Question

A vessel cylindrical in shape and closed at the top and the bottom of the radius R and height H.The vessel is completely filled with water?

If it is rotated about its vertical axis with a speed of w Rad/sec, what is the total pressure force exerted by water on the top and bottom of the vessel?

Answer 1

Let us consider a drop of water of mass `m` at the top of cylinder as depicted in the figure above, we see that two forces are acting on the drop. Weight of drop `mg` and normal reaction `N`. The direction of normal reaction is tilted due to rotation of water taken as incompressible and settled into rigid body rotation.

The `costheta` component of normal reaction creates pressure on the top. It can be shown`"^$` that surface of water at the top takes the shape of parabola and total pressure force exerted by the liquid on the top of the vessel.

`P_"top"=pi/4ρomega^2R^4`
where `omega` is angular speed of cylinder about its central vertical axis, `rho` density of water and `R` radius of cylinder.

We know Newton's Third Law of Motion that for every action there is there is equal and opposite reaction. As such this pressure generated on the top has a reaction. The pressure on the bottom thus is

`P_"bottom"=P_"top"+"Pressure due to weight of water"`
`P_"bottom"=pi/4ρomega^2R^4+piR^2rhoHg`
where `H` is height of water in the cylinder

.-.-.-.-.-.-.-.-.-.

`"^$`
`Ncostheta=mg` .....(1)
`Nsintheta=mromega^2` ....(2)
Dividing (2) with (1) we get
`tantheta=(romega^2)/g`

Taking

`tantheta=(dz)/(dr)`
where `r` and `z` are cylindrical coordinates.

Above expression becomes

`(romega^2)/g=(dz)/(dr)`

This separates and integrates into the parabolic surface profile

`z=(omega^2r^2)/(2g)+C`
where `C` is constant of integration.

1. The cylinder is closed at the top. As such water in the cylinder does not spill over.
2. For mass of water below the center of paraboloid formed by free surface, reference `z=0`, normal reaction `R=mg`. Therefore, mass of water contributing to the upwards pressure is only which is contained within the paraboloid surface and the walls of the cylinder.

At `r=0`, taking `z=0`, we get `C=0`. Therefore, above expression becomes

`z=(omega^2r^2)/(2g)`........(3)

Mass of rotating water creating pressure towards top

`M=int_0^R\ rhoz(2pir)dr`

Using (3)

`M=int_0^R\ rho(omega^2r^2)/(2g)(2pir)dr`
`=>M=|(rhopiomega^2r^4)/(4g)|_0^R`
`=>M=(rhopiomega^2R^4)/(4g)`

Pressure on the top

`P_"top"=(rhopiomega^2R^4)/(4g)g`
`=>P_"top"=pi/4rhoomega^2R^4`

Note that as expected it is independent of `g`