Answer:
Explanation:
To get the angular velocity with which the bottom of the tank first be exposed - The careful and detailed step by step analysis is as shown in the attached file
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) = [tex]\rho(\frac{r^{2} \omega^{2} }{2}-gz)+c[/tex]
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 + [tex]\rho(\frac{r^{2} \omega^{2} }{2}-g(z-z_{0} ))[/tex]
When P(r, z) = P₀ we have [tex]\rho(\frac{r^{2} \omega^{2} }{2}-g(z-z_{0} ))[/tex] = 0 from which
[tex]\frac{r^{2} \omega^{2} }{2} = g(z-z_{0} )[/tex] and
[tex]z-z_{0} = \frac{r^{2} \omega^{2} }{g2}[/tex] which is the equation for paraboloid revolution
The fluid height at the wall of the cylinder where r = R is then
[tex]z_{R} = z_{0} + \frac{R^{2} \omega^{2} }{g2}[/tex]
For a paraboloid circumscribed in a circle, its volume = half the volume of the cylinder thus
[tex]\frac{1}{2}(z_{R}-z_{0})R^{2} \pi= (h-z_{0})R^{2}.\pi[/tex] from which, we have by combining with the previous equation
[tex]z_{0} = h-\frac{R^{2} \omega^{2} }{g2}[/tex] therefore [tex]z_{R} = h+\frac{R^{2} \omega^{2} }{g2}[/tex]
At the point the fluid spills from the cylinder we have
Critical condition, [tex]\omega =\omega_{l} , z_{R} = H[/tex] this results in
[tex]\omega_{l} = \frac{2}{R}\sqrt{g(H-h)}[/tex] the condition when the bottom of the tank will be exposed is given by h = H/2 hence
[tex]\omega_{l} = \frac{2}{R}\sqrt{g(H-\frac{H}{2} )}[/tex] =[tex]\omega_{l} = \frac{2}{R}\sqrt{(\frac{gH}{2} )}[/tex]
Solving we have for R = 0.5 m, H = 0.7 m and g = 9.81 m/s²
[tex]\omega_{l}[/tex] = 7.4106 s⁻¹
