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- #1

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- #2

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I'm guessing we aren't allowed to assume constant acceleration and just say ##9.8\text{ ms}^{-2}##, right? :tongue:

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- #4

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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.

- #5

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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 :(.

- #6

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

- #7

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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?

- #8

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It is interesting that the acceleration does not depend on the density.

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.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

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.

- #9

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- #11

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- #12

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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.

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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.

- #15

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

- #16

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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.

- #17

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[tex]

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

[/tex]

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..

- #18

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We start with this:

##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.

- #19

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- #20

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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.

Last edited:

- #21

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Sorry to bother you again but I forgot to ask this before. Is the above usually taken as aMore 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.

- #22

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- #23

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

We start with this:

##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.

- #24

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- #25

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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.

- #26

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[tex]

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

[/tex]

is relevant for this problem. I found that

[tex]

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

[/tex]

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

- #27

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

- #28

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- #29

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- #30

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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

[tex]

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

[/tex]

with some a and b, then a relation

[tex]

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

[/tex]

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

- #31

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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).

- #32

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(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.

- #33

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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.It seems that a DE

[tex]

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

[/tex]

is relevant for this problem.

- #34

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

[tex]

m(t) = \frac{4\pi \mu r(t)^3}{3}

[/tex]

[tex]

\dot{m}(t) = \rho\pi r(t)^2 v(t)

[/tex]

[tex]

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

[/tex]

If [itex]m(t)[/itex] is eliminated, we get the following two equations:

[tex]

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

[/tex]

[tex]

\dot{r}(t) = \frac{\rho}{4\mu} v(t)

[/tex]

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

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

- #35

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Hmm...

(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.

##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"

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.

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