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Classical mechanics with time dependent force

  1. Sep 2, 2011 #1
    1. The problem statement, all variables and given/known data

    A point mass m is exposed to a time dependent force F(t). Determine the position r(t) of the point mass for the initial conditions r(0) = [itex]r_{0}[/itex]and v(0) = [itex]v_{0}[/itex]

    2. Relevant equations



    3. The attempt at a solution

    [itex]\sum[/itex]F= ma
    [itex]F_{z}[/itex](t) - mg = ma

    a = 1/m [itex]F_{z}[/itex](t) - g
    v = 1/m [itex]\int[/itex][itex]F_{z}[/itex](t) dt - [itex]\int[/itex]g dt
    but how do I start to int F(t) without knowing the equation for F itself?
     
  2. jcsd
  3. Sep 3, 2011 #2

    I like Serena

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    Hi shyta! :smile:

    Since you don't know the actual F(t), what you will have to do is to find a generic formula for r(t) that holds true for any F(t).
    Basically that's what you've already been doing, so that's good! :smile:

    Some comments on your attempt:

    Do you have a reason to assume F(t) is vertical, and furthermore that gravity is not included in F(t) itself?

    You have been given the boundary condition v = v0.
    You should incorporate that in your formula.

    And you haven't calculated r(t) yet.
     
  4. Sep 3, 2011 #3
    Consider how much impact you would add to the mass when


    m*Vi + integral ( F dt ) [from 0 to t] = m*v(t)

    but after you find v(t), find r(t) by integration and please note that you are working with vectors here !
     
  5. Sep 3, 2011 #4
    Hi both! thanks for the replies.

    iloveserena: I was trying to simplify things by considering the z-direction forces first, maybe i was over-simplifying/

    Now assuming I follow stallionx's equation,

    then v(t) = v_0 + 1/m [itex]\int[/itex]F(t) dt from 0 to T
    v(t) = v_0 + [itex]\int[/itex]dv/dt dt from 0 to T
    v(t) = v_0 + [itex]\int[/itex] dv from v to v+aT

    am I in the correct direction?
     
  6. Sep 3, 2011 #5

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    Yep. That's the formula for v as far as you can define it from the generic F(t).


    No. These formulas basically only say that v=v.
    But you need v based on F(t), which is what you already had.

    You need to take the next step, by trying to define r(t).



    Btw, for these formulas it is irrelevant whether they are vectors or scalars.
    The formulas are true for both.
     
  7. Sep 3, 2011 #6
    v(t) = v_0 + 1/m ∫F(t) dt from 0 to T

    r(t) = v_0T + 1/m ∫∫ F(t) dtdt from 0 to T


    But it makes not much sense from this equation because there is a last part that says check that you get the familiar results when F(t) = mg is constant in time
     
  8. Sep 3, 2011 #7

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    Well, you should use different symbols when integrating, and use "t" consistently.
    Let me rewrite your equation:
    [tex]r(t) = v_0 t + {1 \over m} \int_0^t \int_0^{\tau'} F(\tau) d\tau d\tau'[/tex]

    Can you work this out for F(t)=mg?
     
  9. Sep 3, 2011 #8
    I got the equation :) thanks!

    however just a quick query, why is there a need for the different symbols when integrating if it's the same closed intervals?
     
  10. Sep 3, 2011 #9

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    I was not sure if the integration confused you, so I made it more explicit.
    I guess in this case the difference is not all to obvious, since the result will come out the same.
    To illustrate the difference consider the following 2 integrals:
    [tex]\int_0^t t \cdot \tau d\tau = {1 \over 2}t^3[/tex]
    versus
    [tex]\int_0^t t \cdot t dt = {1 \over 3}t^3[/tex]
    See the difference?



    Btw, your equation for r(t) is not right yet.
    You missed something.
     
  11. Sep 3, 2011 #10
    oh i see i see.

    for the r(t) equation do you mean the initial conditions?
     
  12. Sep 3, 2011 #11

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  13. Sep 3, 2011 #12
    Hehe I got that! Thanks for the reminder and help! :)
     
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