- #1
largich
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I have a cone filled with liqid with radius R and height H rotating with \omega. Where do we have to drill a hole that the water would spray to the maximum distance from the cone?
I used the Bernoulli equation obtainig
p_0+0.5 \rho {v_1}^2=p_0+0.5 \rho v^2
v is the speed at the hole, getting
v^2=2g(H-h-h^2-r^2\frac{\omega^2}{2g})=2g(H-h-h^2\frac{(tg{\alpha})^2}\omega^2}{2g}),
where tg{\alpha}=R/H.
I taught using Lagrange multiplicator, where the constraint is the water falling on the floor prom the upward cone:
\psi=v sin{\alpha} t+gt^2/2-h=0.
Further more:
F=v_x t+\lambda(v sin{\alpha} t+gt^2/2-h)
=v cos{\alpha}+\lambda(v sin{\alpha} t+gt^2/2-h)
Solution should be obtained by
\frac{\partial F}{partial t} and
\frac{\partial F}{partial v}, v=v(h),
but i can't solve it.
Did I make the concept wrong? Any ideas would be helpfull.
PS: The cone is standing on its tip and it is opened at the top.
I used the Bernoulli equation obtainig
p_0+0.5 \rho {v_1}^2=p_0+0.5 \rho v^2
v is the speed at the hole, getting
v^2=2g(H-h-h^2-r^2\frac{\omega^2}{2g})=2g(H-h-h^2\frac{(tg{\alpha})^2}\omega^2}{2g}),
where tg{\alpha}=R/H.
I taught using Lagrange multiplicator, where the constraint is the water falling on the floor prom the upward cone:
\psi=v sin{\alpha} t+gt^2/2-h=0.
Further more:
F=v_x t+\lambda(v sin{\alpha} t+gt^2/2-h)
=v cos{\alpha}+\lambda(v sin{\alpha} t+gt^2/2-h)
Solution should be obtained by
\frac{\partial F}{partial t} and
\frac{\partial F}{partial v}, v=v(h),
but i can't solve it.
Did I make the concept wrong? Any ideas would be helpfull.
PS: The cone is standing on its tip and it is opened at the top.