Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Falling Bodies with air resitance

  1. Jan 11, 2005 #1
    I know this may be seen as a physics question and it is, but i am just looking for an explicit solution to falling bodies with air resistance.

    OK many of you are probably familiar with the equation for a body falling with air resistance.

    mdv/dt=mg-kv^2
    I am not sure exactly how to solve for v in terms of t, It seems to be seperable but i have problems with the integration.
    mdv=(mg-kv^2)dt

    If an explicit solution is indeed possible then i would like to know because I am unsure of my results on this one. Thanks! :smile:
     
  2. jcsd
  3. Jan 11, 2005 #2

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    [tex] \frac{dv}{mg-kv^{2}} = \frac{dt}{m} [/tex], now integrate this function and apply the boundary conditions.
     
  4. Jan 11, 2005 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Hint: mg- kv2 can be factored as [itex](\sqrt{mg}-\sqrt{k}v)(\sqrt{mg}+\sqrt{k}v)[/itex]. You can use "partial fractions" to integrate it.
     
    Last edited: Jan 12, 2005
  5. Jan 11, 2005 #4
    Sorry guys i should have clarified the integration is what is getting me, help would be greatly appreciated..... :smile:
     
  6. Jan 12, 2005 #5

    [tex]\frac{1}{mg-kv^2} = \frac{A}{\sqrt{mg} - \sqrt{k}v} + \frac{B}{\sqrt{mg} + \sqrt{k}v}[/tex]

    Determin A and B and then integrate these two fractions...

    marlon
     
  7. Jan 12, 2005 #6

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    It would have been nicer,if we had some Stokes force there.It would have given it a little spice... :tongue2:

    Daniel.
     
  8. Jan 12, 2005 #7

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    the integral is [tex] \int \frac{dv}{\sqrt{mg - kv^{2}}} = \frac{1}{\sqrt{k}}\arcsin\left(\sqrt{\frac{ k}{mg}}v\right) [/tex]

    There is no need to break the fraction into partial fractions to integrate.
     
  9. Jan 12, 2005 #8
    Drtransport.... where did the squareroot come from :confused:
     
  10. Jan 12, 2005 #9

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Nope,wrong formula i'm afraid.no square root over the denominator,no harmonic motion whatsoever...


    Daniel.
     
  11. Jan 13, 2005 #10

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    OOOPPPS,

    [tex] \int \frac{dv}{ a^{2} - v^{2}} = \ln \left ( \frac{ v + a}{ v - a } \right ) + c [/tex]

    extra square root, my mistake (senior moment). You can get this from parital fractions also....... an online list of integrals is at


    http://www.mathwords.com/i/integral_table.htm
     
  12. Jan 13, 2005 #11

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Why didn't you post the correct formula??Where's that [itex] \frac{1}{2a}[/itex] ??

    Daniel.
     
    Last edited: Jan 13, 2005
  13. Jan 13, 2005 #12

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    I gave the general formula, just factor out the necessary terms and go from there.
     
  14. Jan 13, 2005 #13

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    If this is the "general formula":
    [tex] \int \frac{dv}{a^{2}-v^{2}}=\ln(\frac{v+a}{v-a})+C [/tex]

    ,then it's deadly wrong since,for the formula to stand,it assumes a particular value for the integration constant (viz.
    [tex] C=\ln(\frac{1}{2a}) [/tex]

    ),which would flagrantly contradict the rule of expressing the "family" of antiderivatives using ARBITRARY constants.


    Daniel.
     
  15. Jan 13, 2005 #14

    Dr Transport

    User Avatar
    Science Advisor
    Gold Member

    [tex] \int \frac{dv}{a^{2}-v^{2}}=\frac{1}{2a}\ln(\frac{v+a}{v-a})+C [/tex]

    constant factor included.......
     
  16. Jan 13, 2005 #15

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    That's more like it... :approve: Now it looks goooooood... :tongue2:

    Daniel.
     
  17. Jan 13, 2005 #16
    Hey guys this problem is getting to me and i want to be able to work at this to get v explicitly so im gonna keep at it.

    [tex] \int \frac{dv}{mg-kv^{2}}=\int \frac{dt}{m} [/tex]

    following the general integration rule i took [tex] \sqrt{\frac{mg}{k}}[/tex] as a so to fit the form. Since k is a constant and is also a constant i rewrote that as k. I am not sure if all this simplifying of k is valid so if it isnt let me know.

    The resulting integral is simply

    [tex] \frac{1}{2k}\ln{\frac{v+k}{v-k}=\frac{t}{m}+C[/tex]


    Now let me know if this is all valid cuz i always make careless mistakes.

    from there simplifying that one i get

    [tex] \frac{v+k}{v-k}=Ce^{\frac{2kt}{m}}[/tex]


    Let me know if this is right?
     
  18. Jan 13, 2005 #17

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Pay attention with relabeling constants.In this case,i'm afraid you did it badly.You forkot one "k" (the "original one") when u factored it out from the denominator so u could put the latter as a difference of squares.

    Daniel.
     
  19. Jan 13, 2005 #18
    Thanks man that helps alot.... but now im stuck on how to relabel so you can make the equation a bit nicer to work with any ideas? By the way i like your icon......
     
  20. Jan 13, 2005 #19

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    Yes.Forget about the labeling this time.It's not helpful since it would mean the same number of constants:eek:ne the "k",one the relabeling (the sq.root),one from the integration;it would be better to carry arond that radical.

    :blushing: Thenx.I am an icon... :tongue2: :approve:

    Daniel.
     
  21. Jan 14, 2005 #20
    Man leave it to a physicist to have an ego.... oh well if you know your stuff no point in pretending you dont right....?
    Alright well so it should actually be....

    [tex] \frac{1}{2\sqrt\frac{mg}{k}}\ln{\frac{v+sqrt\frac{mg}{k}}{v-sqrt\frac{mg}{k}}=\frac{t}{m}+C[/tex]
    Is this right? Any help appreciated...... :smile:

    Remember humility is good for even the most brilliant of minds.... :biggrin:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Falling Bodies with air resitance
  1. Air drag equasion (Replies: 4)

Loading...