Subfield: Mechanics

## String Cheese
Context question:
a. When a faucet is turned on, a stream of water flows down with initial speed $v_{0}$ at the spout. For this problem, we define $y$ to be the vertical coordinate with its positive direction pointing up.

Assuming the water speed is only affected by gravity as the water falls, find the speed of water $v(y)$ at height $y$. Define the zero of $y$ such that the equation for $v^{2}$ has only one term and find $y_{0}$, the height of the spout.
Context answer:
\boxed{$y_{0}=\frac{-v_{0}^{2}}{2 g}$ ,$v=\sqrt{-2 g y}$}


Context question:
b. Assume that the stream of water falling from the faucet is cylindrically symmetric about a vertical axis through the center of the stream. Also assume that the volume of water per unit time exiting the spout is constant, and that the shape of the stream of water is constant over time.
In this case, the radius $r$ of the stream of water is a function of vertical position $y$. Let the radius at the faucet be $r_{0}$. Using your result from part (a), find $r(y)$.


If $r(y)$ is not constant, it implies that the water has some radial velocity during its fall, in contradiction to our assumptions in part (a) that the motion is purely vertical. You may assume throughout the problem that any such radial velocity is negligibly small.
Context answer:
\boxed{$r=r_{0} \sqrt[4]{\frac{v_{0}^{2}}{-2 g y}}$}


Question: c. The water-air interface has some surface tension, $\sigma$. The effect of surface tension is to change the pressure in the stream according to the Young-Laplace equation,
$$

\Delta P=\sigma\left(\frac{1}{r}+\frac{1}{R}\right)

$$

where $\Delta P$ is the difference in pressure between the stream and the atmosphere and $R$ is the radius of curvature of the vertical profile of the stream, visualized below. ( $R<0$ for the stream of water; the radius of curvature would be positive only if the stream profile curved inwards.)



<img_4431>



For this part of the problem, we assume that $|R| \gg|r|$, so that the curvature of the vertical profile of the stream can be ignored. Also assume that water is incompressible.



Accounting for the pressure in the stream, find a new equation relating for $r(y)$ in terms of $\sigma, r_{0}, v_{0}$, and $\rho$, the density of water. You do not need to solve the equation for $r$.

Solution:
Our conservation of energy approach from part (b) needs to be modified to account for the work done against pressure. As we look further down in the stream, the radius is smaller. This means the pressure is higher there, and the water is slowed compared to when we assumed only gravity acted on the water.



The result of accounting for changes in pressure in a flow where no energy is dissipated is the Bernoulli equation,



$$

\frac{1}{2} \rho v^{2}+\rho g y+P=\frac{1}{2} \rho v_{0}^{2}+\rho g y_{0}+P_{0}

$$



where $P_{0}$ is the pressure in the stream at the spout.



Using the Young-Laplace equation to replace $P$ and $P_{0}$, we have



$$

\frac{1}{2} \rho v^{2}+\rho g y+\frac{\sigma}{r}=\frac{1}{2} \rho v_{0}^{2}+\rho g y_{0}+\frac{\sigma}{r_{0}}

$$



If we substitute in $y_{0}=-\frac{v_{0}^{2}}{2 g}$ and $v=v_{0} \frac{r_{0}^{2}}{r^{2}}$, this becomes



$$

\frac{1}{2} \rho v_{0}^{2} \frac{r_{0}^{4}}{r^{4}}+\rho g y+\frac{\sigma}{r}=\frac{1}{2} \rho v_{0}^{2}-\rho g \frac{v_{0}^{2}}{2 g}+\frac{\sigma}{r_{0}}

$$



This may be simplified to



$$

\frac{1}{2} \rho v_{0}^{2} \frac{r_{0}^{4}}{r^{4}}+\rho g y=\sigma\left(\frac{1}{r_{0}}-\frac{1}{r}\right)

$$

Final answer: $\frac{1}{2} \rho v_{0}^{2} \frac{r_{0}^{4}}{r^{4}}+\rho g y=\sigma\left(\frac{1}{r_{0}}-\frac{1}{r}\right)$