Cray My Blackboard-WL Bla x y VA Homework 18-SP17 x C Two Disks Are Initially At
ID: 1550482 • Letter: C
Question
Cray My Blackboard-WL Bla x y VA Homework 18-SP17 x C Two Disks Are Initially At x C Secure h ps://www.webassign.n /web/Student/Assignment-Responses/submit?dep 5321260 4. 012 points I Previous Answers MI4 9.3.033. My Notes A hoop of mass M 2 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 since in that case the instantaneous speed is zero for the part of the hoop that is in CM contact with the ground (v 0). Therefore, the angular speed of the rotating hoop is a) vcM/R. v of rim center of mass relative to center of mass (a) The initial speed of the hoop is vi 4 m/s, and the hill has a height h 3.7 m. What is the speed vf at the bottom of the hill? x m/s Vf (b) Replace the hoop with a bicycle wheel whose rim has mass M 2 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 m/s Additional Materials Section 9.3 634 PM Ask me anything 3/26/2017Explanation / Answer
initial energy at top Ei = (M)*g*h + (1/2)*I*w^2 + (1/2)*M*V^2 = (M)*g*h + (1/2)*M*R^2*wi^2 + (1/2)*M*V^2
R*wi = vi
Ei = (M)*g*h + (1/2)*M*Vi^2 + (1/2)*M*Vi^2
final energy at the bottom Ef = (1/2)*I*w^2 + (1/2)*M*V^2
I = moment of inertia about the point in contact with the ground = M*R^2
w = Vcm/R
from energy conservation
Ef = Ei
(1/2)*M*R^2*Vcm^2/R^2 + (1/2)*M*Vcm^2 = M*g*h + (1/2)*M*Vi^2 + (1/2)*M*Vi^2
Vcm = sqrt(gh + vi^2)
Vcm = sqrt(9.8*3.7+16) = 7.23 m/s <<<===============ANSWER
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initial energy at top Ei = (M+m)*g*h+ (1/2)*M*Vi^2 + (1/2)*M*Vi^2 + (1/2)*m*Vi^2
final energy at the bottom Ef = (1/2)*I*w^2 + (1/2)*M*Vf^2 + (1/2)*m*Vf^2
I = moment of inertia about the point in contact with the ground = M*R^2 + m*R^2
w = Vcm/R
from energy conservation
Ef = Ei
(1/2)*M*R^2*Vf^2/R^2 + (1/2)*M*Vf^2 + (1/2)*m*Vf^2 = (M+m)*g*h+ (1/2)*M*Vi^2 + (1/2)*M*Vi^2 + (1/2)*m*Vi^2
(1/2)*(2M+m)*Vf^2 = (M+m)*g*h + (1/2)*M*Vi^2 + (1/2)*M*Vi^2 + (1/2)*m*Vi^2
(1/2)*((2*2)+1.5)*Vf^2 = (2+1.5)*9.8*3.7 + ((1/2)*2*4^2) + ((1/2)*2*4^2) +( (1/2)*1.5*4^2)
Vcm = 7.88m/s <<<===============ANSWER