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I'm reviewing some mechanics, and having an issue where my choice of coordinate system results in an incorrect derivation, one that is clearly incorrect. I suspect there's a missing step in my reasoning, but I'm not sure exactly why that is. It might just be a math issue, so I'll show the work as well. Here's the situation:

We drop an object, treated like a particle, into a container of fluid ($v_y$ = 0 at the top of the container). Assume the only forces on the object are from gravity and from a "fluid resistance" with magnitude $kv$; this resistance force is in the opposite direction of the motion of the particle. We are trying to find the velocity and position of the particle as a function of time. We'll only look at vertical movement, and so will leave off the $y$ subscript for $F_y$, $a_y$, $v_y$, and so on.

Let's consider two coordinate systems and solve with each one to see the issue arise:

^ -y                                 ^ +y
|                                    |
|____> +x             And            |____> +x

Thus the particle is moving in the $+\hat{y}$ direction in the left system, and the $-\hat{y}$ direction in the right system.

In the left system, the force of gravity is $mg$, and the force of fluid resistance is $-kv$.

In the right system, the force of gravity is $-mg$, and the force of fluid resistance is $kv$. (note that I suspect it's here that the issue is, since the lack of a negative on the kv term causes the issue, and experimentally adding it fixes results in the correct solution, but I don't know why it would be added).

Thus:

Left System                  Right System

$\sum{F} = ma = mg - kv$            $\sum{F} = ma = kv - mg$

Divide by the right side, then mutliply by $-\frac{k}{m}$ or $\frac{k}{m}$ to simplify integration.

$\frac{-ak}{mg - kv} = \frac{-k}{m}$                 $\frac{ak}{kv - mg} = \frac{k}{m}$

$\int_{0}^{v}\frac{-ak}{mg - kv} = \int_{0}^{t}\frac{-k}{m}$              $\int_{0}^{v}\frac{ak}{kv - mg} = \int_{0}^{t}\frac{k}{m}$

Integrate, then combine the logarithms, and simplify the negative in the right logarithm.

$\ln|\frac{mg-kv}{mg}| = \frac{-kt}{m}$               $\ln|\frac{mg-kv}{mg}| = \frac{kt}{m}$

Exponent, then simplify the fraction.

$1 - \frac{kv}{mg} = e^{\frac{-kt}{m}}$                $1 - \frac{kv}{mg} = e^{\frac{kt}{m}}$

Solve for v.

$v = \frac{mg}{k}(1 - e^{\frac{-kt}{m}})$               $v = \frac{mg}{k}(1 - e^{\frac{kt}{m}})$

And we can already see the issue here, but let's find the acceleration as a function of time by differentiating, just to make it clearer (we can integrate to find the position as a function of time instead).

$a = ge^{\frac{-kt}{m}}$                   $a = -ge^{\frac{kt}{m}}$

Obviously, as time goes on, the acceleration should tend to zero as time increases, which is the case for the left equation. But for the right equation, it tends to infinity as time increases, and I'm not sure why the change in coord system causes this. I suspect it's just something to do with the direction of $v$, but I can't quite figure it out. If you just make $kv$ into $-kv$ in the right equation, you get the right answer, further suggesting it's just a direction issue.

I'm interested in modeling these kinds of things, so understanding the difference and making sure it's correct in diff coordinate systems is important to me.

Thanks for your time!

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  • $\begingroup$ If $k$ is positive, the force of fluid resistance is $-kv$ in both coordinate systems. The force is in the opposite direction to $v$. It doesn't matter whether you call the positive direction "up" or "down". In one system you have $v$ positive and the force negative, in the other $v$ is negative and the force is positive. $\endgroup$
    – alephzero
    Jan 14, 2019 at 22:01
  • $\begingroup$ The equations of motion can’t depend on the choice of the coordinate system. To get a „stable“ result the equation must be $ma+kv=\pm mg$ Important is +v $\endgroup$
    – Eli
    Jan 14, 2019 at 22:35

3 Answers 3

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Let the up unit vector be $\hat u$ and the down unit vector be $\hat d$.

