Question 6.44: A closed vertical cylinder of 30 cm diameter and 100 cm heig......

Given data:
Diameter of cylinder                     D = 30 cm = 0.3 m
Radius of cylinder                         R = 0.15 m
Height of cylinder                         H =100 cm =1 m
Initial height of water                H_{\mathrm{in}}=80 \mathrm{~cm}=0.8 \mathrm{~m}
Rotational speed                             N = 200 rpm
Angular velocity is given by

\omega=\frac{2 \pi N}{60}=\frac{2 \pi \times 200}{60}=20.94 \mathrm{rad} / \mathrm{s}

Volume of air present in the cylinder before rotation

=\frac{\pi}{4} D^2\left(H-H_{\mathrm{in}}\right)

=\frac{\pi}{4}(0.3)^2(1-0.8)=0.014 \mathrm{~m}^3

Let Z and r be the height of paraboloid and radius of paraboloid formed when the cylinder is rotating (Fig 6.29 (b)).
Volume of air present in the cylinder after rotation

=\pi r^2 \frac{Z}{2}

Since the volume of air present in the cylinder before the rotation must be same as that of after rotation,
we have

\pi r^2 \frac{Z}{2}=0.014               (6.61)

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

or                                    r^2=\frac{2 \times 9.81 \times Z}{20.94^2}=0.0447 Z                   (6.62)

From Eqs. (6.61) and (6.62), we have

\pi \times 0.0447 Z \times \frac{Z}{2}=0.014

or                  Z^2=\frac{2 \times 0.014}{\pi \times 0.0447}=0.199

or                    Z=\sqrt{0.199}=0.446 \mathrm{~m}=44.6 \mathrm{~cm}

The height of paraboloid formed is 44.6 cm.