The falling raindrop, revisited Available to Purchase

K. S.

 

Krane

, “,”  , – ().

I.

 

Adawi

, “,”  , – ().

B. G.

 

Dick

, “,”  , – ().

M. H.

 

Partovi

and

D. R.

 

Aston

, “,”  , – ().

As a recent article put it, “undergraduate mechanics students are sometimes [my emphasis] able to solve the nonlinear dynamical equations of motion to find the deceptively simple acceleration $g/7$” [

B. F.

Edwards

,

J. W.

Wilder

, and

E. E.

Scime

, “,”  , – ()].

A physics student, answering another student’s query on an online forum, was blunter: “This is a very old problem. Unfortunately, I remember the answer, $g/7$⁠, but I don’t remember how you get it. It has an unusual solution. There [is] a special substitution that you need to make for the mass; otherwise the problem is insoluble” ⟨www.physicsforums.com/showthread.php?t=198859⟩.

For instance, if $F(x,v)=g(x)h(v)$⁠, then Eq. (3) is separable [in particular, when $F(x,v)=g(x)$⁠, the solution to Eq. (3) gives the usual conservation-of-energy equation]. If $F(x,v)=g(x)v+h(x)v2$⁠, then Eq. (3) is a first-order linear equation with nonconstant coefficients for the function $v(x)$⁠; and more generally, if $F(x,v)=g(x)v2−α+h(x)v2$⁠, then Eq. (3) is a first-order linear equation for the function $v(x)α$⁠. Likewise, if $F(x,v)=v/[g(v)+h(v)x]$⁠, then Eq. (3) can be turned upside-down to obtain a first-order linear equation with nonconstant coefficients for the inverse function $x(v)$⁠; and more generally, if $F(x,v)=v/[g(v)x1−β+h(v)x]$⁠, then Eq. (3) gives a first-order linear equation for the function $x(v)β$⁠.

At least two other cases of Eq. (6) correspond to physically realizable (albeit highly artificial) situations: namely, $(α,β)=(0,1)$ and $(1,1)$ arise when the raindrop is constrained (for example, by a massless container) to be a cylinder of fixed base and growing height (respectively, fixed height and growing base). Note that the cylinder’s base can be of arbitrary shape and need not be circular.

The case $β=−1$ needs to be treated separately and yields $v=Cm−1exp[{g/[(1−α)λ]}m1−α]$⁠. Of course, all cases $β<0$ are unphysical.

The case $α=2+β$ needs to be treated separately and yields $v=({[(1+β)g]/λ}{[log(m/m0)]/m1+β})1/(1+β)$⁠. Of course, all cases $α>1$ are probably unphysical.

This step assumes $β>−1$⁠.

This step assumes $β>−1$ and $α<1$⁠.