Air pressure holds coaster on glass: What if only half full?

[greypilgrim]

Hi.

If I fill a glass with water up to the top and close it with a coaster, I can turn it upside down without the coaster falling off and spilling everything. Usually, this is explained with the atmospheric air pressure being bigger than the pressure of the water inside the glass (hence, it should also work with a tube up to a height of about 10 m).

However, I noted that the glass doesn't need to be full up to the top at all, in fact it works with any water level. Why is that? Shouldn't the pressure of the air-filled part still be atmospheric and add up with the pressure of the water-filled part to something bigger than the outside atmospheric pressure?

A wet coaster also sticks to a completely empty glass, I assume due forces from the surface tension of the water. But I don't think they are strong enough to explain the above.

 

[jbriggs444]

However, I noted that the glass doesn't need to be full up to the top at all, in fact it works with any water level. Why is that?

As long as the height of the liquid in the glass is small compared with 10 meters, the air inside the glass is approximately incompressible. It does not expand much under the weight of only a half-glass of water. For best results (in our house we typically used playing cards), I would always dish the card a little bit into the glass before releasing. The resulting suction allows one to deal with imperfections like a glass being only half full.

The pressure in the head space in an inverted glass will definitely not be atmospheric. It has to be enough below atmospheric to support the column of water remaining in the glass -- minus whatever surface tension can hold by itself.

 

[greypilgrim]

The pressure in the head space in an inverted glass will definitely not be atmospheric.

Why not? I used a rigid coaster to cover and invert the glass. The air has the same space and hence volume as before, why should the pressure change?

 

[Nugatory]

I noted that the glass doesn't need to be full up to the top at all, in fact it works with any water level.

Is the maximum weight that can be supported (that is, the sum of the weight of the water and the coaster) the same for all water levels?

Why not? I used a rigid coaster to cover and invert the glass. The air has the same space and hence volume as before, why should the pressure change?

The assumption of perfect rigidity is valid when you're holding the coaster in place, and it is valid after you've released the coaster. However, these cannot be the same state (because the forces acting on the coaster are different). Is the assumption of perfect rigidity consistent with the transition between the two states?

 

[greypilgrim]

(that is, the sum of the weight of the water and the coaster)

Why is this the maximum weight? If the atmospheric pressure is much greater than all pressures acting from the inside, one should be able to add additional weights to the coaster. Anyway, how does this matter?

The assumption of perfect rigidity is valid when you're holding the coaster in place, and it is valid after you've released the coaster. However, these cannot be the same state (because the forces acting on the coaster are different). Is the assumption of perfect rigidity consistent with the transition between the two states?

What if I only hold the coaster opposite the edge of the glass?

 

[Nugatory]

Why is this the maximum weight?

Maybe I wasn't clear... The weight that is supported is, by definition, the sum of the weight of the coaster and the water. For any given amount of water in the glass, what is the maximum value of that sum before the coaster will no longer be supported by the air pressure from below? Adding more water to the glass increases the sum, as does using a heavier coaster. There is a maximum value for that sum, beyond which the coaster will separate from the glass.

 

[Nugatory]

What if I only hold the coaster opposite the edge of the glass?

That makes the problem harder to analyze without providing any new insight into the physics. There's not much point in taking on that problem until you've worked through the case in which the force from your hand is acting directly on the center of mass of the coaster.

 

[greypilgrim]

For any given amount of water in the glass, what is the maximum value of that sum [...]

I might be particularly slow at the moment, but if the amount of water is given (and the mass of the coaster), what degree of freedom is there left such that the sum of their weights can have a maximum value?

Or are you talking about the 10 m (minus something if we consider the weight of the coaster) limit due to atmospheric pressure?

 

[Nugatory]

I might be particularly slow at the moment, but if the amount of water is given (and the mass of the coaster), what degree of freedom is there left such that the sum of their weights can have a maximum value?

You get to choose the coaster, so you get to choose the weight of the coaster. That's a degree of freedom.
You get to choose the amount of water you put into the glass. That's another degree of freedom.

You said "in fact it works with any water level". Does that mean that no matter what choices you make for these two degrees of freedom, the coaster will stick to the glass? No, because (obviously) if the coaster is made out of a slab of 10 cm cast iron it will fall. So what combinations of coaster weight and water in the glass will allow the coaster to remain stuck to the glass?

Answer that question, and answer the question about the transition between the "coaster supported by your hand" state and the "coaster supported by air pressure" state, and you'll see what's going on here. Answering the second question will also help you see what @jbriggs444 was getting at when he (correctly) stated that the pressure inside the glass cannot be atmospheric after you've removed your hand.

 

[greypilgrim]

If the glass is completely full with height $h$ and lower cross section $A$, then the coaster must not weigh more than
$$m=\frac{(p_0-\rho_W\cdot g\cdot h)\cdot A}{g}\enspace.$$

For the second question, I assume the water makes the coaster bulge slightly outwards. The air inside expands a little bit, such that its pressure plus the pressure of the water plus the pressure of the coaster is lower than the outside atmospheric pressure.