How to perfect the bottle flip?

We make use of the formula

$\bar{h} =\cfrac{\sum_{k=1} ^{n}m_{k} h_{k}}{\sum_{k=1}^{n}m_{k}}$

Here,

$\bar{h} =\cfrac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}$

Let the water level be $h$

$\bar{h} =\cfrac{12\cdot\cfrac{20} {2}+1\cdot36\cdot h\cdot \cfrac{h}{2}} {12+36h}=\cfrac{120+18h^2}{12+36h}=\cfrac{20+3h^2}{2+6h}$

To make ourselves adapt to the problem-solving here better, we can let $\bar{h}=y, h=x$

Our function becomes $y=\cfrac{20+3x^2}{2+6x}$

Now, we are trying to find the value of $x$ when $y$ is minimum. You can use calculus(differentiation) to find it but I personally think algebra will do the trick here.

$\Rightarrow y(2+6x)=20+3x^2$

$3x^2+(-6y)x+20-2y=0$

We know that value of $x$ is real.

$\therefore (-6y)^2-4(3)(20-2y)\geq0$

$36y^2-12(20-2y)\geq0$

$3y^2-(20-2y)\geq 0$

$3y^2+2y-20\geq0$

When $3y^2+2y-20=0$

$y=\cfrac{-2\pm\sqrt{4+4(3)(20)} }{6}$

We neglect the negative value of $y$

We have $y\geq\cfrac{-2+\sqrt{244}}{6}$

Minimum value of y $=\cfrac{-2+\sqrt{244}}{6}$

$x=\cfrac{6y+0}{6}=y\approx2.270$

Water level sought after $=h=x\approx2. 270cm$

To find the total centre of gravity $C_t$ , we can take the centre of gravity of just the water $C_w$ , the centre of gravity of just the bottle $C_b$ , and find the point where the distance from each centre is the ratio between the mass of the bottle $M_b$ and the mass of the water $M_w$ . To do this, we find the distance between the two centres $C_b - C_w$ , divide this by $M_w+M_b$ , and then multiply it by $M_b$ . We can then add this to $C_w$ to find $C_t$ . $C_t=C_w+(C_b-C_w)\times\frac{M_b}{M_b+M_w}$ We can now substitute the information we're given in the question to get: $C_t=\frac{h}{2}+\left(10-\frac{h}{2}\right)\times\frac{12}{12+36\times h}$ Simplify: $C_t=\frac{h}{2}+\left(10-\frac{h}{2}\right)\times\frac{12}{12+36\times h}=\frac{h}{2}+\frac{20-h}{2+6h}=\frac{3 h^2 + 20}{6 h + 2}$ To find the local minimum, We need to differentiate this, using the quotient rule: $\frac{d}{dh}\left(\frac{3 h^2 + 20}{6 h + 2}\right)=\frac{(6h+2)(6h)-(3h^2+20)(6)}{(6h+2)^2}=\frac{3(3h^2+2h-20)}{2(3h+1)^2}$ We then set it to equal $0$ in order to find the local minimum, which we then solve: $\frac{3(3h^2+2h-20)}{2(3h+1)^2}=0$ Divide both sides by $\frac{3}{2(3h+1)^2}$ : $3h^2+2h-20=0$ Divide both sides by $3$ again, then add $\frac{20}{3}$ : $h^2+\frac{2h}{3}=\frac{20}{3}$ Then add $\frac{1}{9}$ so the left side of the question can be expressed as a square: $h^2+\frac{2h}{3}+\frac{1}{9}=\frac{61}{9}$ $(h+\frac{1}{3})^2=\frac{61}{9}$ And now we can easily solve for $h$ : $h+\frac{1}{3}=\frac{\sqrt{61}}{3}$ $h=\frac{\sqrt{61}}{3}-\frac{1}{3}\approx\boxed{2.2701}$

This is not a solution, the others are fine, but puzzlement that the jar, like Bottom's Dream, "hath no bottom" (MSND, Act IV, end of scene i). If we assume that it hath a bottom, of density equal to that of the sides, then the optimal depth is slightly lower: 2.221487551 cm. However if the base has the same mass as the sides, then the optimal depth is 2 cm exactly.

NO CALCULUS: If we imagine the process of adding water into the bottle, we realize that the center of mass will drop as water is initially added and then rise as more water is added above the center of mass. The problem then becomes finding the "turning point". Logically, this point happens when the center of mass is equal to the water line because additional water brings the center of mass up. To calculate the center of mass, we can simplify the bottle to a point mass of $12$ g in the center of the bottle. Similarly, the water can be represented as a point mass. Let $h$ represent the height of filled water. The mass of water is then $36h$ and the point is at a height of $\frac{h}{2}$ . To find the weighted average of the points (the center of mass), we write $\frac{10 \cdot 12+36h \cdot \frac{h}{2}}{36h+12}$ . The denominator represents the total mass. Since we are finding the point the center of mass is equal to the water line, we can set this equal to $h$ . Multiplying out, we get $36h^2+12h=18h^2+120$ , $18h^2+12h-120=0$ , $3h^2+2h-10=0$ , which yields the solution $\boxed{h=\frac{\sqrt{61}-1}{3}=2.270083225}$

I solved this using Solver feature found in most spreadsheet programs. (Libreoffice in my case)

先把重心关于h的函数表示出来，然后求导，导数为0时的x值即为重心最小时的h值。

If added water's height $h$ exaclly equal to height of mass center, it should have lowest center of mass. So we can make this equations: $\cfrac{\cfrac{20}{2}(12)+1(36h(\cfrac{h}{2}))}{12+36h}=h$ , simplification get $\cfrac{20+3h^2}{2+6h}=h$ . then, $20+3h^2=2h+6h^2$ , $-3h^2-2h+20=0$ . use quadratic formula we get $h=\cfrac{2\pm\sqrt{4-4(-3)(20)}}{2(3)}$ . simplification get $h=\cfrac{1+\pm\sqrt{61}}{3}$ . we don't need negative value, so final answer is $h=\cfrac{1+\sqrt{61}}{3}\approx\boxed{2.270}$ .

First, let's consider the center of mass of the bottle and water separately. Since the mass is evenly distributed throughout the length, the center of mass of the cylinder is at its mid-height. The center of mass of the water is also at its mid-height since the density of water is uniform throughout. Now, since the center of mass are directly on top of each other vertically, we will only consider the vertical component. The center of mass is essentially the weighted average of each center of mass, weighted with the mass at its location. Thus, the center of mass with a given water height of $h$ would be $P=\frac{\frac{h}{2}\cdot 36h + 10\cdot 12}{36h+12}=\frac{3h^2+20}{6h+2}$ $P'=\frac{36h^2+12h-18h^2-120}{(6h+2)^2}=0$ $18h^2+12h-120=0 \implies h\approx 2.27$