Question

A tube of length $L$ is filled 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 at the other end will be

• A. $\frac{1}{2}{M}^2{\omega }^{2}L$
• B. $\frac{1}{2}M{\omega }^{2}L$
• C. $\frac{1}{2}M{\omega }^2{L}^2$
• D. None of the above

Answer 1

Answer: (B)

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

Let us take an infinitesimal small length $dx$ situated at distance $x$ on road from the axis of rotation at one of its ends.
Mass of liquid of length $dx$
$$dm=M\frac{dx}{L}$$
Centrifugal force acting on this mass
$$dF=dm\times x\times {\omega }^2=\frac{M{\omega }^{2}xdx}{L}$$
Integrating from $0$ to $L$ we get $F=$$\frac{M{\omega }^{2}L}{2}$
This force is balanced by the force exerted at the other end by the liquid.