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Finding the equation for Velocity with Drag

  • Thread starter Theorγ
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  • #1
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Homework Statement


Given a situation where a ball of mass m is dropped off at a height of h, find the equation that would give the velocity of the ball with respect to time given that gravity and drag are intrinsic factors.


Homework Equations


[tex]F_{g} = m g[/tex]
[tex]F_{D} = \frac{1}{2}pCAv^2[/tex]


The Attempt at a Solution



[tex] F_{net}= F_{g} - F_{D}[/tex]
[tex] F_{net} = m g - \frac{1}{2}pCAv^2[/tex]
[tex] a_{net} = g - \frac{1}{2m}pCAv^2[/tex]
[tex] \frac{dv}{dt} = g - \frac{1}{2m}pCAv^2[/tex]
[tex] \int \frac{1}{g - \frac{1}{2m}pCAv^2} dv = \int dt[/tex]

Now I'm stuck; what in the world would you do to integrate the left side? I tried relating [tex]\int\frac{1}{1 - x^2} dx = \int\frac{-1}{(x-1)(x+1)} dx = \frac{1}{2}ln(x + 1) - \frac{1}{2}ln(x - 1) + C[/tex] to the left side and that integrates out by factoring and using partial fractions. However, I can't find any way to easily factor, if the left side is even factorable to begin with...

_________________________
[tex] \int \frac{1}{(\sqrt{g})^2 - (\sqrt{\frac{1}{2m}pCA}v)^2} dv = t + Constant[/tex]
[tex] \int \frac{1}{(\sqrt{g} + (\sqrt{\frac{1}{2m}pCA}v))(\sqrt{g} - (\sqrt{\frac{1}{2m}pCA}v))} dv = t + Constant[/tex]
[tex] \int \frac{\alpha}{(\sqrt{g} + (\sqrt{\frac{1}{2m}pCA}v))} dv + \int \frac{\beta}{(\sqrt{g} - (\sqrt{\frac{1}{2m}pCA}v))} dv = t + Constant[/tex]
[tex] \int \frac{\frac{1}{2\sqrt{g}}}{(\sqrt{g} + (\sqrt{\frac{1}{2m}pCA}v))} dv + \int \frac{\frac{1}{2\sqrt{g}}}{(\sqrt{g} - (\sqrt{\frac{1}{2m}pCA}v))} dv = t + Constant[/tex]
[tex] u_{1} = \sqrt{g} + \sqrt{\frac{1}{2m}pCA}v[/tex]
[tex] \frac{du_{1}}{dv_{1}} = \sqrt{\frac{1}{2m}pCA}[/tex]
[tex] u_{2} = \sqrt{g} - \sqrt{\frac{1}{2m}pCA}v[/tex]
[tex] \frac{du_{2}}{dv_{2}} = -\sqrt{\frac{1}{2m}pCA}[/tex]
[tex] \int \frac{\frac{1}{2\sqrt{g}}}{u_{1}} \frac{du}{\sqrt{\frac{1}{2m}pCA}} - \int \frac{\frac{1}{2\sqrt{g}}}{u_{2}} \frac{du}{\sqrt{\frac{1}{2m}pCA}} = t + Constant[/tex]
[tex] \frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} ln(\sqrt{g} + \sqrt{\frac{1}{2m}pCA}v) - \frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} ln(\sqrt{g} - \sqrt{\frac{1}{2m}pCA}v) = t + Constant[/tex]
 
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Answers and Replies

  • #2
137
5
You're on the right track. :smile:
If you substitute ½pCA/m for e.g. k, you can easily use the identity a2 - b2 = (a-b)(a+b), from which you can integrate with partial fractions.

It's going to get messy, though.
 
  • #3
Filip Larsen
Gold Member
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  • #4
57
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So now that I've solved the velocity function, is there a way to integrate to find the position function of time?
 
  • #5
137
5
Well, v = ds/dt, so if you simplify the formula you can integrate it directly.

You're probably going to need to pull out some hyperbolic trigonometry, though.

