To solve the exercise it is necessary to apply the concepts related to kinetic energy by rotation and the moment of rotational inertia.
Rotational energy is defined as
[tex]KE = \frac{1}{2}I\omega^2[/tex]
Where,
I = Inertia moment
[tex]\omega =[/tex] Angular velocity
While the Rotational inertia of each blade is given as
[tex]I = \frac{1}{3}ml^2[/tex]
Where,
m= mass
l = length
We have also that the assembly of the motor has three blade, then the total rotational inertia is
[tex]I_m = 3*\frac{1}{3}ml^2[/tex]
[tex]I_m = ml^2[/tex]
Replacing with our values
[tex]I_m = 45*4^2[/tex]
[tex]I_m = 720Kgm^2[/tex]
We have the angular velocity in rev per minute then in rad per second is
[tex]\omega= 240rpm (\frac{2\pi rad}{1rev})(\frac{1min}{60s})[/tex]
[tex]\omega=25.13rad/s[/tex]
Then the total Kinetic Energy at the system is
[tex]KE = \frac{1}{2}720*(25.13)^2[/tex]
[tex]KE= 2.27*10^5J[/tex]
Therefore the total Kinetic Energy at the system is [tex]2.27*10^5J[/tex]
