Rotating liquid in a closed, completely filled cylindrical vessel In a closed cylindrical vessel completely filled with liquid and rotated about its vertical axis at constant speed, the total resultant force on the bottom equals the sum of the total centrifugal-pressure resultant and the weight of liquid in the vessel. Is this statement correct?

Introduction:
Fluids in rigid-body rotation develop a characteristic pressure field. This question probes whether the integrated force on the vessel bottom accounts both for the hydrostatic weight and for the additional pressure due to rotation (often nicknamed centrifugal pressure) when the container is completely filled and sealed.

Given Data / Assumptions:

• Cylindrical vessel of plan area A and liquid height H, completely filled.
• Rotation at angular speed omega about the vertical axis; steady rigid-body rotation (no relative shear).
• Density rho, gravity g, no free surface deformation due to the closed top.

Concept / Approach:

The pressure field in rigid-body rotation is p(r, z) = p0 + rho * g * (H − z) + rho * omega^2 * r^2 / 2, where r is radial distance and z is measured upward from the bottom. The force on the bottom is F_bottom = ∫ p dA evaluated at z = 0 over the circular area. The first vertical term integrates to rho * g * H * A, equal to the liquid's weight. The rotational term integrates to rho * omega^2 / 2 * ∫ r^2 dA, the resultant of the so-called centrifugal-pressure distribution. Their sum (plus any uniform reference pressure p0 * A that cancels with external support) is the total bottom force, validating the statement.

Step-by-Step Solution:

Verification / Alternative check:

If the top were open, a parabolic free surface would form; still, integrating p on the bottom gives W plus a rotational contribution relative to the free-surface reference. Thus, the inclusion of both components is physically consistent.

Why Other Options Are Wrong:

Incorrect / side-walls only: Although radial pressure acts on side walls, the rotational term also appears in bottom pressure because p depends on r^2 at the bottom.Conditional options: The relation does not require very small rotation or an open top for its validity.

Common Pitfalls:

Assuming that centrifugal effects do not influence vertical forces; they do via the r^2 dependence of pressure at the bottom surface.

Final Answer:

Correct