A horizontally oriented tube $AB$ of length $l$ rotates with a constant angular velocity $\omega$ ab

Question

A horizontally oriented tube $AB$ of length $l$ rotates with a constant angular velocity $\omega$ about a stationary vertical axis passing through the open end $A$. The tube is partially filled with an ideal incompressible fluid. The closed end $B$ has a very small orifice. If the length of the fluid column in the tube is $h$ at a given instant, what is the velocity $v$ of the fluid jet relative to the tube as it emerges from the orifice?

Professor PC Thomas Classes · Answer

Applying Bernoulli's equation in the rotating frame of reference attached to the tube: $P + \frac{1}{2}ho v_{	ext{rel}}^2 - \frac{1}{2}ho\omega^2 r^2 = 	ext{constant}$. Let us apply this between the free surface of the fluid inside the tube (at distance $r_1 = l-h$ from the axis) and the orifice at end $B$ (at distance $r_2 = l$). The pressure at both points is atmospheric ($P_0$). Since the orifice is very small, the relative velocity of the free surface is negligible ($v_1 \approx 0$). Thus, $P_0 + 0 - \frac{1}{2}ho\omega^2(l-h)^2 = P_0 + \frac{1}{2}ho v^2 - \frac{1}{2}ho\omega^2 l^2$. Solving for $v$, we get $v^2 = \omega^2[l^2 - (l-h)^2] = \omega^2(2lh - h^2)$. Therefore, $v = \omega\sqrt{h(2l - h)}$.