a) The speed vf at the bottom of the hill is given by the equation vf = √2gh, where g is the acceleration due to gravity (9.8 m/s2). Therefore, vf = √2(9.8)(3.7) = 10.2 m/s.
b) The speed of the bicycle wheel at the bottom of the hill is given by the equation vf = √2gh + (m/M)vCM, where m is the mass of the hub, M is the mass of the rim, and vCM is the speed of the center of mass. Therefore, vf = √2(9.8)(3.7) + (1.3/4)2 = 11.2 m/s.
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 ω = vCM/R.
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?
(b) Replace the hoop with a bicycle wheel whose rim has mass M = 4 kg and radius R = 0.4 m, and whose hub has mass m = 1.3 kg, as shown in the figure. The spokes have negligible mass. What would the bicycle wheel's speed be at the bottom of the hill? (Assume that the wheel has the same initial speed and start at the same height as the hoop in part (a)).
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