But There Would Be No Mass!

Relevant wiki: Moment of Inertia

First of all, since the pool of water can be modeled as the area between the $x$ -axis and a function, $f\left( x \right)$ , one can flip the model on its side as so: In this case, the center of mass we are looking for is the $y$ -coordinate of the center of mass, since in the original model, this would be along the horizontal. The formula for this is $Y_\text{cm} = \frac{\int y\, dm\,} {\int \, dm}$ .

For this two-dimensional setting, $\rho =\frac { M }{ A }$ , where $\rho$ is density, $M$ is total mass, and $A$ is total area. Writing this in differential form, $\rho =\frac { dm }{ da }$ , and this can be solved for $dm$ : $\rho da=dm$ . Substituting $dm$ into the center of mass formula (and noting that the water is of uniform density, so $\rho$ can be factored out), we get the following: $\frac { \int { y\rho da } }{ \int { \rho da } } =\frac { \rho \int { yda } }{ \rho \int { da } } =\frac { \int { yda } }{ \int { da } }$ From here, since the water is level with the top of the tub, the area under $f\left( x \right)$ makes a right triangle. Thus, $\tan { \theta =\frac { f\left( x \right) }{ x } } =slope$ when $x>0$ , and rearranging this equation, $y=f\left( x \right)=\left( \tan { \theta } \right) x$ .

Going back to our modified center of mass equation, since $da=dxdy$ , we can further rewrite this equation by using a double integral and solving it: $\frac { \int { yda } }{ \int { da } } =\frac { \iint _{ 0 }^{ f\left( x \right) }{ ydydx } }{ \iint _{ 0 }^{ f\left( x \right) }{ dydx } } =\frac { \int { \frac { 1 }{ 2 } { \left( f\left( x \right) \right) }^{ 2 }dx } }{ \int { f\left( x \right) dx } }$ Plugging in the necessary variables (and noting that $\tan { \theta }$ does not depend on $x$ and, therefore, can be factored out), we can solve: $\frac { \int _{ 0 }^{ L }{ \frac { 1 }{ 2 } { \left( f\left( x \right) \right) }^{ 2 }dx } }{ \int _{ 0 }^{ L }{ f\left( x \right) dx } } =\frac { \int _{ 0 }^{ L }{ \left( \frac { 1 }{ 2 } \tan ^{ 2 }{ \theta } \right) { x }^{ 2 }dx } }{ \int _{ 0 }^{ L }{ \left( \tan { \theta } \right) xdx } } =\frac { \frac { 1 }{ 2 } \tan ^{ 2 }{ \theta } \int _{ 0 }^{ L }{ { x }^{ 2 }dx } }{ \tan { \theta \int _{ 0 }^{ L }{ xdx } } } =\frac { \tan { \theta } }{ 2 } \frac { \left( \frac { 1 }{ 3 } { L }^{ 3 } \right) }{ \left( \frac { 1 }{ 2 } { L }^{ 2 } \right) } =\frac { L\tan { \theta } }{ 3 }$ By this point, one might be inclined to say that the center of mass would be infinitely far from the pivot, since $\lim _{ \theta \rightarrow { 90° }^{ - } }{ \tan { \theta =\infty } }$ . But once one views the model again, we can easily see that the length between the pivot and the top of the water, $L$ , will also tend to $0$ as $\theta$ approaches $90°$ . Therefore, since $\lim _{ \theta \rightarrow { 90° }^{ - } }{ \frac { L\tan { \theta } }{ 3 } } =\frac { \left( 0 \right) \left( \infty \right) }{ 3 }$ , we have an ambiguous case. However, this problem is not over yet. Since the area under $f\left( x \right)$ makes a right triangle, we can define another property of this triangle to get rid of the singularity: $\cos { \theta } =\frac { L }{ \ell }$ . By rearranging this relation to solve for $L$ , we get $L=\ell \cos { \theta }$ , which can be substituted back into ${ \frac { L\tan { \theta } }{ 3 } }$ to simplify it: ${ \frac { L\tan { \theta } }{ 3 } }=\frac { \left( \ell \cos { \theta } \right) \left( \frac { \sin { \theta } }{ \cos { \theta } } \right) }{ 3 } =\frac { \ell \sin { \theta } }{ 3 }$ The limit of $\sin { \theta }$ at $90°$ is defined, so the limit at $90°$ of the distance to the center of mass is $\lim _{ \theta \rightarrow 90° }{ \frac { \ell \sin { \theta } }{ 3 } = } \boxed { \frac { \ell }{ 3 } }$

The water tube always forms a triangle with one vertex being the pivot point. We know that the distance from any vertex to the center of mass in a triangle is $2/3$ of the median line. Consider the point where the median line from the leftmost point of the triangle intersects the wall and call it $P$ . This point will have distance $1/3$ of the median line to the center of the mass. As the angle tends to $90^\circ$ the triangle looks more and more like a straight line and therefore the median line tends to the length l and the point P tends to be the same as the pivot point. Therefore, the solution is $\frac \ell 3$ ..

This might not be the most elegant solution. Also sorry for the formatting, I'm on my phone.

As $\theta$ approaches $\frac {\pi}{2}$ , the length of the median (i.e. the line joining the center of the platform and the opposite vertex) tends to $l$ (you can draw this). Since we know that the centroid (center of mass) is divided in the ratio $2 : 3$ with the side being bisected (platform) being closer to it, the answer is $\frac {l}{3}$ .