Although you can do it without hyperbolic trigonometry, too.
 
  • #6
57
0
Well I managed to manipulate and simplify the equation into this, but I haven't learned how to use hyperbolic functions, so what do I do?

[tex]a = \frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}}[/tex]
[tex]b = \sqrt{\frac{1}{2m}pCA}[/tex]
[tex]\frac{a}{\sqrt{g}}\frac{ds}{dt} = \frac{e^{bt} - 1}{e^{bt} + 1}[/tex]
[tex]\int\frac{a}{\sqrt{g}} ds= \int\frac{e^{bt} - 1}{e^{bt} + 1} dt[/tex]

____________
EDIT: Okay so I read up briefly on how hyperbolic functions work and this is what I've come up with:
[tex]\int\frac{a}{\sqrt{g}} ds= \int\tanh(\frac{bt}{2}) dt[/tex]
[tex]\frac{a}{\sqrt{g}}s =\frac{2}{b} ln(cosh(\frac{bt}{2})) + Constant[/tex]
 
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  • #7
57
0
Can someone confirm if I manipulated the hyperbolic function correctly and if I integrated correctly?
 
  • #8
137
5
Your integration looks correct. However,
[tex]\frac{a}{\sqrt{g}}\frac{ds}{dt} = \frac{e^{bt} - 1}{e^{bt} + 1}[/tex]
there's an error in your simplification. You've forgotten there's an [tex]\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}}[/tex] before the natural logarithms.

I.e.

[tex]\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}}\times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{|\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t+D[/tex]
 
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  • #9
57
0
So it should be solved like this?:

[tex]\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t+D[/tex]

Because the object is at rest at time 0, then D is:
_____________________________________________________________
[tex]\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+0\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-0\sqrt{\frac{1}{2m}pCA}|}\right) = 0+D[/tex]
[tex]\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}|}{ |\sqrt{g}|}\right) = D[/tex]
[tex]\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln (1) = D = 0[/tex]
_____________________________________________________________
Using these variables as the following expressions:
[tex]\alpha = \frac{1}{2\sqrt{g}}[/tex]
[tex]\beta = \sqrt{\frac{1}{2m}pCA}[/tex]
_____________________________________________________________
The substitution:
[tex]\frac{\frac{1}{2\sqrt{g}}}{\sqrt{\frac{1}{2m}pCA}} \times ln\left(\frac{|\sqrt{g}+v\sqrt{\frac{1}{2m}pCA}|}{ |\sqrt{g}-v\sqrt{\frac{1}{2m}pCA}|}\right)=t[/tex]
[tex]\frac{\alpha}{\beta} \times ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)=t[/tex]
[tex]ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)=\frac{\beta}{\alpha}t[/tex]
[tex]e^{ln\left(\frac{|\sqrt{g}+v\beta|}{ |\sqrt{g}-v\beta|}\right)}=e^{\frac{\beta}{\alpha}t}[/tex]
[tex]\frac{\sqrt{g}+v\beta}{\sqrt{g}-v\beta}=e^{\frac{\beta}{\alpha}t}[/tex]
 
