Answer
The thickness of the wheel should be 5.26 cm.
Work Step by Step
We can find the angular speed at $t = 8.00~s$:
$\omega = \alpha ~t$
$\omega = (3.00~rad/s^2)(8.00~s)$
$\omega = 24.0~rad/s$
The kinetic energy is required to be 800 J. We can find the required mass $M$.
$KE = 800~J$
$\frac{1}{2}I\omega^2 = 800~J$
$\frac{1}{2}(\frac{1}{2}MR^2)\omega^2 = 800~J$
$M = \frac{(4)(800~J)}{R^2~\omega^2}$
$M = \frac{3200~J}{(0.250~m)^2~(24.0~rad/s)^2}$
$M = 88.89~kg$
We can find the thickness $x$ of the wheel.
$\rho \times volume = M$
$\rho\times x\times Area = M$
$x = \frac{M}{\rho ~\pi ~R^2}$
$x = \frac{88.89~kg}{(8600~kg/m^3)(\pi)(0.250~m)^2}$
$x = 0.0526~m = 5.26~cm$
The thickness of the wheel should be 5.26 cm.
