To solve this problem we will apply the concept of rotational kinetic energy. Once this energy is found we will proceed to find the time from the definition of the power, which indicates the change of energy over time. Let's start with the kinetic energy of the rotating flywheel is
[tex]E_r = \frac{1}{2} I\omega^2[/tex]
Here
I = moment of inertia
[tex]\omega =[/tex] Angular velocity
Here we have that,
[tex]\omega = 3000\frac{rev}{min}(\frac{2\pi rad}{1rev})(\frac{1min}{60s})[/tex]
[tex]\omega = 314.159rad/s[/tex]
Replacing the value of the moment of inertia for this object we have,
[tex]E_r = \frac{1}{2} (\frac{MR^2}{2})\omega^2[/tex]
[tex]E_r = \frac{1}{2} (\frac{2000(0.5)^2}{2})(314.159)^2[/tex]
[tex]E_r = 1.233698*10^7J[/tex]
The expression for average power is
[tex]P = \frac{E_r}{\Delta t}[/tex]
[tex]\Delta t = \frac{E_r}{P}[/tex]
[tex]\Delta t = \frac{1.233698*10^7}{20*10^3}[/tex]
[tex]\Delta t = 616.8s \approx 620s[/tex]
Therefore the correct answer is 620s.
