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Fictitious forces and artificial gravity

  1. Aug 30, 2017 #1
    I'd be extremely grateful if anyone could help me with this ... its mainly part iv) that I'm stuck on but the other parts build up to it! Thanks very much!

    1. The problem statement, all variables and given/known data

    I have a wheel shaped space station of radius 100m rotating about its symmetry axis (defined to be the z axis) at angular velocity w in order to create artificial gravity.

    i) how fast does the space station have to rotate in order for artificial gravity at the rim to be the same as that on earth (10m/s^2)

    ii) show that the equations of motion for a free particle in a co-ordinate system fixed to the space station are :

    x'' = 2wy' + w^2 x
    y'' = -2wx' + w^2y
    z'' = 0

    ('' denotes second derivate wrt time, ' denotes first derivative)

    iii) show that by defining the variable u = x + iy the solution is

    u = (a + bt) e^(-iwt)

    iv) there is a shower in the space station of height 2m from the floor.
    Ignoring the initial velocity as it flows out of the shower how long does it take the water to reach the floor?

    How much is the water displaced horizontally when it reaches the floor?


    2. Relevant equations
    Coriolis force = -2w x v
    Centrifugal force = -wxwxr

    3. The attempt at a solution
    i) I just equated w^2 r to 10 and solved to get w = 0.316 rad/s
    ii) and iii) Managed to get these fine

    iv) I'm not really sure whether I am supposed to use part 3 or not here...

    To get the time I just used s = 0.5gt^2 where g = 10 and then solved to get t = 0.632 s

    To get the horizontal distance I just used that the horizontal force is 2wy'. I then subbed in
    y ' = gt and integrated twice to get x = (wgt^3)/3. subbing in for t and w I then got horizontal displacement is 26.6cm

    I feel like I have made too many approximations and should have used part 3?
     

    Attached Files:

  2. jcsd
  3. Aug 30, 2017 #2

    FactChecker

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    I don't know if this will help, but I think Part iv is a little tricky. If you totally ignore velocity at the shower head, then the water will go nowhere and will still be in place when the shower head circles back around. I think they want you to only ignore the radial velocity of the water but keep the velocity tangent to the shower head circle. In that case, the water will travel in a horizontal line in your diagram. So you can figure the distance on the horizontal line from the shower head to the floor circle and the tangential velocity at the shower head and calculate the result. I don't think you would need to do any integrations as you have.
     
  4. Aug 30, 2017 #3
    Thanks ... I think you just ignore the initial velocity. i.e. assume its starts from rest
    Have I worked out the time to hit the ground correctly ?
     
  5. Sep 1, 2017 #4

    TSny

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    The acceleration of the water relative to the station is not really constant as it falls. However, it is approximately constant over the 2 m distance, and so your answer is pretty accurate.

    But, I think they probably want you to answer iv based on the result for iii. You can get a more accurate answer this way.

    Yes, the initial velocity (relative to the station) is zero.
     
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