Upside Down Paper-Glass Water Experiment

[S-Ragnork1729]

Homework Statement

A cylindrical glass was filled with water to a height H. The glass had a total height of h and a radius r. The glass was then covered with a sheet of paper and turned upside down. The volume of water that escaped between the paper and the glass was V . When the paper was no longer held against the glass, it remained attached, while the remaining water stayed inside the glass. What is the maximum mass m of the paper? The air pressure is p, the gravitational acceleration is g, and the water’s density is ρ. The paper used was coated and did not absorb water. At the moment the paper was released, the temperatures of the air and water were the same.

Relevant Equations

$$F_{up} \geq F_{down}$$

$$F_{down} \leq F_{up} $$ $$\Longrightarrow mg \leq \rho g(h-H)\pi r^2 $$ $$\Longrightarrow m \leq \rho (h-H)\pi r^2 $$

 

[Greg Bernhardt]

Show your work. What have you tried?

 

[haruspex]

Homework Statement: A cylindrical glass was filled with water to a height H. The glass had a total height of h and a radius r. The glass was then covered with a sheet of paper and turned upside down. The volume of water that escaped between the paper and the glass was V . When the paper was no longer held against the glass, it remained attached, while the remaining water stayed inside the glass. What is the maximum mass m of the paper? The air pressure is p, the gravitational acceleration is g, and the water’s density is ρ. The paper used was coated and did not absorb water. At the moment the paper was released, the temperatures of the air and water were the same.
Relevant Equations: $$F_{up} \geq F_{down}$$

$$\Longrightarrow mg \leq \rho g(h-H)\pi r^2 $$

Decoding that, “the weight of the paper is less than or equal to the weight of the water that could have been put in the glass initially but wasn’t”.
Eh?

Start by answering this: how is it possible that any water stayed in the glass after turning the glass over and taking away whatever was supporting the paper?

 

[jbriggs444]

I would like to see OP explain the process of inverting the glass and the final resulting equilibrium with words before trying to write down equations.

My understanding of the situation is that upright glass is sealed with a piece of paper at the rim. It is inverted and then some water is allowed to escape at the seal between paper and rim. Importantly, no outside air is allowed to enter as the water escapes.

The answer supplied by OP did not take account of the volume $V$ of water that was allowed to escape. Accordingly, it cannot be correct.

I have not personally attempted to invert a glass in this fashion. When I do it, I am always careful not to allow any air-filled head space. The expansion of air in the head space is undesirable. I generally use a playing card for the paper seal. The task is made easier if one presses gently on the center of the playing card under the inverted glass, squeezing a bit of water out before removing one's hand. That squeezed out water is the $V$ in the problem.