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Calculating a bolt's free fall time in an accelerating elevator

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  1. Jul 13, 2017 #1
    1. The problem statement, all variables and given/known data
    upload_2017-7-13_21-12-24.png

    2. Relevant equations


    3. The attempt at a solution
    A) W.r.t. earth's frame,
    After 2s , velocity of bolt = at = 1.2*2 = 2.4 m/s
    distance traveled = (1/2) 1.2*22 = 2.4 m
    Let's say that the bolt reaches the floor at time t.
    Using s = ut + (1/2)at2
    2.4 = 2.4t +(1/2)(-9.8) t2
    Solving this equation doesn't give proper answer.

    After leaving the car, the bolt goes up till its speed becomes 0 m/s. Let's say that this process takes the time t1 and by this time the bolt travels a distance s1.
    0 = 2.4 - 9.8t 1
    t1 = 2.4/9.8 = 0.245
    s1 = 2.4* 0.245 + (1/2)(-9.8)0.2452 =0.294 m
    After the speed of the bolt becomes 0, the bolt starts falling down.
    Let's say that the bolt takes time t2 to reach the floor from now.
    So,
    s + s1 = (1/2)9.8 t2
    Solving this gives t2 = 0.741s

    What is meant by free fall time ?
    t2 = 0.741 s or t2 + t1 = 0.986 s ?


    b) W.r.t. elevator,Taking t2 as the free fall,
    The displacement is
    d = s + s1 +[ {1.2*(2 + t1 )} t2 + (1/2) 1.2 * t22 = 5.019 m where the bold part is distance traveled by the car in time t2

    The distance is D = d+ s1 = 5.019+0.294= 5.313 m,
     
  2. jcsd
  3. Jul 13, 2017 #2
    Bolt doesn't go up. Here, it is given in the question that the bolt directly comes down.So, the bolt's speed is 0 at the moment it leaves the ceiling.

    So, s = (1/2)9.8 t22
    t2 = 0.7 s , free fall time
     
  4. Jul 13, 2017 #3
    The bolt falls inside the elevator.
    So, after 2s, the ceiling is at a distance 2.7 m + s =2.70+ 2.4 =5.1 m
    The bolt falls from the ceiling and touches the floor of the elevator under free fall and the free fall takes time t.
    distance traveled by the bolt in downward direction + distance traveled by the floor in upward direction = 5.1 m
    ⇒ (1/2)g t2 + 2.4t + (1/2)1.2 t2 + 2.4 m = 5.1 m
    t = 0.52 s
    which one is correct?
     
  5. Jul 13, 2017 #4

    Doc Al

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    Not sure what you mean. At the moment it leaves the ceiling, the bolt's moving at the same speed as the ceiling (and rest of the elevator).
     
  6. Jul 13, 2017 #5
    o.k.
    Now I understood.
    First of all, the bolt falls inside the elevator.
    W.r.t. earth's frame,

    At the moment of leaving, both the bolt and ceiling moves upward with same speed i.e. 2.4 m/s.

    Due to gravity , the speed of the bolt goes on decreasing to 0 and then the bolt falls down while the ceiling goes up with acceleration 1.2 m/s2.
    Distance traveled by the bolt after its speed becomes 0 is d = ( 1/2) g t22

    Now,
    2.7+ 0.294 = ( 1/2) g t22 + 2.4( t1 + t2) + ( 1/2) 1.2( t1 + t2[/SUB)2 , the bold part is distance traveled by the floor during free fall
    2.37 = 4.9 t22 + 2.4 t2 + 0.6 t22 + 0.294 t2
    2.37 = 5.5t222 +2.694 t2
    t2 = 0.456 s

    Time of free fall =t1+ t2 = 0.7 s
     
    Last edited: Jul 13, 2017
  7. Jul 13, 2017 #6
    I didn't know how to solve part (B).
    Frame attached to the shaft is equivalent to a frame attached to the floor for our purpose.
    Now, a person on the floor will see the bolt travelling from ceiling to the floor. So, the displacement is the distance between the ceilling and the floor i.e. 2.7m.

    Is this correct?
     
  8. Jul 13, 2017 #7

    Doc Al

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    That's what I would say. But that seems a strange question.

