Answer:
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