Correction follows. Sorry! The shape of the final equilibrium surface is:
h = (w^2/2g) x R^2 + h0 where h is height, w is angular velocity, g = 9.80665 m/s^2, and R is radius. Using R1 as the radius of the hole, R2 as radius of of tank, we have h = 0 at the radius R1 when equlilbrium is established and no more water can go down the hole. So: 0 = (w^2/2g) x (R1)^2 + h0 h0 = 2g/( w^2 x (R1)^2) The final height Hf of the water at the edge of the tank is thus: Hf = (w^2/2g) x (R2)^2 + 2g/( w^2 x (R1)^2) The above assumes that the initial angular velocity is small. If the angular velocity of the initial condition is high then the initial condition integration of angular momentum and energy also has to be outside the boundary established by h = (w^2/2g) x R^2 + h0 It is possible that in the initial condition all the water will be located outside the radius of the hole, and thus no water can go down the hole at all. Regards, Horace Heffner
