A hoop of mass M = 4 kg and radius R = 0.4 m rolls without slipping down a hill, as shown in the figure. The lack of slipping means that when the center of mass of the hoop has speed v, the tangential speed of the hoop relative to the center of mass is also equal to vCM, since in that case the instantaneous speed is zero for the part of the hoop that is in contact with the ground (v − v = 0). Therefore, the angular speed of the rotating hoop is ù = vCM/R.
(a) The initial speed of the hoop is vi = 2 m/s, and the hill has a height h = 3.7 m. What is the speed vf at the bottom of the hill?
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