Answer
The work done by the torque is $~~1.25\times 10^6~J$
Work Step by Step
We can find the angular speed:
$\omega = (320~rev/min)(\frac{2\pi~rad}{1~rev})(\frac{1~min}{60~s}) = 33.51~rad/s$
We can find the rotational kinetic energy:
$K = \frac{1}{2}I~\omega^2$
$K = (\frac{1}{2})~(\frac{1}{3}ML^2)~(\omega^2)$
$K = (\frac{1}{6})(110~kg)(7.80~m)^2~(33.51~rad/s)^2$
$K = 1.25\times 10^6~J$
The work done by the torque is $~~1.25\times 10^6~J$
