Angular Velocity – Definition, Unit and Formula

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The angular velocity of an object in circular motion is defined as the time rate of change of its angular displacement. It is generally denoted by $\omega$ (omega) and its unit is radians per second (denoted by rad.s$^{-1}$). Its dimensional formula is $[M^0L^0T^{-1}]$.

Table of Contents

Angular Velocity Formula

The angular velocity ($\omega$) formula is

$ \displaystyle \omega =\frac{{\text{angle traced}}}{{\text{time taken}}}$

i.e. $\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle \omega =\frac{{\Delta \theta }}{{\Delta t}}\approx \frac{{d\theta }}{{dt}}}}$

Consider a point object moving along a circular path, with a center (i.e., axis of rotation) at $O$. Let the object move from $P$ to $Q$ in the time interval $\Delta t$, where $\angle POQ = \Delta \theta$.

Now, angular velocity is given by:

\[
\omega = \frac{\text{angle traced}}{\text{time taken}} = \frac{\Delta \theta}{\Delta t} \approx \frac{d\theta}{dt}
\]

$\omega$ (omega) is a vector quantity. Its direction is the same as that of $\Delta \theta$.

Average Angular Velocity

Suppose that a rotating body is at angular position $\displaystyle {{\theta }_{1}}$ at time $\displaystyle {{t}_{1}}$ and at angular position $\displaystyle {{\theta }_{2}}$ at time $\displaystyle {{t}_{2}}$. The average angular velocity in this time interval $\displaystyle {\Delta t}$ from time $\displaystyle {{t}_{1}}$ to $\displaystyle t{}_{2}$ is

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\omega =\dfrac{{{{\theta }_{2}}-{{\theta }_{1}}}}{{{{t}_{2}}-t{}_{1}}}=\dfrac{{\Delta \theta }}{{\Delta t}}}}$

Instantaneous Angular Velocity

The instantaneous angular velocity $\displaystyle \omega $, with which we shall be most concerned, is the limit of the ration of the above equation as $\displaystyle {\Delta t}$ is made to approach zero. Thus

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\omega =\underset{{\Delta t\to 0}}{\mathop{{\lim }}}\,\dfrac{{\Delta \theta }}{{\Delta t}}=\dfrac{{d\theta }}{{dt}}}}$

Angular Velocity Direction

For anticlockwise rotation of the point object on the circular path, the direction of $\omega$, according to the Right-Hand Rule, is along the axis of the circular path directed upwards as shown in the figure.

For clockwise rotation of the point object on the circular path, the direction of $\omega$ is along the axis of the circular path directed downwards as shown in the figure above.

It is important to note that nothing actually moves in the direction of $\displaystyle \overrightarrow{\omega }$. The direction of $\omega$ simply represents that the rotational motion is taking place in a plane perpendicular to it.

Relation between Angular Velocity, Frequency and Time Period

Consider a point object describing a uniform circular motion with frequency $\displaystyle \nu $ and time period T. When the object completes one revolution, the angle traced at its axis of circular motion is $\displaystyle 2\pi $ radians. Therefore, when time $\displaystyle t=T,$ $\displaystyle \theta =2\pi $ radians.

Therefore, angular velocity ($\omega$) is

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol{\displaystyle \omega =\frac{\theta }{t}=\frac{{2\pi }}{T}=2\pi \nu}}$

Angular Velocity of Earth Around its Own Axis

The time period of Earth revolution about its own axis is 24 hours i.e. 86400 seconds.

$\displaystyle \begin{array}{l}\because \omega =\dfrac{{2\pi }}{T}\\\therefore \omega =\dfrac{{2\pi }}{{86400}}\\\Rightarrow \omega =7.27\times {{10}^{{-5}}}\text{ rad/s}\end{array}$

Example: Which is greater: the angular velocity of the hour hand of a watch or the angular velocity of earth around its own axis?

Solution:

For the hour hand of the watch, time period of revolution,

$\displaystyle T=12\text{ hours}$

As,

$\displaystyle \begin{array}{l}\omega =\dfrac{{2\pi }}{T}=\dfrac{{2\pi }}{{12}}\\\Rightarrow \omega =\dfrac{\pi }{6}\text{ rad}\text{./h}\end{array}$

For earth, time period of revolution = 24h

So, $\displaystyle {\omega }’=\dfrac{{2\pi }}{{24}}=\dfrac{\pi }{{12}}$

$\displaystyle \therefore \omega >{\omega }’$

Relation between Linear Velocity and Angular Velocity

The relation between linear velocity and angular velocity in scalar form is given by

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle v=\omega r}}$

In vector notation it is given by

$\displaystyle \overrightarrow{v}=\overrightarrow{\omega }\times \overrightarrow{r}$

Consider a point object moving with a uniform angular velocity $\omega$ and linear speed $\displaystyle v$, on a circulr path of radius $\displaystyle r$ with center at O. Let at any time $\displaystyle t$, the object e at Q, where $\displaystyle \angle POQ=\theta $, and $\displaystyle \overrightarrow{{OQ}}=\overrightarrow{r}+\Delta \overrightarrow{r}$. It means an object describes an arc PQ of length $\displaystyle \Delta l$ at time interval $latex \displaystyle \Delta t$.

$\displaystyle \therefore v=\frac{{\Delta l}}{{\Delta t}}$ or $\displaystyle \Delta l=v\Delta t$

and

$\displaystyle \omega =\frac{{\Delta \theta }}{{\Delta t}}$ $\displaystyle \Delta \theta =\omega \Delta t$

Since, $\displaystyle \text{angle }=\text{ }\frac{{\text{arc}}}{{\text{radius}}}$

$\displaystyle \therefore \Delta \theta =\frac{{\Delta l}}{r}$ or $\displaystyle \omega \Delta t=\frac{{v\Delta t}}{r}$

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle v=\omega r}}$