Your left equation of motion you have written as $ma \,\hat d = mg \,\hat d - kv \,\hat d$ with the acceleration as $a \,\hat d$ and the velocity as $v\,\hat d$ where $a$ and $v$ are the components of the acceleration in the $\hat d$ direction.

Why did you write $mg\,\hat d$? It was because you decided that the gravitational field was downwards and $g$ to be a positive number.

Why did you write $-kv\,\hat d$?
It was because if the velocity of the mass was in the $\hat d$ direction then the component of velocity $v$ would be a positive number and if $k$ is a positive constant the drag $kv \, (-\hat d)$ would be in the negative $\hat d$ direction ie upwards.
Note that if the mass was moving upwards $v$ would be negative and $-kv\,\hat d$ would be in the downward direction because $-kv$ is positive.
So the drag term works for masses falling downwards and moving upwards.

Now lets consider the motion using the up unit vector $\hat u$.
What you have to remember is that when the mass is falling down $v$, the component of velocity in the upward direction, will be negative.
This means that the drag term is now $-kv \,\hat u$ and not$+kv \,\hat u$ as you have assumed in your right hand system.
My equation assigns the correct direction (opposite to the direction of the velocity) to the direction of the drag.
For the mass falling down $v$ is negative so $-kv$ is positive ie in the upward $\hat u$ direction.
You can check it also works for the mass moving upwards.

Now what about the equation of motion?
You can write it as $ma\,\hat u = -mg \,\hat u-kv\,\hat u$ and solve this equation to get $v = -\frac{mg}{k}(1 - e^{\frac{-kt}{m}})$ for the component of velocity in the upward $\hat u$ direction.

The only difference from your other equation is that a negative sign has appeared because of the reversal of the direction of the unit vector from $\hat d$ to $\hat u$.

Note if you leave the assignment of the direction gravitational field right to the end the equation of motion is $ma \,\hat d = mg \,\hat d - kv \,\hat d$, the same as you had before.
All that you will have to remember is that when numerical values are substituted into the equation for the component of the velocity in the upward $\hat u$ direction, the numerical value of the gravitational field strength $g$ would be negative.

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  • $\begingroup$ Thanks! Very thorough explanation, and using unit vectors in the equations is a good way to avoid this in the future. $\endgroup$
    – Brandon
    Jan 15, 2019 at 2:43
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In both coordinate systems, the force is $-kv$. The force is always opposite the direction of the velocity. Flipping the positive-y direction can EITHER be taken care of by sending $mg\rightarrow-mg$ or by sending $v\rightarrow -v$ and $a\rightarrow -a$. It depends on the convention you choose. But you don't flip the sign of $v$ and $mg$ and ignore $a$.

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  • $\begingroup$ The note about ignoring a is I think the critical part I missed when flipping these, and utilizing unit vectors (as in the accepted answer) is a way of making sure you don't ignore these by accident. Thanks! $\endgroup$
    – Brandon
    Jan 15, 2019 at 2:46
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The equation of motion can't depend on the choice of a coordinate system.

the correct equation is

$$m\,\ddot{x}+k\,\dot{x}=-m\,g\tag 1$$

why ?

To solve the equation you make this ansatz:

$x(t)=A\,e^{\lambda\,t}$

$\dot{x}(t)=A\,\lambda\,e^{\lambda\,t}$

$\ddot{x}(t)=A\,\lambda^2\,e^{\lambda\,t}$

in equation (1), with $m\,g=0$

$A\,e^{\lambda\,t}\left(\lambda^2+k\,\lambda\right)=0$

$\Rightarrow\quad \lambda=-k \quad$ and $\quad x(t)=A\,e^{-k\,t}$, so for $t\mapsto\inf\quad$ you get $\quad x(t)\mapsto 0$

But if you write this equation:

$$m\,\ddot{x}-k\,\dot{x}=-m\,g\tag 2$$

with the same ansatz , you get for $t\mapsto\inf\quad$ $x(t)\mapsto \inf$

this is not waht you want?

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