Physics problem: viscous flow in a cylindrical pipe

This problem is related to the previous one, where we discussed a setup with a fluid flow between two horizontal plates. The solution method is very similar, with a difference that here we select a cylindrical element and consider forces acting on it.

Problem statement

A liquid is pumped from one container to another through a long pipe of radius $R$ and length $l$. Determine the flow velocity in the pipe as a function of distance to its axis (i.e. to the center of the pipe). Given that the pressure drop between the two ends of the pipe is $\Delta P$ and the dynamic viscosity of the liquid is $\eta$.

Problem statement is taken from Savchenko et al (1981) book and translated by the post author.

Exploiting the symmetry around the axis of the pipe we select a coaxial cylinder of radius $r$ in the flow and investigate forces acting on it.

Figure 1: Flow in pipe of length $l$ and radius $R$.

If we denote the viscosity stress along the cylinder's sides and parallel to its axis as $\tau_x(r)$, then we can write the balance of the viscous and pressure gradient forces acting on it, along the $x$-axis, as:

\begin{align*} & \tau_x(r) \cdot 2\pi r l + \Delta P \cdot \pi r^2 = 0 & \Rightarrow \\ & \tau_x(r) = -\frac{\Delta P r}{2 l} & \end{align*}

Now we can use the expression linking viscous stress and flow velocity gradient through the dynamic viscosity to determine the flow velocity as follows:

\begin{align*} & \tau_x(r) = -\frac{\Delta P r}{2 l} = \eta \frac{dv}{dr} & \Rightarrow \\ & \frac{dv}{dr} = -\frac{\Delta P r}{2 l \eta} & \Rightarrow \\ & \int\limits_{R}^{r} \frac{dv}{dr}\,dr = -\int\limits_{R}^{r}\frac{\Delta P r}{2 l \eta}\,dr & \Rightarrow \\ & v(r) - v(R) = -\frac{\Delta P \left( r^{2} - R^{2} \right) }{4 l \eta} \\ \end{align*}

Now using the non-slip condition at the pipe walls (i.e. $v(R) = 0$), we get the final expression for the flow velocity at a distance $r$ from the pipe axis: \begin{align*} \boxed{ v(r) = -\frac{\Delta P \left( r^{2} - R^{2} \right) }{4 l \eta} }. \end{align*}

Physics problem: viscous flow between two horizontal plates

Problem statement

Determine velocity of fluid in a stationary flow between two fixed horizontal plates as a function of position between the two plates. The distance between the plates is $h$ and the dynamic viscosity of the fluid is $\eta$. The pressure drop in the direction of the flow per unit length is $\frac{dP}{dx}=-\Delta P$. Additionally, compute the fluid discharge per unit of horizontal length perpendicular to the flow.

Problem statement is taken from Savchenko et al (1981) book and translated by the post author.

Solution

Tihs is not a praticularly complicated problem, but I find it important for understanding the momentum exchanges between layers of viscous fluids and it does not require very deep knowledge of the process but allows us to draw some interesting insights.

Let's introduce a coordinate system with $x$-axis along and pointing in the direction of flow, $z$-axis in the vertical direction upwards, and $y$-axis in the horizontal direction transversal to the flow and perpendicular to the plane formed by $x$ and $z$ axes. The origin of the coordinate system $O$ is on the lower plate (see Figure 1).

Figure 1: Vertical cross-section of the viscous flow between to horizontal plates. $y$-axis is perpendicular and points into the figure plane.

Since the flow is stationary and horizontally uniform the flow velocity is only a function of the vertical coordinate $z$: $v(z)$.

We assume a non-slip boundary condition near the horizontal plates: $v(z=0)=v(z=h)=0$. Additionally, we assume that the stress (friction force per unit of contact area between horizontal layers of the fluid, $\tau_x$) due to viscosity decreases towards the plane of symmetry of the flow: $z=h/2$, i.e the layers of the fluid closer to the middle have higher velocities. Therefore, we can determine $v(z)$ for $z \in [0, h/2]$ and then use the symmetry of the flow to find the velocity for $z \in [h/2, h]$.

Let's consider an element of fluid with dimensions $(\Delta x, \Delta y, \Delta z = h - 2z)$ with it's lower face at $z \in [0, h/2]$ and parallel to the axes $x$ and $y$. Assuming that the viscous stress $x$-coordinate $\tau_x(z)$ is negative, we can express the balance of the forces acting on the selected element of fluid along the $x$-axis as follows:

\begin{align*} \tau_x(z) \cdot \Delta x \cdot \Delta y + \tau_x(z + \Delta z) \cdot \Delta x \cdot \Delta y + \Delta P \cdot \Delta x \cdot \Delta y \cdot \Delta z = 0 \end{align*}

It is easy to see that with $\Delta z = h - 2z$ the upper and lower faces of the selected element are symmetric with respect to the middle plane of the flow ($z=h/2$), and are at the same distance from the upper and lower plates respectively (this is by design of our imaginary flow element) which means that $\tau_x(z + \Delta z) = \tau_x(z)$. Then we can simplify the above balance of forces as:

\begin{align*} & 2 \tau_x(z) + \Delta P \cdot \Delta z = 0 & \Rightarrow \\ & 2 \tau_x(z) + \Delta P \cdot (h - 2z) = 0 & \Rightarrow^{\,} \\ & \tau_x(z) = -\frac{\Delta P}{2} (h - 2z) & \end{align*}

The above expression for the stress is valid for $z \in [0, h/2]$.

Let's now determine $v(z)$ for $z \in [0, h/2]$ and then we'll use the symmetry of the flow to extend the solution to the whole range $z \in [0, h]$.

We can now use the relation between the velocity and the stress through the dynamic viscosity $\eta$:

\begin{align*} \tau_x(z) = -\eta\frac{dv}{dz}, \; z \in [0, h/2] \end{align*}

Using previously derived expression for $\tau_x(z)$ and then integrating we get:

\begin{align*} & -\frac{\Delta P}{2} (h - 2z) = -\eta\frac{dv}{dz} & \Rightarrow \\ & v(z) = \frac{\Delta P}{2\eta} \left( hz -z^{2} \right) + C & \end{align*}

where $C$ is the constant of integration.

