Sinking Object Motion Equations

[Johnnnnnnnn]

Hi guys! I am currently learning about fluid dynamics, and I am stuck on a certain equation derivation. It's about sinking motion which considers only gravity force, buoyant force, and viscous resistance. The link attached has the details.

http://hyperphysics.phy-astr.gsu.edu/hbase/lindrg.html#c2
The problem I have is how the equation for velocity was found. I understand that Fnet = mg - pVg - bv, which is basically net force = weight - buoyant force - viscous resistance, but I don't get how the velocity (equation below) was derived in terms of time, terminal velocity (Vt), and initial velocity (V0).

Is there a mathematical proof for this? Or was the equation created based on trial and error? If you could, could you also please explain how the equation for distance (image below) was also obtained? Thanks!

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[BvU]

Hello Joh8n,

The Newton equation of motion $F_{\rm net} = mg'- bv$ is a differential equation
($F_{\rm net} = {dv\over dt}$)

The solution is discussed here (6.4.10 onwards; they don't express in terms of $v_0$ and $v_t$, but I hope you can follow the reasoning)

 

[Johnnnnnnnn]

Oh I see... Thanks for the help!

 

[Johnnnnnnnn]

Hello Joh8n,

The Newton equation of motion $F_{\rm net} = mg'- bv$ is a differential equation
($F_{\rm net} = {dv\over dt}$)

The solution is discussed here (6.4.10 onwards; they don't express in terms of $v_0$ and $v_t$, but I hope you can follow the reasoning)

So after seeing the derivation of sinking motion, I was wondering if this method of derivation is also correct.

Since $F_{\rm net} = ma = mg'- bv$,
$a = g' - (b/m)v$

From here, could I set a = d²y/ dt², and v = dy/dt, letting y be the vertical distance the object travels, and use second order linear differential equation to solve it?

So in another words, (using y' as the notation for dy/dt)

y'' = g - (b/m) y'
y'' + (b/m)y' = g

Would solving for y with this second order linear differential equation be a correct way to solve for y? If not, could you also please explain why such way is not applicable? Thanks!

 

[BvU]

It's still an equation in y', not in y ...

 

[Johnnnnnnnn]

But I thought second order linear differential equations are in the form of y'' + p(x)y' + q(x)y = g(x), where q(x) would be 0 in this case. I'm guessing you mean that it's unnecessary to change v into y' as it just complicates the calculation?

 

[BvU]

Correct. You have an equation where y does not appear, but y' does. So you solve for y' and then integrate once to get y.