Contractor 6

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# Contractor 6 - PowerPoint PPT Presentation

## Contractor 6

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1. Contractor 6 Homework #2 Cody Bagnall Sarah Canterbury Nallely Davila Keith Jackson AinurKushaliyeva Jose A. Perez Zach Taylor

2.  = 1.5 rpm r=100 m cable Spacecraft segment M = 4.5 MT Problem 2 In this problem, we explore an approach to providing artificial gravity for the lengthy mission to the asteroid. We assume that the spacecraft has two habitable sections, each of mass 4.5MT. During the cruise portions of the flight, the two sections are separated and a 200m long, connecting cable is deployed. The cable is stretched straight and the attitude control jets are used to spin up the whole assemblage to an angular velocity of  =1.5 rpm with the center of rotation midway between the two segments. The final arrangement is as illustrated in the figure above. Assume that the cable is 1 cm in diameter with elastic modulus and volumetric density:

3. Part A What is the magnitude of the artificial gravity experienced by the crew? If a crew member walks at 2 miles per hour, what is the maximum value of the Coriolis acceleration? How fast would a crew member have to run anti-spinward to become “weightless”?

4. FBD Solution A) Centripetal acceleration Equal and opposite Force Tangential Velocity V [m/s] R Coriolis Acceleration To be “weightless”, Velocity of module relative to the crew’s velocity =0

5. Part B Crew motion or equipment vibration will excite the axial vibration modes of the cable. For the values of the parameters given above, and neglecting the mass of the cable, what is the lowest natural frequency involving only axial deformation of the cable/spacecraft system? Do you think this will degrade crew comfort?

6. Solution E- Elastic Modulus A-Cable cross-sectional area D-Cable’s Diameter L-Cable’s length m-Module’s mass Cable Axial Stiffness, Natural Frequency, This could affect crew comfort because it’s in the upper limit of 0.8Hertz.

7. Part C There is another low-frequency oscillation mode that involves the rocking of the crew modules about their centers of mass as illustrated in the following diagram. Assume that the distance, , from the center of mass to the point of cable attachment is 6m and that the moment of inertia, I, is 72,000 kg*m^2. Considering motions that leave the center of rotation stationary, and again neglecting the mass of the cable, calculate the oscillation frequency of this “pendulum” motion. What do you think the impact on crew comfort might be?

8. Torque due to CM τ – Torque due to center of mass. α – Angular acceleration. δ – Distance to CM. –– Acceleration. Solution δ NEWTON’S SECOND LAW OF ROTATION EQUATION OF MOTION Small angle approximation yields: Since maximum sensitivity occurs at 0.167Hertz, it is likely to make crew member nauseous.