Question

A cylinder of diameter $0.4$ $m$ is rotated at an angular velocity 25 radian/second about its axis. Now consider two causes, one when closed and other when open from top. What will be the difference in total force at the bottom of cylinder for two causes? (Take ${\gamma }_w=10$ $kN/m^3$)

• A. $1.57$ $kN$
• B. $15.7$ $kN$
• C. $25.12$ $kN$
• D. $80$ $kN$

Answer 1

The correct option is A $1.57$ $kN$

Equivalent water profile in two cases will be:
It is clear from diagram that the pressure in closed case is

$\frac{{\omega }^2{r}^2}{2g}\cdot \rho gwhichismorethanthatofopenfromtop$

$\therefore$ Thus total difference of force at bottom =

$\frac{{\omega }^2{r}^2}{2g}\cdot {\gamma }_w\cdot \pi r^2$

$\frac{{\omega }^2{r}^4\cdot \rho \pi }{2\times 10}=\frac{(25{)}^2\times (0.2{)}^4\times 10\times \pi }{2\times 10}=1.57kN$