A spinning green liquid

Let the bottom center of the column be $(0, 0)$ , half the width of the column be $L$ , the vertex of the parabola be at $(0, k)$ , and the rightmost fixed red point be at $(p, q)$ .

Then the equation of the parabola has a general format of $y = ax^2 + k$ , which means the rightmost point on the parabola is $(L, aL^2 + k)$ . Since $(p, q)$ is on the parabola, $q = ap^2 + k$ or $q - k = ap^2$ .

Since the area inside a parabola is $\frac{2}{3}$ the area of the rectangle it is contained in, the area of the liquid under the curve is $\frac{1}{3}aL^3 + kL$ . This must be the same as the area of the liquid at rest, which is $qL$ . Therefore, $\frac{1}{3}aL^3 + kL = qL$ , which simplifies to $q - k = \frac{1}{3}aL^2$ .

Now $q - k = ap^2 = \frac{1}{3}aL^2$ , which solves to $L = \sqrt{3}p$ .

Therefore, the ratio of the distance between the red points to the length of the container is $\frac{2p}{2L} = \frac{2p}{2(\sqrt{3}p)} = \frac{1}{\sqrt{3}} \approx \boxed{0.577}$ .

I'm going to cheat a little bit and use the general result derived in this video . The video does not specifically address the question posed in this problem.

The water surface has the form of a parabola:

$y = \frac{2 \pi^2}{g T^2} x^2 - \frac{2 \pi^2 \, L^2}{3 g T^2}$

In the above equation, $T$ is the rotational period, $g$ is the ambient gravitational acceleration, $x$ is the distance from the center, and $L$ is the distance from the center to either tank edge. It also important to note that $y = 0$ corresponds to the water level when the tank is not spinning.

This problem is saying that there is some value of $x$ , say $x'$ , such that $y(x')$ is the same, regardless of the rotational speed. Since this must also hold true when the container is not spinning, it follows that $y(x') = 0$ .

$0 = \frac{2 \pi^2}{g T^2} x'^2 - \frac{2 \pi^2 \, L^2}{3 g T^2} \\ \frac{2 \pi^2}{g T^2} x'^2 = \frac{2 \pi^2 \, L^2}{3 g T^2} \\ x'^2 = \frac{ L^2}{3 } \\ \frac{x'}{L} = \frac{1}{\sqrt{3}}$

We'd like to set the volume at d = to the volume when the column is still, so: