A hoop of mass M = 4 kg and radius R = 0.4 m rolls without slipping down a hill,

Question

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 = vCM/R.

(a) The initial speed of the hoop is vi = 2 m/s, and the hill has a height h = 4.3 m. What is the speed vf at the bottom of the hill?

vf = ? m/s

(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 = 1.7 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)).

vf = ? m/s

Explanation / Answer

The potential energy PE + kinetic energy KE at the top of the hill is equal to the KE + PE at the bottom of the hill plus any frictional losses on the way down.In most cases at the "bottom" of the hill means height = 0 and therefore PE = 0. Thus all of the energy is not KE. For rolling objects, the PE + KE at the top of the hill will be linear and rotational KE at the bottom of the hill
At the top of the hill KE = 1/2 *m*V^2 + 1/2 *I*w^2
I = m*R^2 = 4*0.4^2 = 0.64, = V/R

a) KE = 1/2 *4*2^2 + 1/2 *0.64*(2/0.4)^2 = 8+8=16 J PE = m*g*h = 4*9.8*4.3 = 168J
PE + KE = 16+168=184 J at the top of the hill
At the bottom of the hill 184 = 1/2 *4*V^2 + 1/2 *0.64*(V/0.4)^2
Vf^2(2 + 2) =184 => Vf = sqrt[157/4] = 6.78m/s = vf <---------------------- (a)

(b)The wheel has a hub but you haven't provided the hub's radius. The figure is not visible.All that could change is the moment of the wheel and the mass of the wheel.Assume the moment doesn't change. Mass = 4+1.7 = 5.7kg
PE + KE at hilltop is 198 J
Vf^2(5.7/2+ 2) = 198
Vf^2(2.85+ 2) = 198 => vf = 6.38m/s

calculation may be wrong but concept is right

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