Then using the non-slip condition at the plates we determine the integration constant as $C = v(z=0)=0 \Rightarrow C=0$ and the final expression for the velocity profile is:

\begin{align*} v(z) = \frac{\Delta P}{2\eta} z (h - z), \; z \in [0, h/2]. \end{align*}

Now for $z \in [h/2, h]$ we can use the flow symmetry to express the flow velocity as:

\begin{align*} \hat{v}(z) = v(h - z) = \frac{\Delta P}{2\eta} (h - z) z, \; z \in [h/2, h]. \end{align*} which, curiously, is the same expression as for $z \in [0, h/2]$, therefore finally the flow velocity profile can be expressed using a single formula:

\begin{align*} \boxed{v(z) = \frac{\Delta P}{2\eta} (h - z) z, \; z \in [0, h]} \end{align*}

In the second part of the problem we are asked to compute the flux density per unit of length in $y$ direction, i.e. $q=\frac{dQ}{dy}$. This is just a simplification for us to use a simpler flux expression ($\int{v\,dz}$) instead of the more general one ($\iint{v\,dy\,dz}$).

Therefore we obtain the flux density by integrating the above profile:

\begin{align*} &q=\int\limits_{0}^{h}v(z)\,dz & \Rightarrow \\ &q=\int\limits_{0}^{h} \frac{\Delta P}{2\eta} (h - z) z \,dz &\Rightarrow \\ &q= \frac{\Delta P}{2\eta} \left(\frac{h^{3}}{2} - \frac{h^{3}}{3} \right) & \Rightarrow \\ &\boxed{q= \frac{\Delta P h^{3}}{12\eta}} & \end{align*}

Physics problem: Collapsing spherical cavity (bubble) in a water pond

Consider a collapsing spherical cavity in a large pond of water. The initial pressure of water in the pond is $P_0$, the initial radius of the cavity is $R_0$. Determine the speed of the edge of the cavity when its radius reaches $r_0$ ($r_0 < R_0$). The water density is $\rho$.

Note: The problem statement is translated (and slightly modified) from the original text in Savchenko et al 1981.

Discussion

I did not plan on writing up this problem as it seemed trivial to me at first. I thought, ok there is some kind of differential equation to compose and solve.

I tried to use the Newton's law and the energy conservation equation approaches but could not really get it.

I asked ChatGPT, and it came up with a solution based on the energy balance equation, the solution did resemble the correct one but it was off by a constant factor. ChatGPT mentioned the Rayleigh-Plesset equation which helped me to come up with my solution below. But here I decided to base the solution on the Newton's equation of motion to be able to present a self-contained solution, at least from my stand point.

Solution

Let's consider a water parcel at a distance $x$ from the center of the cavity $C$ (see the figure below).

Figure 1: Cross-section of collapsing cavity in a large pond. Dashed circles show positions of the cavity's edge when its sizes are $R_0, r, r_0$.

The pressure gradient will act to accelerate the parcel towards the point $C$.

\begin{equation} \rho \frac{dv}{dt} = -\frac{\partial P}{\partial x} \end{equation}

where $v=v(x, t)$ - is velocity of the water parcel at the distance $x$ from the center of the collapsing hollow cavity at time $t$.

We can express the derivative of $v(x(t), t)$ in the above equation by assuming that the parcell travels along a tragectory $x(t)$ and employing the compound function differentiation formula:

\begin{align*} \frac{dv}{dt} &= \frac{\partial v}{\partial t} + \frac{\partial v}{\partial x} \frac{d x}{d t} \\ &= \frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x} \end{align*}

Therefore, the equation of motion takes the following form (this is also known as an Euler form of the equation of motion): \begin{align} \boxed{\frac{\partial v}{\partial t} + v\frac{\partial v}{\partial x} = -\frac{1}{\rho}\frac{\partial P}{\partial x}} \end{align}

Let's consider the continuity equation for the spheres with radii $x$ and $R_0$ and centered around the point $C$:

\begin{align*} & v(x, t) \cdot 4\pi x^2 = v(R_0, t) \cdot 4\pi R_0^2 & \Rightarrow \\ & v(x, t) \cdot x^2 = v(R_0, t) \cdot R_0^2 = F(t) & \Rightarrow \\ & \boxed{v(x, t) = \frac{F(t)}{x^2}} & \end{align*}

Now let's plug the above expression for $v(x, t)$ into the equation of motion:

\begin{align*} & \frac{1}{x^2} \frac{d F(t)}{d t} + \frac{F(t)^2}{x^2}\left(-\frac{2}{x^3} \right) = -\frac{1}{\rho}\frac{\partial P}{\partial x} & \Rightarrow \\ & \frac{1}{x^2} \frac{d F(t)}{d t} - \frac{2F(t)^2}{x^5} = -\frac{1}{\rho}\frac{\partial P}{\partial x} \end{align*}

Let's consider the above equation at a fixed moment in time when the radius of the hollow cavity is $r(t)$. Then integrating the above equation from $r(t)$ to $+\infty$ and using that $P(r) = 0$ and $P(+\infty)=P_0$ we get:

\begin{align*} & \frac{d F(t)}{d t}\int\limits_r^{+\infty}\frac{dx}{x^2} - 2F(t)^2\int\limits_r^{+\infty}\frac{dx}{x^5} = -\frac{1}{\rho}\int\limits_r^{+\infty}\frac{\partial P}{\partial x} dx & \Rightarrow \\ & \frac{1}{r}\frac{d F(t)}{d t} - \frac{1}{2} \frac{F(t)^2}{r^4} = -\frac{P_0}{\rho} & \end{align*}

We can express $F(t)$ using flux continuity at the edge of the cavity as follows:

\begin{align*} F(t) = u(t)r^2 \end{align*}

where $u(t) = dr/dt=\dot{r}$ - is the speed of the edge of the cavity at time $t$ when its radius is $r=r(t)$.

After plugging in the above expression for $F(t)$ into the integrated equation of motion, we get:

\begin{align*} \boxed{\dot{u}r + \frac{3}{2} u^2 = -\frac{P_0}{\rho}} \end{align*}

The above is a particular case of an existing named equation (Rayleigh-Plesset equation: $\ddot{r} r + \frac{3}{2} \dot{r}^2 = -P_0/\rho$, where $\dot{r}=u$).

