## College Physics (4th Edition)

(a) The work done on the flywheel is $5524~J$ (b) The applied torque on the flywheel is $29.3~N \cdot m$
(a) We can express the final angular velocity in units of rad/s: $120~rpm \times \frac{2\pi~rad}{1~rev}\times \frac{1~min}{60~s} = 4\pi~rad/s$ We can find the final rotational kinetic energy of the flywheel: $KE_{rot} = \frac{1}{2}I~\omega^2$ $KE_{rot} = \frac{1}{2}MR^2~\omega^2$ $KE_{rot} = \frac{1}{2}(182~kg)(0.62~m)^2~(4\pi~rad/s)^2$ $KE_{rot} = 5524~J$ Since the change in rotational kinetic energy of the flywheel is $5524~J$, the work done on the flywheel is $5524~J$ (b) We can find the angular displacement of the flywheel: $\Delta \theta = \omega_{ave}~t$ $\Delta \theta = \frac{\omega_f-\omega_0}{2}~t$ $\Delta \theta = \frac{4\pi~rad/s-0}{2}~(30.0~s)$ $\Delta \theta = 188.5~rad$ We can find the applied torque on the flywheel: $\tau \Delta \theta = Work$ $\tau = \frac{Work}{\Delta \theta}$ $\tau = \frac{5524~J}{188.5~rad}$ $\tau = 29.3~N \cdot m$ The applied torque on the flywheel is $29.3~N \cdot m$.