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`. Determine the force exerted by the liquid at the other end.

Options

A. `(M omega^2 L)/(2)`
B. `M omega^2 L`
C. `(M omega^2 L)/(4)`
D. `(M omega^2 L^2)/(2)`

Answer

A

Explanation

(a) The force acting on the mass of liquid dm of length dx
at a distance x from the axis of rotation O.

`dF = (dm) x omega^2`
`:. dF = (M)/(L) dx xx xomega^2`
Where (M)/(L) is mass of liquid in unit length.
`:.` The force acting at the other end is for the whole liquied
in tube.
`F = int_0^L(M)/(L) omega^2 x dx = (M)/(L) omega^2 int_0^L x dx`
`=(M)/(L) omega^2 [(x^2)/(2)]_0^L = (M)/(L) omega^2 [(L^2)/(2) - 0] = (MLomega^2)/(2).`