Answer:
Approximately [tex]232[/tex] radians per second.
Explanation:
At an angular speed of [tex]\omega[/tex], the rotational kinetic energy of the flywheel would be:
[tex]\displaystyle (\text{KE}) = \frac{1}{2}\, I\, \omega^{2}[/tex],
Where [tex]I[/tex] is the moment of inertia. Rearrange this equation to find an expression for angular speed [tex]\omega[/tex]:
[tex]\displaystyle \omega = \sqrt{\frac{2\, (\text{KE})}{I}}[/tex].
If the flywheel is to be considered as a disk of uniform density revolving around its diameter, the moment of inertia would be:
[tex]\displaystyle I = \frac{1}{2}\, m\, r^{2}[/tex],
Where:
Given that [tex](\text{KE}) = 3.2 \times 10^{7}\; {\rm J}[/tex], substitute the expression for moment of inertia into the expression for angular speed [tex]\omega[/tex] to find the value of angular speed:
[tex]\begin{aligned} \omega &= \sqrt{\frac{2\, (\text{KE})}{I}} \\ &= \sqrt{\frac{2\, (\text{KE})}{\displaystyle \frac{1}{2}\, m\, r^{2}}} \\ &= \sqrt{\frac{2\, (3.2 \times 10^{7}\; {\rm J})}{\displaystyle\frac{1}{2}\, (303\; {\rm kg})\, (2.8\; {\rm m})^{2}}} \\ &\approx 232\; {\rm s^{-1}}\end{aligned}[/tex].
In other words, the angular speed of the flywheel should be approximately [tex]232[/tex] radians per second.