Answer
The fraction of the total kinetic energy that is the rotational kinetic energy of the wheels is $0.0194$
Work Step by Step
We can find an expression for the total rotational kinetic energy of the two wheels:
$KE_{rot} = 2\times \frac{1}{2}I\omega^2$
$KE_{rot} = (I)(\frac{v}{r})^2$
$KE_{rot} = (0.080~kg~m^2)(\frac{v}{0.32~m})^2$
$KE_{rot} = (0.78125~kg)~v^2$
We can find an expression for the translational kinetic energy:
$KE_{tr} = \frac{1}{2}mv^2$
$KE_{tr} = \frac{1}{2}(79~kg)~v^2$
$KE_{tr} = (39.5~kg)~v^2$
We can find the fraction of the total kinetic energy that is the rotational kinetic energy of the wheels:
$\frac{KE_{rot}}{KE} = \frac{KE_{rot}}{KE_{tr}+KE_{rot}}$
$\frac{KE_{rot}}{KE} = \frac{(0.78125~kg)~v^2}{(39.5~kg)~v^2+(0.78125~kg)~v^2}$
$\frac{KE_{rot}}{KE} = 0.0194$
The fraction of the total kinetic energy that is the rotational kinetic energy of the wheels is $0.0194$.
