How Does Viscosity Affect Motion in a Fluid?

[dykuma]

Homework Statement

A mass M falls under gravity (force mg) through a liquid with decreasing viscosity so that the retarding force is -2mv/(1+t). If it starts from rest, what is the speed, acceleration, and distance fallen at time t=1.

Homework Equations

F=ma

The Attempt at a Solution

F = 2mv/(1+t) - mg
F = m[- g + 2v/(1+t)]
ma = m[-g + 2v/(1+t)]
a = -g + 2v/(1+ t)

At this point, I see two paths to a solution, and I don't know which is correct.
I can let a=x'', v=x'
x'' - (2/(1+ t)) x' = -g
where the solution according to mathmatica is (I'm not sure how to solve this exactly, my guess is to use frobenius)
x =1/3 - C1 T^3 + 1/2(2 C1 + g) T^2 + g T + C1 T + C2

or I can let
a=v'
v' = -g + 2v/(1+ t)
v' - 2v/(1+ t)= -g
Where the solution would be
v = (1 + t) (C1 t + C1 + g) - Edit (I see this solution might be wrong, checking now)

I'm not sure which is the correct result, and I am concerned about the units on each,

 

[cnh1995]

a=v'
v' = -g + 2v/(1+ t)

I would prefer this.
But I think the equation should be
F=mg-2v/(1+t).
Negative sign of the retarding force indicates that it is directed opposite to the gravity.

 

[dykuma]

I would prefer this.
But I think the equation should be
F=mg-2v/(1+t).
Negative sign of the retarding force indicates that it is directed opposite to the gravity.

I see, I was thinking in terms of gravity pulling down, retarding force pushing up.

 

[dykuma]

Okay, flipping the signs around I get the following:
a = g - 2v/(1+ t)
let a=v'
v' = g - 2v/(1+ t)
v' + 2v/(1+ t) = g

Does this look correct? And what should I do with this constant C?

 

[cnh1995]

I see, I was thinking in terms of gravity pulling down, retarding force pushing up.

I think that won't make any difference as long as both the forces are directed opposite to each other.

 

[cnh1995]

v' + 2v/(1+ t) = g

I believe this could be solved using the integrating factor method.
If dy/dx+Py=Q, the integrating factor I=e^∫Pdx. Look up this method.

 

[dykuma]

I believe this could be solved using the integrating factor method.
If dy/dx+Py=Q, the integrating factor I=e^∫Pdx. Look up this method.

I did that, and I got a solution that matched the more complicated case of us x''=a, x'=v.

The problem now Is that I am unsure of what to do with those constants.

 

[cnh1995]

I did that, and I got a solution that matched the more complicated case of us x''=a, x'=v.
View attachment 108277

The problem now Is that I am unsure of what to do with those constants.

First find the solution for v as a function of time. Here, you will have only one constant. Use the initial condition i.e. v=0 at t=0 to get this constant. So you'll have a complete v vs t relationship. Now differentiate it w.r.t time and put t=1 to get the acceleration at t=1 and integrate v(t) w.r.t time from t=0 to t=1 to get the displacement. I don't think you need three constants for this.

 

[dykuma]

First find the solution for v as a function of time. Here, you will have only one constant. Use the initial condition i.e. v=0 at t=0 to get this constant. So you'll have a complete v vs t relationship. Now differentiate it w.r.t time and put t=1 to get the acceleration at t=1 and integrate v(t) w.r.t time from t=0 to t=1 to get the displacement. I don't think you need three constants for this.

Oh my god, I'm such and Idiot. I don't know why I didn't see that before!

Thanks!