Finally, to solve the equation we will invert the $r(t)$ function and will consider the speed of the edge of the cavity as a function of the radius of the cavity.

\begin{align*} \dot{u} = \frac{d u}{d r} \frac{1}{\dot{r}} \end{align*}

Replacing the temporal derivative $\dot{u}$, using the above formula, in the initial differential equation we get an easily separable first order differential equation with respect to $u=u(r)$:

\begin{align*} \frac{du}{dr}ur + \frac{3}{2} u^2 = -\frac{P_0}{\rho} \end{align*}

Further, we can separate $du$ and $dr$ and integrate the equation from $R_0$ to $r_0$, taking into account that $u(R_0) = 0$ (i.e. the velocity of the cavity edge was 0 in the beginning when its radius was $R_0$):

\begin{align*} & -\int\limits_{u(R_0)}^{u(r_0)} \frac{u du}{\frac{P_0}{\rho} + \frac{3}{2}u^2} = \int\limits_{R_0}^{r_0}\frac{dr}{r} & \Rightarrow \\ & \ln\left(\frac{\frac{P_0}{\rho} + \frac{3}{2}u_(r_0)^2}{\frac{P_0}{\rho}}\right) = -3 \ln\left(\frac{r_0}{R_0}\right) & \Rightarrow \\ & \frac{P_0}{\rho} + \frac{3}{2}u(r_0)^2 = \frac{P_0}{\rho}\left(\frac{R_0}{r_0}\right)^3 & \Rightarrow \\ & \boxed{\left| u(r_0)\right| = \sqrt{\frac{2}{3}\frac{P_0}{\rho}\frac{R_0^3 - r_0^3}{r_0^3}} } & \end{align*}

So the above expression is the formula for the speed of the edge of the collapsing cavity at the moment when its radius reaches $r_0$ from $R_0$. It is interesting to note that there is a singularity at $r_0 \rightarrow 0$, the speed of the cavity edge increases to infinity.

Physics problem: acceleration of water level in a container due to draining from an opening in its bottom

Problem statement

From an opening at the bottom of a tall container, water drains out. The cross-section area of the container is $S$, and the cross-section area of the draining stream of water is $\sigma$. The water level in the vessel moves downward with constant acceleration. Determine this acceleration.

Solution

I'll describe two methods to solve the problem: the first one is the one I used and it is based on Bernoulli's law (energy conservation) and the second one is using the Toricelli's formula and accompanying assumptions from the get go. The second one I would guess a more experimented physicist would use and be done with it (see the approach 2). If I were to think about the second approach right away, this post would not happen.

Approach 1

Let's denote the speed of the water at the surface of the container and at the draining hole as $v_0$ and $v_1$ respectively. Then we can write energy and mass conservation expressions for the two cross-sections as follows:

\begin{align*} \begin{cases} \frac{v_0^2}{2} + gh = \frac{v_1^2}{2} \\ v_1\sigma = v_0 S \end{cases} \end{align*}

where $h$ is the current water level in the container, ang $g$ is the acceleration due to gravity.

Now, if we eliminate $v_1$ from the energy conservation equation and apply a time derivative to the both sides of the equation we will obtain the following relation (we use an upper dot to denote a time derivative, $\dot{x}=\frac{dx}{dt}$):

\begin{equation*} 2 v_0 \dot{v_0} + 2g\dot{h}=\left(\frac{S}{\sigma}\right)^2 2 v_0 \dot{v_0} \end{equation*}

Let's denote the acceleration of the water level as $a=\dot{v_0}$, and also note that $v_0 = -\dot{h}$ (the minus sign here is added because the level $h$ is decreasing with time and I prefer to consider $v_0$ as speed value without direction). Using this notation, the previous equation can be written as:

\begin{equation*} 2 v_0 a - 2gv_0=\left(\frac{S}{\sigma}\right)^2 2 v_0 a \end{equation*}

Simplifying we get the following expression for the water level acceleration in the container:

\begin{equation*} a = \frac{g}{1 - \left(\frac{S}{\sigma}\right)^2} \end{equation*}

Now we could consider that the conditions given in the problem statement imply $S\gg\sigma \rightarrow S/\sigma \gg 1$, therefore we can approximate the above expression as:

\begin{equation*} a \approx -g\left(\frac{\sigma}{S}\right)^2 \end{equation*}

Note that $a = \dot{v_0} < 0$, meaning that the water level descent in the container is slowing down with time.

Approach 2

For this approach we recognize right away that the problem conditions allow the use of Toricelli's formula for the speed $v_1$ of draining water from a large container with the water level at $h$:

\begin{equation*} v_1=\sqrt{2gh} \end{equation*}

The draining water will cause the change to the water level as follows: \begin{align*} \dot{h}=-v_1\sigma/S=-\frac{\sigma}{S}\sqrt{2gh} \end{align*}

If we apply a time derivative to the above equation and substitute $\dot{h}$, we obtain:

\begin{align*} \ddot{h} & =-\frac{\sigma}{S}\sqrt{2g}\frac{1}{2\sqrt{h}}\dot{h} \\ & = g\left(\frac{\sigma}{S}\right)^2 \end{align*}

Since the water level acceleration is $a = \dot{v_0} = d_t \left( -\dot{h} \right) = -\ddot{h}$, we obtain the final expression using the above relation for the second derivative of the water level as:

\begin{align*} a = -g\left(\frac{\sigma}{S}\right)^2 \end{align*}

Note that in this case we did not have to make any approximations as they were already applied for the Toricelli's formula to be valid (mainly that the speed of the water level $v_0$ is much lower than the draining speed $v_1$).

Physics problem: cumulative shell colliding with an armoured wall

Introduction and problem statement

This problem, as many others appearing here, took me a lot of time to wrap my head around. I almost lost faith, even tried to ask ChatGPT to solve it for me few times, but ChatGPT's solution was wrong or completely obscure for me.

Here is the problem statement (translated by ChatGPT from the Savchenko et al 1981 book).

Problem statement: In 1941, the Germans invented a cumulative anti-tank shell. The shell has a fuse on its front part, which, upon impact, causes detonation and ignites the entire charge. The shell penetrates the armor. In 1944, such German shells came into the hands of both the Soviets and their allies. Extensive experiments began. Various additional effects and paradoxes were discovered. The researchers started to investigate what exactly was penetrating the armor. At first, they thought it was a jet of hot gas that burns through the armor. However, it turned out that a metal jet is piercing through the armor traveling in a very puzzling way: in front of the plate at a speed of $v_{0} = {\rm 8\;km/s}$, inside the plate at ${\rm 4\;km/s}$, and behind the armor plate again at ${\rm 8\;km/s}$.

Explain this phenomenon and determine the speed of the metal wall of the conical cavity covering the charge, given that the cone’s vertex angle is 30°.

I have to admit that the translation is pretty accurate.

• You might prefer to read the notebook version of the post here.
• Latex sources for the plots are available here.

Figure 1: Cross-section of cumulative shell heading towards the armoured wall.

Solution

The problem consists of two parts. The first one asks to explain the phenomenon and the second one asks to determine the speed of the collapsing conical wall that is separating the fuse and the explosive charge of the shell, this collapsing wall is forming the metal jet piercing through the armor. I'll proceed in the order I solved it initially, starting with the part 2.

