An inflatable ball is made of a thin, inelastic, but bendable material. When fully inflated, it has a radius $R$. If the ball is filled half full with air, and is then continuously pushed underwater by a large, flat surface. What shape does the ball assume when fully underwater? Approach the shape should be such that surface area remains the same and volume too and so maybe the energy minimization by lagrangian give the required surface?
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$\begingroup$ I'm guessing the problem means that you'll have a dome-shape and the radius of the dome is such that the pressure of the enclosed air equals the mean water pressure. Since you started with essentially one-half atmosphere pressure, the rest should follow. $\endgroup$– Carl WitthoftCommented Feb 4, 2022 at 12:36
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$\begingroup$ We need to prove it the shape too @Carl Witthoft i dont know if that shape is optimal $\endgroup$– ProblemDestroyerCommented Feb 4, 2022 at 12:49
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$\begingroup$ Without the ball, the air would form a thin flat layer under the surface as it tries to rise as far as possible. The ball will keep it from spreading. The air can spread out as much as possible if it occupies a flat region near the equator of the ball. $\endgroup$– mmesser314Commented Feb 4, 2022 at 13:48
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$\begingroup$ May you give a answer with a bit calculation so that we can get yhe exact shape , although your reasoning for the type of shape is pretty much what i think should be @mmesser314 $\endgroup$– ProblemDestroyerCommented Feb 4, 2022 at 13:56
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$\begingroup$ This exact question has already been posted here physics.stackexchange.com/q/671881/23615 $\endgroup$– TriatticusCommented Feb 4, 2022 at 22:39
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1 Answer
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I'm guessing homework, so I won't give a complete answer, just ask a lot of suggestive questions.
As a warmup exercise consider this. You take a rectangular solid box with five sides solid, and the top side open. Then turn it upside down and push it down into water. There will be air trapped inside. What will be the shape of the surface where the air contacts the water? Suppose instead of a rectangular box you had a cylindrical box. What will be the shape of the surface then?
What can you do about the deformation of the ball producing something that isn't spherical. You will have folds and bends and wrinkles. Can you ignore them as a low order approximation?
Consider the upper surface of the ball. What shape will it be? That's hopefully an easy one. It will conform to the flat surface. Suppose there was some small part that poked down into the air inside the ball. Air pressure will push it back up and squeeze the water out.
Consider the bottom surface of the air. Suppose there is a small portion that pokes down into the water. How will the pressure pushing on it compare to its nearby neighbor portions? How will the air pressure compare?
What will the surface tension of the ball be as a function of the shape of the ball under water? Consider a place in the water where the water pressure is greater than the pressure of the air inside the ball. The water pushes that part of the ball upward. But water pressure only depends on depth. So what will the bottom surface of the ball look like?
Finally, what has to be minimized? The air in the ball "wants" to rise out of the water. What parameter will be the controlling parameter you need to minimize to get the final shape?
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Some more hints.
In the rectangular box the air will form a flat surface. The water will find a single level where the pressure of the water trying to rise equals the (very slightly compressed) air in the box. If the surface was not flat then the raised portions would feel less water pressure than the lower portions, but the same air pressure (very nearly) and so would try to equalize.
If you wanted to include the small effect of compressing the air you could probably get quite accurate by treating it as an ideal gas. PV=nRT etc.
The controlling parameter will be depth of the bottom of the air-filled region. You want to minimize this.
There is a symmetry available that will get you a very obvious answer.
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$\begingroup$ And sure i would like to answer your questions : first one it would be like a hemispherical type but not exact spherical because sides are a and b which will make the shape a bit different , contact angle is 0° ideal case, water wouldnt rise any because of same pressure inside and outside .in case of cylinder it will be perfect hemisphere i think , second one i dont know about this .. , third one isnt the inside pressure P° of atmosphere inside the ball and so when it goes inside water the pressure may change assuming ideal gas behavior , temp. Might too change ?? even if these stuff dont $\endgroup$ Commented Feb 4, 2022 at 15:50
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$\begingroup$ how much will the gas compress how do we know ? And shouldnt near the sides the water should be elevated due to surface tension ? In rectangular box case and "then the raised portion would feel less water pressure" is it due to Pfrom air = weight of rised portion + Pressurefrom water below ? Why the controlling parameter is the depth please explain in more detail . $\endgroup$ Commented Feb 4, 2022 at 19:00