(A) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation.The mass of this small element is $dm = (M/L) dx

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(A) Consider a small element of the liquid of length $dx$ at a distance $x$ from the axis of rotation.The mass of this small element is $dm = (M/L) dx$.The centrifugal force $dF$ acting on this element is $dF = (dm) x \omega^2 = (M/L) x \omega^2 dx$.To find the total force exerted by the liquid on the other end,we integrate this expression from $x = 0$ to $x = L$:$F = \int_0^L \frac{M}{L} \omega^2 x dx = \frac{M \omega^2}{L} \left[ \frac{x^2}{2} \right]_0^L = \frac{M \omega^2}{L} \cdot \frac{L^2}{2} = \frac{ML\omega^2}{2}$.

Class 11 Physics System of Particles and Rotational Motion Rotation Motion Basic, Motion of Connected Mass

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$A$ 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 $\omega$. The force exerted by the liquid on the tube at the other end is

A
$\frac{ML\omega^2}{2}$
B
$ML\omega^2$
C
$\frac{M\omega^2L^2}{2}$
D
$\frac{M\omega L^2}{2}$

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System of Particles and Rotational Motion

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Rotation Motion Basic, Motion of Connected Mass

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$A$ 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 $\omega$. The force exerted by the liquid on the tube at the other end is

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The direction of the angular velocity vector is along which of the following?

$A$ solid cylinder of mass $M$ and radius $R$ is placed in a horizontal position. Two strings are wrapped around the cylinder. Find the tension in the strings and the acceleration of the cylinder as the strings unwind.

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