Making suspend work on an arch-flavoured linux

I have a custom installation of linux (EndeavourOS if you are curious) that I use with the i3 window manager. But whenever I was doing

systemctl suspend

it was going to suspend but then turned back on, without any error message... Recently, I've guessed that the reason for this was the missing swap file. If you find yourself in a similar situation there is a nice howto on the creation and activation of a swapfile on Linux: https://btrfs.readthedocs.io/en/latest/Swapfile.html

Physics problem: Air-filled balloon floating in a rotating cylindrical tank partially filled with water

I want to discuss this moderate complexity question about a balloon, filled with air, floating in a rotating tank, which is partially filled with water, as shown in the plot. This is a nice closure for 2023 as it happens to be the last problem on Archimedes' force in the edition of the Savchenko et al problem book that I have. I like this one as its key hint to the solution is in the illustration to the problem.

Problem statement In a cylindrical container, partially filled with water, an air-filled balloon is floating attached to the side wall with a rope as shown in the figure. The container is rotating and the rope is deviated from the side wall towards the center by an angle $\alpha$. The length of the rope is $l$ and the radius of the container is $R$. The radius of the balloon is $r$. Given the information above, determine the angular speed $\omega$ of rotation of the cylinder.

Latex code for the illustration can be found here.

Solution

Before I start describing the solution, I would like to draw your attention to the fact that the balloon is deviated towards the axis of rotation. What would be the reason for this? We need to have a force pulling it towards the center to provide the acceleration towards the rotation axis. It is clear that the rope, attaching the balloon to the wall, pulls the balloon away from the center. Therefore, in principle, I would expect the balloon to be pressed to the side wall of the tank just above the point where the rope is attached to the wall, due to buoyancy of air. And that would probably be true if we had a flat water surface. But here, judging from the illustration, the water surface is not level, but curved, actually its shape is parabolic. This creates a pressure gradient force pushing the air balloon towards the rotation axis.

We can easily compute the pressure gradient at the distance $x$ due to the curvature of the water surface, let's denote the horizontal profile of the water surface as $h(x)$, where $x$ is the distance from the rotation axis on the water surface.

Then

\begin{equation} \frac{\partial P(x)}{\partial x} = \rho_0 g \frac{\partial h(x)}{\partial x} \end{equation}

where $\rho_0$ is the water density, and $g$ is the acceleration due to gravity.

Now, to determine $\frac{\partial h(x)}{\partial x}$ we consider a water parcel at the surface at the distance $x$ from the rotation axis. There are two forces acting on it resulting in the acceleration directed towards the center of rotation: the first one is the reaction of the rest of the water, perpendicular to the water surface as there is no tangent acceleration, and the second one is the force of gravity. It is easy to show that the $\tan$ of the angle between the force of gravity ($dm \vec{g}$) and the reaction force $d\vec{N}$ is:

\begin{equation} \tan\sigma(x) = \frac{\partial h(x)}{\partial x} = \frac{\omega ^ 2 x}{g} \end{equation}

We can use the above to calculate the horizontal pressure gradient as a function of $x$ as follows

\begin{align*} \frac{\partial P(x)}{\partial x} = \rho_0 \cdot \omega ^ 2 x \Rightarrow \\ P(x) = P_0 + \frac{\rho_0 \cdot \omega ^ 2 x ^ 2}{2} \end{align*} where $P_0$ is the pressure at the center of the tank.

So now we would have to find the resulting force, actually the horizontal component of the force, acting on the balloon due to the above pressure field in the water as:

\begin{align*} F_P = \int\limits_{B} P(\beta, \gamma) \cdot \cos \phi(\beta, \gamma) dA \end{align*}

Where $B$ denotes the surface of the balloon, $dA$ is an element of its surface area, $(\beta, \gamma)$ are the angles defining a position of an element $dA$ on the surface of the balloon, and $\phi$ is the angle between the horizontal axis and the normal to the surface of the balloon pointing outside of the balloon (just to have the force positive when it is pushing the balloon towards the center).

