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B A doubt about tangential and normal aceleration, pilot g forces

  1. Dec 7, 2016 #1
    so a pilot is spining in a circle at a constant speed, the g forces he will feel will be the same than normal acceleration dependant on wr

    but now you put him inside a tube where he can move along varying the radius of the circle

    now the normal acceleration will be zero because he is in the equivalent of centrifugal gravity free fall

    but he will feel as g forces the tangential acceleration as tangential velocity is function of wr and r is increasing as the pilot moves along the tube

    but now you propel the spining tube in such a way than as the pilot doubles radius the w velocity of the tube halves, on this way the tangential velocity is constant and hence tangential acceleration is zero

    but how can the pilot be in this last case in zero tangential acceleration and zero normal acceleration which implies he endures zero g forces if he is making a spiral trajectory?
     
    Last edited: Dec 7, 2016
  2. jcsd
  3. Dec 7, 2016 #2

    A.T.

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    Define "tangential" and "normal" precisely.
     
  4. Dec 7, 2016 #3
    i thought they were universal terms:

    normal or tangential to the pilot trajectory
     
  5. Dec 7, 2016 #4

    Dale

    Staff: Mentor

    Exactly how would you do that? What motion of the tube would make this happen?
     
  6. Dec 7, 2016 #5
    well you put a precise gps on the pilot and connect the data with the engine of the tube

    alternatively you could calculate an spiral tube in such a way that the spiral of the tube made the w half as the radius doubles with the right shape of the spiral tube
     
  7. Dec 7, 2016 #6

    Dale

    Staff: Mentor

    And what function would you program into the GPS? What motion of the tube could you program which would result in the desired motion of the pilot?

    What shape spiral is the right shape?
     
    Last edited: Dec 7, 2016
  8. Dec 7, 2016 #7
    well i would take as the center of reference the center of spin of the tube and from this reference the pilot will make a perfect spiral for which i dont know the formula of position but i guess someone does
     
  9. Dec 7, 2016 #8

    A.T.

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    Only initially. It doesn't stay zero if angular velocity of the tube is constant.

    If you spin the tube such that it exerts no force on the pilot, then he moves on a straight line.
     
  10. Dec 7, 2016 #9

    Dale

    Staff: Mentor

    The requirement that you proposed in your original post is for the motion of the tube that would produce the specified motion of the pilot. I dont think that the motion you proposed is even possible, and it seems that you are simply assuming that it is possible without any knowledge of the details.

    If you assume that something impossible is possible then clearly you will wind up with contradictions.
     
    Last edited: Dec 7, 2016
  11. Dec 7, 2016 #10
    i think i figure it out with your help guys, thanks

    the pilot makes a spiral trajectory and this is the key

    but the center of curvature in a given instant is not the center of the spiral but its somewhere offset from the center

    so you just have to calculate this center and obtain the normal aceleration with this new instant temporal center that will be the g forces the pilot will experiment as he makes an spiral trajectory
     
  12. Dec 7, 2016 #11

    Nidum

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    (1) Draw us a picture .

    (2) Aircraft and pilot g forces have been well studied over many years . Have you done a search for any existing information about your problem ?
     
  13. Dec 7, 2016 #12
    IMG_20161207_160421_zpswq8fsjnw.jpg

    g forces a pilot feels for what i know go in function of At and An

    as a curiosity on free fall on vacuum you feel zero g

    edit:

    where i wrote At i meant An
     
    Last edited: Dec 7, 2016
  14. Dec 7, 2016 #13
    Your analysis is incorrect. Increasing the tube speed proportional to radius is NOT the condition that keeps the tube from touching him.

    Start with the idea that with no acceleration he must move in a straight line. For there to be no acceleration the tube must be positioned around him at all times but not touch him

    One might question how it came about that the pilot started moving in a straight line in the first place, but I didn't make up the problem. You did.

    So if the pilot moves parallel to a radial, the initial tube velocity is 0 and dividing by the pilots radial position still gives 0. The tube does not move.

    Now say the pilot moves at a constant velocity v starting at t0 on a tangent at radius r0. The pilots position at time t is angle = atan(v t / r0) and radius = sqrt((v t)^2 + r0 ^2)
    For small angles atan(v t / r0) ~= v t /r0 and the angular speed increases linearly with time, but not linearly with the pilots radius. In fact at small times the sqrt approximates to r0 + (v t)^2 / (2 r0). For the tube not to touch him and keep the acceleration at zero the tube moves proportional to the square root of the radial position. This is because in addition to the radius becoming larger the fraction of the velocity which is tangential becomes less and less. This is where your logic failed. Any other motion of the tube, say angular velocity proportional to radius as you propose WILL accelerate the pilot.
     
  15. Dec 7, 2016 #14
    well obviously as soon as the wall of the tube pushes the pilot he will feel g forces and if he goes in a staright line the tube is not spining which is a condition

    what i initially failed to see is that the center of the trajectory at a given instant is not the center of the spiral but the center of curvature of that portion of the spiral where the pilot is at the given instant

    if instead you wrongly take the center of curvature in the center of the spiral thats when the contradiction appears
     
  16. Dec 9, 2016 #15
    i still have some trouble with this case problem:

    imagine the case there are two counterrotating tubes in space:

    the tubes are massless and frictionless, i can make this assumption for being them counterrotaing they still can apply a torque

    there are two astronauts in the center of each tube and exactly at the same moment they start a centrifugal gravity free fall

    we measure their trajectory exactly in the moment both astronauts intersect

    the trajectories will be V which added will be a I aiming up

    if initially the astronauts vectors in an isolated system was zero, how can it the isolated system go to a vector I aiming up without counterpart since the tubes are massless and hence can not have a vector

    i hope you dont mind me asking questions of things i dont see and setting my self physics problems in order to solve them for entertainment
     
  17. Dec 9, 2016 #16

    A.T.

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    If they were massless they would have an infinite acceleration from any unbalanced force, so you already found the flaw in your assumptions.
     
  18. Dec 9, 2016 #17
    i think what would be going on is that actually none of the astronauts would be moving forward or backwards just lateraly, it would be the massles platform which would be rocking back and fordward :)
     
  19. Dec 9, 2016 #18
    now a last case that will help me understand vectors and their arms:

    now theres a single masless frictionless tube in space and two astronauts that jump to free fall artificial centrifugal gravity in opposite sense

    IMG_20161209_115711_zps2ylew3p4.jpg

    initially theres a vector aiming upwards from the surface of the paper where the drawing is

    how did it mathematically go to those two opposite vectors and the growing arm?
     
    Last edited: Dec 9, 2016
  20. Dec 9, 2016 #19

    Dale

    Staff: Mentor

    What you describe cannot happen without some external torque.
     
  21. Dec 9, 2016 #20
    well then make the tube have the same mass than the both astronauts and be 10 m long, then the torque will be provided by the tubes inertia to keep spinning

    consider than the tube with the astronauts inside its spinning at w=10

    the initial separation between the astronauts cog would be 10 cm
     
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