \dot r \cos\theta - r\dot \theta \sin\theta$

(1)

$\dot y = \frac{dy}{dt} = \dot r \sin\theta + r\dot \theta \cos\theta$

(2)

We can write the tangential component of velocity, $v_T$ , which is the component of velocity perpendicular to $OP$ , as:

$v_T = \dot y \cos\theta - \dot x \sin\theta$

and, by using the expressions of $\dot x$ and $\dot y$ from equations (1) and (2) respectively, we obtain:

$v_T = r \dot \theta$

(3)

We define the angular velocity $\omega$ as the rate of change of $\theta$ with respect to time:

$\omega = \frac{d\theta}{dt}$

(4)

Using (4) in (3) we get that:

$v_T = r\omega$

from which the angular velocity becomes:

$\omega = \frac{v_T}{r}$

Angular Acceleration

In order to define the acceleration, we first have to calculate $\ddot x$ and $\ddot y$ . To do this we differentiate equations (1) and (2) with respect to time, when we obtain that:

$\ddot x = \ddot r \cos\theta - \dot r \sin\theta - r\ddot \theta^2 \cos\theta - r \ddot \theta \sin\theta$

(5)

$\ddot y = \ddot r \sin\theta + \dot r \dot \theta \cos\theta + r \dot \theta^2 \sin\theta +r \ddot \theta \cos\theta$

(6)

We can write the radial component of acceleration, $a_r$ , which is the component of acceleration in the direction of $OP$ , as:

$a_r = \ddot x \cos\theta + \ddot y \sin\theta$

and, by using the expressions of $\ddot x$ and $\ddot y$ from equations (5) and (6) respectively, we get:

$a_r = \ddot r - r \dot \theta^2$

(7)

We can also write (7), by using (4), as:

$a_r = \dot v - r\omega^2$

where $\dot v$ is the rate of change of velocity, while the $r\omega^2$ term is called the Centripetal Acceleration ( $a_c$ ).

The tangential component of acceleration on the other hand, $a_T$ , can be written as:

$a_T = \ddot y \cos\theta - \ddot x \sin\theta$

and, again by replacing $\ddot x$ and $\ddot y$ from equations (5) and (6) respectively, we obtain:

$a_T = r \ddot \theta + 2 \dot r \dot \theta$

(8)

We define the angular acceleration $\alpha$ as the rate of change of the angular velocity $\omega$ with respect to time:

$\alpha = \frac{d\omega}{dt} = \ddot \theta$

(9)

Equation (8) can also be written, by taking into account (9), as:

$a_T = r \alpha + 2v\omega$

where the $2v\omega$ term is called the compound supplementary acceleration, or the Coriolis component of acceleration $a_C$ .

Example:

[imperial]

Example - Linear velocity

Problem

Consider that the hard disk of a computer is circular and rotates with an angular velocity of 7200 revolutions per minute.

Calculate the linear velocity (in miles per hour) of a particle which is found 2 inches away from the center of the hard disk.

Workings

The angular velocity of the hard disk expressed in SI units is:

$$\omega$ = 7200 \cdot \frac{2\pi}{60} = 240\pi \; \frac{rad}{s}$

As the linear velocity can be defined as $v=\omega r$ , we obtain the linear velocity of the particle as:

$v = 240\pi\cdot 2\ = 1507.2 \;\frac{inch}{s}$

or, by converting it in miles per hour:

Solution

$v = 85.63 \; mph$

Reference

http://www.algebralab.org

For an application of angular velocity to mechanics, also see the reference page on Velocity and Acceleration of a Piston .