The angular speed of an automobile engine is increased at a constant rate from $\mathbf{1200}\mathbf{}\raisebox{1ex}{${\displaystyle \mathbf{\text{rev}}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{min}}$}\right.$ to$\mathbf{3000}\raisebox{1ex}{${\displaystyle \text{rev}}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{\text{min}}$}\right.$ in $\mathbf{12}\mathbf{\text{s}}$. (a) What is its angular acceleration in revolutions per minute-squared? (b) How many revolutions does the engine make during this$\mathbf{12}\mathbf{\text{s}}$ interval?

Initial angular speed,${\omega }_0=1200\raisebox{1ex}{$\text{rev}$}\!\left/ \!\raisebox{-1ex}{$\text{min}$}\right.$

Final angular speed, $\omega =3000\raisebox{1ex}{$\text{rev}$}\!\left/ \!\raisebox{-1ex}{$\text{min}$}\right.$

Time,$t=12\text{\hspace{0.17em}s}$

The problem deals with the calculation of angular acceleration. It is the time rate of change of angular velocity. Also, it involves kinematic equation of motion in which the motion of an object is described at constant acceleration.

Formulae:

$\omega ={\omega }_0+\alpha t$ ..... (i)

$\theta =\frac{1}{2}\left({\omega }_0+\omega \right)t$ ..... (ii)

Here, $\alpha$is the angular acceleration.

Here, assuming the rotation is positive. Thus, using equation (i),

$$\begin{gathered} \omega ={\omega }_0+\alpha t \\ \alpha =\frac{\omega -{\omega }_0}{t} \end{gathered}$$

Substitute given values in the above equation.

$\begin{array}{rcl}\alpha & =& \frac{3000\raisebox{1ex}{$\text{rev}$}\!\left/ \!\raisebox{-1ex}{$\text{min}$}\right.\text{-12}00\raisebox{1ex}{$\text{rev}$}\!\left/ \!\raisebox{-1ex}{$\text{min}$}\right.}{{\displaystyle \raisebox{1ex}{$12\text{s}$}\!\left/ \!\raisebox{-1ex}{$60\text{min}$}\right.}}\\ & =& 9\times {10}^3\raisebox{1ex}{$\text{rev}$}\!\left/ \!\raisebox{-1ex}{${\min }^2$}\right.\end{array}$

Using equation (ii) you can find the revolution as below.

$\begin{array}{rcl}\theta & =& \frac{1}{2}({\omega }_0+\omega )t\\ & =& \frac{1}{2}\left(1200\text{rev}+3000\raisebox{1ex}{${\displaystyle \text{\hspace{0.17em}rev}}$}\!\left/ \!\raisebox{-1ex}{$\min $}\right.\right)\left(\frac{{\displaystyle 12}}{{\displaystyle 60}}\min \right)\\ & =& 4.2\times {10}^2\text{rev}\end{array}$

Hence, the revolution of an engine is$\begin{array}{rcl}& & 4.2\times {10}^2\text{rev}\end{array}$.