Part 2

Let's determine the speed of the conical wall collapsing into the jet flowing onto the armor at a speed of $v_{0} = {\rm 8\;km/s}$ as shown in Fig. 2. We focus here only on the upper part of the conical wall cross-section ($AB$, see Fig. 1 as well).

The breaking moment for me here was to realize that the velocity of the conical wall is orthogonal to its surface ($\vec{v}$ in Fig. 2). Since it is collapsing under the pressure force of the gas/liquid created to the left of the cone surface after the explosion of the charge.

Figure 2: Cross-section of the upper part of the collapsing conical wall.

Let's consider a reference frame in which the cone material flows along the cone surface at speed $\vec{v}_1$. This reference frame is moving to the right at a horizontal speed $\vec{u}$, so that the following relation is satisfied:

\begin{equation} \vec{v} - \vec{u} = \vec{v}_1 \end{equation}

In this reference frame the horizontal part of the jet is moving at speed $\left|\vec{v}_1\right| = v_1$ (here and below we ommit the arrow to denote scalar variables). This is relatively easy to prove using energy conservation. In summary we have the following system of equations:

\begin{align} \begin{cases} \vec{v} - \vec{u} = \vec{v}_1\\ v_0 - u = v_1\\ \end{cases} \end{align}

From the triangle formed by the speed vectors in Fig. 2 we get: \begin{align} \begin{cases} v_1 = v / \tan\left(\frac{\alpha}{2}\right)\\ u = v / \sin\left(\frac{\alpha}{2}\right) \end{cases} \end{align}

Now using the last two equations and the relation $v_0 - u = v_1$ we get an equation with respect to $v$:

\begin{align*} v_0 - u = v_1 & \Rightarrow v_0 - v / \sin\left(\frac{\alpha}{2}\right) = v / \tan\left(\frac{\alpha}{2}\right) \\ &\\ & \Rightarrow \boxed{v = \frac{v_0 \sin\left(\alpha/2\right)}{1 + \cos\left(\alpha/2\right)}}\\ &\\ & \Rightarrow v\approx 1.05\:{\rm km/s} \end{align*}

Part 1

The first part of the problem is to explain the phenomenon, mainly why the speed decreases by a factor of two in the armor. It is more or less clear that the jet will slow down due to the friction with the armor and then will accelerate back behind the armoured wall to the initial speed as it is being pushed forward. The factor of two for the velocity can be explained by the size of the hole pierced by the jet in the armoured wall (as shown in Fig. 3) which is two times larger than the diameter of the jet.

Figure 3: Metallic jet formed from the collapsed conic wall after piercing through the armoured wall, after the stationary flow regime is established.

So now we need to explain why the hole would be approximately twice the size of the metal jet piercing through the armor. For this we need to consider the moments before the hole is pierced completely. At this point there is a part of the jet that flows backwards reflected from the armor as shown in Fig.4 below, which expands the hole. The symmetry and mass conservation would hint us towards the factor of two for the hole diameter.

There is however a question, why would the backflow velocity be directed exactly in the opposite direction with respect to the incident jet. Indeed, this is not obvious.

Figure 4: Metallic jet formed from the collapsed conic wall prior to piercing through the armoured wall.

To find the angle at which the backward flow is reflected we can consider two colliding jets of liquid: smaller with a radius $r$ and a larger one with a radius $R$, moving towards each other at equal and parallel speeds (we can always switch to the reference frame where the metallic jet and the armoured wall move at equal speeds towards each other). As a result of the collision of the jets, a conical backflow in the direction of the smaller jet is formed. If we denote the angle of the cone as $2\beta$, then using momentum, mass and energy conseration it is relatively easy to show that the angle of the cone can be computed using dimensions of the colliding jets as follows: \begin{equation} \cos\beta = \frac{R^2-r^2}{R^2 + r^2} \end{equation}

Now, for our problem if we take the radius of the metallic jet to be $r$, it is much smaller than the effective radius of the armoured wall:

\begin{align} r \ll R & \Rightarrow \cos\beta = \frac{R^2-r^2}{R^2 + r^2} \approx 1 \\ & \Rightarrow \beta \approx 0 \end{align}

which means that indeed in our case the backflow is directed antiparallel to the incident jet velocity.

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: calculating the tension of ropes tying together timbers of a raft

This problem comes from my usual source of inspiration, the book by Savchenko et al., from the section on floating bodies and Archimedes' force. I spent a considerable amount of time figuring this one out, mostly because the answers in the edition I have and in the newer edition do not match. Another reason I could not get the correct answer is that I did not think that we should account that there are actually two ropes one at the front and the second is at the back of the raft.

Here is the problem statement: Consider two rows of wooden timbers tied together into a raft, as shown in the figure. Half of the top row of the timbers is above the water surface. Determine the minimum tension of the two ropes holding the timbers together, if the mass of each timber is $m$. The raft is very wide.

Solution

Let's denote water density as $\rho$ and the volume of each timber as $V$. Let's refer to forces, acting between the timbers and between a tibmer and a rope as $N$ and $F$ respectively. See the figure below (the Latex code for the fiigure can be found here).

We can express the equillibrium of a bottom and of a top timbers separately using vertical components of forces acting on them:

$$\tag{1} \begin{cases} 2F - N\sqrt{3} - mg + \rho V g = 0 \\ -2F + N\sqrt{3} - mg + \frac{1}{2} \rho V g = 0 \end{cases}\\ \Rightarrow -6F + 3N\sqrt{3} - mg = 0 $$

Note that we have $2F$ in each of the equations above for the force the ropes push timber downwards or upwards because we have two ropes one at the front and another at the back of the raft.

Let's consider vertical components of the forces acting on the upper part of one of the ropes. Assuming that we have around $n$ timbers in each row and that the mass of the rope is negligible, we get:

\begin{equation}\tag{2} n \cdot F = 2 T\cos\alpha \end{equation}

Where $\alpha$ - is the angle of the tension force with the vertical at the left and right edges of the raft. The tension arises from the bottom part of the rope pulling the upper part down. Since the $n$ is large, according to the problem statement, we can deduce:

\begin{equation}\tag{3} F = 2 \frac{T\cos\alpha}{n} \rightarrow 0 \end{equation}

So the equation (1) simplifies to the following: \begin{equation}\tag{4} 3 N\sqrt{3} - mg = 0 \end{equation}

Now let's consider the right part of the timbers as a system and write down the equillibrium condition for horizontal components of the forces acting on it. I highlighted in red the timbers in the selected system as well as relevant forces:

\begin{equation}\tag{5} \frac{N}{2} + P_1 + P_2 - 4T = 0\\ \Rightarrow 4T = \frac{N}{2} + P_1 + P_2 \end{equation}

where $P_1 \geq 0$, and $P_2 \geq 0$ - horizontal forces exerted by the neighbour timbers from the top and the bottom rows respectively onto the above selected system of the timbers.

