Drawback of dimensional analysis - motion in a resistive medium

[brotherbobby]

TL;DR Summary

The author of a problem book on introductory mechanics discusses, in his first chapter, some of the drawbacks of dimensional analysis. Among others, he mentions the case of the distance travelled by a particle with some initial velocity in a resistive medium. I copy and paste the portion of the text below.

I hope someone can explain to me the meaning of the sentence I underline. I fail to see how $l_{max}$, the total distance covered by the particle, can be indefinitely large.

Statement from the text : I copy and paste the portion of the text that am struggling to understand and underline in red the claim the author makes which I can't believe to be true.

Doubt : As you can read in the first line of the paragraph and in the one I underlined, the author believes that $l_{max}$ "blows up to infinity". But isn't it the case that a particle travelling with some initial velocity $v_0$ would come to a stop in a resistive medium after covering a finite distance?

A hint or help would be appreciated. Thank you for your time.

 

[Frabjous]

Solving dv/dt = -v² gives Δt ~ (1/v_final)-1 which gives Δx ~ ln(1+Δt).

 

[brotherbobby]

Thank you @Frabjous . Writing $F = m\frac{dv}{dt} = -kv^2\Rightarrow \frac{dv}{dt}=-\frac{k}{m}v^2$, which, upon solving first for $v$ and then for $x$ gives what you wrote : $\boxed{\Delta x = \frac{1}{\alpha}\ln\left(1+\frac{\alpha}{\beta}\Delta t\right)}$, where $\Delta x$ is the total distance travelled by the particle for a time $t$ and the constants $\alpha = \frac{k}{m}$ and $\beta = \frac{1}{v_0}$, $v_0$ being the initial velocity.

So the conclusion is that for a length of time $\Delta t$ large enough, the distance travelled $\Delta x$ goes up indefinitely?

Also, doesn't it violate intuition? Don't you expect the particle to come to rest at some stage?

 

[Swamp Thing]

doesn't it violate intuition? Don't you expect the particle to come to rest at some stage?

I too was taken aback for a few seconds. But in fact, this intuition is based on the fact that in the real world, there is always some static friction that kicks in below a certain velocity, bringing the object to a dead stop. In a situation without static friction the velocity only approaches zero without ever becoming actually zero.

 

[brotherbobby]

yes @Swamp Thing , but there is one point that I failed to consider. The resistive force varies as a square of the velocity. Hence the smaller the velocity, the less the resistance. In high school physics, we speak of static friction ($\mu_S$) which has some maximum value and kinetic friction ($\mu_K$) which is lower than static friction and acts for as long as the body moves. Both these frictions have constant values. In fact, we don't talk of a velocity-dependent kinetic friction $\mu_K(v)$. I am not sure frictional coefficients vary with velocity for moving bodies when both surfaces are solid. But surely they do when bodies move in fluids.

Here, as you said, the velocity will fall, which will make the frictional force fall, which in turn will make the velocity fall but less so. With the velocity never becoming zero, it is conceivable that the particle keeps moving for good.

I plot below the distance-time graph for the function $x(t) = 0.5\ln (1+0.5t)$. I don't know if the coefficients are realistic. Upon zooming and going as far along the time axis as I could, I found that the distance covered doesn't end in a horizontal line.

 

[Frabjous]

ln(infinity)=infinity

If memory serves, there is a closed form solution for F=a+bv^2

 

[Haborix]

I'm not really sure the author makes a very clear point. We usually use dimensional analysis to how things will scale in the physical constants. You can actually see from the full solution in post #3 that it does scale like dimensional analysis suggests, but there is an additional logarithmic factor. I think the author of the book simply asks too much of dimensional analysis.

 

[A.T.]

In high school physics, we speak of static friction ($\mu_S$) which has some maximum value and kinetic friction ($\mu_K$) which is lower than static friction and acts for as long as the body moves. Both these frictions have constant values.

Static friction doesn't have to be constant, the constant coefficient just tells you its maximal value, as you wrote.

The transition between sliding (kinetic friction) and grip (static friction) is complex. When you consider the macroscopic deformation of an object, static friction can be a function of that deformation and raises briefly to stop the object within the distance given by the limits of deformability, then goes to back zero once the deformation ceases.

 

[brotherbobby]

I'm not really sure the author makes a very clear point. We usually use dimensional analysis to how things will scale in the physical constants. You can actually see from the full solution in post #3 that it does scale like dimensional analysis suggests, but there is an additional logarithmic factor. I think the author of the book simply asks too much of dimensional analysis.

The author is showing where dimensional analysis fails. This is one example in that it cannot predict the factor under logarithmic (or exponential/trigonometric) expressions.

 

[Haborix]

It even does that in this case, dimensional analysis would give you a time scale $\beta/\alpha$, using post #3 notation. The author already goes off the tracks by putting an equal sign between $\ell_{max}$ and the constants of equivalent dimension. I'd ignore this section of the book and maybe the whole book if it is full of this kind of thing.