  • #10
57
0
Continued from above:
[tex]\sqrt{g}+v\beta=e^{\frac{\beta}{\alpha}t} \times (\sqrt{g}-v\beta)[/tex]
[tex]\sqrt{g}+v\beta=e^{\frac{\beta}{\alpha}t}\sqrt{g}-e^{\frac{\beta}{\alpha}t}v\beta[/tex]
[tex]e^{\frac{\beta}{\alpha}t}v\beta+v\beta=e^{\frac{ \beta }{\alpha}t} \sqrt{g}-\sqrt{g}[/tex]
[tex]v\beta \times (e^{\frac{\beta}{\alpha}t}+1)=\sqrt{g} \times (e^{\frac{\beta}{\alpha}t}-1)[/tex]
[tex]\frac{\beta}{\sqrt{g}}v=\frac{(e^{\frac{\beta}{ \alpha }t}-1)}{(e^{\frac{\beta}{\alpha}t}+1)}[/tex]
[tex]\frac{\beta}{\sqrt{g}}\frac{dx}{dt}=\frac{(e^{ \frac {\beta}{\alpha}t}-1)}{(e^{\frac{\beta}{\alpha}t}+1)}[/tex]
[tex]\int\frac{\beta}{\sqrt{g}}dx=\int\frac{(e^{\frac{ \beta }{ \alpha }t}-1)}{(e^{\frac{\beta}{\alpha}t}+1)}dt[/tex]
____________________________________________
Hyperbolic Substitution:
[tex]\frac{e^{2t}-1}{e^{2t}+1}=tanh(t)[/tex]
[tex]\frac{e^{\frac{ \beta }{ \alpha }t}-1}{e^{\frac{\beta}{\alpha}t}+1}=tanh(\frac{\beta}{2\alpha}t)[/tex]
[tex]\int\frac{\beta}{\sqrt{g}}dx=\int\tanh(\frac{\beta}{2\alpha}t)dt[/tex]
[tex]\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|cosh(\frac{\beta}{2 \alpha }t)| + C[/tex]
[tex] cosh(t)=\frac{e^{2t}+1}{2e^t}[/tex]
[tex] cosh(\frac{\beta}{2 \alpha }t)=\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}[/tex]
[tex]\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}| + C[/tex]
 
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  • #11
137
5
Looks fine to me. :smile:
 
  • #12
57
0
Because the object is at rest at time 0, then the object has moved a distance of 0:
[tex]\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}| + C[/tex]
[tex]\frac{\beta}{\sqrt{g}}0=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }0}+1}{2e^{\frac{\beta}{2 \alpha }0}}| + C[/tex]
[tex]0=\frac{2 \alpha }{\beta} \times ln|\frac{2}{2}| + C[/tex]
[tex]0=0 + C[/tex]
[tex]C = 0[/tex]
______________________________________________________
Completing the equation:
[tex]\frac{\beta}{\sqrt{g}}x=\frac{2 \alpha }{\beta} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}|[/tex]
[tex]x=\frac{2 \alpha \sqrt{g}}{\beta^2} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}|[/tex]
____________________________________
Previous Variable Denotation:
[tex]\alpha = \frac{1}{2\sqrt{g}}[/tex]
[tex]\beta = \sqrt{\frac{1}{2m}pCA}[/tex]
____________________________________
Simplification:
[tex]x=\frac{2 \alpha \sqrt{g}}{\beta^2} \times ln|\frac{e^{\frac{\beta}{ \alpha }t}+1}{2e^{\frac{\beta}{2 \alpha }t}}|[/tex]
[tex]x=\frac{2 \frac{1}{2\sqrt{g}} \sqrt{g}}{\sqrt{\frac{1}{2m}pCA}^2} \times ln|\frac{e^{\frac{\sqrt{\frac{1}{2m}pCA}}{ \frac{1}{2\sqrt{g}} }t}+1}{2e^{\frac{ \sqrt{\frac{1}{2m}pCA}}{2 \frac{1}{2\sqrt{g}} }t}}|[/tex]
[tex]x=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{2m}}t}}|[/tex]
____________________________________
Okay so can someone confirm the validity of my steps now, and confirm whether the below equation represents an object with mass m, dropped at a height of h, with respect to time t, given that gravity and drag are the only forces acting on it:

[tex]h=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{2m}}t}}|[/tex]
 
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  • #13
57
0
I redid the problem on paper this time and I got a slightly different answer:

[tex]h=\frac{2mg}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}|[/tex]

Compared to my original one from the above post:

[tex]h=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}|[/tex]

Can anyone confirm which one is right? My new one contains an extra g, but my other one doesn't...
 
  • #14
137
5
[tex]h=\frac{2m}{pCA} \times ln|\frac{e^{2 \sqrt{\frac{pCAg}{2m}}t}+1}{2e^{\sqrt{\frac{pCAg}{ 2m}}t}}|[/tex]
It's this one. If you check the units, you get meter.
 

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