Answer
The motor needs to do $0.0512~J$ of work.
Work Step by Step
We can find the angular velocity $\omega$:
$\omega = \frac{v}{r} = \frac{1.20~m/s}{0.020~m} = 60~rad/s$
We can find the rotational kinetic energy:
$KE_{rot} = \frac{1}{2}I~\omega^2$
$KE_{rot} = \frac{1}{2}(\frac{1}{2})MR^2~\omega^2$
$KE_{rot} = \frac{1}{4}(0.0158~kg)(0.060~m)^2(60~rad/s)^2$
$KE_{rot} = 0.0512~J$
Since the change in energy of the disk is $0.0512~J$, the motor needs to do $0.0512~J$ of work.
