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 $6$ cm 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 $6$ cm.
Assuming the density of soda is $1$ gram per cubic cm and the density of air is $0.001$ gram per cubic cm.
Find the depth of soda in the can for which the center of mass is at it's lowest point.
I really do not know where to begin on this mess.
Thank you for the help.
Best Answer
Hint:
When the surface is at height $h$ from the bottom, hence height $H-h$ from the top, the gravity center of the soda is at $\dfrac h2$ and that of air at $h+\dfrac{H-h}2=\dfrac{H+h}2$.
The height of the global gravity center is the average of these two heights, weighted by the weights of the two parts.
$$h_g=\frac{d_sh}{d_sh+d_a(H-h)}\frac h2+\frac{d_a(H-h)}{d_sh+d_a(H-h)}\frac{H+h}2=\frac12\frac{(d_s-d_a)h^2+d_aH^2}{(d_s-d_a)h+d_aH}\\ =\frac12\frac{\delta\left(\dfrac hH\right)^2+1}{\delta\dfrac hH+1}=\frac12\left(\delta\dfrac hH-1\right)+\frac1{\delta\dfrac hH+1}.$$
Find the root of the derivative of this expression.