Basic stuff - velocity dependant friction

1. Sep 28, 2004

abertram28

please help! recently we went over velocity dependant frictional forces in my engineering physics I class. we all went through the standard model using

F(friction) = -bv

we applied F=ma and got

mg-bv = m (dv/dt)

i realize how this works. im a little fuzzy when it comes to integrating these and using "dummy variables" but i understand how to work around them.
we derived a few equations for t and an equation for v and position using more integration and some simplification.

more or less, i understood this. then we decided to create problems where the F(friction) had some factor of v in it. we then traded with partners. mine was pretty easy, Fr = -(bmv). my partner made one a bit harder. he chose Fr = -mk(v)^2 unluck for me, we traded, then solved. he had a bit of trouble with conceptulizing the mass being a factor in each term, having it cancel out... it wasnt that bad though.

for some reason when i tried to do this problem myself, i ran into some kind of error setting it up or integrating it. please show me step by step how to get a v equation.

my teacher looked at this for about 20 mins and then told me when i have an equation that is using Vo then i am on the right track. she says she got
V=(Vo)/(1 + (Vo)kt) she said we need Vo because of the V^2. please show me how to start this problem at least.

i did this

looking at mg-bv = m (dv/dt)
i decided that im using -mkv^2 instead of -bv. we are theorizing a solution for a ball falling through a liquid. i know the units dont make sense, we made these problems up.
so mg-mkv^2 = m (dv/dt) (F=ma)
Terminal velocity is mg=mkv^2 with v= (g/k)^(1/2)

so, i rearranged my initial equation to be with v and dv on the left and dt on the right.
m factors out and divides. im left with

(dv)/(g-kv^2)=(dt)
integrate both sides. left from 0 to v using vprime as variable. right from 0 to t with tprime as variable. my calculus II class skipped hyperbolic functions.

isnt this a hyperbolic solution? anyhow, my TI-92 gives me

(gk)^(-1/2) *ArctanH(v*(k/g)^(-1/2))

i guess i would have done a factoring of -1/k out of the left side and used arctan rule. i guess i would have been wrong. anyhow. i looked up the definitions of hyperbolics. using the arctanH x = (1/2) ln ((1+x)/(1-x))
AHH! out of boredom i find that the original ingrand fits the
Int(du/(a^2-x^2)) = 1/2a ln(abs((a+u)/(a-u)))+C from the hyperbolic chapter of my calc text.

so i can replace a with (g/k)^(1/2) and pull a 1/k out like i thought would be good... i suppose i could have done this by hand if my damn calc teacher had bothered to go over these. so anyways. im lost after this, if im correct or not, i dont know. i tried to differentiate my answer, buts its too hard for me to do by hand. im tired. do i just solve for v? how do i go about doing that? where the heck does Vo come into play?

2. Sep 28, 2004

HallsofIvy

Arctangent is the anti-derivative of $\frac{1}{1+x^2}$, not $\frac{1}{1-x^2}$. Since you are doing this with real numbers, the difference is important! Of course, the difference between tangent and hyperbolic tangent is that x is replaced by ix.

To integrate $\frac{dv}{g-kv^2}$ "by hand", use partial fractions. g-kv2 can be factored as $(\sqrt{g}- \sqrt{k}v)(\sqrt{g}+\sqrt{k}v)$, of course.

3. Sep 28, 2004

quasar987

I can't follow your explanation but I did a similar problem last week and I recognized many of the things you mentionned as part of the way I proceeded to solve the problem so you must be very near from the answer.

you have mg - bv² = m*dv/dt

You simply put all the v and dv on the same side and the dt on the other side so that you get a big messy function of v on one side and a constant function on the other side. Like this:

dt/m = dv/(mg - bv²)

You integrate both sides from t=0 to t=t. You will need to take b out of there though so you can use the formula I think you mentionned (int [] is for integral)

1/m int[dt] = 1/b int[dv/(mg/b - v²)]

Setting v(0) = 0 (since the ball is stading still in your hand before then you drop it into water), you get (abs[] is for absolute value)

t/m = 1/2(mg/b)½ ln(abs[v+(mg/b)½/v-(mg/b)½])

e^((mg/b)½ * 2t/m) = v+(mg/b)½/v-(mg/b)½

Ok, if you permit I will use easier notation. We will call...

A = e^((mg/b)½ * 2t/m) and B = (mg/b)½

Now all we gotta do is isolate v

A = v+B/v-B

A(v-B) = v+B

Av - AB = v+B

Av - AB - v = B

v(A - 1) - AB = B

v(A - 1) = B + AB

v(A - 1) = B(A + 1)

v = B (A + 1)/(A - 1)

and you can figure out that (A + 1)/(A - 1) is tanh[(mgb)½ * t/m]

and that's your final answer. There are no Vo because Vo = V(t=0) = 0

Last edited: Sep 28, 2004
4. Sep 28, 2004

abertram28

halls>

my antiderivitive up there is the hyperbolic function arctanH not arctan. my mention of the arctan was just that it was my original direction. to integrate
du/(a^2 - u^2) you most certainly do not need to use partial fractions. though it is an accepted method, the inverse hyperbolic functions yield this integral and my calc ii class hasnt gotten to partial fractions by the time i need this integral in physics! how silly! my understanding wasnt that in hyperbolic functions x was replaced by ix. i learned that they are defined by using e^x and e^-x and division by 2. sin is minus, cos is add. is that wrong? there are no i or imaginary numbers in that.

quasar> my masses cancelled out, but i see how to do that algebra since you cleaned it up for me.

thanks guys.

5. Sep 28, 2004

abertram28

sorry to double post...

im screwed up where i get to the V = Vo/(1+Vokt)

where do Vo come from, the C from integration?

6. Sep 29, 2004

abertram28

ok, i figured out my problem with this problem.
first of all, i misinterpreted the problem. f=-mkv^2 the opposing force is a constant and not needed to find the v and a equations.

my problem is after i integrate my dv/-kv^2=dt i need to solve the definate integral from 0 to V, right? if i solve 1/kv for v=0, i get 1/0.

how do i work through this?