Question 1.5.4: A closed cylindrical vessel, 0.3 m diameter and 0.6 m high, ......

Given for case (a): \mathrm{z_1 = 0.6 × 2/3 = 0.4  m,  z_R = 0.6  m} (see Fig. E1.5.4 a), and R = 0.15 m.

Using Eq. 1.5.8, we get ω = 18.68 rad/s. The liquid free surface is shown in Fig. E1.5.1.

\mathrm{z_R=z_1+\frac{\omega ^2R^2}{4g}and z_0=z_1-\frac{\omega ^2R^2}{4g} }                 (1.5.8)

(b) In this case the liquid will collect more and more near the wall as it cannot spill out. Let R be the radius at the cylinder brim when the liquid has touched the bottom of the cylinder. Initially, 1/3 rd cylinder is not filled with liquid. Therefore, volume of the paraboloid of revolution should be equal to it. This gives Volume of paraboloid of revolution = volume of unfilled space initially.
or,

1/2 (πR² × 0.6) = 1/3 {π(0.3/2)² × 0.6

→

R = 0.122 m.

Now
\mathrm{z_R = 0.6  m,  z_0 = 0 m  and  R = 0.122  m,}

Eq. 1.5.8 gives ω = 28.1 rad/s.

If the cylinder is open then liquid will spill out as ω increases beyond 18.68 rad/s. When the liquid touches the bottom of the cylinder then \mathrm{z_0 = 0  m, z_R = 0.6  m  and  R = 0.15  m.} Eq. 1.5.8 gives ω = 22.9 rad/s. Further, the volume of the oil left in the cylinder = ½ A × 0.6 = 0.3 A, here A is the cross sectional area of the cylinder. In the initial condition, the volume occupied is A × 0.6 × 2/3 = 0.4 A. Therefore, percentage of oil left in the cylinder = 0.3 A/(0.4A) × 100 = 75%. Thus the effect of removing the lid from the cylinder top is that the oil remaining in the cylinder is decreased by 25% at the time liquid touches cylinder bottom.