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Physics › Kinematics ›

Angular Velocity and Acceleration

Angular velocity and acceleration, including centripetal and coriolis acceleration.

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Contents

Overview
Angular Velocity
Angular Acceleration
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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

moving in a

reference plane, as diagramed in Figure 1. Let

be the projection of

on the

axis,

the projection of

on the

axis,

the length of

, 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,

, which is the component of velocity perpendicular to

, 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,

, which is the component of acceleration in the direction of

, 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 (

The tangential component of acceleration on the other hand,

, 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

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 .

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