The flywheel of an engine is rotating at$\mathbf{25}\mathbf{.}\mathbf{0}\raisebox{1ex}{$\mathbf{r}\mathbf{a}\mathbf{d}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{s}$}\right.$ . When the engine is turned off, the flywheel slows at a constant rate and stops in$\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{s}$ . Calculate

(a) The angular acceleration of the flywheel,

(b) The angle through which the flywheel rotates in stopping, and

(c) The number of revolutions made by the flywheel in stopping

1. Flywheel rotation is $25\text{rad}/\text{sec}$
2. Flywheel stops in $20\text{sec}$

Here, initial and final angular velocity and time are given. Using the angular kinematic equations, find angular acceleration and angle. Convert the angle into revolutions by using the relationship between the radians traced in one rotation.

Formulae are as follow:

$\omega ={\omega }_0+\alpha \text{t}$

$\theta =\frac{\omega +{\omega }_0}{2}\times \text{t}$

Where,

t is time, $\omega , {\omega }_0$ are final and initial angular velocities, is angular acceleration and $\theta$ is displacement.

Using the angular kinematic equation,

$$\begin{gathered} \omega ={\omega }_0+\alpha \text{t} \\ 0=25+\alpha \times 20 \\ \alpha =-1.25\text{rad}/s^2 \end{gathered}$$

Hence, angular acceleration of the flywheel is $\text{- 1.25}\raisebox{1ex}{$\text{rad}$}\!\left/ \!\raisebox{-1ex}{${\text{s}}^{\text{2}}$}\right.$

The angular displacement can be found as,

$$\begin{gathered} \theta =\left(\frac{\omega +{\omega }_0}{2}\right)\text{t} \\ \theta =\frac{25}{2}\times 2 \\ \theta =250\text{rad} \end{gathered}$$

Hence, angle through which the flywheel rotates is $250\text{rad}$

Convertangular displacement into number of rotations as,

$$\begin{gathered} \text{n}=250\text{rad}\times \frac{1\text{rev}}{2\times \pi } \\ \text{n}=39.8\text{rev} \end{gathered}$$

Hence, number of revolutions made by the flywheel is$39.8\text{rev}$.

Therefore, basic formulas of rotational kinematics can be used to find angular acceleration, angle, and number of revolutions.