## College Physics (4th Edition)

(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.
(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.