An open 1- -m-diameter tank contains water at a depth of 0.7 m when at rest. As the tank is rotated about its vertical axis the center of the fluid surface is depressed. At what angular velocity will the bottom of the tank first be exposed? No water is spilled from the tank.

The answer to the question is

At an angular velocity of 7.4106 s⁻¹ the bottom of the tank of water will be first  exposed with no water spilling from the tank

Explanation:

The rotational motion of the fluid in the tank is

P(r, z) =

where r = radius,

g = Acceleration due to gravity

P = Pressure

ρ = Fluid density

c = constant

We have the boundary condition giving by

at r = 0,  z = z₀ and P = P₀

Therefore as we move from  the center outwards towards the wall of the cylinder we have

P(r, z) = P₀ +

When P(r, z) = P₀ we have   = 0 from which

and

which is the equation for paraboloid revolution

The fluid height at the wall of the cylinder where r = R is then

For a paraboloid circumscribed in a circle, its volume = half the volume of the cylinder thus

from which, we have by combining with the previous equation

therefore

At the point the fluid spills from the cylinder we have

Critical condition, this results in

the condition when the bottom of the tank will be exposed is given by h = H/2 hence

=

Solving we have for R = 0.5 m, H = 0.7 m and g = 9.81 m/s²

= 7.4106 s⁻¹

answer:

i dont know the answer but good luck

explanation: