IPhO 2023 · Theory problem 3

Water and Objects

Download PDFIPhO 2023

In this problem, we consider the phenomena caused by the interaction between water and objects, related to surface tension. Part A treats motion, while Parts B and C are regarding static situations.

If necessary, you can use the fact that if the function \(y(x)\) satisfies the differential equation \(y''(x)=ay(x)\) (\(a\) is a positive constant), then its general solution is \(y(x)=Ae^{\sqrt{a}x}+Be^{-\sqrt{a}x}\), where \(A\) and \(B\) are arbitrary constants.

Part A. Merger of water drops (2.0 points)

As shown in Fig.1, we consider two stationary, spherical water drops on the surface of a super-hydrophobic material, i.e., very strong repulsive force exists between material and water.

Initially neighboring two identical spherical water drops are placed on the surface; then these two drops are merged after touching each other and form a larger spherical water drop, which suddenly jumps up.

A.1 2.0pt

The radius \(a\) of both water drops before the merger is \(100\,\mu\mathrm{m}\). The density of water \(\rho\) is \(1.00\times10^3\,\mathrm{kg/m^3}\). The surface tension \(\gamma\) is \(7.27\times10^{-2}\,\mathrm{J/m^2}\).

A portion \(k\) of the difference of the surface energy before and after the merger, \(\Delta E\), is transformed into the kinetic energy of the jumped water drop. Then, determine the initial jump-up velocity, \(v\), of the merged water drop in two significant digits under the following assumptions:

  • \(k=0.06\)
  • Before and after merger, the total volume of water is conserved.
Figure 1

Fig. 1: Merger of two water drops and jump of the merged water drop.

Part B. A vertically placed board (4.5 points)

A flat board is immersed vertically in water. Figures 2(a) and 2(b) respectively show water surface forms for the hydrophilic (attractive) and hydrophobic board materials. We neglect the thickness of the board.

The board surface is on the \(yz\) plane, and the horizontal water surface far away from the board is on the \(xy\) plane with \(z=0\). The surface shape does not depend on the \(y\)-coordinate. Let \(\theta(x)\) be the angle between the water surface and the horizontal plane at a point \((x,z)\) on the water surface in the \(xz\) plane. Here \(\theta(x)\) is measured with respect to the positive \(x\) axis and the counterclockwise rotation is taken as positive. Let \(\theta(x)\) be \(\theta_0\) at the point of contact between the board and the water surface \((x=0)\). In the following, \(\theta_0\) is fixed by the properties of the board material.

Water density \(\rho\) is constant and water surface tension \(\gamma\) is uniform. The gravitational acceleration constant is given by \(g\). The atmospheric pressure, \(P_0\), is assumed to be always uniform. Let us determine the water surface form in the following steps. Note that the unit of surface tension is \(\mathrm{J/m^2}\) as well as \(\mathrm{N/m}\).

Figure 2

Fig. 2: Boards vertically immersed in the water. (a) hydrophilic board case; (b) hydrophobic board case.

B.1 0.6pt

We consider a hydrophilic board case, as shown in Fig.2(a). We note that the water pressure, \(P\), satisfies the conditions \(P<P_0\) for \(z>0\) and \(P=P_0\) for \(z=0\). Then, express \(P\) at \(z\) in terms of \(\rho\), \(g\), \(z\), and \(P_0\).

B.2 0.8pt

We consider a water block whose cutout is shown as shaded in Fig.3(a). Its \(xz\) plane cross-section is shown in a hatched area in Fig.3(b). Let \(z_1\) and \(z_2\) respectively be the left and right edge coordinates of the boundary (water surface) between the water block and the air.

Obtain a horizontal component (\(x\) component) of the net force per unit length along the \(y\)-axis, \(f_x\), which is exerted on the water block due to the pressure, in terms of \(\rho\), \(g\), \(z_1\), and \(z_2\). Note that \(P_0\) results in no net horizontal force on the water block.

Figure 3

Fig. 3: Cutout form of water block on the water surface. (a) Bird's eye view and (b) cross-sectional view.

B.3 0.8pt

Surface tension acting on the water block is balanced with the force \(f_x\) discussed in B.2. We respectively define \(\theta_1\) and \(\theta_2\) as the angles between the water surface and the horizontal plane at the left and right edges. Express \(f_x\) in terms of \(\gamma\), \(\theta_1\), and \(\theta_2\).

