A circular cylinder of height $h_{0} = 10$ cm and radius $r_{0} = 2$ cm is opened at the top and filled with liquid. It is rotated about its vertical axis. Determine the speed of rotation so that half the area of the bottom gets exposed (g = 10 $m / s e c^{2}$ ).

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Solution

Area of bottom $= π r_{0}^{2}$
$r = \frac{r_{0}}{s q r t 2}$
Applying Bernoulli's equation between points (1) and (2) -
$P_{a t m} + \frac{1}{2} p v_{1}^{2} - p g H = O_{a t m} + \frac{1}{2} p v_{2}^{2} - p g (H - h_{0})$
$- p g h_{0} = \frac{1}{2} p (v_{2}^{2} - v_{1}^{2}) \Rightarrow 2 g h_{0} = [v_{1}^{2} - v_{2}^{2}] = [w^{2} r_{0}^{2} - w^{2} r^{2}]$
$r_{0} = 2 \times 10^{- 2} m \Rightarrow 2 g h_{0} = w^{2} [r_{0}^{2} - r^{2}]$
$w = \frac{2}{r_{0}} \sqrt{g h} = \frac{2}{2 \times 10^{- 2}} \sqrt{10 \times 0.1} = 100$ radian / sec.

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A circular cylinder of height h0=10 cm and radius r0=2 cm is opened at the top and filled with liquid. It is rotated about its vertical axis. Determine the speed of rotation so that half the area of the bottom gets exposed (g = 10 m/sec2).

Area of bottom =πr20r=r0sqrt2Applying Bernoulli's equation between points (1) and (2) - Patm+12pv21−pgH=Oatm+12pv22−pg(H−h0)−pgh0=12p(v22−v21)⇒2gh0=[v21−v22]=[w2r20−w2r2]r0=2×10−2m⇒2gh0=w2[r20−r2]w=2r0√gh=22×10−2√10×0.1=100 radian / sec.

A circular cylinder of height $h_{0} = 10$ cm and radius $r_{0} = 2$ cm is opened at the top and filled with liquid. It is rotated about its vertical axis. Determine the speed of rotation so that half the area of the bottom gets exposed (g = 10 $m / s e c^{2}$ ).