Drag exerted on condensing raindrop

Question

Newton's second law of motion $F = ma$ = $m{dv\over dt}$can be written in the form $F = {d\over dt}(mv)$ in terms of the momentum $mv$ of a particle of mass $m$ and velocity $v$, and remains valid in this form even if $m$ is not constant, as assumed so far. Suppose a spherical raindrop falls through air saturated with water vapor, and assume that by condensation the mass of the raindrop increases at a rate proportional to its surface area, with $c$ the constant of proportionality. If the initial radius and velocity of the raindrop are both zero, show that the drag exerted by the condensation of the water vapor has the effect of making the raindrop fall with acceleration ${1\over 4}g$.

• Hint: Show that ${d\over dr}(r^3v) = ({\delta \over c})r^3g$, where $r$ is the radius of the raindrop and $\delta$ is its density.

This is a question from a math textbook from the chapter integration and differential equations

My approach

Given that, ${dm\over dt} \propto 4\pi r^2$

${dm\over dt} = c\cdot 4\pi r^2\cdot \delta \cdot{dr\over dt}$

since mass = density × volume

$m = {4\over 3}\pi r^3\cdot\delta$

I can't write the equation for $v$.

Also I am totally stucked at this step.

Plus I can't reach the hint given in question.

Any kind of help or suggestion or worked out steps would be appreciated.

Answer 1

Substitute $m=\frac{4\pi\delta}3r^3$ into the Newton’s law $F=mg = \frac{d}{dt}(mv)$ to get $$r^3g = \frac{d}{dt}(r^3v)\tag1$$ and also into the proportional condensation $\frac{dm}{dt}= 4\pi r^2 c$ to get $$\frac{dr}{dt}=\frac c\delta$$ Then, plug the resulting solution $r= \frac c\delta t$ into (1) to get $$t^3g =\frac{d}{dt}(t^3v)$$ Integrate both sides to obtain $\frac14 t^4g = t^3 v$, or $v=\frac14gt$, i.e. accelerating at $\frac14g$.

Answer 2

SEVERAL HINTS (Try to use as few of them as possible!)

You are told that ${dm\over dt} = c\cdot 4\pi r^2$ s0 your first equation needs correcting.

From $m=\frac{4}{3}\pi r^3\delta$ you should now be able to get ${dr\over dt} = \frac{c}{\delta}.$

You now need to use $mg={d\over dt} (mv)={d\over dr} (mv){dr\over dt}$. Substituting $m=\frac{4}{3}\pi r^3\delta$ and ${dr\over dt} = \frac{c}{\delta}$ will get you to the hint in the textbook.

You are now nearly finished! Integrate the hint equation with respect to $r$ and simplify the expression this gives you for $v$. Finally differentiate $v$ with respect to $t$ to obtain the required answer. (Remember that you know the value of (${dr\over dt}$).

Answer 3

An alternative solution is to explicitly calculate $m(t)$: $$\frac{dm}{dt}=c4\pi r^2\delta$$ With $$m=\frac{4\pi}3r^3\delta$$you get $$\frac d{dt}\left(\frac{4\pi}3r^3\delta\right)=4\delta\pi r^2\frac{dr}{dt}=c4\pi r^2\delta$$ This simplifies to $$\frac{dr}{dt}=c$$Knowing that $r(0)=0$, you get $$r(t)=ct$$ Then $$m(t)=\frac{4\pi}3\delta r^3(t)=\frac{4\pi}3\delta c^3t^3$$ Then $$\frac d{dt}(mv)=mg$$can be written as $$v=g\frac1{m(t)}\int_0^t m(\tau)d\tau$$ or $$a=\frac {dv}{dt}=g\frac d{dt}\left[\frac1{m(t)}\int_0^t m(\tau)d\tau\right]\\=g\left[\frac{-1}{m(t)^2}\frac{dm}{dt}\int_0^t m(\tau)d\tau+\frac1{m(t)}m(t)\right]$$ I've used here the product rule for derivation and the fundamental theorem of calculus. Then $$m(t)^2=\left(\frac{4\pi}3\right)^2\delta^2c^6t^6\\\frac{dm}{dt}=4\pi r^2c\delta=4\pi c^3\delta t^2\\\int_0^tm(\tau)d\tau=\frac{4\pi}3 \delta c^3\frac{t^4}4$$ Use these values in the equation for acceleration and you get $$a=g\left[-\frac34+1\right]=\frac g4$$