Q: A tube of length L is filled completely with an incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity ω. The force exerted by the liquid at the other end is
(a)(Mω2 L)/2
(b)Mω2 L
(c) (Mω2 L)/4
(d) (Mω2 L2)/2
Ans: (a)
Sol: Taking element of length dx at a distance x from the axis of rotation .
Mass of element dx is $\large m = \frac{M}{L}dx$
For rotation , this element require centripetal force .
$\large dF = m \omega^2 x$
$\large dF = (\frac{M}{L}dx) \omega^2 x$
$\large F = \int_{0}^{L}dF $
$\large F = \int_{0}^{L} \frac{M}{L} \omega^2 x dx $
$\large F = \frac{M}{L} \omega^2 \int_{0}^{L} x dx $
$\large F = \frac{M}{L} \omega^2 (\frac{L^2}{2})$
$\large F = \frac{M\omega^2 L}{2} $
The force exerted by the liquid at the other end is
$\large F = \frac{M\omega^2 L}{2} $