Therefore,

\begin{equation}\tag{6} 4 T \geq \frac{N}{2} \Rightarrow T_{\min} = \frac{N}{8} \Rightarrow N = 8 T_{\min} \end{equation}

Then substituting the above expression for $N$ into equation (4) we get:

\begin{equation}\tag{7} 24 T_{\min}\sqrt{3} - mg = 0 \\ \Rightarrow \boxed{T_{\min} = \frac{mg}{24\sqrt{3}} = \frac{mg\sqrt{3}}{72}} \end{equation}

The above result is the same as in the newer edition of the book. At this point, I am not sure how the answer in the old edition of the book was obtained, because if we forget about one rope, the $T_{\min}'$ would be two times greater then the correct one, but in the older edition the tension is $4 T_{\min}$ for some reason....

Stability of a rectangular box floating on a surface of a liquid

This problem, as many of the previous discussed here, is coming from the Savchenko et al book. It took me around one month to wrap my head around this one. I have to admit that I improved (or refreshed) my knowledge of Archimedes' force thanks to this one. Specifically, this problem highlights the fact that Archimedes' force is applied to the center of mass of the water replaced by a floating object. Therefore, even if the object's density is uniform, it can happen that the torques of the gravity force and the Archimedes' force are not compensated even though the modules of the forces are equal. This fact makes more interesting the question about stability of a floating object in a liquid.

Problem statement

Given a rectangular body of length $a$, width $a$, and height $b$. Its density is $\rho$ and it floats in a liquid of density $\rho_0$. Determine the ratio $\frac{a}{b}$ for which the object's equillibrium state is stable. It is assumed that the side $b$ of the box is vertical in the equillibrium position.

Discussion

It is obvious that the object's equllibrium is stable with respect to vertical displacements, because if we push it down, the Archimedes' force increases and pushes it back. Similarly, if we pull the object up, the Archimedes' force becomes smaller and the gravity pushes it back to its initial equillibrium position where the gravity force is compensated by the Archimedes' force. Therefore, further down we consider only stability of the object with respect to rotations.

We won't consider the case with the box completely submerged into the liquid as in this case the center of mass of the box coincides with the center of mass of the liquid pushed out by the box, due to the uniform density of the box, and therefore the net torque (exerted by the gravity and Archimedes' forces) will always be zero. I guess this case is not stable with respect to rotations, as the box won't be restored to its original vertical position. Zero gravity, like in the outer space, all orientations are equivalent.

Here we can speculate a bit. Intuitively we would expect that the larger base (i.e. greater $a$) with respect to the height $b$ would give us a more stable equillibrium. Hence we can expect the result to be in a form $\frac{a}{b} > r$ or $\frac{a}{b} \geq r$.

Let's denote the height of submerged part of the box as $h$ and the pivot point as $O$. The pivot point is the intersection of the vertical axis of symmetry of the box and the horizontal plane at the level of the surface of the liquid (the updated level after the box is submerged), see the figure below. We consider rotations at a small angle $\varphi$ around this pivot point $O$.

The latex source for the plot can be found here.

There are two ways to find out if the vertical upright orientation of the box is stable: using the change in the height of the center of mass of the system liquid and box, i.e., if the above rotations lower the center of mass of the system then its vertical orientation is not stable, otherwise it is stbale, and using the net torque exerted by the gravity and Archimedes' forces, i.e., in the stable case the torque would act in the direction opposite to the direction of rotation,otherwise it is not stable. We'll use both methods and compare the results.

But first let's prepare a few useful expressions. If we denote as $A$ the center of mass of the liquid pushed out by the box (this is also the point where the Archimedes' force is applied to the box). Let's determine its coordinates $\left(x_{A\varphi}, y_{A\varphi}\right)$ when the box is rotated, in the system of coordinates with the center at the point $O$ and the $y$-axis pointing downward and the $x$-axis pointing to the right. It is easy to see that the mass ($m_A=\rho_0 a^2 h$) of the liquid pushed out by the box does not change during the rotation as well as the module of the Archimede's force $F_A=m_A g$. By definition, expressions for the coordinates of the center of mass of the liquid pushed out by the box can be written as:

\begin{cases} m_A x_{A\varphi} = m_A x_{A'\varphi} + m_L x_L + m_R x_R \\ m_A y_{A\varphi} = m_A y_{A'\varphi} + m_L y_L + m_R y_R \end{cases}

where

\begin{align} m_L = -m_R\\ x_L = -x_R\\ y_L = -y_R\\ \end{align}

and the $A'\varphi$ is the center of mass of the fictious liquid volume when it is rotated from the upright position as a solid body, $m_L$ and $m_R$ are the masses of the prisms (triangles in 2D, see the above figure) to be removed from and added to the fictious liquid body respectively to get the real shape of the liquid pushed out by the box during rotation. $\left(x_R, y_R\right)$ and $\left(x_L, y_L\right)$ are the centers of mass of the prisms of masses $m_R$ and $m_L$ respectively (red and blue triangles in the figure, the blue one has negative mass and the red one has positive mass). Since $m_L$ is the mass to be removed, it is negative in the equations to avoid confusion with the signs. We use these triangles (prismes in 3D) to facilitate the calculation of the position of the center of mass of the liquid pushed out by the box $A\varphi$. It is easy to compute coordinates of the center of mass of the fictious rectangular volume $B_1B_2RL$ and then to find coordinates of $A\varphi$ (of the complex shape of the liquid pushed out by the rotated box) we compute the center of mass of a system consisting of $B_1B_2RL$, $m_R$ and the negative mass $m_L$. The center of mass of such a system is $A\varphi$. If we substitude the relations between $m_L$ and $m_R$ into the equations for $x_{A\varphi}$, $y_{A\varphi}$ we get the following expressions:

\begin{cases} x_{A\varphi} = x_{A'\varphi} + 2 \frac{m_R}{m_A} x_R \\ y_{A\varphi} = y_{A'\varphi} + 2 \frac{m_R}{m_A} y_R \end{cases}

It is easy to see that

\begin{align} \frac{m_R}{m_A} & = \frac{\rho_0 \frac{1}{2}\cdot \frac{1}{2} a \cdot \frac{1}{2} a \tan\varphi\cdot a}{\rho_0 a^2 h} = \ \frac{a}{8h}\tan\varphi \approx \frac{a}{8h}\varphi \\ x_{A'\varphi} & = -\frac{h}{2}\sin\phi \approx -\frac{h}{2}\phi\\ y_{A'\varphi} & = \frac{h}{2}\cos\phi \end{align}

We keep exact values for the $y$ coordinates here as the changes in $y$ due to the rotation are on the order of $\sim\varphi^2$ and the variations of $x$ coordinates are on the order of $\sim\varphi$.

