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(B) Consider a small element of the liquid of mass $dm$ at a distance $r$ from the axis of rotation. The length of this element is $dr$. Since the tub

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$\frac{1}{2}M{\omega ^2}L$

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(B) Consider a small element of the liquid of mass $dm$ at a distance $r$ from the axis of rotation. The length of this element is $dr$. Since the tube has length $L$ and mass $M$,the linear mass density is $\lambda = M/L$.Thus,$dm = \lambda dr = (M/L) dr$.The centripetal force $dF$ required for this element to rotate is $dF = (dm) \omega^2 r = (M/L) \omega^2 r dr$.To find the total force exerted by the liquid at the axis of rotation,we integrate from $r = 0$ to $r = L$:$F = \int_0^L \frac{M}{L} \omega^2 r dr = \frac{M \omega^2}{L} \left[ \frac{r^2}{2} \right]_0^L = \frac{M \omega^2}{L} \cdot \frac{L^2}{2} = \frac{1}{2} M \omega^2 L$.This force represents the total centripetal force required to keep the liquid in circular motion,which is effectively the force exerted by the liquid at the pivot point.

$A$ tube of length $L$ is filled completely with an incompressible liquid of mass $M$ and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity $\omega$. Then the force exerted by the liquid at the other end is

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