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Help Explaining Physics of freefall tower

  1. May 24, 2007 #1
    1. The problem statement, all variables and given/known data

    I need to explain the physics of a free fall drop tower ride, 250 feet high, car holds 35 people and I'm assuming you would need the mass of the car so help me out there. Also, It is not supposed to exceed 3 g's when braking. It would be awesome if someone could help me out with this, I can try to help you out with anything else needed to do the math.
     
  2. jcsd
  3. May 24, 2007 #2

    berkeman

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    Staff: Mentor

    Are you familiar with the algebraic form of the kinematic equations of motion in a gravitational field?

    http://en.wikipedia.org/wiki/Kinematics

    For free fall, you can calculate the velocity as a function of time (since you know the acceleration is constant at a = g = 9.8m/s^2). Then you need to do some algebra to see at what time you need to start 3g braking in order to stop at the bottom. Can you show us how you would set up those equations?
     
  4. May 24, 2007 #3
    Hmm, I'm not sure of what you mean by velocity as a function of time but I think I need...

    V=v+(9.8)t where V is final and v is initial

    I'm assuming the velocity as a function of time would help me fill the voids in the equation
     
  5. May 24, 2007 #4

    berkeman

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    That's the correct kinematic equation for the velocity as a function of time in a constant force field (like a gravitational field).

    [tex]V = V_0 + a t = V_0 + (9.8 \frac{m}{s^2})[/tex]

    If the ride starts at rest at the top, then Vo = 0, right? So this equation lets you figure out what the speed is as the ride falls.

    Now write the equation for the stop part. The stop deceleration force on the passengers is supposed to be limited to "3g's", which means what in terms of the force on the ride? Remember, if there is no deceleration, then the people on the ride feel "0g". If there is a "1g" acceleration up and gravity's "1g" acceleration down, the passengers feel no net acceleration and what does their velocity do? If there is a "2g" acceleration up and gravity's "1g" acceleration down, what do the passengers feel? Etc., etc.

    So what is the equation for the velocity as a function of time for the deceleration phase? Now can you use some algebra to determine when the deceleration has to start in order to reach V = 0 at the bottom of the ride?

    I have to bail for a few hours. Keep on calculating, and others will try to help out.
     
  6. May 24, 2007 #5
    Do I have the right idea for the time of deceleration?

    0=9.8+(-29.4)t

    That would be for a total of 2 g's, not 3? because I am adding the force of gravity?
     
    Last edited: May 24, 2007
  7. May 25, 2007 #6

    berkeman

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    First of all, I doubt the highest velocity before starting to decelerate is 9.8m/s (which is what you seem to be showing in your equation for the deceleration phase). I do think you have the net deceleration correct at 2g, since you will be pushing up with the 3g that the people will feel, and gravity will still be acting down with 1g.

    For the deceleration phase, you need to write the same form of the equation that I posted above, with the Vo being the variable that you don't know (that's the fastest that the ride was falling before you started the 3g deceleration).
     
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