A hoop of mass M = 3 kg and radius R = 0.4 m rolls without slipping down a hill,

Question

A hoop of mass M = 3 kg and radius R = 0.4 m rolls without slipping down a hill, as shown in the figure below. The lack of slipping means that when they enter of mass of the hoop has speed v, the tangential speed of the hoop relative to the center of mass is also equal to nu_cm, since in that 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 omega = v_CM/R. The initial speed of the hoop is v_1 = 4 m/s, the hill has a height h = 3.6 m. What is the speed v_t at the bottom of the hill? v_y = m/s Replace the hoop with a bicycle wheel whose rim has mass M = 3 kg and radius R = 0.4 m, and whose hub has mass m = 1.5 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)).

Explanation / Answer

PE + KE at the top of the hill = KE + PE at the bottom of the hill

h = 0 , PE = 0

At the top of the hill KE = 1/2 m V2 + 1/2 I w2

I = m R2 = 3 * 0.42 = 0.48

a) KE = 1/2 * 3 * 42 + 1/2 * 0.48 * (4 / 0.4)2 = 48 J

PE = m * g * h = 3 * 9.8 * 3.6 = 105.84 J

PE + KE = 153.84 J [at the top of the hill ]

At the bottom of the hill 153.84 J = 1/2 * 3 * V2 + 1/2 * 0.48 * (V/0.4)2

Vf2 (1.5 + 1.5) = 153.84

Vf = 7.16 m/s

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b) Mass = 3 + 1.5 = 4.5 kg

PE + KE = 230 J

Vf2(2.25 + 2.25) = 230

vf = 7 m/s

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