In a free-fall scenario, a bucket filled with water and featuring a side hole will not allow water to flow out due to the absence of gravitational force acting on the water. According to Bernoulli's equation, the pressure remains constant, and the cohesive forces among water molecules dominate. While a stationary observer would perceive the water extruding in a parabolic trajectory, a free-falling observer would see the water moving in a straight line. The cohesive forces lead to the formation of spherical shapes, potentially resulting in 'blobs' of water being expelled if the hole size and bucket shape permit.
PREREQUISITESPhysics students, fluid dynamics researchers, and anyone interested in the behavior of liquids under gravitational and free-fall conditions.
Rakshit Joshi said:There is a situation that a bucket which is full of water and has a hole at the side is let to fall freely. Will the water flow out of the bucket? And if it does, will its path be parabolic or linear? Assume that there is no air and hydrodynamics is working. Full detailed answer would be appreciated.
Why would it continue to be extruded, rather than just produce a hemispherical bulge?NTW said:Gravitational force disappears in a free-falling frame of reference, and in the absence of gravitational force, only one force remain: the cohesive force of attraction between water molecules. Hence, the mass of water will tend to acquire a spherical form. In so doing, constrained by the cylindrical or conical walls of the bucket, and its flat bottom, some water might be 'extruded' through the hole... For a stationary, non free-falling observer, the trajectory of the 'extruded jet' would be parabolic. For a free-falling observer, the extruded jet would move in straight line...