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Integral of non-constant acceleration

  1. May 6, 2013 #1
    1. The problem statement, all variables and given/known data
    A stone falls from a high tower. what is the distance, in the x direction, parallel to the ground, that the stone reaches, caused only by the x component, the tangential component, of the gravity, after t seconds of free fall.

    2. Relevant equations
    The constant acceleration formula:
    [tex]x=\frac{1}{2}a\cdot t^2[/tex]

    3. The attempt at a solution
    According to the drawing attached, the acceleration in the x direction is:
    [tex]a=g\cdot\sin\theta=g\sin(\omega t)[/tex]
    Where ω is the angular velocity of the earth.
    I insert this a into the constant acceleration formula, since i don't know anything else, and, maybe, in short intervals of time the acceleration is approximately constant.
    So, i get:
    [tex]dx=\frac{1}{2}g\sin(\omega dt)\cdot dt^2[/tex]
    And:
    [tex]x=\frac{g}{2}\int_{0}^{4} \sin(\omega t)\cdot t^2[/tex]
    Is this approach correct? since, numerically, the results aren't.
     

    Attached Files:

  2. jcsd
  3. May 6, 2013 #2

    SammyS

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    Since the numerical result isn't correct, I doubt that the approach is correct either.


    Kind of a strange problem.

    I guess that the tower is so high that stone doesn't land at a point directly below the position it was dropped from.

    It also looks like you were to to assume that the tower is short enough to consider g to be a constant value.


    What class is this for?
     
  4. May 7, 2013 #3

    rude man

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    Yes, weird. Looks like a Coriolis problem except the latitude is not given, nor the tower height.
     
  5. May 7, 2013 #4

    haruspex

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    I agree, strange question. What is 'tangential component of gravity', unless caused by a gravitational anomaly? What does the diagram represent? Perhaps theta therein is supposed to be pi/2 - latitude, but the rest of the picture still doesn't make sense, and as Rude man says you still need the tower height. (I get ωht sin(θ), where h is the tower height and ω = 2π/24hours.)

    Edit: What rubbish, haruspex :blushing:. The tower height is not needed. My h above should be the vertical distance fallen in time t, so it reduces to ωgt3 sin(θ)/2 = 2.3 cm at the equator.
    But I don't see that you need any of that fancy non-inertial frame calculation - at a distance h below the release point, the tangential velocity difference between the stone and the tower is ωh sin(θ), and that is constant, so the horizontal separation is ωht sin(θ).
     
    Last edited: May 8, 2013
  6. May 7, 2013 #5

    TSny

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  7. May 8, 2013 #6

    haruspex

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