B.4 0.8pt

The following equation holds at an arbitrary point \((x,z)\) on the water surface,

\[ \frac{1}{2}\left(\frac{z}{\ell}\right)^a+\cos\theta(x) =\text{constant}. \tag{1} \]

Determine the exponent \(a\) and express the constant \(\ell\) in terms of \(\gamma\) and \(\rho\). Note that this equation holds regardless of hydrophilic or hydrophobic board materials.

B.5 1.5pt

In Eq. (1) in B.4, we assume that variation of the water surface is slow, i.e., \(\lvert z'(x)\rvert\ll1\), so that we can expand \(\cos\theta(x)\) with respect to \(z'(x)\) up to the second order. Then, differentiating the resultant equation with respect to \(x\), we obtain the differential equation satisfied by \(z(x)\). Solve this differential equation and determine \(z(x)\) for \(x\geq0\) in terms of \(\tan\theta_0\) and \(\ell\). Note that the vertical directions of Figs. 2 and 3 are exaggerated for better view and they do not satisfy the condition, \(\lvert z'(x)\rvert\ll1\).

Part C. Interaction between two rods (3.5 points)

The identical rods A and B made of the same material floating in parallel on the water surface are placed at the same distance away from the \(y\)-axis (Fig.4).

Figure 4

Fig. 4: Two rods A and B floating on the water surface.

C.1 1.0pt

At the contact points of the rod B and the water surface, we define the \(z\)-coordinates \(z_a\) and \(z_b\), and the angles \(\theta_a\) and \(\theta_b\), as shown in Fig.5. Determine the horizontal force component, \(F_x\), on the rod B per unit length along the \(y\)-axis in terms of \(\theta_a\), \(\theta_b\), \(z_a\), \(z_b\), \(\rho\), \(g\), and \(\gamma\).

Figure 5

Fig. 5: Vertical cross-sectional view of two rods floating on the water surface.

C.2 1.5pt

We define the \(z\)-coordinate of the water surface, \(z_0\), at the midpoint of two rods in the \(xz\) plane. Express the force \(F_x\) obtained in C.1 without using \(\theta_a\), \(\theta_b\), \(z_a\), and \(z_b\).

C.3 1.0pt

Let \(x_a\) be the \(x\)-coordinate of the contact point between the water surface and the left side of the rod B. Using the differential equation obtained in B.5, express the water level coordinate \(z_0\) of the midpoint of these two rods A and B in terms of \(x_a\) and \(z_a\). You can use the constant \(\ell\) introduced in B.4.

Official English text of the problem.

Answers

Positive horizontal forces point along \(+x\). Both \(f_x\) and \(F_x\) are forces per unit length along \(y\), with units \(\mathrm{N/m}\).

  • A.1 \[ \boxed{v=\sqrt{\frac{3k\gamma}{\rho a}\left(2-2^{2/3}\right)} =0.23\,\mathrm{m/s}\quad\text{upward}.} \]

  • B.1 \[ \boxed{P(z)=P_0-\rho gz.} \]

  • B.2 \[ \boxed{f_x=\frac{\rho g}{2}\left(z_2^2-z_1^2\right).} \]

  • B.3 \[ \boxed{f_x=\gamma\left(\cos\theta_1-\cos\theta_2\right).} \]

  • B.4 \[ \boxed{a=2,\qquad \ell=\sqrt{\frac{\gamma}{\rho g}}.} \] The constant on the right of Eq. (1) is \(1\) for the board’s meniscus.

  • B.5, in the small-slope approximation: \[ \boxed{z''(x)=\frac{z(x)}{\ell^2},\qquad z(x)=-\ell\tan\theta_0\,e^{-x/\ell},\quad x\geq0.} \]

  • C.1 \[ \boxed{F_x=\gamma\left(\cos\theta_b-\cos\theta_a\right) +\frac{\rho g}{2}\left(z_b^2-z_a^2\right).} \]

  • C.2 \[ \boxed{F_x=-\frac{\rho g}{2}z_0^2.} \] The force on B is toward A when \(z_0\neq0\).