From the geometry and the properties of the centre of mass of a triangle we can determine $x_R$ and $y_R$: \begin{cases} x_R & = \frac{a}{2\cos\varphi}\cdot\frac{1}{2} + \frac{1}{3} \frac{a}{2\cos\varphi} \cos2\varphi = \frac{a}{4\cos\varphi}\left(1 + \frac{\cos2\varphi}{3}\right)\\ y_R & = \frac{1}{3}\left(\frac{a}{2\cdot2\cos\varphi}\right)\sin2\varphi = \frac{1}{6}a\sin\varphi \end{cases}

For small angles $\varphi$ (which is the case in the problem at hand), we get the following expressions for $x_R$ and $y_R$:

\begin{cases} x_R & \approx \frac{a}{3} \\ y_R & \approx \frac{a}{6}\varphi \end{cases}

Using the above relation we can express $\left(x_{A\varphi}, y_{A\varphi}\right)$ as functions of $h$, $a$, and $\varphi$:

\begin{cases} x_{A\varphi} = x_{A'\varphi} + 2 \frac{m_R}{m_A} x_R \approx -\frac{h}{2}\phi + 2 \frac{a}{8h}\varphi \cdot \frac{a}{3} = \frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right)\\ y_{A\varphi} = y_{A'\varphi} + 2 \frac{m_R}{m_A} y_R \approx \frac{h}{2}\cos\varphi + 2 \frac{a}{8h}\varphi \cdot \frac{a}{6}\varphi = \ \frac{h}{2}\cos\varphi + \frac{a^2}{24h}\varphi^2 \end{cases}

We did not replace $\cos\varphi$ with one for the $y$ coordinate since the accuracy of the terms is $\varphi ^ 2$.

Let's note the coordinates of the center of mass of the pushed out liquid for $\varphi=0$:

\begin{cases} x_{A0} = 0\\ y_{A0} = \frac{h}{2} \end{cases}

It is easy to derive how the coordinates of the center of mass of the box ($C$) change during rotation, as it is always in the middle of the box:

\begin{cases} x_{C\varphi} = \left( \frac{b}{2} - h \right)\sin\varphi\\ y_{C\varphi} = \left( h - \frac{b}{2} \right)\cos\varphi \end{cases}

Finally, let's express $h$ using equillibrium between the Archimedes' ($F_A$) and the gravity ($F_C = m_C g$) forces:

\begin{align} F_C-F_A=0 \Rightarrow \rho a^2 b - \rho_0 a^2 h = 0 \Rightarrow h=\frac{\rho}{\rho_0}b \end{align}

Method 1: analysis of torques

Without the loss of generality we can consider clockwise rotation as shown in the plot. Torques acting in the counter-clockwise direction are positive and torques acting in the clockwise direction are negative. With this convention, for stability it is required that the net torque ($\tau$) should act in the direction opposite to the rotation (i.e. anti-clockwise). Therefore the condition for stability is $\tau > 0$.

Let's consider two cases and impose the stability condition:

a) the center of mass is below the liquid surface $h\geq b/2$

If $x_{A\varphi} \geq 0$, then $\tau = F_C |x_{C\varphi}| + F_A |x_{A\varphi}|$, which is always positive. Both $\tau_C$ and $\tau_A$ act to stabilize the box against rotation.

If $x_{A\varphi} < 0$, then stability condition is:

\begin{align} \tau = F_C |x_{C\varphi}| - F_A |x_{A\varphi}| > 0 \end{align}

Since $F_C = F_A$ (as well as $m_A = m_C$), for stability we need the following:

\begin{align} |x_{C\varphi}| - |x_{A\varphi}| > 0 &\Rightarrow - \left( \frac{b}{2} - h \right)\varphi - (-1)\cdot\frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ & \Rightarrow -\frac{b}{2} + h + \frac{1}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ & \Rightarrow -2bh + 4h^2 + \frac{a^2}{3} - 2 h^2 > 0 \\ & \Rightarrow -6bh + a^2 + 6 h^2 > 0 \\ & \Rightarrow -6b^2\frac{\rho}{\rho_0} + a^2 + 6 \left(\frac{\rho}{\rho_0} b\right)^2 > 0 \\ & \Rightarrow \left(\frac{a}{b}\right)^2 > 6\frac{\rho}{\rho_0} - 6 \left(\frac{\rho}{\rho_0} \right)^2 \\ & \Rightarrow \boxed{\frac{a}{b} >\sqrt{ 6\frac{\rho}{\rho_0}\left(1 - \frac{\rho}{\rho_0} \right) }} \end{align}

b) the center of mass is above the liquid surface $h < b/2$

In this case the gravity force acts to destabilize the box with its torque acting in the direction of rotation. So the stability in this case is possible when $|x_{A\varphi}| > |x_{C\varphi}|$ and $x_{A\varphi} \cdot x_{C\varphi} > 0$ (i.e. A and C are from the same side from the pivot point $O$) which is equivalent to:

\begin{align} \tau = -F_C |x_{C\varphi}| + F_A |x_{A\varphi}| > 0 &\Rightarrow \ - (-1) \cdot \left(h - \frac{b}{2}\right)\varphi + \frac{\varphi}{4h}\left( \frac{a^2}{3} - 2 h^2 \right) > 0 \\ &\Rightarrow \boxed{\frac{a}{b} >\sqrt{ 6\frac{\rho}{\rho_0}\left(1 - \frac{\rho}{\rho_0} \right) }} \end{align}

Let's check if the conditions we imposed do actually make sense.

\begin{align} h < b/2 \Rightarrow \rho/\rho_0 < 1/2 \Rightarrow a/b > \sqrt{3/2} \Rightarrow \frac{a^2}{3} - \frac{b^2}{2} > 0 \Rightarrow x_{A\varphi} > 0 \end{align}

remembering that we consider only $\varphi > 0$.

