The Question
A closed vertical cylinder is filled with an incompressible liquid of mass m completely. The cylinder is now sealed from all sides and rotated about a vertical axis with uniform angular speed. Force applied by liquid on the bottom of cylinder will be:
Choices And Explanation
[A]
mg
[B]
Greater than mg
[C]
Less than mg
[D]
Zero
Explanation:
For an instant, suppose that the cylinder was open from top. Then obviously, the liquid will tend to spill out from the top. If the top lid is now closed, the liquid will push the lid upward with a force (say F0). Lid will push the liquid downward with the same force F0. Hence FBD of liquid will be as shown in figure.
Here F is the force applied by bottom of container on liquid. For equilibrium of liquid,
Hence, (B)
Now this is the solution, but I do not understand the first line itself. Why will the liquid tend to spill out? Which force is responsible for this? Even if we consider the frame of the rotating liquid, it experiences a pseudo force only in the horizontal direction, so which force is acting in the vertical direction?
Best Answer
It took me some time to convince myself that answer (B) is correct. Here is my reasoning:
It is not enough to say "the liquid would spill out of the top if the top wasn't there" - because the middle of the liquid is lower, and you can't decide just from that piece of information whether those two effects balance out, or whether one is bigger than the other.
In the steady state scenario, viscous forces in the liquid have eliminated velocity gradients in the liquid until the velocity of the liquid is strictly proportional to the distance to the center of rotation. Once that happens, the liquid that's not on the axis is experiencing an acceleration - which means that there must be a force on the liquid, and which means there must be a differential pressure across the liquid in the radial direction, and the differential pressure scales with $r$ - the distance to the axis.
Now if we had a cylinder that was a little bit taller than the liquid level, this pressure differential would result in a slope on the surface of the liquid that is a function of $r$ - specifically, the slope would be proportional to $r$ so the surface takes on the shape of a parabola (ignoring surface tension effects). See https://physics.stackexchange.com/a/88344/26969 for a derivation, or http://www.mne.psu.edu/cimbala/Learning/Fluid/Rigid_body/rigid_body.htm for a more complete analysis including pictures.
Here comes the "bit of insight" that I needed:
Imagine you start with the open container. When it is rotating, the surface will be curved but the only force on the bottom of the container is the force of gravity pushing down on the liquid - in other words the total force on the bottom is equal to the weight of the liquid.
Now we start moving the lid down. When we first touch the water, we start to experience a force - we have to push down against the pressure of the water in order to push down the lid. As we push, the water is pushed inwards. Now in pushing on the lid, we are applying an additional external force on the liquid in the vessel - and so there must be an additional upward force by the bottom of the vessel on the liquid to keep things in equilibrium.
This diagram tries to explain it - the numbers are fictitious: