## Task 18/77

Given is a cylindrical beaker of height $H$. The (homogeneous) mantle has the mass $M$, the mass of the bottom is negligible. The glass is filled with a (homogeneous) liquid of mass $m_{H}$ to the brim, but the liquid drips out through an opening in the bottom until it is completely empty.

Required is a function that gives the position of the center of mass as a function of the fill height $h$, as well as the smallest distance of the center of mass from the bottom.
Due to the homogeneity and symmetry, the center of gravity lies on the axis of the cylinder, so that its position is completely determined by the height $s$ above the ground.

Also due to the homogeneity, the center of gravity of the mantle is at a height of 0.5H, and the center of gravity of the liquid is at a height of $0.5 h$ above the ground (where $0 \leq h \leq H$ applies).

The mass of the liquid at the fill height $h$ is $m=\frac{h}{H} m_{H}$. For the two moments $M_{G}$ and $M_{F}$ (referenced to the ground), the following applies:

$$
M_{G}=M \cdot 0.5 H=0.5 H M \quad ; \quad M_{F}=m \cdot 0.5 h=0.5 \frac{h^{2}}{H} \cdot m_{H}
$$

The total moment then results in

$$
M_{G}+M_{F}=0.5 H M+0.5 \frac{h^{2}}{H} m_{H}=\frac{1}{2 H}\left(H^{2} M+h^{2} m_{H}\right)
$$

If the total moment is divided by the total mass, the distance $s$ of the center of gravity from the ground is obtained:

$$
s=f(h)=\frac{\frac{1}{2 H}\left(H^{2} M+h^{2} m_{H}\right)}{M+\frac{h}{H} m_{H}}=0.5 \frac{H^{2} M+h^{2} m_{H}}{H M+h m_{H}}
$$

For $h=0$ and for $h=H$, this yields (as expected due to symmetry)

$$
s_{0}=s_{H}=f(0)=f(H)=0.5 H
$$

Possible extreme value points of the function are obtained by setting the first derivative to zero, where it suffices to set the numerator to zero:

$$
\begin{gathered}
s^{\prime}=f^{\prime}(h)=\frac{m_{H}}{2} \cdot \frac{h^{2} m_{H}+2 h H M-H^{2} M}{\left(H M+h m_{H}\right)^{2}} \\
h^{2} m_{H}+2 h H M-H^{2} M=0 \rightarrow h_{E}=\frac{H}{m_{H}}\left(\sqrt{M\left(M+m_{H}\right)}-M\right)
\end{gathered}
$$

(since a negative fill height is meaningless, the negative root value is discarded). From this it follows

$$
s_{E}=f\left(h_{E}\right)=\frac{H M}{m_{H}}\left(\sqrt{1+\frac{m_{H}}{M}}-1\right)
$$

That this value is a minimum can be confirmed in various ways:

1. by checking the second derivative; it is $f^{\prime \prime}\left(h_{E}\right)>0$;
2. by the observations that a) $s_{E}=f\left(h_{E}\right)<0.5 H=f(0)=f(H)$ and b) $f(h)$ is continuous.