Falling Bodies with air resitance

[BigStelly]

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!

 

[Dr Transport]

$$\frac{dv}{mg-kv^{2}} = \frac{dt}{m}$$, now integrate this function and apply the boundary conditions.

 

[HallsofIvy]

Hint: mg- kv² can be factored as $(\sqrt{mg}-\sqrt{k}v)(\sqrt{mg}+\sqrt{k}v)$. You can use "partial fractions" to integrate it.

 

[BigStelly]

Sorry guys i should have clarified the integration is what is getting me, help would be greatly appreciated...

 

[marlon]

Sorry guys i should have clarified the integration is what is getting me, help would be greatly appreciated...

$$\frac{1}{mg-kv^2} = \frac{A}{\sqrt{mg} - \sqrt{k}v} + \frac{B}{\sqrt{mg} + \sqrt{k}v}$$

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

marlon

 

[dextercioby]

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

Daniel.

 

[Dr Transport]

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

There is no need to break the fraction into partial fractions to integrate.

 

[vincentchan]

Drtransport... where did the squareroot come from

 

[dextercioby]

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


Daniel.

 

[Dr Transport]

OOOPPPS,

$$\int \frac{dv}{ a^{2} - v^{2}} = \ln \left ( \frac{ v + a}{ v - a } \right ) + c$$

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

 

[dextercioby]

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

Daniel.

 

[Dr Transport]

I gave the general formula, just factor out the necessary terms and go from there.

 

[dextercioby]

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

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

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


Daniel.

 

[Dr Transport]

$$\int \frac{dv}{a^{2}-v^{2}}=\frac{1}{2a}\ln(\frac{v+a}{v-a})+C$$

constant factor included...

 

[dextercioby]

That's more like it... Now it looks goooooood...

Daniel.

 

[BigStelly]

Hey guys this problem is getting to me and i want to be able to work at this to get v explicitly so I am going to keep at it.

$$\int \frac{dv}{mg-kv^{2}}=\int \frac{dt}{m}$$

following the general integration rule i took $$\sqrt{\frac{mg}{k}}$$ 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 isn't let me know.

The resulting integral is simply

$$\frac{1}{2k}\ln{\frac{v+k}{v-k}=\frac{t}{m}+C$$


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

from there simplifying that one i get

$$\frac{v+k}{v-k}=Ce^{\frac{2kt}{m}}$$


Let me know if this is right?

 

[dextercioby]

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.

 

[BigStelly]

Thanks man that helps a lot... but now I am 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...

 

[dextercioby]

Thanks man that helps a lot... But now I am stuck on how to relabel so you can make the equation a bit nicer to work with any ideas?

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

By the way i like your icon...

Thenx.I am an icon...

Daniel.

 

[BigStelly]

Man leave it to a physicist to have an ego... oh well if you know your stuff no point in pretending you don't right...?
Alright well so it should actually be...

$$\frac{1}{2\sqrt\frac{mg}{k}}\ln{\frac{v+sqrt\frac{mg}{k}}{v-sqrt\frac{mg}{k}}=\frac{t}{m}+C$$
Is this right? Any help appreciated...

Remember humility is good for even the most brilliant of minds...

 

[dextercioby]

U forgot about the "k" again.
$$\frac{1}{k}\frac{1}{2\sqrt{\frac{mg}{k}}}\ln(\frac{v+\sqrt{\frac{mg}{k}}}{v-\sqrt{\frac{mg}{k}}})=\frac{t}{m}+C$$

Daniel.

 

[BigStelly]

Ok from here...

$$\frac{1}{k}\frac{1}{2\sqrt{\frac{mg}{k}}}\ln(\frac {v+\sqrt{\frac{mg}{k}}}{v-\sqrt{\frac{mg}{k}}})=\frac{t}{m}+C$$

multpyling through i get

$$\ln(\frac {v+\sqrt{\frac{mg}{k}}}{v-\sqrt{\frac{mg}{k}}})=\frac {2k\sqrt\frac{mg}{k}t}{m}+C$$

I thought that because $$2k\sqrt{\frac{mg}{k}}$$ is a constant(being that k,m, and g are all constants i could rewrite that as C) is this valid? if not what are the rules for relabeling constants because if they are different than i recall it could lie in most of my issues with thie problem...
thanks a bunch

 

[dextercioby]

As long as u can keep track of the relations between these constants,u can relabel them in any way possible.The trick is that these problems have "input variables",whether they are letters ('k','m','g'),or numbers.The results have to be expressed in terms of these "input variables".So it makes no sense to present the reader with the solution with other variables than the original ones.It's like when evaluating integrals using substitution methods.U relabel the original parameter,however,in the final result u must "invert the substitution"...

Daniel.

P.S.The last formula is further symplified by noting that in the RHS,par éxample,the "k" and "m" appear both at the numerator and at the denominator.You can do the same thing in the LHS.sqrt of "k" is at the denominator twice...

 

[BigStelly]

Alright thanks for your help, earlier you mentioned stokes forces as part of the problem... well now I am interested i recall hearing something about that before but i don't remember exactly what it is, so please enlighten me how would you deal with it?

 

[dextercioby]

Alright thanks for your help, earlier you mentioned stokes forces as part of the problem... well now I am interested i recall hearing something about that before but i don't remember exactly what it is, so please enlighten me how would you deal with it?

I was talking about the formula (15) from here

Daniel.