Angular Acceleration – Definition, Unit and Formula

Angular acceleration of an object in circular motion is defined as the time rate of change of its angular velocity. It is generally denoted by $\boldsymbol {\displaystyle \alpha}$.

Angular Acceleration Formula

The formula of angular acceleration is given by

$\bbox[15px, #e4e4e4, border: 2px solid #000000]{\boldsymbol {\displaystyle \alpha =\frac{{d\omega }}{{dt}}}}$

Its S.I. Unit is $\boldsymbol {\displaystyle rad\cdot {{s}^{{-2}}}}$ and its dimensional formula is $\displaystyle [{{M}^{0}}{{L}^{0}}{{T}^{{-2}}}]$

Average Angular Acceleration

Consider an object moving in a circular path of radius $\displaystyle r$ with variable angular acceleration.

If $\displaystyle {{\omega }_{1}}$ and $\displaystyle {{\omega }_{2}}$ be the angular velocities at times $\displaystyle t{}_{2}$ and $\displaystyle t{}_{1}$, respectively, then the average angular acceleration of the rotating body in the interval from $\displaystyle {{t}_{1}}$ to $\displaystyle {{t}_{2}}$ is given by

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

Instantaneous Angular Acceleration

Suppose the angular acceleration is not constant and keeps on changing. In that case, the instantaneous angular acceleration $\boldsymbol {\displaystyle \alpha}$, with which we shall be most concerned, is the limit of this quantity as $\displaystyle {\Delta t}$ is made to approach zero. Thus,

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

Relation between Linear Acceleration and Angular Acceleration

We know that linear velocity $\displaystyle v$ is related with angular velocity $\displaystyle \omega $ by the relation as

$\displaystyle v=\omega r$

Differentiating both sides w.r.t. time, we have

$\displaystyle \frac{{dv}}{{dt}}=\frac{d}{{dt}}\left( {\omega r} \right)=\left( {\frac{{d\omega }}{{dt}}} \right)r$

Since, $\displaystyle r$ is constant in circular motion and $\displaystyle \frac{{d\omega }}{{dt}}=\alpha $, therefore

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

where $ \displaystyle a$ is the linear acceleration and $\displaystyle \alpha $ is the angular acceleration.

Example: During the analysis of a helicopter engine you determine that the rotor’s velocity changes from 320 rev/min to 225 rev/min in 1.50 min as the rotor is slowing to a stop.

a. What is the average angular acceleration of the rotor blades during this interval?

Solution:

Since, $\displaystyle \overline{\alpha }=\frac{{{{\omega }_{2}}-{{\omega }_{1}}}}{{\Delta t}}$

$\displaystyle \begin{array}{l}\Rightarrow \overline{\alpha }=\dfrac{{225-320}}{{1.50}}rev/{{\min }^{2}}\\\Rightarrow \overline{\alpha }=-63.3\text{ }rev/{{\min }^{2}}\end{array}$

b. Assume the angular acceleration $\displaystyle \alpha $ is constant at this average value How long will the rotor blades take to stop from their initial angular velocity of 320 rev/min?

Solution:

$\displaystyle \begin{array}{l}\because \omega ={{\omega }_{\circ }}+\alpha t\\\therefore t=\dfrac{{\omega -{{\omega }_{\circ }}}}{\alpha }\\\Rightarrow t=\dfrac{{0-320}}{{-63.3}}\\\Rightarrow t=5.1\text{ }\min \end{array}$