Question
There is some amount of water in the beaker, and a vertical rod is passing through the centre of the circular base of the cylindrical beaker which is rotated with angular velocity ω , then water takes a particular shape(miniscus is formed). Find the equation of the miniscus of water when the mid point of the miniscus just touches the bottom surface of the beaker?
Joshi sir comment
after making the diagram, complete the miniscus as a sphere, then following conditions will be obtained
1) h = R(1-cosθ) here h, R and θ are height of miniscus, radius of the sphere and contact angle.
2) R = r/cosθ, r is the radius of beaker
3) tanθ = g/rw2
4) πr2hρ - πR3ρcos2θ = 2πrTcosθ
There are 4 equations, remove R, θ and T and get a relation between h and r in terms of g, w, ρ and π