The moment of inertia of the flywheel is 2.63 kg-[tex]m^{2}[/tex]
It is given that,
The maximum energy stored on the flywheel is given as
E=3.7MJ= 3.7×[tex]10^{6}[/tex] J
Angular velocity of the flywheel is 16000[tex]\frac{rev}{min}[/tex] = 1675.51[tex]\frac{rad}{sec}[/tex]
So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :
E = [tex]\frac{1}{2}[/tex][tex]Iw^{2}[/tex]
By rearranging the equation:
I = [tex]\frac{2E}{w_{2} }[/tex]
I = 2.63 kg-[tex]m^{2}[/tex]
Thus the moment of inertia of the flywheel is 2.63 kg-[tex]m^{2}[/tex].
Learn more about moment of inertia here;
https://brainly.com/question/13449336
#SPJ4