    Another (equivalent) way to solve part (a) from the frame of the earth is to write two equations: one for the position of the floor as a function of time and one for the position of the bolt as a function of time. Solve for when bolt and floor are at the same position.
     
  9. Jul 14, 2017 #8

    TSny

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    I believe the frame of the elevator shaft is the same as the earth frame.
    https://www.vocabulary.com/dictionary/elevator shaft
     
  10. Jul 14, 2017 #9
    From this dictionary, what I understood is elevator shaft is something very long tube like and it is attached to the roof of the elevator.

    Hence, the shaft,too, is moving.
    The shaft frame is equivalent to the elevator's floor frame.
     
  11. Jul 14, 2017 #10
    No no, I agree with TSny... An "elevator shaft" is the shaft in which the elevator moves, not something that moves with the elevator.

    I think Irodov also meant it this way, or else the two parts of (b) are redundant (the displacement would equal the distance).
     
  12. Jul 14, 2017 #11

    Doc Al

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    You're right! I misread that as saying the "frame of the elevator". (Which is why I thought it a strange -- trivial -- question!) D'oh!
     
  13. Jul 18, 2017 #12
    Taking elevator frame as the Earth frame,

    distance traveled by the floor during free fall = 2.4( t1 + t2) + ( 1/2) 1.2( t1 + t2)2 = 1.977 m =2 m

    So, the displacement traveled by the bolt =[( 2.7 -2.0) m = 0.7 m ]toward the floor

    the distance traveled by the bolt = s1 + ( 1/2) g t22 = 0.294 + 1.019 = 1.312 =1.3 m

    So, the problem is solved.

    Thanks to all.
     
  14. Jul 18, 2017 #13

    Ray Vickson

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    An easier way is to note that for somebody making measurements inside the elevator, it looks exactly like a free-fall problem on a new planet with a stronger force of gravity than earth's: on this new planet, the acceleration of "gravity" is ##g_0 = 9.8 + 1.2 = 11.0\: m/s^2##. The bolt starts with 0 "velocity" from a height of 2.7 m.
     
  15. Jul 18, 2017 #14
    Thanks for this insight.


    This helps in solving part a easily as

    w.r.t. floor frame,
    2.7 = (1/2) *11*tfall2
    tfall= 0.7 s

    Now, tfall for both the floor and the earth frame is same i.e. tfall = 0.7 s. I am assuming it so. But how to prove this? It is assumed so under Galilean transformation . But, Galilean transformation deals with only inertial frames. And here, floor frame is non - inertial.
    But, how can it help in solving part (b)?
    W.r.t. the earth frame,

    Knowing that tfall= 0.7 s

    the displacement can be given as
    D = 2.4 tfall - (1/2) g tfall2 = 0.7 m
    But, for calculating distance, I will have to do what I did in the earlier post i.e. post 12.
     
  16. Jul 18, 2017 #15

    Ray Vickson

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    You can transform to non-Galilean frames. In this case the transformation would be something like ##x' = x - v_0 t - \frac{1}{2} a t^2, \; t' = t##. From the behavior of ##x(t)## you can get the behavior of ##x'(t') = x'(t).##

    A more familiar example of transformation to a non-Galilean frame is to transfer to a rotating coordinate system, where such effects as "centrifugal force" and "Coriolis force" appear naturally.
     
  17. Jul 18, 2017 #16
    Are non-Galilean frames nothing but non - inertial frames without considering special relativity?
    What are Galilean and non - Galilean frames?
     
  18. Jul 19, 2017 #17

    Ray Vickson

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    I probably should have said non-inertial frames and non-Galilean transformations.
     
  19. Jul 20, 2017 #18
    Floor frame ⇒ (x, t)
    Earth frame⇒(x', t')


    t' =t, assumption
    x' = x + v0t + (1/2) at2 where v0 is the initial speed of the floor and a is the acceleration of the floor measured from the Earth's frame


    w.r.t. floor frame,
    2.7 = (1/2) *11*tfall[/SUB2
    tfall= 0.7 s

    x = 2.7 - 2.4*0.7 - 0.5*1.2*0.72 = 0.7 m

    This transformation helps in calculating displacement , not distance.
     
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