Answer
$12.27\ kg.m^2$
Work Step by Step
It is given that;
Kinetic energy of wheel KE = 24400 J
Angular velocity $\omega = 602\ \frac{rev}{min} $
Next,
We convert Angular velocity from rev/min to rad/sec:
$\omega = 602 \frac{rev}{min}\times \frac{2\pi \frac{rad}{rev}}{60 \frac{s}{min}}=63.04\ \frac{rad}{s}$
Therefore, relation between Kinetic energy and momentum of inertia is:
$KE= \frac{1}{2}I\omega^2$
$I=\frac{2KE}{\omega^2}$
$I=\frac{2\times 24400 J}{(63.04\ rad/s)^2}$
$I=12.27\ kg.m^2$
