Respuesta :
Answer:
Explanation:
The question is one that examine the physical fundamental of mechanics of a cylindrical vessel .
We would use the Euler' equation and some coriolis and centripetal force formula.
The fig below explains it.
Answer:
(a) The shape of the free surface of the water is a parabola of revolution as follows;
[tex]h(r) = h_0 + \frac{ \Omega^2}{2g} r^2[/tex]
(b) The water pressure distribution over the bottom of the vessel is
[tex]\rho \times g \times (h_0 + \frac{ \Omega^2}{2g} r^2)[/tex] where r is the distance from the axis.
Explanation:
To solve the question, we solve the Euler's equation of the form
[tex]\rho (\frac{\partial u}{\partial t} -\textbf{u} \times \textbf{\omega} ) = -\rho \textbf{u} \times \omega = -\bigtriangledown (p + \frac{1}{2} \rho \parallel \textbf{u} \parallel^2) + \rho \textbf{g}[/tex]ω) = -ρu×ω = [tex]- \nabla[/tex](P + [tex]\frac{1}{2}[/tex]ρ ║u║²) + ρg
When in uniform rotation, we have
u = [tex]u_{\theta}\hat{e}_{\theta}[/tex] , ω = [tex]\omega_z \hat{e}_{z}[/tex] where [tex]u_{\theta}[/tex] = rΩ and [tex]\omega_z[/tex] = 2Ω
Therefore, u × ω = 2·r·Ω²· [tex]\hat{e}_{r}[/tex]
From which the radial component of the vector equation is given as
-2·p·r·Ω² = [tex]\frac{\partial P}{\partial r} - \frac{\rho}{2}\frac{d u_{\theta}^2}{dr} = -\frac{\partial P}{\partial r} - \rho r\Omega^2[/tex]
Therefore,
[tex]\frac{\partial P}{\partial r} = \rho r\Omega^2[/tex] = [tex]\rho \frac{u_{\theta}^2}{r}[/tex]
Integrating gives
P(r, z) = [tex]\frac{ \rho \Omega^2}{2} r^2 +f_1(z)[/tex]
By substituting the above into the z component of the equation of motion, we obtain;
[tex]\frac{dp}{dz} = -\rho g \Rightarrow \frac{df_1}{dz} = -\rho g \Rightarrow f_1(z) = -\rho g z+C_3[/tex]
Therefore
P(r, z) = [tex]\frac{ \rho \Omega^2}{2} r^2 + -\rho g z+C_3[/tex]
From the boundary conditions r = R and z = [tex]z_R[/tex], we find C₃ as follows
P(r = R, z = [tex]z_R[/tex]) = [tex]p_{atm}[/tex]
Therefore [tex]p_{atm}[/tex] = [tex]\frac{ \rho \Omega^2}{2} R^2 + -\rho g z_R+C_3[/tex]
From which we have
P(r, z) - [tex]p_{atm}[/tex] = [tex]\frac{ \rho \Omega^2}{2} r^2 + -\rho g z+C_3[/tex] - [tex](\frac{ \rho \Omega^2}{2} R^2 + -\rho g z_R+C_3)[/tex]
P(r, z) - [tex]p_{atm}[/tex] = [tex]\frac{ \rho \Omega^2}{2} (r^2 -R^2) -\rho g (z-z_R)[/tex]
We note that at the surface, the interface between the air and the liquid
P = [tex]p_{atm}[/tex], the shape of the of the free surface of the water is therefore;
[tex]z_R-z =\frac{ \Omega^2}{2g} (R^2 -r^2)[/tex],
Given that at r = 0 we have the height = h₀
Therefore, [tex]z_R-z =h_0 + \frac{ \Omega^2}{2g} r^2 = h(r)[/tex]
The shape of the of the free surface of the water is a parabola of revolution.
(b) The water pressure distribution over the bottom of the vessel is given by
ρ × g × z
= [tex]\rho \times g \times \frac{ \Omega^2}{2g} (R^2 -r^2)[/tex] = [tex]\rho \times g \times (h_0 + \frac{ \Omega^2}{2g} r^2)[/tex]
