# Physics Challenge I: The Raindrop solved by mfb and voko

• micromass

#### micromass

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Assume that a cloud consists of tiny water droplets suspended (uniformly distributed, and at rest) in air, and consider a raindrop falling through them. What is the acceleration of the raindrop? Assume that the raindrop is initially of negligible size and that when it hits a water droplet, the droplet’s water gets added to it. Also, assume that the raindrop is spherical at all times.

Assume that a cloud consists of tiny water droplets suspended (uniformly distributed, and at rest) in air, and consider a raindrop falling through them. What is the acceleration of the raindrop? Assume that the raindrop is initially of negligible size and that when it hits a water droplet, the droplet’s water gets added to it. Also, assume that the raindrop is spherical at all times.
I'm guessing we aren't allowed to assume constant acceleration and just say ##9.8\text{ ms}^{-2}##, right? :tongue:

Should we assume that the cloud droplets are not affected by gravity at all? Is the raindrop affected by gravity?

Should we assume that the cloud droplets are not affected by gravity at all? Is the raindrop affected by gravity?

The raindrop itself is of course affected by gravity since it is falling to the ground.
The cloud droplets can be assumed not to be affected by gravity, since they are at rest.

Let ##\rho## be the fraction of water in the cloud (by volume). Neglect air resistance and assume that the vertical size of the raindrop does not matter (this is equivalent to the assumption ##\rho \ll 1##).
Let r be the radius of the raindrop, V its volume and v its vertical velocity. Define z as vertical coordinate, with z=0 as the point where our raindrop starts and positive z downwards.

##V=\frac{4}{3}\pi r^3##, therefore ##dV=4\pi r^2 dr##.

Water collection is given by ##\frac{dV}{dz}=\rho \pi r^2##.

Let's try a t-dependence:
Combining both, ##\frac{dr}{dz} = \frac{\rho}{4}##.
With the initial condition r(z=0)=0, ##r=\frac{\rho}{4} z##

The rate of water collection is ##\frac{dV}{dt}=\rho \pi r^2 v##. This leads to a deceleration by
$$a_{growth}=-\frac{v}{V} \frac{dV}{dt} = -\frac{3\rho v^2}{4r} = -\frac{3 v^2}{z}$$

Adding gravity: $$\ddot z = g-\frac{3 \dot z^2}{z}$$
Hmm... looks ugly.

Momentum?
$$Vg=\frac{d(Vv)}{dt}=V \ddot z + \dot V \dot z$$

We know that ##V(z)=\frac{4}{3} \pi r^3 = \pi \frac{\rho^3 z^3}{3 \cdot 4^2}##...
$$\frac{ \dot{V}}{V} = \frac{3 \dot z}{z}$$

$$g = \ddot z + \frac{3 \dot z^2}{z}$$
Same equation :(.

The final step here is to assume ## z = ft^2/2 ##, then ## f = g/7 ##.

I just have a question or two about this. The general assumption seems to be that in a given small amount of time ##dt##, the volume swept out by the cross section of the spherical raindrop (i.e. the volume swept out by the equator) is given by ##dV = \pi r^2 vdt ## hence the mass of the droplets accreted in the interval ##dt## is ##dm = \sigma \pi r^2 vdt## where ##\sigma## is the mass density of the droplets in space. I just wanted to have something clarified. Now assuming that the raindrop falls straight down, it seems valid to make the assumption that all the droplets accreted by the raindrop are directly below the raindrop so that we need only worry about the lower hemisphere as far as the accretion goes right? The volume swept out by the cross section will look approximately like a tube (drawn in 2D) in the time ##dt## whereas the volume swept out by the boundary of the lower hemisphere will look approximately like this (again in 2D): http://postimg.org/image/s2rk79f03/ Sorry for the crude diagrams.

I say approximately because the sides will start tilting out as the raindrop grows but for a small enough interval ##dt## we can ignore that for the purposes of my question. Now technically it is the second swept out volume that contains all the droplets which get accreted onto the raindrop and which add to its mass in the time ##dt##; the volume swept out by the cross section (the tube) does not contain the physical droplets that get accreted onto the raindrop in that interval. Would this be fair to say? Granted the two swept out volumes are equal because with the second one I can just translate the extra part at the bottom to the missing part at the top (an action which preserves the volume) and get the tube back. Because the two volumes are the same, I can say that the mass obtained from the accretion of the physical droplets contained in the second swept out volume is just ##dm = \sigma \pi r^2 vdt## which is what we had above anyways.

The reason I bring this up is, it doesn't seem right to me to say that the mass obtained by the raindrop in the given time interval comes from the volume swept out by the cross section; it seems more correct to say that the mass obtained by the raindrop comes from the volume swept out by the boundary of the lower hemisphere (as drawn above). While mathematically the two are equivalent, physically they wouldn't be equivalent so this irked me a bit. Am I off base here?

@voko: Oh, so obvious.
It is interesting that the acceleration does not depend on the density.

WannabeNewton said:
The reason I bring this up is, it doesn't seem right to me to say that the mass obtained by the raindrop in the given time interval comes from the volume swept out by the cross section
You don't have to assume that this volume is a disk, it can be a very small version of your second shape (and it is). This is valid as long as we can neglect the growth of the raindrop, so we don't aggregate significant amounts of water from the sides.

As an extreme case, consider ##\rho \approx 1##. The drop does not have to fall significantly, it will grow extremely fast as growth leads to an expansion sidewards, leading to even more water collected by the drop.

Hi mfb! I'm afraid I don't quite understand what you mean. Are you saying we can assume the raindrop is so small (very small radius) that the lower hemisphere (and upper hemisphere even though it doesn't matter here) is extremely close to the equator (cross section) so that there is negligible space in between the lower hemisphere and equator hence the lower hemisphere and equator essentially sweep out the same physical volume in space when the raindrop falls a bit, meaning we can essentially just think of the equator of the raindrop as picking up all the droplets (and their mass) as it sweeps out its volume during descent?

Newton, what you say explains why accretion is proportional to ## r^2 v ##, so there is no discrepancy between "physics" and "mathematics" (in fact, this is pure geometry all around).

Hi voko! My confusion was that the cross section of the raindrop by itself doesn't actually pick up any droplets, it is the surface of the lower hemisphere that picks up all the droplets so it didn't make sense to me physically to claim that all the droplets picked up by the raindrop during some time interval would be contained in the volume swept out by the cross section, as opposed to being contained in the volume swept out by the surface of the lower hemisphere. I hope I'm being clear enough with my confusion.

Does the center of mass in Keplerian motion attract anything by itself? Same thing.

Note that we get "mist drag" proportional to ## r^2 v^2 ##, just like with ordinary drag. Yet even in ordinary drag the cross-section does not interact with the fluid. Dimensionally, the thing simply has to depend on some area.

Sorry voko but can you explain why it is ok to treat the system as if all the droplets were being picked up by the cross section, giving us ##dm = \sigma \pi r^2 vdt##?

Newton, you answered this question in your very own #7 here: "Because the two volumes are the same, I can say that the mass obtained from the accretion of the physical droplets contained in the second swept out volume is just ## dm=\sigma \pi r^2 v dt ## which is what we had above anyways."

More generally, in any rectlinear motion, the increase of the "volume swept out" is given by ## A v dt##, where ## A ## is the cross-section normal to velocity. Whether it is correct to say "swept out by the cross-section" rather than "swept out by the surface" is a matter of taste, because the two are indistinguishable physically.

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Ah ok I guess I was using the word "physically" in a different way then. Thanks voko and mfb!

A somewhat different solution. Basic equations: $$m\frac {dv} {dt} + \frac {dm} {dt} v = mg \\ m = \frac 4 3 \alpha \pi r^3 \\ \frac {dm} {dt} = \beta \pi r^2 v$$ Excluding ## r ##: $$r = (\frac {3} {4 \alpha \pi} m)^{1/3} \\ \frac {dm} {dt} = \beta \pi (\frac {3} {4 \alpha \pi} m)^{2/3} v = k m^a v$$ Now the trick part: $$m\frac {dv} {dt} + \frac {dm} {dt} v = m\frac {dv} {dm} \frac {dm} {dt} + \frac {dm} {dt} v = (m\frac {dv} {dm} + v)\frac {dm} {dt} = (m\frac {dv} {dm} + v) k m^a v = mg \\ v \frac {dv} {dm} + \frac {v^2} {m} = \frac g {km^a}$$ Let ## u = v^2 ##, then $$\frac {du} {dm} = 2 v \frac {dv} {dm} \\ \frac {du} {dm} + 2 \frac {u} {m} = 2 \frac g {km^a} \\ \frac {du} {dm} m^2 + 2 u m = 2 \frac g k m^{2 - a} \\ \frac {d} {dm} (u m^2) = 2 \frac g k m^{2 - a} \\ um^2 = 2 \frac g {k(3 - a)} m^{3 - a} + C \\ v = \frac 1 m \sqrt { 2 \frac g {k(3 - a)} m^{3 - a} + C } = p m^{\frac {1 - a} 2} \sqrt { 1 - C m^{a - 3} }$$ When ## v = 0 ##, ## C = m^{3 - a} ##, so we can interpret ## C = m_0^{3 - a} ##, where ## m_0 ## is the initial mass, so $$v = p m^{\frac {1 - a} 2} \sqrt { 1 - (\frac {m_0}{m})^{3 - a} }$$ From this we can immediately see that ## m_0 = 0 ##, as the problem prescribes, simplifies the equation radically in the literal sense: it eliminates the radical expression entirely! $$v = p m^{\frac {1 - a} 2} \Rightarrow m = (\frac {v} {p})^{\frac {2} {1 - a}} \\ m\frac {dv} {dt} + \frac {dm} {dt} v = (\frac {v} {p})^{\frac {2} {1 - a}} \frac {dv} {dt} + {\frac {2} {1 - a}} (\frac {v} {p})^{\frac {2} {1 - a}} \frac {dv} {dt} = (\frac {v} {p})^{\frac {2} {1 - a}} g \\ (1 + \frac {2} {1 - a}) \frac {dv} {dt} = g$$ Now, since ## a = 2/3 ##, we again obtain ## 7 \dot{v} = g ##. From the general solution, it is also seen that ## g/7 ## is the asymptotic acceleration as the mass grows infinitely.

After considering problems such as spaceship shooting two cannons at opposite direction, I have been left under impression, that you should never use equation $F(t)=D_t(m(t)v(t))$ with composite objects. Considering this, I would consider an equation

$$\dot{m}(t)v(t) + m(t)\dot{v}(t) = gm(t)$$

a little suspicious. However, I just came up with a complicated way of proving this by taking into account the internal energy related to the continuous inelastic collision between the drop and the mist.

So this equation seems to be correct IMO. Peculiar stuff, anyway..

I see a problem with this idea: ##m' = σ π r^2 v##

##m = \frac{4}{3} π r^3##
##m' = 4 π r^2 r'##

##4 π r^2 r' = σ π r^2 v##
##r' = \frac{σ}{4} v##
##r = r_0 + \frac{σ}{4} x##

where x is just the distance the drop has travelled.

Now here is the problem, suppose the drop starts out very small with negligible volume. At time t, it has swept out a region that is larger than the cone swept by the cross section. This is because the intersection of the sphere and the cone is not a great circle. We in fact have a conical shell unaccounted for, as well as the hemisphere at the end of the cone. So the approximation made utterly fails.