It is easy to see that:

$$ dA = r^2\cos\gamma d\gamma d\beta $$

Let's denote the position of the center of the air ballon as $x_0$. We can split the balloon into two equal parts: one closer to the rotation axis and the other one further away from the rotation axis, let's denote water pressure on these parts as $P_{-}(\beta, \gamma)$ and $P_{+}(\beta, \gamma)$.

To fully and independently cover these parts the angles should be $\gamma \in [-\pi/2, \pi/2]$ and $\beta \in [-\pi/2, \pi/2]$. Then we can use $\beta, \gamma$ pair to parameterize the pressure on those sides of the balloon as follows.

$$ P_{+}(\beta, \gamma) = P_0 + \frac{\rho_0 \omega ^ 2}{2}\cdot(x_0 + r \cos\gamma \cdot \cos\beta)^2 $$

And for the half closer to the rotation axis:

$$ P_{-}(\beta, \gamma) = P_0 + \frac{\rho_0 \omega ^ 2}{2}\cdot(x_0 - r \cos\gamma \cdot \cos\beta)^2 $$

Now we pair opposite points with the same $\beta, \gamma$ on both hemispheres to get the contribution to the pressure gradient across the balloon:

$$ \Delta P(\beta, \gamma) = P_{+} - P_{-} = 2 \rho_0\omega^2 r x_0 \cos\gamma \cdot \cos\beta $$

Then the resulting horizontal component of the force acting on the balloon from the two parts is calculated as follows:

\begin{align*} dF_{P} &= \Delta P(\beta, \gamma) dA \cdot \cos\gamma \cos\beta \\ &= 2\rho_0 \omega^2 r^3 x_0 \cos^3\gamma \cos^2\beta d\gamma d\beta \end{align*}

To get the net force we integrate the above expression:

\begin{align*} F_{P} & = 2\rho_0 \omega^2 r^3 x_0 \int\limits_{-\pi/2}^{\pi/2} \cos^3\gamma d\gamma \int\limits_{-\pi/2}^{\pi/2} \cos^2\beta d\beta \\ & = \frac{4}{3} \pi r ^3 \rho_0 \omega^2 x_0 \end{align*}

Now we are ready to write down the second Newton's law projected on the vertical and horizontal axes:

\begin{cases} F_{A} - mg - T \cos\alpha = 0 \\ F_{P} - T \sin\alpha = m \omega ^ 2 x_0 \end{cases}

where $F_{A}$ - is the Archimedes' force acting on the balloon, $mg$ - is the force of gravity pulling the balloon down, and $T$ - is the tension of the rope attaching the balloon to the side wall of the tank.

Let's denote the volume of the balloon as $V$ and the air density as $\rho$, then using the Newton's law expression projected on the vertical axis we can express the tension force as follows:

$$ T = Vg (\rho_0 - \rho) \frac{1}{\cos\alpha} $$

Then we plug the above expression for $T$ in to the equation for the horizontal force components and obtain the following eqaution with respect to the rotation frequency $\omega$:

\begin{align*} T\sin\alpha & = Vg (\rho_0 - \rho) \tan\alpha \\ & = F_{P} - m \omega ^ 2 x_0 = (\rho_0 - \rho) \omega ^ 2 x_0 V \end{align*}

Simplifying the above we get: \begin{align*} g\tan\alpha = \omega ^ 2 x_0 \Rightarrow \omega = \sqrt{\frac{g\tan\alpha}{x_0}} \end{align*}

Now we need to express the distance from the rotation axis to the center of the balloon ($x_0$) through the quantities given in the problem statement. From the geometry considerations it is easy to obtain the following: $$ x_0 = R - (l + r) \sin\alpha $$

Therefore, the final answer to the problem question is:

$$ \boxed{\omega = \sqrt{\frac{g\tan\alpha}{R - (l + r) \sin\alpha}}}. $$