Rotation of a sphere and cube full with water

[Gh778]

Hi,

A sphere and a cube full at 99.99% with water are put in symetrical position to the center C. Sphere at right and cube at left for example. When they turn together at the center C, the centripetal forces are not the same. The cube has not water all around faces, especially the face closed to the center C and this face can't put pressure (we arrange the shape for that). The sphere can put pressure in the direction of the center even the sphere is at 99.99% with water.

Example:

R=100m
side of cube = 1m
Radius of sphere = 0.62 m
Volume of sphere = volume of cube = 1m
Mass of sphere = mass of cube = 1000 kg
Rotational speed = 10rd/s

For the cube F=1000*100*100.5 = 10.05 e+6 N

For the sphere F=1000*100*(100.3-99.7) = 60000N

Is this result is correct ? Or if not have you the method ?

Edit: we can turn or make oscillation +/- 30 ° around, but this create a force and I think the result is not good but why ?

 

[Gh778]

I added a drawing.

I added result for a cube and a triangle full with water

With:

A rotational speed of w=10rd/s
A radius of R=100m
A cube of 1 m of side full with water, m=1000kg
A isoscele right rectangle of base = 1.5m , h=0.75m, proof = 1.77m => volume =1 m³, m=1000kg (full with water)

The centripetal force for the cube is like m*w²*R = 1000*100*100 = 10e+6 N

The centripetal force for the triangle is like m*w²*100.25-m*w²*99.97*0.28 = 7.2e+6N
0.28 because the weight increase of x² in the triangle
100.25 m and 99.97 m are where the force is apply

Like this the result is right ?

 

[Gh778]

Nobody ? I can't find something with internet. Maybe you have a link to explain this ?

 

[Doc Al]

I'm not understanding the issue. If the two containers of water have the same mass, distance from the axis, and speed why wouldn't the net centripetal force be the same for both?