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
Best Answer
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 $[0,1]$.