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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 $\omega$. The force exerted by the liquid at the other end is :-
A
$\dfrac{ML\omega^2}{2}$
B
$\dfrac{ML^2\omega}{2}$
C
$ML\omega^2$
D
$\dfrac{ML^2\omega^2}{2}$
Detailed Solution
Consider a small element of liquid of mass $dm = \dfrac{M}{L}dx$ at a distance $x$ from the axis of rotation.
Centripetal force needed for this element: $dF = (dm)(\omega^2)(x)$
The total centripetal force for the whole liquid is supplied by the far end of the tube, so the liquid exerts an equal force on that end.
Required force $F = \int dF = \int_0^L \dfrac{M}{L}\omega^2 x\,dx$
$F = \dfrac{M\omega^2}{L}\left[\dfrac{x^2}{2}\right]_0^L = \dfrac{M\omega^2}{L} \times \dfrac{L^2}{2}$
$F = \dfrac{M\omega^2 L}{2}$
(The same result follows by treating the whole mass as concentrated at the centre of mass, at a distance $\dfrac{L}{2}$ from the axis.)
Centripetal force needed for this element: $dF = (dm)(\omega^2)(x)$
The total centripetal force for the whole liquid is supplied by the far end of the tube, so the liquid exerts an equal force on that end.
Required force $F = \int dF = \int_0^L \dfrac{M}{L}\omega^2 x\,dx$
$F = \dfrac{M\omega^2}{L}\left[\dfrac{x^2}{2}\right]_0^L = \dfrac{M\omega^2}{L} \times \dfrac{L^2}{2}$
$F = \dfrac{M\omega^2 L}{2}$
(The same result follows by treating the whole mass as concentrated at the centre of mass, at a distance $\dfrac{L}{2}$ from the axis.)