Answer:
[tex]\dot n = 748178.306\,rpm[/tex]
Explanation:
A flywheel stores mechanical energy in the form of rotational kinetic energy:
[tex]K = \frac{1}{2}\cdot I \cdot \omega^{2}[/tex]
The expression is simplified by considering the flywheel as a solid disk:
[tex]K = \frac{1}{4}\cdot m\cdot r^{2}\cdot \omega^{2}[/tex]
The final speed of the flywheel is:
[tex]\Delta E = \frac{1}{4}\cdot m \cdot r^{2}\cdot \omega^{2}[/tex]
[tex]\omega = \sqrt{\frac{4\cdot \Delta E}{m\cdot r^{2}} }[/tex]
[tex]\omega = \frac{2}{r}\cdot \sqrt{\frac{\Delta E}{m} }[/tex]
[tex]\omega = \frac{2}{0.437\,m}\cdot \sqrt{\frac{2.96\times 10^{9}\,J}{10.1\,kg} }[/tex]
[tex]\omega \approx 78349.049\,\frac{rad}{s}[/tex]
The final speed in revolutions per minute is:
[tex]\dot n = \frac{60\cdot \omega}{2\pi}[/tex]
[tex]\dot n = \frac{60\cdot (78349.049\,\frac{rad}{s} )}{2\pi}[/tex]
[tex]\dot n = 748178.306\,rpm[/tex]
