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 v_CM, 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 omega = v_CM/R. (a) The initial speed of the hoop is v_j = 3 m/s, and the hill has a height h = 4.2 m. What is the speed v_f at the bottom of the hill? v_f = 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 = 1.9 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)). v_f = m/s

Explanation / Answer

a) using the conservation of energy

Initial PE + Initial KE + Initial Rotational KE = Final KE + Final rotational KE

mgh + 1/2 mv^2 + 1/2 Iw^2 = 1/2 mVf^2 + 1/2 I wf^2

I = mr^2 , w= v/r

mgh + 1/2 mv^2 + 1/2 ( mr^2) ( v^2/r^2) = 1/2 mvf^2 + 1/ 2 ( m) ( vf^2)

gh + 1/2 ( v^2) + 1/2 ( v^2) = vf^2

gh + vi^2 = vf^2

vf = sqroot (9.8x 4.2 +9) = 7.08 m/s apprx

b) ( M+m) gh + 1/2 (M+m) vi^2 + 1/2 Iwi^2 = 1/2 (M+m)vf^2 + 1/2 I wf^2 ( as we don't have the radius of hub, we can assume that's it's too small and can be neglected)

Therefore I = MR^2 ( apprx)

( 5.9) (9.8) ( 4.2) + 1/2 ( 5.9) ( 9) + 1/2 ( 4) ( 9) = 1/2 ( 5.9) ( vf^2) + 1/2 ( 4) vf^2

242.844 +26.55 + 18 = 4.95vf^2

vf = 7.619 m/s apprx

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