Question

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 ω. The force exerted by the liquid at the other end is

• A. $\frac{1}{2}M\omega^{2}L$
• B. $M\omega^{2}L$
• C. $\frac{1}{4}M\omega^{2}L$
• D. $\frac{1}{2}M\omega^{2}L^{2}$

Answer 1

Answer: (A)

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

Answer 2

The mass of element = dm = (m/L)dx.

The force on the element towards the axis = (F + dF) – F = dF.

∴ dF = (dm) ω²x = $\left( \frac{M}{L}dx \right)$ω²x.

F = $\frac{1}{2}.\frac{M\omega^{2}}{L}x^{2} + cons ⥂ \tan ⥂ t.$

For x = 0, F = 0. ∴ the constant = 0.

For x = L, F = 1/2 Mω²L.