Rotating Fluid

The cylinder is "full": I interpret that as the liquid standing the full height $2R$ at the edge of the cylinder.

One quarter of the bottom is visible: thus the liquid reaches height zero at a distance $R/2$ from the center.

I use an energy analysis. At the surface the rotating fluid, the total mechanical energy (per unit of mass) is constant. This total energy is a combination of gravitational potential energy $\epsilon_g$ and kinetic energy $\epsilon_k$ .

If at a distance $r$ from the axis, the liquid reaches height $y$ , the energy of a unit mass at the surface is $\epsilon = \epsilon_g + \epsilon_k = -gy + \tfrac{1}{2}r^2\omega^2.$

We apply this equation to the liquid at the edge ( $r = R; y = 2R$ ) and at the bottom ( $r = R/2; y = 0$ ): $\epsilon_{\text{edge}} = -2gR + \tfrac{1}{2}R^2\omega^2;$ $\epsilon_{\text{bottom}} = 0 + \tfrac{1}{2}\left(\frac{R}{2}\right)^2\omega^2.$ Equating these gives $-2gR = -\tfrac{3}{8}R^2\omega^2,$ and solving for $\omega$ results in $\omega = \sqrt{\frac{16g}{3R}}.$ Clearly, $a = 16, b = 3$ , and $a + b = 19$ .