Question #8aca8

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Question #8aca8

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Answer 1 · 2014-12-17T02:53:27

Let's start by stating a couple of assumptions:
1. Water is generally incompressible.
2. The amount of water flowing through any cross-section of the stream will be the same at any height.
3. After coming out of the faucet, the water is influenced only by gravity.

Combining assumptions 1 and 2 we can say that the total flux of water $Φ$ $Phi$ at any height $h$ $h$ is always equal. And it is equal to the velocity at that height $v_{h}$ $v_h$ times the cross-section of the stream at that height $S_{h}$ $S_h$ .

$Φ = v_{0} S_{0} = v_{h} S_{h}$ $Phi = v_0S_0 = v_hS_h$

Assumption 3 allows us to describe the velocity at any height as a function of the initial velocity and the height.

$v_{h}^{2} = v_{0}^{2} + 2 g h$ $v_h^2 = v_0^2 + 2gh$

A little algebra allows us to solve for the cross sectional area as a function of these variables:

$S_{h} = S_{0} \frac{v_{0}}{\sqrt{v_{0}^{2} + 2 g h}}$ $S_h = S_0v_0/sqrt(v_0^2 + 2gh)$

The radius can be calculated from the area using the relation
$S = π r^{2}$ $S = pir^2$

The equation then becomes:
$r_{h} = r_{0} \sqrt{\frac{v_{0}}{\sqrt{v_{0}^{2} + 2 g h}}}$ $r_h = r_0sqrt(v_0/sqrt(v_0^2 + 2gh))$

enter image source here
http://www-e.uni-magdeburg.de/mertens/teaching/seminar/themen/ajp_drop_formation.pdf

Answer 2 · 2014-12-17T05:06:54

The cohesive force between water molecules binds them together as a liquid. Because of this we can assume that droplets of water are not breaking off from the main flow, so the shape of the water stream would pinch tighter and tighter as it fell. This would produce a horn shape.

The mathematical shape would be a volume of rotation, where we would take a graph of the radius of the stream as a function of height and rotate it around the axis. To find this function, we'll start with the equation of continuity. If you have ever put your thumb over the end of a hose to make the water come out faster, you have demonstrated the equation of continuity;

$A_{1} v_{1} = A_{2} v_{2}$ $A_1 v_1 = A_2 v_2$

Where $A$ $A$ is the cross sectional area of a fluid stream, and $v$ $v$ is the velocity of the fluid. It tells us that the amount of fluid passing through a cross section of a stream is the same at all points in the stream, regardless of the area. This means we can calculate the diameter of the stream at different points in time.

We can consider $A_{1}$ $A_1$ to be the area of the faucet, and $v_{1}$ $v_1$ to be the initial velocity, $v_{o}$ $v_o$ , of the water. $v_{2}$ $v_2$ is the velocity at some time after the water drops, $v (t)$ $v(t)$ . When we substitute these values into the equation of continuity, we get;

$(π r_{o}^{2}) v_{o}^{2} = (π r^{2}) (g t)$ $(pi r_o^2) v_o^2= (pi r^2) (g t)$

which rearranges to;

$r (t) = \frac{r_{o} v_{o}}{\sqrt{g t}}$ $r(t) = (r_o v_o)/sqrt(g t)$

Where $g$ $g$ is the acceleration of gravity. We want to know $r (x)$ $r(x)$ though, instead of $r (t)$ $r(t)$ . We can find this by rearranging the equation for constant acceleration.

$x (t) = \frac{1}{2} g t^{2} + v_{o} t + x_{o}$ $x(t)= 1/2 g t^2 +v_ot + x_o$

We can simplify this by setting $x_{o}$ $x_o$ and $v_{o}$ $v_o$ equal to 0. Then we rearrange to find $t$ $t$ in terms of $x$ $x$ .

$t (x) = \sqrt{\frac{2 x}{g}}$ $t(x) = sqrt((2x)/g )$

We can plug this into $r (t)$ $r(t)$ to get $r (x)$ $r(x)$ . After some simplification, we get;

$r (x) = \frac{r_{o} v_{o}}{{(2 g x)}^{\frac{1}{4}}}$ $r(x) = (r_o v_o)/ (2gx)^(1/4)$

A graph of $x^{- (\frac{1}{4})}$ $x^-(1/4)$ looks like this;
Geogebra
and if we rotate it around the x axis we get the shape of the water falling from a faucet. Accounting for the fact that the water actually has a starting velocity would move the graph to the left, making the slope shallower, and including the constants from $r (x)$ $r(x)$ would adjust the amplitude (height)of the graph.

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