Question 6.43: An open cylinder of 20 cm diameter and 30 cm height is fille......

Given data:
Diameter of cylinder            D = 20 cm = 0.2 m
Radius of cylinder                 R = 0.1 m
Height of cylinder                 H = 30 cm = 0.3 m

Let x be the radius of the exposed portion of the bottom of the tank (Fig. 6.28). Let us also consider that the height of the imaginary parabola DOE is z.

Now,                        \pi x^2=\frac{\pi R^2}{3}

or                                \frac{x^2}{R^2}=\frac{1}{3}=0.333

Using the relation z=\frac{\omega^2 r^2}{2 g} , for the imaginary parabola

DOE, we have

z=\frac{\omega^2 x^2}{2 g}                     (6.59)

Using the relation z=\frac{\omega^2 r^2}{2 g} , for the parabola AOB, we have

z + 0.3 = \frac{\omega^2 R^2}{2 g}     (6.6o)

From Eqs. (6.59) and (6.60), we have

\frac{z}{z+0.3}=\frac{x^2}{R^2}=0.333

or                          z = 0.333z + 0.099

or                          z=\frac{0.099}{0.667}=0.1484 \mathrm{~m}

Using the relation z=\frac{\omega^2 r^2}{2 g} ,for the parabola AOB, we have

0.1484+0.3=\frac{\omega^2 \times 0.1^2}{2 \times 9.81}

or                          ω² = 879.76

or                          \omega=\sqrt{879.76}=29.66 \mathrm{rad} / \mathrm{s}

Rotational speed is given by

N=\frac{60 \omega}{2 \pi}                     \left[\because \omega=\frac{2 \pi N}{60}\right]

=\frac{60 \times 29.64}{2 \pi}=283.04 \mathrm{rpm}