Problems with modelling the vertical motion of a dust particle

[Ax_xiom]

TL;DR

I am getting confusing results whenever I use differential equations to model the vertical motion of a dust particle

So I was experimenting with using differential equations to model motion and I wanted to use one to model the motion of dust being projected vertically upwards, and the differential equation I got was this $$\frac {dv} {dt} = -g-cv^2$$ where g is the acceleration due to gravity and c is a constant formed from air density, mass of the particle, drag coefficient and surface area which turns out to be about 0.787

When I solve the equation (with the boundary condition that the velocity at 0 is some set intial velocity) I get this: $$v = \frac {\sqrt{g} \tan\{\tan^{-1}\{\frac {\sqrt{c}} {\sqrt{g}} V_i\} - \sqrt{cg} t\}} {\sqrt{c}}$$

But when I plot and intergrate this I get weird results. If I set the inital velocity to the speed of light the model predicts it would only travel 23m upwards which makes no sense. So what happened and what did I do wrong here?

 

[anuttarasammyak]

\frac {dv} {dt} = -g-cv|v|
seems better. The second term should be opposite to the direction of v.

 

[hutchphd]

Why do you expect a simple 1 dimensional model to give an interesting result? I am asking you to explain the form of the equation you have written.....

 

[erobz]

What do you find for max height after you change the independent variable to height via $$ \frac{dv}{dt}= \frac{dv}{dh}v $$ and solve for it directly?

 

[Ax_xiom]

\frac {dv} {dt} = -g-cv|v|
seems better. The second term should be opposite to the direction of v.

This is only for the dust particle moving upwards so drag and gravity are always moving in the same direction

Why do you expect a simple 1 dimensional model to give an interesting result?

I thought the differential equation would work

Why do you expect a simple 1 dimensional model to give an interesting result? I am asking you to explain the form of the equation you have written.....

So the particle is moving upwards with a velocity V and forces W (weight) and D (drag) acting against it. $$
F = -WD $$
$$W = mg $$
$$D = \frac {C_d \rho A V^2}{2} $$
$$F = -mg-\frac {C_d \rho A V^2}{2} $$
$$\frac {F}{m} = a $$
$$a = -g -\frac {C_d \rho A V^2}{2m} $$
$$c = \frac {C_d \rho A}{2m} $$
$$a = -g - cV^2 $$
$$a = \frac {dV}{dt} $$
$$\frac {dV}{dt} = -g -cV^2 \\$$

What do you find for max height after you change the independent variable to height via dvdt=dvdhv and solve for it directly?

The issue is that the resulting equation results in the natural log of height which breaks the boundary condition (when h=0, V = V_i). Also where does that equation come from?

 

[erobz]

The issue is that the resulting equation results in the natural log of height which breaks the boundary condition (when h=0, V = V_i). Also where does that equation come from?

Im trying to manipulate in my head, but are you seeing to invert the limits because of the negative sign in front of the log. The solution should look like $$ h= k \ln \left[ \frac{g + c v_o^2}{ g + c v^2}\right] $$

 

[A.T.]

So I was experimenting with using differential equations to model motion and I wanted to use one to model the motion of dust being projected vertically upwards, and the differential equation I got was this $$\frac {dv} {dt} = -g-cv^2$$ where g is the acceleration due to gravity and c is a constant formed from air density, mass of the particle, drag coefficient and surface area which turns out to be about 0.787

When I solve the equation (with the boundary condition that the velocity at 0 is some set intial velocity) I get this: $$v = \frac {\sqrt{g} \tan\{\tan^{-1}\{\frac {\sqrt{c}} {\sqrt{g}} V_i\} - \sqrt{cg} t\}} {\sqrt{c}}$$

Have you compred your results with online material dealing with the same problem? For example:



Also, note that you can solve the problem in a free-falling frame, where only the quadratic drag is acting, just like described for horizontal motion in the first part of this document:

https://www.physics.udel.edu/~szalewic/teach/419/cm08ln_quad-drag.pdf

Once you have velocity and position as function of time in the free-falling frame, transforming them to the ground fixed frame should be easy: You just subtract the velocity and distance gained in free-fall after a given time.

 

[Ax_xiom]

Im trying to manipulate in my head, but are you seeing to invert the limits because of the negative sign in front of the log. The solution should look like $$ h= k \ln \left[ \frac{g + c v_o^2}{ g + c v^2}\right] $$

How did you get there? I'm new to solving differential equations so this is what I do:

\begin{align}
\frac {dv}{dh} h = -g-cv^2 \\
\frac {dv}{-g-cv^2} = \frac {dh}{h} \text{ Moving variables as it is a seperable differential equation} \\
-\frac {\arctan(\frac{\sqrt{c}}{\sqrt{g}}v)}{\sqrt{cg}} = \ln(h)+C \text{ solved integrals}
\end{align}

And then I get stuck as ln(h) isn't defined when h = 0

Have you compred your results with online material dealing with the same problem? For example:

Our solutions seem to be equivalent and still results in the odd result of a dust particle being projected upwards at the speed of light only moving upwards 23 meters before stopping. So what is happening?

 

[anuttarasammyak]

@Ax_xiom By integrating your v I got maximum height
h=\frac{1}{2c}\log(1+\frac{c}{g}V_i^2)h diverges for infinite $\frac{c}{g}V_i^2$. For small $\frac{c}{g}V_i^2 << 1$
h \approx \frac{V_i^2}{2g} - \frac{1}{4c} [\frac{c}{g}V_i^2]^2
RHS first term is a usual result of projectile. The second term reduces the heigt.
Does it coincide with your solution ?

 

[A.T.]

still results in the odd result of a dust particle being projected upwards at the speed of light only moving upwards 23 meters before stopping. So what is happening?

Is it an odd result? If your c is based on a tiny dust particle in a dense gas, then maybe that's what Newtonian mechanics (which is of course unrealistic at relativistic speeds) predicts.

Have you tried integrating your initial equation (a = -mg - cv²) numerically, with a sufficiently small time step?

 

[erobz]

How did you get there? I'm new to solving differential equations so this is what I do:

\begin{align}
\frac {dv}{dh} h = -g-cv^2 \\
\frac {dv}{-g-cv^2} = \frac {dh}{h} \text{ Moving variables as it is a seperable differential equation} \\
-\frac {\arctan(\frac{\sqrt{c}}{\sqrt{g}}v)}{\sqrt{cg}} = \ln(h)+C \text{ solved integrals}
\end{align}

You are going off course. Scrap it and try again.

The acceleration transforms with the chain rule like:

$$ \frac{dv}{dt} = \frac{dv}{dh}\frac{dh}{dt} = \frac{dv}{dh}v $$

Once you make the substitution you make another substitution on the other side:

$$ U = g + c v^2 $$

Now take the derivative of $U$ with respect to $h$. You should be looking for something interesting about this result and the last substitution you made with the derivative. The goal is to transform everything in the equation into a simple ODE in terms of $U$ and $h$.

Once you get that evaluate integral at the limits for $U$ as you would in a definite integral.

 

[Ax_xiom]

Is it an odd result? If your c is based on a tiny dust particle in a dense gas, then maybe that's what Newtonian mechanics (which is of course unrealistic at relativistic speeds) predicts.

I'm not necessarily interested in the relativistic case but more what it implies. I'm pretty sure that dust from blasting gets projected close or above the 23m height predicted from the speed of light case without being anywhere close to the speed of light. And Volcanic eruptions can project volcanic ash kilometers into the air which would require speeds far above the speed of light

You are going off course. Scrap it and try again.

I used $\frac {dv}{dh}h$ instead of $\frac{dv}{dh}v$

Using the right one and the substitutions you used I got the solution you posted

 

[A.T.]

I'm not necessarily interested in the relativistic case but more what it implies. I'm pretty sure that dust from blasting gets projected close or above the 23m height predicted from the speed of light case without being anywhere close to the speed of light. And Volcanic eruptions can project volcanic ash kilometers into the air which would require speeds far above the speed of light

You are conflating a single particle moving realtive to a stationary gas surrounding it, with many particles exchanging momentum and moving along with moving gas surrounding them.

 

[Ax_xiom]

You are conflating a single particle moving realtive to a stationary gas surrounding it, with many particles exchanging momentum and moving along with moving gas surrounding them.

That makes a lot of sense, I thought there was something wrong with my model but didn't know what, thanks!

 

[erobz]

That makes a lot of sense, I thought there was something wrong with my model but didn't know what, thanks!

It is interesting, the first out of the top will have resistance, but flow is going to initiate there after locally making the relative velocity between the atmosphere and ejecta go toward zero.

Also, as much as we would like to apply it at “c” as an “∞” the model surely breaks down well before. I don’t think it was obvious unlike the others seem to, and thus a good sanity check demonstrating critical thinking.

 

[Ax_xiom]

It is interesting, the first out of the top will have resistance, but flow is going to initiate there after locally making the relative velocity between the atmosphere and ejecta go toward zero.

I would assume that happens quite quickly due to dust being small so very affected by the movement of the atmosphere

I don’t think it was obvious unlike the others seem to, and thus a good sanity check demonstrating critical thinking.

I think one of the skills needed in physics is being able to tell if the result you're getting makes sense, and if it doesn't being able to figure out why