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How to make an object submerged under water? Neutrally buoyant?

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  • $\begingroup$ I recommend an anchor... on second thought make it 2 $\endgroup$
    – user6760
    Apr 14 '19 at 9:43
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    $\begingroup$ Be aware that neutral buoyancy is an unstable condition. If the average density of your bottle is even the tiniest bit greater than the density of the surrounding water, it will start to sink. In sinking, it will be exposed to higher pressure, which will squeeze the flexible walls and compress the air within. The average density of your bottle will increase as it sinks, and it will sink faster. The opposite will happen if its average density is less than the water, and it starts to rise toward the surface. $\endgroup$ Apr 14 '19 at 13:35
  • $\begingroup$ Without some active means to control its depth, it must eventually either land on the bottom, or float at the surface. $\endgroup$ Apr 14 '19 at 13:37
  • $\begingroup$ I added some water so now it’s almost submerged but it’s not balancing because of the air bubble in the bottle how can I fix that $\endgroup$
    – Amanda_C
    Apr 15 '19 at 0:11
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enter image description hereIf density of submarine is less then water,it will go to the and if density is higher ,it will stay on top. When density of submarine is exactly equal to that of water,then wherever the submarine is in the water ,it will not be affected by gravity or buoyancy force(except if another factors come in way) and will not experience any upward or downward force at any place.But it is difficult to make the submarine of exact density as water.

What you can do is,you can put a balloon inside the submarine and can control the air in it to make it go up or down inside the water. Now you will be able to control its vertical motion.

well I don't think the fan will move freely because of frictional resistance.

In first image,it is water in balloon and in second it is air

enter image description here

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Here is another way that I have used to ensure that the submarine dives- and also returns to the surface when the rubber power is spent.

As described by others here, you "ballast" the sub, but not for neutral buoyancy: you add slightly too much air so it barely floats. Then you add fins to the body of the sub, and bend them slightly down so that when it is being propelled, the sub experiences a downforce which makes it sink.

Down it goes- until it runs out of rubber, and then it rises back to the surface.

While we are on the subject, having a single propeller at the rear of the sub means that the rubber band applies a countertorque to the body of the sub which urges it to rotate in the opposite direction of the prop. This reduces the propulsive efficiency of the rubber motor. This torque reaction can be easily cancelled by installing a second prop in the nose of the sub, connected to the opposite end of the rubber band. You pitch the blades of this prop backwards relative to the rear prop so it produces forward thrust while rotating in the opposite sense of the rear prop. No torque reaction!

Finally, note that a small diameter prop will spin up to great speed in water and furnish a lot of thrust for a short period of time. the slower the prop turns, the less thrust it will develop but the longer the sub will run. But trimming the prop diameter, you can obtain tradeoffs between maximum speed and maximum range.

One last thing: the plastic propellers used in rubber-powered model airplanes work well for sub propulsion- as long as you can get a pair of them in left- and right-hand rotation. Hobby shops sell these. And be sure to have as much fun as possible in your experiments!

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You can calculate the amount of water to put in the submarine. The buoyant force on the submarine will be $$ F_b = V \rho _w g $$ Where V is the volume of submarine submerged in the water and $\rho _w$ is the density of water. This buoyant force must balance all the weight of the submarine ( ie weight of submarine plus the weight of water poured inside it.) This means $$ V \rho_w g = m_{sub} g + m _{w}g $$ Where $m_w$ is mass of water poured into the submarine and $m_{sub}$ is the mass of submarine without any water in it. Since $ density = \frac{mass}{volume} \Rightarrow mass = density × volume $ $$ V \rho_w g = m_{sub} g + V_w \rho _w g $$ Where $V_m$ is the volume of water to be poured in the submarine. $$ V_w = \frac{V \rho _w - m_{sub}}{ \rho _w} $$ The volume of the submarine ie the bottle can be measured easily. Pour water in the submarine until it is full and then measure this water you get the volume of submarine. Be careful to place all the quantities in correct units . For instance if you place the volume of in ml then you must put its mass in grams and the density of water as 1gm/ ml and you will receive the volume of water to be poured in ml. Similarly if you place volume of bottle in $m^3$ (1 $m^3 = 10^3$ litres) you must put mass in kg and the density of water as $10^3$ kg/$m^3$ and you will receive the volume to be poured in litres. Put the water slightly less than the value you arrived at . As we have not considered other resistive forces.

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  • $\begingroup$ So would I put 𝑚𝑠𝑢𝑏 in kg? Also by Vw do you mean Vm? $\endgroup$
    – Amanda_C
    Apr 14 '19 at 21:55
  • $\begingroup$ also what does this mean 1gm/ ml $\endgroup$
    – Amanda_C
    Apr 14 '19 at 22:01
  • $\begingroup$ The msub = 4 grams (0.004kg), density of water = 10^3 kgm^3, volume of sub = 750mL = 0.75L = 0.00075 m^3, when I put all those number in the calculation I get 0.000746 L, which is 0.746 ml... $\endgroup$
    – Amanda_C
    Apr 14 '19 at 22:47
  • $\begingroup$ does that mean it is 746 ml? how doI get rid of the air bubble in the bottle that keeps moving the bottle. I want the bottle to be stable in the water $\endgroup$
    – Amanda_C
    Apr 15 '19 at 0:47
  • $\begingroup$ is there any reference on google where i can read upon that final equation? $\endgroup$
    – Amanda_C
    Apr 16 '19 at 0:16
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I believe it is possible to make a neutral buoyancy object stably submerged under water. The density of the object should be a bit greater than the typical density of water and the bulk modulus should be greater than that of water. As a result, the object would be stably buoyant at some depth because water density increases with depth due to compressibility of water.

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