I should also note that the final equilibrium is given by: h = (w^2/2g) x R^2 - 2g/( w^2 x (R1)^2)
where w is the final angular velocity of the water and tank, R1 is the drain radius, and R is a given radius, *provided* there is enough water in the tank initially to fill the volume under that surface. If not then the radius R1 is not the radius of the drain hole, but is a larger radius, and in that case, assuming the tank is rotating at w, no water can go down the drain at all. Regards, Horace Heffner
