Answer
(a) $53,023~J$ of work is done by the motor.
(b) The motor needs to provide a torque of $~1.515\times 10^6~N \cdot m~$ to the wheel.
Work Step by Step
(a) We can find the final rotational kinetic energy of the flywheel:
$KE_{rot} = \frac{1}{2}I~\omega^2$
$KE_{rot} = \frac{1}{2}MR^2~\omega^2$
$KE_{rot} = \frac{1}{2}(1.90\times 10^6~kg)(67.5~m)^2~(3.50\times 10^{-3}~rad/s)^2$
$KE_{rot} = 53,023~J$
Since the change in rotational kinetic energy of the wheel is $53,023~J$, the work done by the motor is $53,023~J$
(b) We can find the angular displacement of the flywheel:
$\Delta \theta = \omega_{ave}~t$
$\Delta \theta = \frac{\omega_f-\omega_0}{2}~t$
$\Delta \theta = (\frac{3.50\times 10^{-3}~rad/s-0}{2})~(20.0~s)$
$\Delta \theta = 0.035~rad$
We can find the applied torque on the flywheel:
$\tau \Delta \theta = Work$
$\tau = \frac{Work}{\Delta \theta}$
$\tau = \frac{53,023~J}{0.035~rad}$
$\tau = 1.515\times 10^6~N \cdot m$
The motor needs to provide a torque of $~1.515\times 10^6~N \cdot m~$ to the wheel.
