Chapter 2.5, Problem 6E

Interpretation Introduction

Interpretation:

For a water bucket with a hole in the bottom, it is to be shown that $\text{av(t) = A}\dot{\text{h}}\text{(t)}$, where $\text{a}$ is the area of the hole, $\text{A}$ is the cross-sectional area of the bucket, $\text{v(t)}$ is the velocity of the water passing through the hole, and $\text{h(t)}$ is the height of the water remaining in the bucket at time $\text{t}$. The equation ${\text{v}}^{\text{2}}\text{ = 2gh}$ is to be derived. It is to be shown that $\dot{\text{h}}\text{ = -C}\sqrt{\text{h}}$, where $\text{C = }\sqrt{\text{2g}}\left(\frac{\text{a}}{\text{A}}\right)$. Given $\text{h(0) = 0}$, it is to be shown that the solution of $\text{h(t)}$ is non-unique for $\text{t < 0}$.

Concept Introduction:

By invoking the law of conservation of mass, the equation $\text{av(t) = A}\dot{\text{h}}\text{(t)}$ can be proved.

By using the law of conservation of energy, the equation ${\text{v}}^{\text{2}}\text{ = 2gh}$ can be derived.

By using the equations $\text{av(t) = A}\dot{\text{h}}\text{(t)}$ and ${\text{v}}^{\text{2}}\text{ = 2gh}$, it can be shown that $\dot{\text{h}}\text{ = -C}\sqrt{\text{h}}$, where $\text{C = }\sqrt{\text{2g}}\left(\frac{\text{a}}{\text{A}}\right)$.

It can be shown that the solution of $\text{h(t)}$ is non-unique for $\text{t < 0}$ by integrating the equation $\dot{\text{h}}\text{ = -C}\sqrt{\text{h}}$ and using the initial condition $\text{h(0) = 0}$.