# Misunderstanding of Gravitational Potential Energy

• Rsch613
In summary, the conversation discusses a problem involving the tilt of water in a lake and its potential energy. The potential energy is described in terms of ##x## and ##y##, and the change in potential energy after the tilt is calculated using double integrals. The person is unsure of their understanding of gravitational potential energy and is seeking clarification on why the magnitude of potential energy changes at different reference points. This discussion may be more suitable for the Introductory Physics Homework Help forum.f

#### Rsch613

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?

Last edited:
• Steve4Physics and Delta2
However, I chose not to do this because I could not see how there is any change in potential from this reference point.
From this reference point, as you go deeper down, the potential energy decreases, i.e. it becomes negative. Please post the statement of the problem as was given to you and post your attempt at a solution. This will help us to help you better.

For future reference: Homework problems like this must be posted in the Introductory Physics Homework Help forum and conform to the Homework Help Guidelines for Students and Helpers. I will report this posting and perhaps a mentor will move it where it belongs.

• vanhees71
It doesn't matter where you take the zero of the potential, because it doesn't change the physics. I'd put the ##z##-axis pointing up, such that ##\vec{g}=-g \vec{e}_z##. Then the potential for the gravitational force density is
$$V(\vec{r})=-\rho \vec{g} \cdot \vec{r}=\rho g z,$$
where ##\rho=\text{const}## is the density of water, which can be assumed to be incompressible.

The hydrostatic equations then tell you that
$$\vec{\nabla}(P+V)=0 \; \Rightarrow\; P=P_0-\rho g z.$$
If the surface of the lake is at ##z_0## you have ##P|_{z=z_0}=P_{\text{air}}## and thus ##P_0=P_{\text{air}}+\rho g z_0## or
$$P=P_{\text{air}} -\rho g(z-z_0).$$

From this reference point, as you go deeper down, the potential energy decreases, i.e. it becomes negative. Please post the statement of the problem as was given to you and post your attempt at a solution. This will help us to help you better.

For future reference: Homework problems like this must be posted in the Introductory Physics Homework Help forum and conform to the Homework Help Guidelines for Students and Helpers. I will report this posting and perhaps a mentor will move it where it belongs.
I am not asking for the solution to the problem. I got the correct answer, just in a much more difficult way than was necessary. I am simply asking for a correction to what is clearly a faulty understanding of gravitational potential energy. Does this still belong in the Introductory Physics Homework Help forum?
An equivalent way (I think) of stating the problem that causes my confusion would involve a rod with some gravitational potential energy. Suppose it is parallel to the Earth at some distance from the ground. Then tilt the rod so that it is no longer parallel to the earth. Then it makes sense to me that it has gained some potential energy with respect to the ground. But if I were to have chosen the height of the rod as the 0 of the gravitational potential and then tilted the rod, It seems like each tiny piece of mass on the side of the rod that rises above the 0 potential line has a counterpart piece of mass on the other side of the rod that has a potential energy of opposite sign but the same magnitude. Since they both travel the same distance from the zero line, but in opposite directions.

If you tilt the rod but keep its center of mass at the same height, its gravitational potential energy will not change. Parts of the rod gain potential energy and other parts lose potential energy but the net change is zero.

• Rsch613 and jbriggs444
If you tilt the rod but keep its center of mass at the same height, its gravitational potential energy will not change. Parts of the rod gain potential energy and other parts lose potential energy but the net change is zero.
Okay, yes then that does make sense. Maybe the problem reduced by way of some integral manipulation then. I will edit the post and add my solution so maybe somebody can show me an easy way to make the problem less calculation intensive. Thank you for your help.

What precisely do you want to calculate? The force density due to gravity of the Earth acting on your fluid elements is ##\vec{f}=\rho \vec{g}##, where ##\rho=\text{const}## (incompressibility of water). Now you want a potential, such that ##\vec{f}=-\vec \nabla{V}##, from which, because also ##\vec{g}=\text{const}## you get ##V=-\rho \vec{g} \cdot \vec{x}+V_0##, where ##V_0## is an arbitrary constant, which doesn't play any role. You can choose it arbitrarily.

• Rsch613
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.
Unless I have misunderstood, you have in fact used the double integral! So it is not clear what the problem is.

But a couple of points:

The tilted rod analogy is not a good one because
a) the rod's length (presumably correspondng to length of water-surface) changes;
b) when rotated about its centre, the rod’s potential energy is unchanged (because the height of the rod’s centre of gravity is unchanged).

An alternative approach would be to consider the height-change of the water's centre of gravity. In cross-section, the water changes from a rectangle to a right-trapezium. So the height of the centre of gravity changes (by ##\Delta y##). The change in potential energy is then simply ##mg\Delta y## where m is the total mass of water. This approach requires you to look-up (or derive) the formula for the height of the trapezium’s centroid.