Method 2: minimum of the potential energy of the system

I like this method more as it requires less logic branching. Let's denote as $y$ the $y$-coordinate of the center of mass of the system liquid and box. Since the positive direction of the axis is downwards, the change $\Delta y < 0$ during the rotation of the box would mean that the center of mass of the system is lifted and therefore the system's potential energy would be increasing and the system itself will tend to minimize its potential energy, therefore $\Delta y < 0$ is our stability condition. By the definition of the center of mass we have:

\begin{align} (M - m_A + m_C) y = m_C y_{C\varphi} - m_A y_{A\varphi} + M y_0 \end{align}

where $M$ is the mass the liquid would have when the immersed part of the box is replaced by the liquid. $y_0$ is the $y$-coordinate of the center of mass of $M$.

Let's examine the change in the $y$-coordinate of the center of mass of the system during rotation:

\begin{align} \Delta y = \frac{1}{M - m_A + m_C}\left ( m_C \Delta y_{C} - m_A \Delta y_{A} \right) \Rightarrow \Delta y \propto m_C \Delta y_{C} - m_A \Delta y_{A} \end{align}

Noting that $m_C = m_A$, we get:

\begin{align} \Delta y \propto \Delta y_{C} - \Delta y_{A} \end{align}

Therefore the condition for the system stability is equivalent to:

\begin{align*} \Delta y < 0 \Rightarrow \Delta y_{C} - \Delta y_{A} < 0 \end{align*}

From the relations prepared above we can express $\Delta y_{C}$ and $\Delta y_{A}$ as follows:

\begin{align*} \Delta y_{A} & = y_{A\varphi} - y_{A0} = \frac{h}{2}\cos\varphi + \frac{a^2}{24h}\varphi^2 - \frac{h}{2} = \frac{a^2}{24h}\varphi^2 - \frac{h}{4} \varphi^2 \\ \Delta y_{C} & = y_{C\varphi} - y_{C0} = \left(h - \frac{b}{2}\right)\left(\cos\varphi - 1 \right) \approx -\left(h - \frac{b}{2}\right)\frac{2\varphi^2}{4} \end{align*}

Now plugging the above expressions for $\Delta y_{C}$ and $\Delta y_{A}$ into the stability inequality, we get:

\begin{align*} & -\left(h - \frac{b}{2}\right)\frac{2\varphi^2}{4} - \frac{a^2}{24h}\varphi^2 + \frac{h}{4} \varphi^2 < 0 \Rightarrow \\ & \Rightarrow \frac{b}{4}\varphi^2 - \frac{h}{4} \varphi^2 - \frac{a^2}{24h}\varphi^2 < 0 \Rightarrow \\ & \Rightarrow 6 (b - h) h - a^2 < 0 \Rightarrow 6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right) b^2 - a^2 < 0 \Rightarrow \\ & \Rightarrow \boxed{\frac{a}{b} > \sqrt{6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right)}} \end{align*}

In conclusion, we are happy that the minimum potential energy and the torque methods give the same result.

\begin{equation*} \boxed{\frac{a}{b} > \sqrt{6 \frac{\rho}{\rho_0} \left( 1 - \frac{\rho}{\rho_0} \right)}} \end{equation*}

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.

Proving that $\sqrt{2}$ is irrational

In this post I would like to tell you about a math problem that I encountered for the first time during my 9th grade math exam. I guess I was 15 years old back then. The question, if I remember correctly, was to proove that $\sqrt{2}$ is irrational. At the exam I was trying to proove it assuming that the inverse statement is true, i.e. assuming that $\sqrt{2}$ is rational. So basically I did the following:

$$ \sqrt{2} \in Q \Rightarrow \text{ exist } m, n \in Z \text{ such that } \sqrt{2} = \frac{m}{n}\Rightarrow m = n \sqrt{2} $$

And stopped there saying that this is a contradiction because of the whole number on the left and real numer on the right, but the fact that $n\sqrt{2}$ is a real number requires proof not less than the initial problem statement. I guess I got 0 points for the proof like that. I remember that I did very bad at that exam, and the mark was low (almost failed). It nevertheless taught me to never leave an exam even if completed in advance, so at all my following exams at school and the university I was double checking my solutions If I had time and had never left an exam before the alocated time elapsed.

But, surprisingly, I have never got back to this problem afterwards. Until, 20 years later, my wife got the same question as homework. At that point I decided to revisit it. Please see my solution below, still using the contrary assertion:

\begin{equation} \text{Let } \sqrt{2} \in Q \Rightarrow \text{ exist } m, n \in Z \text{ such that } \sqrt{2} = \frac{m}{n}\Rightarrow m = n \sqrt{2} \Rightarrow \boxed{m^2 = 2 n^2} \end{equation}$$ \text{Exist } k, l = 0, 1, 2, ..., +\infty \text{ such that } \left\{\begin{array}{l} &m=2^k\cdot m_1\\ &n=2^l\cdot n_1\\ \end{array}\right. $$

where $m_1, n_1$ are odd.

Substituting the above expressions for $m$ and $n$ into the last part of the equation $m^2 = 2 n^2$, we get:

$$ 2^{2k} m_1^2 = 2^{2l + 1} n_1^2, $$

where $m_1^2, n_1^2$ are odd, because $m_1$ and $n_1$ are odd.

In order to be equal the left and right parts should have equal powers of each divisor. Therefore for the divisor 2 it is necessary that $2k = 2l + 1$, which is not possible, since the sets of odd and even numbers do not intersect. Since our assumption that $\sqrt{2} \in Q$ leads to a contradiction we conclude that the inverse to our initial assumption is true, i.e. $\sqrt{2} \notin Q$.

Physics problem: Superheroes and sticks

Some time ago, when I was around 17 years old, I have started solving physics problems from the book "Physical Problems" edition 1981 written by Savchenko O. Y. I was solving them one by one consecutively. But I did not finish them at the time...

I still have the book. So, last year, after I broke my leg at a soccer game, I had a bit more time, when I could only read, think or watch television. At that point, I had an idea to resume the problem-solving activity, and still solve some problems occasionally, even though my leg is OK now.

Recently, I was talking with a friend about one of the problems during lunch. That one was about a satellite that was losing height due to air friction (this one is interesting to me since the speed of the satellite is actually increasing). The friend suggested that I could write up my solutions in Latex. At that point, I did not feel like any of the solutions I have come up with was good enough to spend time writing them up. But today, I think, I have a good one. I would like to share it here. Who knows, maybe someone will point me to an even easier and nicer solution.

Here is my interpretation of the problem statement (it is Problem 5.42 in my edition of the book):

Superman and Ironman throw heavy sticks upwards at a pole of the Earth. The Superman's stick fell back after 30 days and the Ironman's stick fell back after 7 days. What is the difference between the initial speeds of the sticks?

