Question 6.42: An open cylinder 20 cm in diameter and 30 cm high is full of......

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

Total volume of water in the cylinder ∀ = πR² H = π × (0.1)² × 0.3 = 0.0094 m³

Volume of water spilled out due to rotation =\frac{1}{3} \forall=\frac{1}{3} \times 0.0094=0.00313 \mathrm{~m}^3

Let h be the height of the paraboloid formed due to rotation.

=\frac{\pi(.1)^2}{2} \times h=0.0157 h \mathrm{~m}^3

Since the volume of water spilled is same as the volume of paraboloid, we have

0.0157h = 0.00313

or                                                        h=\frac{0.00313}{0.0157}=0.1994 \mathrm{~m}

Using the relation  z=\frac{\omega^2 r^2}{2 g} \text { (Eq. 6.56), we have }

0.1994=\frac{\omega^2 \times 0.1^2}{2 \times 9.81}

or                                     ω² = 391.22

or                                    \omega=\sqrt{391.22}=19.779 \mathrm{rad} / \mathrm{s}

Rotational speed is found to be

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

=\frac{60 \times 19.74}{2 \pi}=188.5 \mathrm{rpm}