- #1

Rsch613

- 4

- 2

- Homework Statement
- A lake has a normal depth of h across it's whole length. Imagining that at some instant the water level at the extreme ends is at ##\pm y_0## with respect to the normal level. Show that the increased gravitational potential energy of the whole mass of water is given by: ##U = \frac{1}{6}b\rho gLy_{0}^2## where ##b## is the width of the lake. You get this result by finding the increased potential energy of a slice a distance x from the center and integrating.

- Relevant Equations
- ##U = \intF(y)dy##

##dm = \rho b dx dy##

Here is my solution, which is correct.

The tilt of the water at the top can be described in terms of ##x## and ##y## as ##y = \frac{2y_0}{L}x##. The height of the water at any given x is then equal to ##h + \frac{2y_0}{L}x## where ##x \in [-\frac{L}{2}, \frac{L}{2}]##.

So the potential energy in the system after the tilt of the water is

$$\rho g b \int_{-\frac{L}{2}}^{\frac{L}{2}}\int_0^{h + \frac{2y_0}{L}x}ydydx = \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{\left(h + \frac{2y_0x}{L}\right)^2}{2}dx$$

$$= \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{h^2 + \frac{4y_0h}{L}x + \frac{4y_0^2}{L^2}x^2}{2}dx$$

The potential energy of the lake before the tilt is given by

$$\rho b g \int_{-\frac{L}{2}}^{\frac{L}{2}}\int_0^hydydx = \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{h^2}{2}dx$$

The change in potential energy across the whole lake is then,

$$U = \rho g b \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{h^2 + \frac{4y_0h}{L}x + \frac{4y_0^2}{L^2}x^2}{2} - \frac{h^2}{2}dx = \rho g b \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{\frac{4y_0h}{L}x + \frac{4y_0^2}{L^2}x^2}{2}$$

$$= 2 \rho g b\left(\frac{y_0h}{2L}

\left(\frac{L^2}{4} - \frac{L^2}{4}\right) + \frac{y_0^2}{3L^2}\left(\frac{L^3}{8} + \frac{L^3}{8}\right)\right) = \frac{1}{6} \rho g b L y_0^2$$

I know this problem can also be done by the much simpler double integral $$\rho g b \int_{-\frac{L}{2}}^\frac{L}{2}\int_0^{\frac{2y_0}{L}x}ydydx.$$ But I cannot see how to convert what I have into this integral.

I assumed initially that my non-understanding of the easier way to solve the problem was due to a misunderstanding of gravitational potential energy. That may not be the case. However, in case that that is the problem, I will still explain my current understanding of the potential energy in the problem.

If I take the bottom of the lake as the 0 potential line, then when the lake tilts I have a decreased potential energy from one half of the lake and an increased potential energy in the other. But the magnitude of the increased potential is larger because it is further from the 0 potential energy line. When I instead take h as the 0 potential line, I can avoid integrating over the majority of the lake. My bounds of integration in the y direction would just become 0 and the linear function of x that describes the surface depth. There is no need to include the initial depth h of the lake into any of my constants of integration. When the water level changes, one side of the lake will have risen above my 0 potential line, while the other will have fallen below it. My issue is that I cannot see why the magnitude of these is different. It seems like they should have the same magnitude but opposite signs and therefore, there should be no change in potential from this reference point. Of course, the answer does not reflect this.

Is this a flawed understanding of gravitational potential energy on my part? Or is this some integration manipulation? Or is it something else?

The tilt of the water at the top can be described in terms of ##x## and ##y## as ##y = \frac{2y_0}{L}x##. The height of the water at any given x is then equal to ##h + \frac{2y_0}{L}x## where ##x \in [-\frac{L}{2}, \frac{L}{2}]##.

So the potential energy in the system after the tilt of the water is

$$\rho g b \int_{-\frac{L}{2}}^{\frac{L}{2}}\int_0^{h + \frac{2y_0}{L}x}ydydx = \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{\left(h + \frac{2y_0x}{L}\right)^2}{2}dx$$

$$= \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{h^2 + \frac{4y_0h}{L}x + \frac{4y_0^2}{L^2}x^2}{2}dx$$

The potential energy of the lake before the tilt is given by

$$\rho b g \int_{-\frac{L}{2}}^{\frac{L}{2}}\int_0^hydydx = \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{h^2}{2}dx$$

The change in potential energy across the whole lake is then,

$$U = \rho g b \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{h^2 + \frac{4y_0h}{L}x + \frac{4y_0^2}{L^2}x^2}{2} - \frac{h^2}{2}dx = \rho g b \int_{-\frac{L}{2}}^{\frac{L}{2}}\frac{\frac{4y_0h}{L}x + \frac{4y_0^2}{L^2}x^2}{2}$$

$$= 2 \rho g b\left(\frac{y_0h}{2L}

\left(\frac{L^2}{4} - \frac{L^2}{4}\right) + \frac{y_0^2}{3L^2}\left(\frac{L^3}{8} + \frac{L^3}{8}\right)\right) = \frac{1}{6} \rho g b L y_0^2$$

I know this problem can also be done by the much simpler double integral $$\rho g b \int_{-\frac{L}{2}}^\frac{L}{2}\int_0^{\frac{2y_0}{L}x}ydydx.$$ But I cannot see how to convert what I have into this integral.

I assumed initially that my non-understanding of the easier way to solve the problem was due to a misunderstanding of gravitational potential energy. That may not be the case. However, in case that that is the problem, I will still explain my current understanding of the potential energy in the problem.

If I take the bottom of the lake as the 0 potential line, then when the lake tilts I have a decreased potential energy from one half of the lake and an increased potential energy in the other. But the magnitude of the increased potential is larger because it is further from the 0 potential energy line. When I instead take h as the 0 potential line, I can avoid integrating over the majority of the lake. My bounds of integration in the y direction would just become 0 and the linear function of x that describes the surface depth. There is no need to include the initial depth h of the lake into any of my constants of integration. When the water level changes, one side of the lake will have risen above my 0 potential line, while the other will have fallen below it. My issue is that I cannot see why the magnitude of these is different. It seems like they should have the same magnitude but opposite signs and therefore, there should be no change in potential from this reference point. Of course, the answer does not reflect this.

Is this a flawed understanding of gravitational potential energy on my part? Or is this some integration manipulation? Or is it something else?

Last edited: