I'd like to ask about using dimensional analysis to model the projectile of a mass that's thrown vertically upward. In many similar problems, air resistance is ignored- but in this particular case, the air resistance is considered, and its force is proportional to the square of the velocity. The coefficient for the air resistance is $a$, and its dimension is $\{a\}= ML^{-1}.$

We are also given the following differential equation:

$$m\cdot x'(t)=-mg-a[x'(t)]^2, \quad x(0) = 0, \quad x'(0) = v_0$$ Doing dimensional analysis, and calculating the dimensionless products $\Pi_1, \Pi_2, \Pi_3$ such that $\Pi_1 = F(\Pi_2, \Pi_3),$ I calculated:

$$x(t) \frac{g}{v_0^2} = f\bigg(\frac{t}{v_0/g} , \frac{mg}{v_0^2a}\bigg).$$

Now, a similarity solution/dimension reduction will allow us to simplify the differential equation to one variable, which will allow for a more straightforward diff eq solution.

I applied substitutions $u=\frac{x(t)}{v_0^2/g}$ and $\tau = \frac{t}{v_0/g}$. I wasn't sure how to incorporate a third substitution to address $\Pi_3$ (assuming one is even needed). The most I was able to reduce the differential equation above was to: $$\frac{d^2u}{d\tau^2} = -\frac{g}{v_0} - \frac{av_0}{m}\bigg[\frac{du}{d\tau}\bigg]^2.$$

Basically, I'm just asking if anyone would have insight on how to find a proper substitution, so we could end up with a similarity solution for the differential equation above.


With the substitutions to the dimensionless variables, $$u=\frac{gx(t)}{v_0^2},\>\>\>\>\tau = \frac{gt}{v_0}$$

You should get

$$\frac{d^2u}{d\tau^2} = -1 - \frac{av_0^2}{mg}\left(\frac{du}{d\tau}\right)^2.$$

which is indeed dimensionless as desired. (The differential equation you derived is not quite right.)

  • $\begingroup$ This helped me find my error. Thank you so much! $\endgroup$ – daOnlyBG Sep 14 at 19:19
  • $\begingroup$ Glad could help $\endgroup$ – Quanto Sep 14 at 19:25

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