Answer
160 hp.
Work Step by Step
For a motor to maintain a constant angular speed of 3.8 rev/s, its torque will have the same magnitude as the frictional torque. Consider magnitudes only.
$$\tau_{friction}=\tau_{motor}=I\alpha_{friction}=\frac{1}{2}MR^2(\frac{\Delta \omega}{t})$$
The motor’s power is $\tau_{motor}\omega$.
$$P_{m}= \frac{1}{2}MR^2(\frac{\Delta \omega}{t})\omega$$
$$P_{m}= \frac{1}{2}(220kg)(5.5m)^2(\frac{(3.8)(2\pi)rad/s}{16s})((3.8)(2\pi)rad/s)$$
$$P_{m}= 1.186\times10^5W\approx 160hp $$
