Physics problem: Horizontal cylindric water pump

First, let me give you the problem statement so you can decide for yourself if the post is worth your time.

Problem statement

A pump consists of a cylinder with a piston. The cylinder is positioned horizontally. The piston's active pushing area is $A$. There is a hole of area $a$, in the middle of the piston. The piston is pushing water to the left and the water is flowing out to the right through the hole. The force $F$ is applied to the piston to make it move to the left with a constant velocity. Assuming that the flow is laminar and stationary ($\frac{\partial X}{\partial t} = 0$), determine the flow velocity of the liquid through the hole.

[Source: Savchenko et al 1981 physics problems book]

Latex source for the above plot is here: link. In this one I used layers to draw the pump on top of water.

Solution

One thing that we need to realize is that we are asked to compute the outflow speed of the water relative to the piston.

Another one, which is important to get the same answer as in the source physics problem book, is to realize that the total cross-section area of the cylinder is $A+a$.

Now that these two points are clarified we can switch to the reference system moving to the left with the piston. In this reference system the water flows to the right and through the hole. Let's denote the speed of water to the left from the piston and to the right from the piston as $u$ and $v$ respectively. Therefore the speed $v$ is the actual outflow speed we are asked to compute here. Let's denote the pressure to the left and to the right from the piston as $P_1$ and $P_2$ respectively.

Let's write down the following three constraints imposed onto the pump system in the form of equations.

1. The second Newton's law applied to the piston yields:

   $$ F = \left( P_1 - P_2 \right) \cdot A $$

2. The energy conservation law applied to a unit volume of the water (or else known as Bernoulli's law) yields:

   $$ P_1 + \rho \frac{u^2}{2} = P_2 + \rho \frac{v^2}{2} $$

3. The continuity equation yields:

   $$ u \cdot A = v \cdot a $$

Above, we have three equations and three unknowns to be found are $u$, $v$, and $P_1 - P_2$. By solving the above system of equations we obtain the following expression for the outflow speed $v$:

$$ \boxed{v = \sqrt{\frac{2FA}{\rho (A^2 - a^2)}}} $$

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:

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

Physics problem: A mass on an oscillating inclined plane

This problem comes from the problem book by Savchenko et al. which I mentioned in the blog before. It is the last problem on harmonic oscillations in the edition I have (3.30*). Since I am posting it here, it means that the solution took me quite long time and I have still some points that are not completely clear to me (hoping someone will comment and clarify it for me :-)).

Problem statement

Consider an inclined plane at an angle $\alpha$ with respect to the horizon. The plane is oscillating harmonically at high frequency along the axis ($x$) parallel to its surface and making the same angle $\alpha$ with the horizon (as shown in Fig. 1). The maximum speed of the oscillating plane is $v_0$. An object is then placed on the inclined plane. The friction coefficient between the plane and the object is $\mu$. Assuming that $\mu \gg \tan\alpha$ determine how would the object move and its mean velocity after a long period of time.

Fig. 1: Oscillating inclined plane with the mass on it, forces are shown at the moment when the object is sliding up the plane with the inertia force directed upwards.

Link to the latex source of the figure

Solution

We will consider the object in the reference frame tight to the oscillating plane. In this reference frame there would be the inertial force (varying harmonically) acting on the object and the object will oscillate up and down the inclined plane. Since the frequency of the oscillations of the plane is high, then we can consider that there are only two states of the object: it is moving upwards and downwards with respect to the plane. So the object will be oscillating on the plane up and down.

Let's take $T$ to be the period of the oscillating motion of the object. Which is equal to the period of oscillations of the plane.

Let's denote:

• $t_1$ - time it takes the object to travel up with respect to the plane
• $t_2=T-t_1$ - time it takes the object to travel down with respect to the plane

Let's consider projections of forces acting upon the object on the $x$-axis parallel to the inclined plane:

$$ F_x = \pm\,\mu m g \cos\alpha - m g \sin\alpha - m\omega v_0 \cos\left( \omega t + \phi_0 \right) $$

The last term of the equation above is the force of inertia, assuming that the velocity of the inclined plane varies according to the formula $v(t)=v_0\sin\left( \omega t + \phi_0 \right)$. Since the object is sliding, there is the friction force $F_f=\pm\,\mu N = \pm\,\mu m g\cos\alpha$ acting on the object in the direction opposite to its velocity.

Eventually, after a long time the change of momentum during each period $T$ would be 0 (stationarity condition):

$$ \Delta p = \int_{0}^{T}dp = \int_{0}^{T}{F_x}{dt} = 0 \tag{1} $$

Substituting the $F_x$ into the expression above:

$$ -m g T \sin\alpha - \mu m g t_1 \cos\alpha + \mu m g t_2 \cos\alpha = 0 $$$$ - T \sin\alpha - \mu t_1 \cos\alpha + \mu \left( T - t_1 \right) \cos\alpha = 0 \Rightarrow \boxed{t_1 = \frac{T}{2} \left( 1 - \frac{\tan\alpha}{\mu} \right)} $$

By definition the mean velocity is:

$$ \overline{v} \;\equiv\; \frac{1}{T}\left[\int_{0}^{t_1}{v_1(t)}{dt} + \int_{t_1}^{T}{v_2(t)}{dt} \right] \tag{2} $$

Let's determine the functions $v_1(t)$ and $v_2(t)$.

