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Height at which pilot must pull out of dive before hitting 5g

  1. Jan 30, 2014 #1
    1. The problem statement, all variables and given/known data

    A jet is diving vertically downwards at 1200kmh^-1. If the pilot can withstand a max acceleration of 5g before losing consciousness, at what height must the plane start a quarter turn to pull out of the dive.
    Assume the speed remains constant.

    3. The attempt at a solution

    I do not understand the question fully. Could someone give me a leg-up?
     
  2. jcsd
  3. Jan 30, 2014 #2

    rcgldr

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    I would assume the problem is asking what is the lowest height so that the pilot can pull ouf of the dive without hitting the ground or exceeding 5 g's. I would assume the 5 g limit needs to take the effect of gravity into account.
     
  4. Jan 30, 2014 #3

    BvU

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    Oh boy, a lot of parallel questions. What do you think is meant when you read "quarter" turn ?
     
  5. Jan 30, 2014 #4
    That the pilot makes a pi/2 turn?
     
  6. Jan 30, 2014 #5
    I drew a circle for this and take the radius (the -y component) as the height along which the pilot travels before making the quarter turn.
    But beyond which, I have no idea.
     
  7. Jan 30, 2014 #6

    BvU

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    Good. Maybe, because ambiguous. Turning can be done many ways with a jet plane. With a side view and a top view, which kind of turn do you mean ?
     

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  8. Jan 30, 2014 #7

    The side view.
     
  9. Jan 30, 2014 #8

    BvU

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    I see no turn, but you made a good choice. It was more or less stipulated in the word "pull" in the OP. Which way do you want the guy to turn ?
     
  10. Jan 30, 2014 #9

    BvU

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    If he does his thing right, what g does he feel after this scary emergency measure ?
     
  11. Jan 30, 2014 #10

    HallsofIvy

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    I presume you know that in making a circular turn of radius r, at speed v, the inward acceleration has magnitude [itex]\frac{v^2}{r}[/itex] so "centrifugal force" on the pilot will be [itex]\frac{v^2m}{r}[/itex] where m is the mass of the pilot. I'm not clear whether the "5gs" acceleration the pilot can stand is to include the pilots weight or not. If it is not, then set [itex]\frac{v^2m}{r}= 5mg[/itex] and solve for r. If it is, set [itex]\frac{v^2m}{r}= 4mg[/itex] (subtracting the pilots weight, mg) and solve for r.


    Of course, if he is making a vertical circle of radius r, he had better be at least "r" above the ground to avoid hitting the ground!


    We can write a circle of radius r, as parametric equations x= r cos(\omega t), y= r sin(\omega t). differentiating, v_x= -r\omega sin(\omega t), v_y= r\omega cos(\omega t). The acceleration is given by, differentiating again, a_x= -r\omega^2 cos(\omega t), a_y= -r\omega^2 sin(\omega t).

    The speed at each instant is given by \sqrt{v_x^2+ v_y^2}= \sqrt{r^2\omega^2}= r\omega= v so we must have \omega= v/r. Then the acceleration is given by a_x= -(v^2/r) cos(\omega t), a_y= -(v
     
  12. Jan 30, 2014 #11

    Let's suppose he turns in the -x direction
     
  13. Jan 30, 2014 #12

    HallsofIvy

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    I presume you know that making a circle of radius r, at speed v, there must be a centripetal force of magnitude [itex]mv^2/r[/itex]. I don't know whether the pilots own weight is included in the "5 gs" or not. If it is not then we must have [itex]mv^2/r= 5mg[/itex]. If it is, then subtract 1 mg for his weight to get [itex]mv^2/r= 4mg[/itex].

    Solve one of those for r. Of course, his initial height, before making that loop must be at least "r" to avoid hitting the ground!
     
  14. Jan 30, 2014 #13

    BvU

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    Which way is that ? Or was there a coordinate system present already ?
     
  15. Jan 30, 2014 #14

    BvU

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    @HallsofIvy, don't spoil this! With all due respect, of course...
     
    Last edited: Jan 30, 2014
  16. Jan 30, 2014 #15

    BvU

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    If I were at the controls, I'd prefer the one on the right. Flying upside down at height zero is riskier than head up. Agree?
     

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  17. Jan 30, 2014 #16

    HallsofIvy

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    I'm sorry, I foolishly thought we were being serious. I clearly did not read the responses carefully enough.
     
  18. Jan 30, 2014 #17

    BvU

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    Negation, are you comfortable with the equations for circular motion at constant speed?
    All you asked for was a leg-up and now we already have 17 posts!
     
  19. Jan 30, 2014 #18
    To some degree, I am.

    I conceive of a circle with the origin (0,0) and of radius, r. The plane begins from (r,0) and takes the path of an arc length in quadrant 4. At point (-r,0), it begins moving horizontally.
    In moving along the arc length, the acceleration changes and I right? And we want the g value to be lesser to or equal to 5g along the arc length.

    a = g = v^2/ r

    5(9.8) = 333^2/r
    r = 2263m

    r represents the height at which the pilot must make the turn along the arc length cojoining (-r,0) and (0-r).
    Is this it?
     
    Last edited: Jan 30, 2014
  20. Jan 30, 2014 #19
    Yes, this was what I have in mind.
    Although, the left one was what I had exactly in mind. But both diagrams are essentially parallel model of one another.
     
  21. Jan 31, 2014 #20

    BvU

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    You are absolutely right: concerning the physics exercise they are equivalent. But pilot's g-resistance applies to downward g forces: you survive/stay conscious longer with those than with the blood running to your head. (I think -- will ask my son who is in the business)

    My battle plan for you was to focus on the worst case: the end of the quarter circle. Then centripetal force and upward force from the gravity pushing the pilot down into the ejector seat align and therefore the magnitude of the sum is at a maximum. This would leave 4g for the centripetal force to execute the 'turn', and the expressions etc. were already posted.

    You will do fine on this one, I am sure. Look forward to the next encounter.
     
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