Consider now the centripetal force \({F}_{c}(x)\) at depth \(x ≤ h\),
\({F}_{c}(x) = Fc - (h-x)\rho .g.\triangle A\)
where \(\rho\) is the density of the fluid and \(g\) is gravity and \(\triangle A\), an elemental area at depth \((h-x)\) and radius \(y\). As depth increases, \({F}{c}(x)\) decreases and the opening of the vortex closes up, this models the vortex as a inverted cone. We know that \(F_c(0) = 0\), therefore,
\(F_c = h. \rho .g.\triangle A\)
\({F}_{c}(x) = \rho .g.\triangle A.x = \triangle m.y.w^2\),
\(w^2 = \rho .g.\triangle A.\frac { x }{\triangle m.y}\)
where \(w\) is the angular velocity of an elemental volume at depth \((h-x)\) just on the slant surface of the cone at a radius of \(y\) and \(\triangle m\) an elemental mass of fluid in circular motion.
\(\triangle A = y.\triangle \theta .\triangle x\), \(\triangle m = y\triangle \theta . \triangle x. \triangle y. \rho\)
where \(\theta\) is the angle at the axis of rotation. If we consider a elemental ring of radius \( y\) and thickness \(\triangle y\) and integrate over \(d\theta\) for \(2\pi\) then,
\(\triangle A = 2\pi.y.\triangle x\), \(\triangle m = 2\pi.y.\triangle x .\triangle y. \rho\)
\(w^2 = g.\frac { x}{y.\triangle y}\)
The angular kinetic energy of such a ring is,
\(\triangle VKE_y = \frac { 1 }{ 2 }I(y)w^{ 2 }\)
where \(I(y)\) is the moment of Inertia of a thin ring of mass \( m\), radius y spinning at its center axis.
\(I(y)= m.y^2\), where \(m = 2\pi.y.\triangle y . \triangle x.\rho\)
\(I(y)=2\pi.\rho.y^3.\triangle y . \triangle x\)
So,
\(VKE_y = \pi.\rho.g.{ x }.y^{ 2 }. \triangle x\)
For total VKE over all depth \(x\), we consider,
\(\triangle VKE = VKE_y = \pi.\rho .g.{ x }. y^2 .\triangle x \)
\(VKE = \pi.\rho .g.\int^{h}_{0}{ x }y^2 dx\), since \(y=\frac{r}{h}x\)
\(VKE = \pi. \rho .g.\int^{h}_{0}{ x }{(\frac{r}{h}x)^2} dx\),
\(VKE = \pi. \rho .g.\frac{r^2}{h^2}|^{h}_{0}\frac {x^4}{4}\)
\(VKE = \frac{1}{4} \pi \rho g{r^2}{h^2}\)
This is the vortex kinetic enegry.
