Darn it, we're gonna need more soda.

Calculus Level 3

A full can of soda sits on the table. Suddenly, a cosmic ray punches a tiny hole in the bottom of the can, allowing the contents of the soda can to slowly drain away!!

Assuming the can is a perfect uniform cylinder with top and bottom of identical mass, and is perfectly full of soda, the center of mass of the can and the soda inside is initially exactly in the middle of the can .

As the soda drains away, the center of mass begins to drop downwards.

When the soda is completely gone, the center of mass of the can is once again exactly in the middle of the can .

At some point during this process, the center of mass must have attained a lowest height above the table before going back up! Find the height of the level of the soda which causes the height of the center of mass of the can and the soda inside to be at a minimum using the following values:

Mass of empty can: $2~g$
Density of soda: $1~g/cm^3$
Area of the base of the can: $2~cm^2$
Height of the can: $3~cm$

Credit: I found this neat little problem somewhere on the internet, but I can't remember where! If someone knows the source, I'll credit it.

The answer is 1.

3 solutions

Sid 2108
Feb 25, 2015

I used common sense to solve the question. Let the answer be $x$ The height of the can $3~cm$ so $x$ is less than $3$ .The answer must be more than $0$ also it can't be $0$ because when the height of soda is $0$ the can is empty.Thus $0<x<3$ . Now when the answer is in decimal the answer input box shows a message "decimals are OK" which didn't appear in this question thus the answer $x$ is an integer. So possible integral values for $x$ are $1,2$ so I decided to try both the options since there are three tries I got the answer correct in the first try which is $1$ .

Haha, I didn't know that trick about "decimals are ok".

Brian Kardon - 6 years, 3 months ago

ya it was a perfect common sense by @Sid 2108

shreyas garkhedkar - 6 years, 3 months ago

Aditya Pappula
Feb 23, 2015

Using symmetry, we can deduce that it happens when the mass of coke is equal to the mass of the can, which happens when height of the soda is 1cm

That's an interesting argument! Could you explain your symmetry reasoning?

Brian Kardon - 6 years, 3 months ago

When the mass of the soda is larger than the mass of the can, then the center of mass of the system is closer to the center of mass of the soda. Since the center of mass of the soda is go downward, then the center of mass the system is also go downward.

When the mass of the soda is same with the mass of the can, then the center of mass of the system is exactly in the middle of the center of mass of the can and the center of mass of the soda

When the mass of the soda is smaller than the mass of the can, then the center of mass of the system must be closer to the center of mass of the can. Hence the center of mass of the system will be go upward.

So, the minimum height of the center of mass of the system will happen when the mass of the can is same with the mass of the soda.

$Mcan = Msoda = \rho A y$ then $y = \frac{Mcan}{\rho A} = 1 cm$

Fiki Akbar - 6 years, 3 months ago

Nice! Much simpler than my reasoning.

Brian Kardon - 6 years, 3 months ago

I do not see how your third part of the argument works - surely the center of mass is closer to the can, but it does not imply the center of mass must be going up (although in this case it is but we would see otherwise below).

Your reasoning should be valid if we changed the mass of the can, call it $m$ . Using Brian's method (which is what I did as well), the center of mass with height of soda $h$ should be $\frac{h^2+1.5m}{2h+m}$ . Equating the derivative to be $0$ gives you $2x^2+2mx-3m=0$ , taking the positive root of that gives $\frac{1}{2}(\sqrt{m^2+6m}-m)$ . Clearly this has no direct correlation to $m$ in fact in every other case of positive $m\neq 2$ , the mass of the soda at the minimum center of mass, $\sqrt{m^2+6m}-m\neq m$ by simple algebra. The above is no more than a coincidence.

Yong See Foo - 6 years, 3 months ago

@Yong See Foo – As Brian said, the center of mass can be written as a weighted average. As the mass of the soda drops, the can accounts for more of the total weight and therefore the center of mass will get closer and closer to the center of the can as the soda level drops.

Tan Li Xuan - 6 years, 3 months ago

Thank you.. I could not have explained it in a better way. As you mentioned, I was tracking the movement of the center of mass of the total system as soda gets dripped out of the can.

Aditya Pappula - 6 years, 3 months ago

Brian Kardon
Feb 22, 2015

The height of the center of mass of the can and the liquid inside can be written as a weighted average of the center of mass of the can and the liquid:

$h_{cm} = \frac{M_c \frac{h_c}{2} + M_s \frac{h_s}{2}}{M_c + M_s}$

Substituting for the mass of the soda, which is a function of the height of the soda ( $M_s = \rho_s A h_s$ ):

$h_{cm} = \frac{M_c \frac{h_c}{2} + \rho_s A \frac{h^2_s}{2}}{M_c + \rho_s A h_s}$

Since we want to find the minimum value of the center of mass height with respect to the soda height, we need to find $\frac{dh_{cm}}{dh_s}$ , which is below, after the application of the chain rule:

$\frac{dh_{cm}}{dh_s} = \frac{\rho_s A h_s(M_c + \rho_s A h_s) - \rho_s A M_c \frac{h_c}{2} - \rho^2_s A^2 \frac{h^2_s}{2}}{(M_c + \rho_s A h_s)^2}$

Setting this derivative equal to zero allows us to solve for the minimum value of $h_{cm}$ .

$\frac{dh_{cm}}{dh_s} = 0$

After simplification, this leads to a quadratic equation:

$h_s^2 + \frac{2 M_c}{A \rho_s} h - \frac{M_c h_c}{A \rho_s} = 0$

For which the solution is:

$h_s = \frac{-M_c \pm \sqrt{M^2_c + M_c h_c A \rho_s}}{A \rho_s}$

We will choose the solution that gives us a positive soda height rather than a negative one. Plugging in the values $M_c = 2~g$ , $\rho_s = 1~g/cm^3$ , $A = 2~cm^2$ , and $h_c = 3~cm$ gives us that the height of the soda which causes the center of mass of the system to be at a minimum is

$h_s = 1~cm$

Darn it, we're gonna need more soda.

The answer is 1.

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