Question

The force on the bottom of a cylinder which is rotating contains water

Answer 1

The shape of the empty space in the cylinder is a paraboloid.

If h is the height of the empty space, volume of liqiuid spilled out is

$\frac{1}{2} {\pi}^{2} \: h$
. r is the radius of the cylinder.


Therefore, if half the volume is spilled out, then height of the paraboloid is h. Above the center of the bottom there is no liquid .

And the pressure due to liquid is zero.

Answer 2

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I would like to add to Carltons answer and go a little more in depth on how to approach problems like that in general.

Hydrostatics or Hydrodynamics

There are two possible solutions to the problem. You can either choose a reference frame that rotates with the fluids or that stays still and observes the system from the outside. This will determine if hydrostatics are allowed or not.

Rotating reference frame

The origin of the coordinate system (polar coordinates) is at the top of the cylinder, positive z axis direction is upward, against the gravity.

Since there is no movement in our system as we rotate the same way the fluid does, hydrostatics can be used.

∇p=ϱf⃗

∇p=ϱf→

Due to the rotation of our reference frame we have to add the centrifugal force as an external force.

f⃗ =⎛⎝⎜Ω2r0−g⎞⎠⎟

f→=(Ω2r0−g)

Hence

⎛⎝⎜⎜⎜⎜∂p∂r1r∂p∂φ∂p∂z⎞⎠⎟⎟⎟⎟=ϱ⎛⎝⎜Ω2r0−g⎞⎠⎟

(∂p∂r1r∂p∂φ∂p∂z)=ϱ(Ω2r0−g)

Successively Integration yields

p(r,φ,z)=ϱΩ2r22+C(φ,z)(1)(1)p(r,φ,z)=ϱΩ2r22+C(φ,z)

Inserting in the second row

1r∂ϱΩ2r22+C(φ,z)∂φ=01r∂ϱΩ2r22+C(φ,z)∂φ=0

C′(φ,z)=0C′(φ,z)=0

Hence C(z)C(z)

Inserting in the third row yields

C′(z)=∂C(z)∂z=−ϱgC′(z)=∂C(z)∂z=−ϱg

C(z)=−ϱgz+DC(z)=−ϱgz+D

Hence

p(r,z)=ϱΩ2r22−ϱgz+D

p(r,z)=ϱΩ2r22−ϱgz+D

With the boundary condition that p(r=0,z=0)=pap(r=0,z=0)=pa it follows that

p(r,z)=pa−ϱgz+ϱΩ2r22(2)

(2)p(r,z)=pa−ϱgz+ϱΩ2r22

This approach is more flexible as you can account for any changes in the problemset without relying on a formula that is just 1 solution for 1 specific problem.

Fixed reference frame

In this case we observe the system from the outside and we can no longer use hydrostatics for our problem.

Here Navier-Stokes comes in handy.

In polar coordinates for an inviscid, stationary flow the equation for the radial direction is

u∂u∂r+vr∂u∂φ+w∂u∂z−v2r=−1ϱ∂p∂r+fru∂u∂r+vr∂u∂φ+w∂u∂z−v2r=−1ϱ∂p∂r+fr

with v⃗ =⎛⎝⎜uvw⎞⎠⎟v→=(uvw)

As there are now no external forces acting on the system fr=0fr=0 and we omit all zero terms that follow from u=0u=0 and z=0z=0 as we only have a velocity in vv direction, it follows that.

−v2r=−1ϱ∂p∂r−v2r=−1ϱ∂p∂r

with Ω=vrΩ=vr

Ω2r=1ϱ∂p∂r

Ω2r=1ϱ∂p∂r

which equals (1).

Repeat with v and z direction and the same boundary condition you get to (2).

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I prefer general solutions over specific solutions that why I felt the need to add to this answer.

Calculating force

Now you can integrate over the walls of the cylinder.

F=∫Ap(r=R,z)Rdφdz