Answer
The maximum kinetic energy that can be stored in the flywheel is 73500 J.
Work Step by Step
Let's assume the radial acceleration is $3500~m/s^2$. We can find the angular speed.
$a_{rad} = \omega^2 ~r = 3500~m/s^2$
$\omega = \sqrt{\frac{3500~m/s^2}{r}}$
$\omega = \sqrt{\frac{3500~m/s^2}{1.20~m}}$
$\omega = 54.0~rad/s$
We can find the kinetic energy stored in the flywheel.
$KE = \frac{1}{2}I\omega^2$
$KE = \frac{1}{2}(\frac{1}{2}mr^2)\omega^2$
$KE = \frac{1}{4}(mr^2)\omega^2$
$KE = \frac{1}{4}(70.0~kg)(1.20~m)^2~(54.0~rad/s)^2$
$KE = 73500~J$
The maximum kinetic energy that can be stored in the flywheel is 73500 J.