Let ##r = r(x)##, some unknown function such that ## r(0) = r_0 ##. Let us consider the volume of the figure of revolution formed by rotating the curve ## r(x) ## about the ##x## axis, where ##x## varies from 0 to X. This volume is ## V = \pi \int_0^X r^2(x) dx ##, and it contains ## v = \sigma V ## water, and the latter must be equal to the volume of a hemisphere whose radius is ## r(X) ##: $$4 \pi r^3(X) /3 = \sigma \pi \int_0^X r^2(x) dx$$ Differentiating, $$4 \pi r^2 r' = \sigma \pi r^2 \Rightarrow r' = \sigma /4 \Rightarrow r = r_0 + \sigma / 4 x$$ Which is the same result as we got from the accretion law given by ## \sigma \pi r^2 v ##, so it must be correct.

Let ##r = r(x)##, some unknown function such that ## r(0) = r_0 ##. Let us consider the volume of the figure of revolution formed by rotating the curve ## r(x) ## about the ##x## axis, where ##x## varies from 0 to X. This volume is ## V = \pi \int_0^X r^2(x) dx ##, and it contains ## v = \sigma V ## water, and the latter must be equal to the volume of a hemisphere whose radius is ## r(X) ##: $$4 \pi r^3(X) /3 = \sigma \pi \int_0^X r^2(x) dx$$ Differentiating, $$4 \pi r^2 r' = \sigma \pi r^2 \Rightarrow r' = \sigma /4 \Rightarrow r = r_0 + \sigma / 4 x$$ Which is the same result as we got from the accretion law given by ## \sigma \pi r^2 v ##, so it must be correct.

I agree it is difficult to see the problem.

One way to look at it is to notice that the volume is wrong. If we add in Riemann style cylinders of width ds (I meant dxs here) and radius equal to r(t), the volume is too small. The reason is that, as the sphere moves, it gets bigger. The surface of the trumpet carves through the sphere because the tangent to the sphere at its cross section is horizontal but the trumpet surface is not. The sphere sweeps over a slightly bigger shape, and there is the unaccounted hemisphere at the mouth of the trumpet.

Another way to look at it is, for each Riemann cylinder, the volume of the cylinder is instrumental in deciding what the radius of the next cylinder should be. If the first cylinder is too small, this error will bias all the future cylinders. It is almost like feedback with a microphone, the errors in the summation are feeding back, building to a crescendo.

I can't give a good explanation for why this happens, I don't know that much about it myself.

-- correction -- In this post, I should have said dx instead of ds.

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More generally, in any rectlinear motion, the increase of the "volume swept out" is given by ## A v dt##, where ## A ## is the cross-section normal to velocity. Whether it is correct to say "swept out by the cross-section" rather than "swept out by the surface" is a matter of taste, because the two are indistinguishable physically.
Sorry to bother you again but I forgot to ask this before. Is the above usually taken as a definition or derived from some more fundamental relations? Thanks.

Regarding what I wrote before on the accruing of errors, I think one can fix it by using cylinders of width ds instead of dx. But now I'll leave this, back to trying to solve this challenge.

I see a problem with this idea: ##m' = σ π r^2 v##

##m = \frac{4}{3} π r^3##
##m' = 4 π r^2 r'##

##4 π r^2 r' = σ π r^2 v##
##r' = \frac{σ}{4} v##
##r = r_0 + \frac{σ}{4} x##

where x is just the distance the drop has travelled.

Now here is the problem, suppose the drop starts out very small with negligible volume. At time t, it has swept out a region that is larger than the cone swept by the cross section. This is because the intersection of the sphere and the cone is not a great circle. We in fact have a conical shell unaccounted for, as well as the hemisphere at the end of the cone. So the approximation made utterly fails.
That is exactly the reason for the approximation I used in post #5.

So you were imagining the raindrop as essentially just a disk added with really tiny hemispheres ("assume that the vertical size of the raindrop does not matter") given that the droplets sparsely populate the space?

That is exactly the reason for the approximation I used in post #5.

I'm starting to think this problem is infeasible without that assumption, that the trumpet is almost cylindrical. The volume formula is just too complicated otherwise. Or at least, one would have to write a program to solve it in that case.