  • C.3, in the small-slope approximation: \[ \boxed{z_0=\frac{z_a}{\cosh(x_a/\ell)} =\frac{2z_a}{e^{x_a/\ell}+e^{-x_a/\ell}}.} \]

Solution

A.1

Let \(R\) be the radius of the merged drop. Conservation of volume gives

\[ \frac{4\pi R^3}{3} =2\frac{4\pi a^3}{3}, \qquad R=2^{1/3}a. \]

The mass of the merged drop is therefore

\[ m=\rho\frac{4\pi R^3}{3} =\frac{8\pi}{3}\rho a^3. \]

Surface energy equals surface tension multiplied by surface area. Thus the decrease in surface energy is

\[ \begin{aligned} \Delta E &=\gamma\left(2\cdot4\pi a^2-4\pi R^2\right)\\ &=4\pi\gamma a^2\left(2-2^{2/3}\right). \end{aligned} \]

Using the stipulated conversion of a fraction \(k\) into translational kinetic energy,

\[ \frac12mv^2=k\Delta E. \]

Consequently,

\[ v^2 =\frac{3k\gamma}{\rho a}\left(2-2^{2/3}\right). \]

With \(a=1.00\times10^{-4}\,\mathrm m\),

\[ m=8.3776\times10^{-9}\,\mathrm{kg}, \qquad \Delta E=3.7694\times10^{-9}\,\mathrm J, \]

and

\[ v=0.23236\ldots\,\mathrm{m/s}. \]

To two significant digits, the initial velocity is

\[ \boxed{v=0.23\,\mathrm{m/s}\ \text{upward}.} \]

B.1

In hydrostatic equilibrium, the upward pressure force balances gravity, giving

\[ \frac{dP}{dz}=-\rho g. \]

Since \(\rho\) is constant and \(P(0)=P_0\), integration yields

\[ \boxed{P(z)=P_0-\rho gz.} \]

In particular, water above \(z=0\) has pressure below atmospheric pressure, as stated.

B.2

A uniform pressure \(P_0\) applied over the entire closed boundary produces zero resultant force. We can therefore calculate using the gauge pressure

\[ p(z)=P(z)-P_0=-\rho gz. \]

The gauge pressure on the air-facing surface is zero. Let \(z_*\) be the common height of the block’s bottom. Pressure on the left vertical face pushes toward \(+x\), and pressure on the right face pushes toward \(-x\). The bottom contributes no horizontal force.

Per unit length along \(y\),

\[ \begin{aligned} f_x &=\int_{z_*}^{z_1}p(z)\,dz -\int_{z_*}^{z_2}p(z)\,dz\\ &=-\rho g\int_{z_2}^{z_1}z\,dz\\ &=\frac{\rho g}{2}\left(z_2^2-z_1^2\right). \end{aligned} \]

Hence

\[ \boxed{f_x=\frac{\rho g}{2}\left(z_2^2-z_1^2\right).} \]

For the illustrated hydrophilic meniscus, \(z_1>z_2>0\), so this pressure force points toward \(-x\).

B.3

Surface tension pulls tangentially away from the block at each edge. Its horizontal components per unit length are

\[ -\gamma\cos\theta_1 \quad\text{at the left edge}, \qquad +\gamma\cos\theta_2 \quad\text{at the right edge}. \]

Horizontal equilibrium requires

\[ f_x-\gamma\cos\theta_1+\gamma\cos\theta_2=0. \]

Therefore,

\[ \boxed{f_x=\gamma\left(\cos\theta_1-\cos\theta_2\right).} \]

B.4

Equating the expressions from B.2 and B.3,

\[ \frac{\rho g}{2}\left(z_2^2-z_1^2\right) =\gamma\left(\cos\theta_1-\cos\theta_2\right). \]

Rearranging gives

\[ \frac{\rho g}{2\gamma}z_1^2+\cos\theta_1 = \frac{\rho g}{2\gamma}z_2^2+\cos\theta_2. \]

Since the two edges can be chosen arbitrarily along the surface,

\[ \frac{\rho g}{2\gamma}z^2+\cos\theta(x)=C. \]

Comparison with Eq. (1) gives

\[ \boxed{a=2,\qquad \ell=\sqrt{\frac{\gamma}{\rho g}}.} \]

The length \(\ell\) is the capillary length. The derivation permits either sign of \(z\), so it applies to both raised and depressed menisci.

Far from the board, \(z\to0\) and \(\theta\to0\). Thus \(C=1\), and the board’s exact surface relation is

\[ \boxed{\frac{z^2}{2\ell^2}+\cos\theta(x)=1.} \]

B.5

The tangent to the surface satisfies

\[ \tan\theta(x)=z'(x). \]

For the surface represented as a graph,

\[ \cos\theta(x)=\frac{1}{\sqrt{1+[z'(x)]^2}}. \]

Under the stated approximation \(\lvert z'(x)\rvert\ll1\), its expansion through second order is

\[ \cos\theta(x) =1-\frac12[z'(x)]^2+O([z'(x)]^4). \]

Substitution into Eq. (1) gives, to this order,

\[ \frac{z^2}{2\ell^2}+1-\frac12(z')^2=C. \]

Differentiating,

\[ \frac{zz'}{\ell^2}-z'z''=0. \]

For a nonflat profile this yields

\[ \boxed{z''=\frac{z}{\ell^2},} \]

with the equation extending to isolated stationary points by continuity. The flat profile \(z=0\) also satisfies it.

