Equation of motion with proportional drag


by Hannibal123
Tags: drag, equation, motion, proportional
Hannibal123
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#1
Dec19-12, 02:47 PM
P: 19
1. The problem statement, all variables and given/known data

the forces will be like this

[tex]m⋅dv/dt=-m⋅g-k⋅v[/tex]

I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0



3. The attempt at a solution
i have found this
[tex]v(t)=mg/k*(-1+e^(-k/m⋅t))+v_0⋅e^(-k/m⋅t)[/tex] or in picture form http://imgur.com/WjksG

But I am not sure it's correct
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ehild
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#2
Dec19-12, 03:39 PM
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Quote Quote by Hannibal123 View Post
1. The problem statement, all variables and given/known data

the forces will be like this

[tex]m⋅dv/dt=-m⋅g-k⋅v[/tex]

I need to find the velocity equation and the place equation, thereby meaning the integral of the velocity equation. for the condition t,v=0,v_0



3. The attempt at a solution
i have found this
[tex]v(t)=\frac{mg}{k}(-1+e^{-\frac{k}{m}t})+v_0⋅e^{-\frac{k}{m}t}[/tex] or in picture form http://imgur.com/WjksG

But I am not sure it's correct
It looks correct, but v0=0 .

Edit: I misread vo, so it is correct.

ehild
haruspex
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#3
Dec19-12, 03:43 PM
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Your answer looks right to me. Why do you doubt it?

haruspex
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#4
Dec19-12, 03:44 PM
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Equation of motion with proportional drag


Quote Quote by ehild View Post
It looks correct, but v0=0 .

ehild
I read the OP condition as initially (t,v) = (0, v0)
ehild
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#5
Dec19-12, 03:53 PM
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Quote Quote by haruspex View Post
I read the OP condition as initially (t,v) = (0, v0)
Sorry, I misread it. Edit my post.

ehild
Hannibal123
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#6
Dec19-12, 04:40 PM
P: 19
Quote Quote by haruspex View Post
I read the OP condition as initially (t,v) = (0, v0)
that is what i meant. I doubt the results because when i draw them in a graph they dosen't seem to be correct. I have drawn them i geogebra if you are familiar with that software.
My values are:
m: 0,145 kg
k: 0,0032
g: (gravity acceleration) 9,82
v_0: 9,93 m/s

this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
if you want to try for yourself (should save you some time). These values are not meant for a proportional drag, but still it seems weird that the graph looks like one of a constant function
http://imgur.com/FFElN (the red one)
Hannibal123
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#7
Dec19-12, 04:42 PM
P: 19
Furthermore when i integrate the velocity equation i get this http://imgur.com/IO6G6
Again not sure if I am correct
haruspex
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#8
Dec19-12, 07:44 PM
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Quote Quote by Hannibal123 View Post
m: 0,145 kg
k: 0,0032
g: (gravity acceleration) 9,82
v_0: 9,93 m/s

this is what i have entered in geogebra: g(x) =(0.145*9.82/0.0032)*(-1 + ℯ^(0x / 0.15)) + 9.93ℯ^(0x / 0.15)
How did -k/m end up as 0 / 0.15? Looks like rounding error, and the red line graph seems to be a consequence.
Hannibal123
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#9
Dec19-12, 07:59 PM
P: 19
Turns out i did make some input mistakes, this looks better
http://imgur.com/eVRwQ

EDIT*
Nope wrong again, this should be the correct equation http://imgur.com/bJ19e
Hannibal123
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#10
Dec19-12, 08:09 PM
P: 19
this is the equation of place (if that is the correct term in english) as far as I am concerned the height is not suposed to be negative to positive time values???
http://imgur.com/CxAYC
haruspex
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#11
Dec19-12, 09:54 PM
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Your velocity time graph shows velocity +ve at time 0, so the distance time graph should show distance increasing at time 0.
Hannibal123
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#12
Dec20-12, 05:29 AM
P: 19
Indeed it should, however it dosen't. Is this a wrong integration of the velocity equation?
http://imgur.com/A6BKT
Based on the gaph it does not seem to be correct.
Hannibal123
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#13
Dec20-12, 09:14 AM
P: 19
Just found the correct place equation i think
http://imgur.com/quiLu
can you see a good way to shoten it?
haruspex
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#14
Dec20-12, 03:21 PM
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Looks right. I'd write it as
[itex]\frac{mg}{k}\left(-t+\left(\frac{m}{k}+\frac{v_0}{g}\right)\left(1-e^{-\frac{kt}{m}}\right)\right)[/itex]


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