In the ride shown, a person A sits in a seat that is attached via a cable of len

Question

In the ride shown, a person A sits in a seat that is attached via a cable of length L to a freely moving trolley B of mass mB. The total mass of the person and the seat is mA . The trolley is constrained by the beam to move only in the horizontal direction. The system is released from rest at the angle theta = theta0 and it is allotted to swing in the vertical plane. Neglect the mass of the cable and treat the person and the scat as a single particle. Determine the velocities of the trolley and the rider the first time that# = 0 degree . Evaluate your solution for WA = 98 lb, WB = 24 lb, L = 13 ft, and theta0 = 75degree. The velocity of the trolley is ft/s and velocity of the rider is ft/s

Explanation / Answer

for momentum conservation

MaVa+MbVb=0

Vb=MaVa/Mb.....................1
for energy conservation at =0
1/2 MaVa2 + 1/2 MbVb2 = MagL(1-cos) ...............2

solving

Va=-(2gl(1-cos)Mb/(Ma+Mb))i

Vb=(Ma/Mb)(2gl(1-cos)Mb/(Ma+Mb))i

For given Values

Va=-11.04 i ft/s

Vb=45.09 i ft/s

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