Rotating liquid in a bucket

Lets try it like this. @Lucas Tell Marchi sir's solution is also good but a bit complicated, lets solve it in easier way.

We know for any point on the curve of fluid we have coordinates say $(x,y)$ taking (0,0) to be at the minima of the curve.

So we can write that $\dfrac{dy}{dx} = \tan \theta$ .

Now lets see what forces are acting on the pt mass at (x,y).

Y direction

• Weight= dm g.

X direction

• Centrepetal force = $m \omega^2 x$ .

Now we know $\dfrac{dy}{dx} = \dfrac{\omega^2 x}{g}$

We can solve this Differential equation and put the limits of $x$ as $0 \rightarrow r$ and y as $0 \rightarrow z$ .

We will get

$\boxed{\boxed{z=\dfrac{\omega^2 x^2}{2g}}}$

$\implies A=1,B=2,C=2,D =2 \implies A+B+C+D=\boxed{7}$

Let $p$ be the pressure at any point inside the liquid. If $u$ is the potential energy density due to all forces acting on the liquid, it is a well-known fact of fluid statics that $\nabla p = \nabla (-u).$ In this case, the liquid spins, but in its own referential, it is static; in this referential, there exists a force density $\overrightarrow{f_c} = \rho \omega^{2}r \hat{r} = -\nabla u_c,$ due to inertial forces. As there's also gravity, the total potential energy density is

$u = \rho g z - \frac{1}{2} \rho \omega^{2} r^{2}.$

As we said before, $\nabla p = \nabla (-u),$ which leads us to the conclusion that

$p = \frac{1}{2} \rho \omega^{2} r^{2} - \rho g z + c,$

being $c$ some constant. If we let $r = z = 0$ we are at the surface of the liquid and the pressure must be $p_0,$ the atmosphere's pressure; this leads to the conclusion that $c = p_0.$ As we want to find the free surface equation, all we need to do now is let $p = p_0,$ which gives us

$z = \frac{ \omega^{2} }{2g} r^{2},$

from where we get the sum $A + B + C + D.$