My Notes 4. 1/2 points | Previous Answers MI4 9.3.033 Total 1/2 Question Part 1

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My Notes 4. 1/2 points | Previous Answers MI4 9.3.033 Total 1/2 Question Part 1 2 Points Submissions Used 2/5 3/5 A hoop of mass M = 3 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 v of v of rim relative to center of mass center of mass (a) The initial speed of the hoop is vi4 m/s, and the hill has a height h 3.9 m. What is the speed vf at the bottom of the hill? Vf7.36 m/s (b) 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 = 2.1 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)) VF8.179 m/s Enter a number Additional Materia A Section 9.3

Explanation / Answer

Applying energy conservation,

PEi + KEi = PEf + KEf

KE = translational KE + rotational KE

KE = m v^2 / 2 + I w^2 / 2

{ I = m r^2 and w = v / r}

KE = m v^2 /2 + m v^2 /2 = m v^2

SO applying energy conservation,

m gh + m v0^2 = m v^2 + 0

v = sqrt(4^2 + (9.8 x 3.9))

v = 7.4 m/s .....Ans

(B) now KE = (M + m)v^2 /2 + (M R^2)(v/R^2)/2

= (M + m)v^2 /2 + M v^2 / 2

= M v^2 + m v^2 /2

(M + m)g h + (2M + m) v0^2 /2 = 0 + (2 M+ m) v^2 / 2

2(3 + 2.1)(9.8 x 3.9) + (6 + 2.1)(4^2) = (6 + 2.1) v^2

v = 8 m/s ....Ans

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