It seems that a DE

$$\ddot{r}(t) = \alpha - \beta\frac{\dot{r}(t)^2}{r(t)}$$

is relevant for this problem. I found that

$$r(t) = \Big(\sqrt{r(0)} + \sqrt{\frac{\alpha}{4\beta + 2}}t\Big)^2$$

solves it, but this apparently isn't the collection of all solutions. What are the other solutions, with different $\dot{r}(0)$?

solves it, but this apparently isn't the collection of all solutions. What are the other solutions, with different $\dot{r}(0)$?
I would expect that they just correspond to a different r(0) together with a shift in time.

Do keep this discussion going. But I want to congratulate mfb and voko for finding the correct solution!

You could create a much simpler model if you assume the distance traveled by the rain drop in 1 second is significantly greater than the mean interparticle distance of the drops. You could then approximate the raindrop as moving through a fluid with a mean density and viscosity. The primary y direction forces will be drag and gravity. Further more, this can be coupled with a conservation of momentum proprtional to the drag. I am on my phone now so i cannot detail further

I would expect that they just correspond to a different r(0) together with a shift in time.

This cannot be true. If we assume

$$r(t) = (a+bt)^2$$

with some a and b, then a relation

$$\dot{r}(0) = 2\sqrt{\frac{\alpha}{4\beta + 2}}\sqrt{r(0)}$$

inevitably follows. There must be solutions of some other form.

I don't see the issue. t=0 is just the point where the second equation is satisfied.

With
$$r(t) = (a+bt)^2$$
i get ##\dot{r}(0)=2ba## and ##r(0)=a^2##
Apart from the special case of r(0)=0, I can find a,b for every pair of ##(r,\dot{r})## (assuming both are positive, to stay in the physical range).

You should find a and b such that the following three condition are satisfied:

(1) r(0) is what we want.

(2) r'(0) is what we want.

(3) r(t) satisfies the ODE.

If you choose a and b so that (1) and (2) hold, the third one will probably not.

It seems that a DE

$$\ddot{r}(t) = \alpha - \beta\frac{\dot{r}(t)^2}{r(t)}$$

is relevant for this problem.
If you want to solve that differential equation, you can start by substituting v = dr/dt, so d^2 r/dt^2 = dv/dt = dv/dr * dr/dt = v* dv/dr. So we have a first-order differential equation of the form v*dv/dx = a - bv^2/r.

Actually I had a two component first order ODE in the beginning, and I was hoping to make things easier by eliminating the other component Now I'm not sure which way would turn out easier...

Here's a summary of equations: First, the physics must be used to justify the following three equations:

$$m(t) = \frac{4\pi \mu r(t)^3}{3}$$
$$\dot{m}(t) = \rho\pi r(t)^2 v(t)$$
$$\dot{m}(t)v(t) + m(t)\dot{v}(t) = gm(t)$$

If $m(t)$ is eliminated, we get the following two equations:

$$\dot{v}(t) = g - \frac{3\rho}{4\mu}\frac{v(t)^2}{r(t)}$$
$$\dot{r}(t) = \frac{\rho}{4\mu} v(t)$$

Which form a two component first order ODE, and which should have a unique solution as long as $r(t)\neq 0$.

If $v(t)$ is eliminated from this, we get a second order DE for $r(t)$ that has the form I was asking about.

You should find a and b such that the following three condition are satisfied:

(1) r(0) is what we want.

(2) r'(0) is what we want.

(3) r(t) satisfies the ODE.

If you choose a and b so that (1) and (2) hold, the third one will probably not.
Hmm...

##r(t) = (a+bt)^2##
##r'(t) = 2b(a+bt)##
##r''(t) = 2b##
$$2b^2=g-3\frac{r'^2}{r} = g-12b^2$$
Good point, b is fixed by the differential equation. Actually, your solutions are all solutions which start with a zero size somewhere, as "a" is a shift in time (by -a/b).

To find the other solutions, we can set r'(0)=0 and look at r(0) > 0.
r''(0)=g and r''(t) approaches g/7 for large t.