Let's note that the velocity should be 0 at the turning points, therefore:

$$ v_1(0) = v_1(t_1) = v_2(t_1) = v_2(T) = 0 $$

Applying the second Newton's law the velocity time derivatives can be expressed as:

$$ \begin{split} d_t v_1(t) & = -\mu g \cos\alpha - \omega v_0 \cos\left( \omega t + \phi_0 \right), t\in[0, t_1] \\ d_t v_2(t) & = \mu g \cos\alpha - \omega v_0 \cos\left( \omega t + \phi_0 \right), t\in[t_1, T] \end{split} $$

Here we neglected the projection of the gravity force ($F_{gx} = -mg\sin\alpha$) as it is much smaller than the force of friction, according to the problem statement.

Integrating the equations for acceleration and using the 0-velocity conditions at turning points, we get:

$$ \begin{split} v_1(t) & = -\mu g t \cos\alpha - v_0 \sin\left(\omega t + \pi \right), t\in[0, t_1] \\ v_2(t) & = \mu g (t - t_1) \cos\alpha - v_0 \left(\sin\left( \omega t + \pi \right) - \sin\left( \omega t_1 + \pi \right)\right), t\in[t_1, T] \end{split} $$

$\phi_0=\pi$ follows from the equations $v_1(0) = 0$ and $v_1(t_1)=0$.

Finally, by substituting the expressions for $v_1(t)$ and $v_2(t)$ we get:

$$ \begin{split} \overline{v}\cdot T & = \int_{0}^{t_1}{-\mu g t \cos\alpha}{dt} + \int_{t_1}^{T}{\mu g (t - t_1) \cos\alpha}{dt} - \int_{0}^{t_1}{v_0 \sin\left( \omega t + \pi \right)}{dt}\\ & - \int_{t_1}^{T}{v_0 \sin\left( \omega t + \pi \right)}{dt}+\int_{t_1}^{T}{v_0 \sin\left( \omega t_1 + \pi \right)}{dt}\\ \end{split} $$

By simplifying and integrating, we get:

$$ \overline{v}\cdot T = -(T-t_1) v_0 \sin\left( \omega t_1 \right) $$

then substituting $t_1$ and using the equation $\omega \cdot T = 2\pi$:

$$ \begin{split} \overline{v} \cdot T & = -\left(\frac{T}{2} + \frac{\tan\alpha}{\mu}\right) v_0 \sin\left(\pi - \frac{\tan\alpha}{\mu} \pi \right) \\ & = -v_0 \left(\frac{T}{2} + \frac{\tan\alpha}{\mu}\right) \sin\left(\frac{\tan\alpha}{\mu} \pi \right) \\ & \approx -v_0 \frac{T}{2} \frac{\tan\alpha}{\mu} \pi \end{split} $$

Then the final expression for the mean velocity is:

$$ \boxed{\overline{v} = -v_0 \frac{\tan\alpha}{2\mu} \pi} $$

Reflections

• Let's double check that $v_1(t_1)=0$, i.e.,

  $$ -\mu g t_1 \cos\alpha - v_0 \sin\left(\omega t_1 + \pi \right) = 0 $$

  Substituting $t_1$:

  $$ -\mu g \frac{T}{2} \left( 1 - \frac{\tan\alpha}{\mu} \right) \cos\alpha + v_0 \sin\left(\omega \frac{T}{2} \left( 1 - \frac{\tan\alpha}{\mu} \right) \right) = 0 $$

  Simplifying: $$ -\mu g \frac{T}{2} \left( 1 - \frac{\tan\alpha}{\mu} \right) \cos\alpha + v_0 \sin\left(\pi\frac{\tan\alpha}{\mu}\right) = 0 $$

  Using the approximation $\frac{\tan\alpha}{\mu} \ll 1$:

  $$ -\mu g \frac{T}{2} \cos\alpha + v_0 \pi\frac{\tan\alpha}{\mu} = 0 $$

  Well, it seems that in order to have $v_1(t_1) = 0$, it is necessary that the period of oscillation be:

  $$ T=v_0 2\pi\frac{\tan\alpha}{\mu^2 g\cos\alpha} $$

  This makes me uncomfortable with the solution and the problem statement. Maybe there is something more to the motion described in the problem than I understand...

• It is very easy to verify as well that $v_2(T) = 0$, assuming that $t_1\approx T/2$.

  $$ \begin{split} v_2(T) & = \mu g (T - t_1) \cos\alpha - v_0 \left(\sin\left( \omega T + \pi \right) - \sin\left( \omega t_1 + \pi \right)\right) \\ & = \mu g (T - t_1) \cos\alpha + v_0\sin\left( \omega t_1 + \pi \right) \end{split} $$

  Then using the condition $v_1(t_1)=0$, we get $$ v_2(T) = \mu g (T - t_1) \cos\alpha - \mu g t_1 \cos\alpha \approx 0 $$

• $\overline{v} < 0$, so the object will slide down from the plane.