Answer:
[tex]I=12.5\ kg.m^2[/tex]
Explanation:
Given that
Stored energy ,[tex]E=1\times 10^6[/tex]
Angular speed ,[tex]\omega =400\ rad/s[/tex]
Lets take moment of inertia of the flywheel = I
As we know that stored energy in the flywheel is given as
[tex]E=\dfrac{1}{2}\omega^2 I[/tex]
[tex]I=\dfrac{2\ttimes E}{\omega^2}[/tex]
Now by putting the values in the above equation we get
[tex]I=\dfrac{2\times 1\times 10^6}{400^2}\ kg.m^2[/tex]
[tex]I=12.5\ kg.m^2[/tex]
Therefore the moment of inertia of the flywheel will be [tex]I=12.5\ kg.m^2[/tex]
The answer will be C.
[tex]I=12.5\ kg.m^2[/tex]
