Answer
The translational kinetic energy is $~~39.2~J$
Work Step by Step
We can find vertical height that the wheel falls:
$h = (2.00~m)~sin~30.0^{\circ} = 1.00~m$
We can use conservation of energy to find the speed:
$mgh = \frac{1}{2}mv^2+\frac{1}{2}I~\omega^2$
$mv^2+I~(\frac{v}{r})^2 = 2mgh$
$v^2 = \frac{2mgh}{m+I/r^2}$
$v = \sqrt{\frac{2mgh}{m+I/r^2}}$
$v = \sqrt{\frac{(2)(10.0~kg)(9.8~m/s^2)(1.00~m)}{(10.0~kg)+0.600~kg~m^2/(0.200~m)^2}}$
$v = 2.80~m/s$
We can find the translational kinetic energy:
$K = \frac{1}{2}m~v^2$
$K = \frac{1}{2}(10.0~kg)~(2.80~m/s)^2$
$K = 39.2~J$
The translational kinetic energy is $~~39.2~J$
