Centre of Mass of a Can

question physics short

Question

When drinking a can of a beverage, how much liquid is in the can when the centre of mass is at its lowest? Assume the can has a constant cross-sectional area, and a height of $2H$. Give your answer in terms of $H$, the mass of the can $m$, and the density of the liquid $\rho$.


Hints

There are two contributions to the centre of mass - the constant weight of the can itself, and the weight due to mass of the beverage which depends on the remaining volume.

Initially, the centre of mass is in the middle of the can. After drinking all of the beverage, the centre of mass is in the middle of the can.


Answer

The equation for the centre of mass, for an object made up of is given by $$\text{total mass}*\bar{x}=\sum_{i=0}^N{m_i*x_i}$$
Where $\bar{x}$ is the distance of the centre of mass from the origin, $m_i$ and $x_i$ is the mass and distance from origin of the $i$th particle respectively.

In this case, $$(m+x\rho)\bar{x}=mH+\frac{1}{2}\rho x^2$$ You can find the minimum value of $\bar{x}$ via differentiation. However, another method is to notice that the minimum occurs at $\bar{x}=x$ - when the top of the fluid is in line with the CoM. This is because, at this point:

  • Increasing $x$ will bring more mass above the CoM, raising $\bar{x}$
  • Decreasing $x$ will reduce the amount of mass that is below the CoM, raising $\bar{x}$

Therefore, just substitute $x=\bar{x}$ into the equation above to get a quadratic in $\bar{x}$, the (positive) solution to which is

$$ x=\frac{-m+\sqrt{m^2 +2\rho mH}}{\rho} $$

This answer can be simplified by considering the ratio of masses $r=\frac{m}{\rho}$, giving

$$ x=-r+\sqrt{r^2 +2rH} $$