Question

A cylindrical vessel of radius R is filled with an ideal liquid of density p up to height h above its flat bottom. It is now set into rotation about its vertical axis with constant angular velocity m. When a steady state is achieved, the liquid does not overflow and no portion of the bottom is dry. Now at the centre of the bottom a small orifice of radius r is made. How much liquid can flow out of the vessel?

Answer 1

π R^2 (h - (ω^2 R^2)/(4g))

Answer 2

The free surface of a liquid in a cylindrical vessel rotating about its vertical axis with angular velocity $\omega$ is a paraboloid of revolution. The equation of the free surface is given by $z = z_0 + \frac{\omega^2 r^2}{2g}$, where $z$ is the height of the liquid surface at a radial distance $r$ from the axis of rotation, and $z_0$ is the height of the liquid surface at the center ($r=0$).

Initially, the vessel contains a volume of liquid $V_{initial} = \pi R^2 h$. When the vessel is set into rotation, the liquid volume is conserved. The volume of liquid in the rotating vessel is given by integrating the height $z(r)$ over the base area: $V_{rotating} = \int_0^R 2\pi r z(r) dr = \int_0^R 2\pi r (z_0 + \frac{\omega^2 r^2}{2g}) dr$ $V_{rotating} = 2\pi \left[ z_0 \frac{r^2}{2} + \frac{\omega^2 r^4}{8g} \right]_0^R = \pi R^2 z_0 + \frac{\pi \omega^2 R^4}{4g}$.

Since the volume is conserved, $V_{initial} = V_{rotating}$: $\pi R^2 h = \pi R^2 z_0 + \frac{\pi \omega^2 R^4}{4g}$. Dividing by $\pi R^2$, we get the relationship between the initial height $h$ and the height at the center $z_0$ in the rotating state: $h = z_0 + \frac{\omega^2 R^2}{4g}$.

The problem states that in the steady state of rotation (before the orifice is made), no portion of the bottom is dry, which means $z_0 \ge 0$.

Now, a small orifice is made at the center of the bottom. Liquid will flow out as long as the pressure at the orifice is greater than atmospheric pressure. The pressure at the orifice (at $r=0$, $z=0$) is $P(r=0, z=0) = P_{atm} + \rho g z_0'(t)$, where $z_0'(t)$ is the instantaneous height of the liquid surface at the center. The flow stops when the height at the center becomes zero, i.e., $z_0' = 0$.

When the flow stops, the height of the liquid surface at the center is $z_0' = 0$. The free surface equation becomes $z(r) = 0 + \frac{\omega^2 r^2}{2g} = \frac{\omega^2 r^2}{2g}$. The volume of the liquid remaining in the vessel at this point is: $V_{final} = \int_0^R 2\pi r z(r) dr = \int_0^R 2\pi r \left(\frac{\omega^2 r^2}{2g}\right) dr = \frac{\pi \omega^2}{g} \int_0^R r^3 dr$ $V_{final} = \frac{\pi \omega^2}{g} \left[ \frac{r^4}{4} \right]_0^R = \frac{\pi \omega^2 R^4}{4g}$.

The amount of liquid that has flowed out of the vessel is the difference between the initial volume and the final volume: $V_{out} = V_{initial} - V_{final} = \pi R^2 h - \frac{\pi \omega^2 R^4}{4g}$.

We can express this amount in terms of $z_0$, the height at the center in the initial steady state. From the equation $h = z_0 + \frac{\omega^2 R^2}{4g}$, we have $\frac{\omega^2 R^2}{4g} = h - z_0$. Substituting this into the expression for $V_{out}$: $V_{out} = \pi R^2 h - \pi R^2 \left(\frac{\omega^2 R^2}{4g}\right) = \pi R^2 h - \pi R^2 (h - z_0) = \pi R^2 h - \pi R^2 h + \pi R^2 z_0$. $V_{out} = \pi R^2 z_0$.

So, the amount of liquid that flows out is $\pi R^2 z_0$, where $z_0$ is the height of the liquid at the center of the base in the initial steady state of rotation.

We can also express the amount flowed out in terms of $h$ and $\omega$ using the derived relationship: Amount flowed out = $\pi R^2 (h - \frac{\omega^2 R^2}{4g})$.

The question asks "How much liquid can flow out of the vessel?". This refers to the volume of liquid that flows out. Both $\pi R^2 z_0$ and $\pi R^2 (h - \frac{\omega^2 R^2}{4g})$ represent this volume. Without knowing $z_0$ explicitly, the answer is best expressed in terms of the given parameters $R$, $h$, and $\omega$.

The amount of liquid that can flow out is $\pi R^2 h - \frac{\pi \omega^2 R^4}{4g}$. Since the problem states that no portion of the bottom is dry in the initial steady state, $z_0 \ge 0$. This implies $h - \frac{\omega^2 R^2}{4g} \ge 0$, so the amount flowed out, $\pi R^2 (h - \frac{\omega^2 R^2}{4g})$, is non-negative.