- #1
abertram28
- 54
- 0
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. I am 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 I am using -mkv^2 instead of -bv. we are theorizing a solution for a ball falling through a liquid. i know the units don't 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. I am 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. I am lost after this, if I am correct or not, i don't know. i tried to differentiate my answer, buts its too hard for me to do by hand. I am tired. do i just solve for v? how do i go about doing that? where the heck does Vo come into play?
F(friction) = -bv
we applied F=ma and got
mg-bv = m (dv/dt)
i realize how this works. I am 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 I am using -mkv^2 instead of -bv. we are theorizing a solution for a ball falling through a liquid. i know the units don't 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. I am 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. I am lost after this, if I am correct or not, i don't know. i tried to differentiate my answer, buts its too hard for me to do by hand. I am tired. do i just solve for v? how do i go about doing that? where the heck does Vo come into play?