Q.1. Maximize exposed wetted area of a disc rotating vertically in a liquid

Question 1:

A circular disk of radius r is used in an evaporator and is rotated in a vertical plane. If it is to be partially submerged in the liquid so as to maximize the exposed wetted area of the disk, show that the center of the disk should be positioned at a height r/sqrt(1+pi^2) above the surface of the liquid.

Answer 1:

Let the center of the circle O make an angle θ with the chord AB cut on the liquid surface.

Exposed wetted area, A

= area of the disc - area of sector OAB + area of triangle OAB - area of circle with center O and touching AB

= πr^2 - (1/2)r^2 * θ + (1/2)r^2 * sinθ - π[rcos(θ/2)]^2

For A to be maximum, dA/dθ = 0

=> - (1/2)r^2 + (1/2)r^2 cosθ + πr^2cos(θ/2)sin(θ/2) = 0

=> - (1/2)r^2 + (1/2)r^2 [1 - 2sin^2 (θ/2)] + πr^2cos(θ/2)sin(θ/2) = 0

=> - sin^2 (θ/2) + πcos(θ/2)sin(θ/2) = 0

=> tan(θ/2) = π

=> height of the center of the disc above the surface of the liquid

= rcos(θ/2)

= r / √(1 + π^2).

Link to YA!