A cylindrical vessel of radius r containing a liqiud is rotating about a vertical axis through the centre of circular base, If the vessel is rotating with angular velocity `omega` then what is the difference of the heights of liquid at centre of vessel and edge?

Explanation

When cylindrical vessel will rotate the shape of surface of liquid in the vessel is as shown in figure. Take two points P and Q in the vessel at same level as shown in figure. From Bernoulli’s theorem
`P_(P)/rho + 1/2V_(P)^(2) + gh_(P) = P_(Q)/rho + 1/2V_(Q)^(2) + gh_(Q)`
As `h_(P) = h_(Q)` and `v_(P) = 0`
`:. P_(P) = P_(Q) + 1/2 rhov_(Q)^(2)`
P_(P) - P_(Q) = 1/2 rhov_(Q)
`rhogh = 1/2 rho r^(2)omega^(2)`
h = `(r^(2) omega^(2))/2g`