The general solution is

\[ z(x)=A e^{x/\ell}+B e^{-x/\ell}. \]

The boundary condition \(z(x)\to0\) as \(x\to+\infty\) requires \(A=0\). At the board,

\[ z'(0)=\tan\theta_0=-\frac{B}{\ell}, \]

so \(B=-\ell\tan\theta_0\). Therefore,

\[ \boxed{z(x)=-\ell\tan\theta_0\,e^{-x/\ell}, \qquad x\geq0.} \]

For a hydrophilic board, \(\theta_0<0\), so \(z>0\). For a hydrophobic board, \(\theta_0>0\), so \(z<0\). In both cases the surface approaches \(z=0\) exponentially over the length scale \(\ell\). Consistency requires \(\lvert\tan\theta_0\rvert\ll1\).

C.1

There are two contributions to the horizontal force on rod B: surface tension at the contact lines and pressure on the rod.

Surface tension pulls the rod toward the adjoining free surface. Its net horizontal component is therefore

\[ F_{\gamma,x} =-\gamma\cos\theta_a+\gamma\cos\theta_b. \]

For the pressure force, subtract the uniform atmospheric pressure, whose resultant over the closed rod surface is zero. The remaining pressure on the wetted surface is \(p(z)=-\rho gz\).

The horizontal pressure force on a curved surface is obtained by integrating pressure over its vertical projection. Let \(z_*\) be the height of the bottom of the rod. The left wetted side contributes toward \(+x\), and the right wetted side toward \(-x\):

\[ \begin{aligned} F_{p,x} &=\int_{z_*}^{z_a}(-\rho gz)\,dz -\int_{z_*}^{z_b}(-\rho gz)\,dz\\ &=\frac{\rho g}{2}\left(z_b^2-z_a^2\right). \end{aligned} \]

Gravity has no horizontal component. Adding the two contributions,

\[ \boxed{ F_x=\gamma\left(\cos\theta_b-\cos\theta_a\right) +\frac{\rho g}{2}\left(z_b^2-z_a^2\right). } \]

C.2

The surface relation from B.4 applies separately to the surface between the rods and to the exterior surface. Their integration constants need not be equal.

On the exterior side of B, the surface becomes horizontal at \(z=0\) far away. Thus

\[ \frac{z_b^2}{2\ell^2}+\cos\theta_b=1. \]

Between the identical rods, reflection symmetry about \(x=0\) makes the surface horizontal at the midpoint. There \(z=z_0\), so the interior relation is

\[ \frac{z_a^2}{2\ell^2}+\cos\theta_a =1+\frac{z_0^2}{2\ell^2}. \]

Since \(\rho g=\gamma/\ell^2\), the force from C.1 can be written as

\[ F_x =\gamma\left[ \left(\cos\theta_b+\frac{z_b^2}{2\ell^2}\right) - \left(\cos\theta_a+\frac{z_a^2}{2\ell^2}\right) \right]. \]

Substitution gives

\[ F_x =\gamma\left[1-\left(1+\frac{z_0^2}{2\ell^2}\right)\right]. \]

Hence

\[ \boxed{F_x=-\frac{\gamma z_0^2}{2\ell^2} =-\frac{\rho g}{2}z_0^2.} \]

Rod B lies on the positive-\(x\) side, so the negative sign means attraction toward rod A. The force vanishes when \(z_0=0\). This result does not require the small-slope approximation.

C.3

Now apply the small-slope equation from B.5 to the surface between the rods:

\[ z''=\frac{z}{\ell^2}. \]

Writing its solution in hyperbolic form,

\[ z(x)=C_1\cosh(x/\ell)+C_2\sinh(x/\ell). \]

Symmetry gives \(z'(0)=0\), which requires \(C_2=0\). Since \(z(0)=z_0\), we have \(C_1=z_0\), and therefore

\[ z(x)=z_0\cosh(x/\ell). \]

At the inner contact with rod B, \(x=x_a\) and \(z=z_a\). Consequently,

\[ z_a=z_0\cosh(x_a/\ell), \]

so

\[ \boxed{ z_0=\frac{z_a}{\cosh(x_a/\ell)} =\frac{2z_a}{e^{x_a/\ell}+e^{-x_a/\ell}}. } \]

This height–separation relation is valid within the small-slope approximation.

The solution was checked against the official solution.