Solution It is probably possible to verify that the sticks would go far enough that the approximation $\vec{g}=const$ is not valid anymore. Although I did not do it myself. I have assumed from the start that it is not valid for the given travel times.

From the energy conservation law we can obtain the expression for the speed of a stick, for example, on its way up at a distance $r$ from the Earth center:

\[ e=\frac{E}{m}=\frac{v^2}{2} - \frac{GM}{r} \]

Then, by equating it to the energy at the highest point, we get:

$$ e=\frac{v^2}{2} - \frac{gR^2}{r}=-\frac{gR^2}{x} $$

$x$ is the distance from the center of the Earth to the highest point of the trajectory of a stick. Here I have used the definition of the free-fall acceleration at the Earth surface:

$$ g = \frac{GM}{R^2} $$

$G$ is the gravitational constant, $R$ is the Earth radius.

Eliminating the speed we get

$$ v = \sqrt{\frac{2gR^2}{x}}\cdot\sqrt{\frac{x}{R} - 1} $$

On the way up $v=\frac{dr}{dt}$, so:

$$ \frac{dr}{dt} = \sqrt{\gamma}\cdot\sqrt{\frac{x}{R} - 1}\\ \gamma\equiv\frac{2gR^2}{x} $$

I have integrated the above equation by using the following substitution: $$ \frac{1}{\cos^2{\alpha}} = \frac{x}{R} $$

and obtained the time ($t$) it takes the stick to travel one way as a function of the maximum distance from the centre of the Earth ($x$):

$$ t = \frac{1}{\sqrt{\gamma}}\left( x \cos^{-1}\left(\frac{R}{x}\right) + \sqrt{R(x-R)} \right) $$

I would prefer to have it the other way around...

The fact that the stick speed should be bounded between the first $v_1=\sqrt{gR}$ and the second $v_2=\sqrt{2gR}$ cosmic speeds inspired me to use a search algorithm to solve for $x$. To include the speed at the Earth surface I have used the expressions for the total energy at the surface and at the highest point to construct the following system of equations:

$$ e_0 = \frac{v_0^2}{2} - gR \\ x = -\frac{gR^2}{e_0}\\ t = \frac{1}{\sqrt{\gamma}}\left( x \cos^{-1}\left(\frac{R}{x}\right) + \sqrt{R(x-R)} \right) \\ \gamma\equiv\frac{2gR^2}{x} $$

Finally, I used Python to implement an iterative search algorithm to find solutions of the system above. At the start, the limits of the speed are set to:

$$ vmin_0 = \sqrt{gR} \\ vmax_0 = \sqrt{2gR} $$

then we substitute $v_n = (vmin_n + vmax_n) / 2$ until the calculated $2t_n$ is close enough to the travel times given in the problem statement. Until this happens the following iterations are perfomed (based on the fact that $v$ grows monotonuously with $t$):

if $2t_n > 2t$, then $vmax_{n + 1} = v_n$, else $vmin_{n+1} = v_n$ and the next trial for $v$ is calculated as:

$$ v_{n+1} = (vmin_{n+1} + vmax_{n + 1}) / 2 $$

Below is the implementation of the algorithm:

In conclusion, the solution gives 11156 m/s for the Superman's stick and 11084 m/s for the Ironman's stick. Therefore the difference is around 72 m/s.

How I tried to derive the formula for the sine of sum on my own …

One evening I just got interested in this formula. But my school notes were not reachable at the moment and I did not want to look for it in the internet. And also maybe I wanted to prove something to myself. Either way I want to show you here the results. I have spent a lot of time solving this and it is a little embarrassing :), but still.


This solution is not finished. The cases when the angles are situated on the different sides from 90 degrees axis should be also considered.

The following is my first attempt to solve this,so if you are not interested, you can skip it ...






And also the cases when one of the cosines or both sines of the angles are zero should be considered separately.

I do not imply in any way that this is my invention, just the way I would use if I had to derive these relations.

Just nostalgy for my geometry classses ...

Today I took a book with English grammar which I used at school just to refresh my knowledge. And I found a home task in geometry which was given to us by our mathematics teacher Loboda Oleksiy Anatoliyovych at 10th grade. Since it is very interesting and dear for me in some way I've decided to translate it into English and post it here for everybody interested and for me to remember and not to loose. Here it is.

Home Task 11, geometry., 10th grade, to 04.12.00. Vectors

1. The points A,B,C and D are given. Points M and N - are the centers of the segments AB and CD, the point O is a center of the segment MN ( if a segment degenerates into a point then the point is the center). Prove that for any point S the following equation is valid:
SA + SB + SC + SD = 4SO.

2. The points M and N are the centers of the edges AB and A₁D₁ of the parallelepiped ABCDA₁B₁C₁D₁. The plane CMN intersects
the lines B₁C₁ and DB₁ respectively at the points P and Q. Express the vectors AP and AQ in terms of the vectors AB = a, AD = b, AA₁ = c.
(My answer: AP = a - 1/2 b + c; AQ = ( 4a + 5b + 4c ) / 9)

3. Different points A, B, C and D are given. The points M and N are taken at AC and BD respectively and in a way that the following relations are true: AM = λ⋅AC, BN = λ⋅BD. Prove that the vectors AB, CD and MN are coplanar and express MN in terms of AB and CD.
(My answer: MN =  λ⋅CD  + (1-λ)⋅AB )

4. Different points A, B, C and D are given. The points M and N are the centers of the segments BC and AD respectively. Prove that if the equation MN = 0.5⋅(AB + CD) is valid, then AB || CD.

5. The skew lines a and b are given and respectively the points A₁, A₂, A₃ and B₁, B₂, B₃ are at the lines (A₂ is between A₁ and A₃, B₂ is between B₁ and B₃ ). And also given that A₁A₂:A₂A₃ = B₁B₂:B₂B₃. Prove that the centers of the segments A₁B₁, A₂B₂, A₃B₃ belong to the same line.

6. ABCDA₁B₁C₁D₁ is a parallelepiped. Let's draw a line which intersects the lines AA₁, BC and C₁D₁ respectively at points M, N and P in a way that MN:MP = 2. What is the value of the relation BN:BC (find all solutions)?
(My answer: BN:BC = 2, could not find any additional cases yet).

6H (Example of how a problem difficulty can increase almost to infinity by changing one word:), I don't think this problem has sufficient data in the statement). ABCDA₁B₁C₁D₁ is a parallelepiped. Let's draw a plane which intersects the lines AA₁, BC and C₁D₁ respectively at points M, N and P in a way that MN:MP = 2. What is the value of the relation BN:BC (find all solutions)?

OK, it took a little longer than I thought, so the rest will appear here later ... I hope soon.