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Help finding center of mass of soda can? If you represent the soda can as a right-circular cylinder

radius=4 cm height =12 cm

We are told to neglect the mass of the can itself.

When the can is full the center of mass is at 6cm above the base, halfway along the axis of the can.

As the can is drained and air replaces the soda, the center of mass descends towards the bottom.

However when the can is completely empty the center of mass is still at 6cm.

Assuming the density of soda is 1 gram per cubic cm and the density of air is 0.001 grams per cubic cm.

Find the depth of soda in the can for which the center of mass is at it's lowest point.

I am not sure where to begin, I was trying to think of this as a linear case where the fulcrum is at a point with masses on either end, but I am extremely lost.

Thank you for the help

Superman
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    This isn't too difficult. Write an equation that shows you what the mass of the can is for any filled volume, i.e. allow the filled volume to change via the soda. Then, notice that the volume of the air filling the can as the soda decreases is inversely proportional, i.e. as the soda goes down the air goes up. The mass of the can itself never changes and therefore may be ignored. Find the balance point between the mass of the air and the mass of the soda -- the point where the mass of air and soda is the same. Could be solved in terms of torque, in Physics terms. – bjd2385 Nov 19 '14 at 04:43
  • If you are to ignore the mass of the can the CM of the empty can is not defined. Usually for this problem you are given the mass of the can, which is rather smaller than the mass of the soda. You are then correct that the CM is halfway up both when the can is full and when it is empty. – Ross Millikan Jun 15 '21 at 01:12

2 Answers2

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Denote by $m_s$ the mass of the remaining soda, by $c_s$ the $z$-coordinate of its center of gravity, similarly for the air in the can, and let $c_*$ be the $z$-coordinate of the overall centroid. Then $c_*$ can be computed from the moment equation $$m_sc_s+m_ac_a=(m_s+m_a)c_*\ .\tag{1}$$ In order to compute the quantities appearing in $(1)$ let $\rho_a$ be the density of air and $\rho_s=1000\rho_a$ the density of soda. Furthermore let $h$ be the height of the can, $A$ the area of its circular base, and let $z:=t\> h$ with $0\leq t\leq1$ be the level of the remaining soda. Then we have $$m_s=\rho_s t h A,\quad c_s={1\over2}th,\quad m_a=\rho_a(1-t)hA,\quad c_a={1+t\over 2}h\ .$$ Now plug this all into $(1)$ and solve for $c_*$. You will get an expression of the form $$c_*=C f(t)\ ,$$ where all data of the problem can be collected into one constant $C$, and $f$ is a "pure" function of $t$. Now determine $\arg\min$ of $f$ in the interval $[0,1]$.

Christian Blatter
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Another approach is to consider that the first bit of soda removed from the can is the top soda and is above the CM, so removing it must lower the CM. As you remove soda the CM will continue to drop until the top of the soda is at the CM. Then the soda you remove will come from below the CM and the CM will rise. Find the height of soda that makes its height match the CM of the can+soda. That will be the minimum height of the CM and the amount of soda to achieve it. If the height of the soda is $h$ the CM of the soda is $\frac h2$. Let $M$ be the mass of soda in a full can and $m$ the mass of the can. The mass of soda in the can is then $\frac h{12}M$ and the cm of the can+soda is $$\frac{\frac h{12}\cdot \frac 12M+6m}{\frac h{12}M+m}$$ which has to equal $h$. You can solve for $h$

Ross Millikan
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