Answer
The amount of energy transferred to thermal energy by friction is $~~45.9~kJ$
Work Step by Step
We can find the rotational inertia:
$I = (\frac{1}{12})(6.40~kg)(1.20~m)^2+(2)(1.06~kg)(0.600~m)^2$
$I = 1.53~kg~m^2$
We can find the initial rotational kinetic energy:
$K = \frac{1}{2}I~\omega^2$
$K = (\frac{1}{2})(1.53~kg~m^2)[(39.0~rev/s)(2\pi~rad)]^2$
$K = 45.9~kJ$
The amount of energy transferred to thermal energy by friction is $~~45.9~kJ$
