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Falling objects, springs

  1. Sep 11, 2004 #1
    hi,

    how can I figure out the peak velocity of 1kg dropped 1m onto a spring? i know the peak velocity occurs when F = -kx + mg = 0.
     
    Last edited: Sep 11, 2004
  2. jcsd
  3. Sep 11, 2004 #2

    Integral

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    The peak velocity will occur when the mass first reaches the spring after its fall.
     
  4. Sep 11, 2004 #3
    you need to rethink that.
     
    Last edited: Sep 11, 2004
  5. Sep 11, 2004 #4

    Tide

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    The maximum speed will occur when the gravitational force is balanced by the spring force: [itex]x = -\frac{mg}{k}[/itex]
    Before that point is reached the object is still accelerating and beyond that point the acceleration is upward.
     
  6. Sep 11, 2004 #5
    thanks. anyone else?
     
  7. Sep 11, 2004 #6

    Integral

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    The instant the mass hits the spring it begins to decelerate. Perhaps we are not speaking of the same speed. I am referring to the mass, and you?
     
  8. Sep 11, 2004 #7

    Tide

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    No, because the spring force is zero at that point and gravity is still acting.
     
  9. Sep 11, 2004 #8
    in my initial post i tried to indicate how far i had gotten. i can get x. i can also get the velocity before the spring starts stretching. what i need is the peak velocity. that's the part i'm having trouble with.
     
    Last edited: Sep 11, 2004
  10. Sep 11, 2004 #9

    Tide

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    Use energy conservation!
     
  11. Sep 11, 2004 #10

    Integral

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    Ok, ok...
    Looks to me like once we hit the spring we must satisfy the DE

    [tex] \ddot {x}m = -mg - kx [/tex]


    Find the velocity expression from this and you are done. :)

    I'm workin on it.

    Edit: inserted a lost factor.
     
    Last edited: Sep 16, 2004
  12. Sep 11, 2004 #11
    how? the spring continues to stretch after kx = mg.
     
  13. Sep 11, 2004 #12

    Tide

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    No, the spring is being compressed. Remember, you're DROPPING the object onto the spring.

    Do you know what the potential energy is for a spring and how to calculate gravitational potential energy?
     
  14. Sep 11, 2004 #13
    a flicker of a light bulb. 1/2kx^2 is the potential energy of the spring, so subtracting that from the change in potential energy of the falling mass should give me the kinetic energy. am i right?
     
    Last edited: Sep 11, 2004
  15. Sep 11, 2004 #14

    Tide

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    That's part of it. Energy conservation tells you that
    [tex]mv^2 + kx^2 + 2mgx = mv_0^2[/tex]
    is a constant. I've taken the zero point of the potential to be at the point of contact.
     
    Last edited by a moderator: Sep 16, 2004
  16. Sep 11, 2004 #15

    Integral

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    From the solution of the DE I get
    Vi = velocity at the top of the spring.
    [tex] x(t) = \frac {mg} k Cos ( \sqrt { \frac k m}t) + V_i \sqrt {\frac m k} Sin (\sqrt {\frac k m } t) -\frac {mg} k [/tex]
    The velocity equation is:

    [tex] \dot {x}(t) =\sqrt {\frac m k}g Sin(\sqrt{\frac k m}t) + V_i Cos{(\sqrt {\frac k m}t)}[/tex]
    The time at which the max velocity occurs will be

    [tex] t_m =\sqrt {\frac m k} Tan^{-1} (\frac g {V_i} \sqrt { \frac m k})[/tex]


    Simply plug this time into the velocity equation to find the Max velocity.

    Edit the edit: All units are now happy!
     
    Last edited: Sep 16, 2004
  17. Sep 11, 2004 #16

    Tide

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    Your units aren't right! :-)
     
  18. Sep 11, 2004 #17

    Integral

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    Thats for sure!

    Got 'em fixed now. A lost constant or 2 can really mess things up!
     
  19. Sep 11, 2004 #18
    ok but shouldn't it be:

    [tex]KE_{mass} + PE_{spring} = \Delta PE_{mass}
    [/tex]

    when [tex]v_{0}=0[/tex].

    so,

    [tex]1/2mv^2 + 1/2kx^2 = mg(h+x)[/tex]

    or

    [tex]mv^2 + kx^2 - 2mg(h+x) = 0[/tex]

    where h is the drop distance and x is the spring compression.

    in other words in your equation the distance in the [tex]kx^2[/tex] term shouldn't be the same as the distance in the [tex]2mgx[/tex] term, and it seems to me the sign of the [tex]2mg(h+x)[/tex] term should be negative. as far as i know, energy isn't a vector quantity, so i'm not sure how this would result in different signs:

    also, these equations seem to express the same thing I said in my previous post:

    (how do you get two lines of equations between a set of tags? i tried \\ at the end of a line and the equations just ran together.)
     
    Last edited by a moderator: Sep 16, 2004
  20. Sep 11, 2004 #19

    Tide

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    The full equation would be
    [tex]mv^2 + kx^2 + 2mgx = mv_0^2 + kx_0^2 + 2 mgx_0[/tex]
    where the subscripts refer to some initial value. I took [itex]x_0[/itex] to be zero which corresponds to the point of initial contact. My [itex]v_0[/itex] is the speed of the object just as it hits the spring which you can calculate if you know the height above the spring when it was released.
     
    Last edited by a moderator: Sep 16, 2004
  21. Sep 12, 2004 #20

    ehild

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    [tex]mv^2 + kx^2 -2mgx = mv_0^2[/tex]
    V is maximum when
    [tex]x=\frac{mg}{k}\mbox{. }
    v_{max}=\sqrt{v_0^2+\frac{mg^2}{k}}.[/tex]

    ehild
     
    Last edited by a moderator: Sep 16, 2004
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