Answer
The work done on the flywheel is $~~-29.9~J$
Work Step by Step
In part (b), we found that the initial angular velocity is $~~\omega_i = 21.429~rad/s$
In part (b), we found that the final angular velocity is $~~\omega_f = 5.714~rad/s$
We can find the initial rotational kinetic energy of the flywheel:
$K_i = \frac{1}{2}I\omega_i^2$
$K_i = \frac{1}{2}(0.140~kg~m^2)(21.429~rad/s)^2$
$K_i = 32.144~J$
We can find the final rotational kinetic energy of the flywheel:
$K_f = \frac{1}{2}I\omega_f^2$
$K_f = \frac{1}{2}(0.140~kg~m^2)(5.714~rad/s)^2$
$K_f = 2.285~J$
The work done on the flywheel is equal to the change in rotational kinetic energy:
$Work = K_f-K_i$
$Work = 2.285~J-32.144~J$
$Work = -29.9~J$
The work done on the flywheel is $~~-29.9~J$
