Closed cylindrical vessel completely filled and rotating about its vertical axis: What is the total resultant pressure on the top circular cover (radius r) due to rotation at angular speed ω, for a liquid of specific weight w?

Introduction / Context:When a completely filled cylindrical vessel rotates as a solid body about its vertical axis, the liquid develops a radial pressure variation. Even though there is no free surface at the top, the pressure at the top cover increases with radius, producing a net resultant force on the cover.

Given Data / Assumptions:

• Liquid of specific weight w (so density ρ = w / g).
• Angular speed ω (rad/s), radius r.
• Solid-body rotation; neglect end effects.

Concept / Approach:In solid-body rotation, radial equilibrium gives dp/dr = ρ * ω^2 * r. Integrate from the axis to radius r to obtain pressure rise relative to the axis: Δp(r) = (1/2) * ρ * ω^2 * r^2. The resultant on the circular top is the area integral of Δp over the disk area, giving a closed-form expression proportional to r^4.

Step-by-Step Solution:

Verification / Alternative check:Dimension check: w has units N/m^3, ω^2 r^4 / g gives m, multiplied by π yields N, consistent with force (total pressure resultant).

Why Other Options Are Wrong:

• π * r^2 * (w * ω^2 * r^2) / (2 * g): Uses edge pressure as uniform—incorrect; average over the area is half of the edge value.
• (w * ω^2 * r^3) / g: Wrong power of r and missing π factor.
• (π * w * ω * r^4) / (2 * g): Uses ω instead of ω^2 and wrong constant.

Common Pitfalls:Assuming uniform pressure equal to rim value; forgetting that pressure varies with r^2, making the average half the edge value and leading to the r^4 dependence after integration.

Final Answer:(w * π * ω^2 * r^4) / (4 * g)