[User]
Q3-1
Theory

                                                                                English (Official)

Theoretical Problem 3: Cavitation [10.0 points]
Introduction

Cavitation is the phenomenon of vapour bubbles or ``cavities'' occurring in a liquid medium due to drop
in pressure. This is in contrast to boiling, where vapour bubbles are created due to rise in temperature.
Since the vapour bubbles collapse and generate shock waves as well as supersonic jets when the dropped
pressure is restored, cavitation is a constant source of damage and even of catastrophe in hydraulic
machines, ships, and more generally in any device involving liquid flow. On the other hand, it has found
many positive applications, for example in chemical industry, cleaning, and in treatment of kidney stones.

         Figure 1. (a) Cavitating propeller (b) Cavitation damage (Source: Wikimedia Commons)

It is understood that cavitation generally grow out of microscopic bubbles, called nuclei, that preexisted in
the liquid. These micro-bubbles are a few microns in size and contain both vapour and non-condensable
gas (the latter is just air when ordinary water is under consideration). If the pressure in the liquid be-
comes sufficiently low, the nuclei grow into a macroscopic size, initiating cavitation. Liquid purified of
such nuclei can even withstand negative pressure without cavitating. One usually compares this with
solid under tension, which does not rupture easily if there are no preexisting pockets or cracks in it.
In this problem, we will be concerned with various idealized scenarios related to cavitation. As is often
the case, we can glean some nontrivial information from simple dimensional analysis. However, we will
need differential equations embodying fundamental laws such as Newton's second law of motion and
Fick's law of diffusion, if we want to conduct a more precise study.
One of the first things we want to know is the so called critical (or threshold) pressure, that is the minimum
value of the water pressure so that the nuclei remain microscopic without growing into macroscopic
bubbles. The critical pressure is roughly equal to the vapour pressure at the given temperature, but the
exact value is slightly lower due to surface tension and the air content of the nucleus.

                                                                                Q3-2
Theory

                                                                                English (Official)

     Figure 2. (a) Cavitation (down arrow) and boiling (right arrow) on a phase diagram (b) Typical
     bubble (see Table 1 for notations)

If the external pressure suddenly drops below the critical pressure of a nucleus, then the nucleus starts
expanding and the expansion rate quickly reaches a stable value. In practice, after the bubble becomes
macroscopic in size, the pressure is typically restored to its original value, and the bubble starts collaps-
ing. We will model this situation by considering a macroscopic bubble in equilibrium, whose external
pressure is then suddenly risen. The collapsing bubble will rebound after reaching a minimum size if
the bubble had air in it. On the other hand, a pure vapour bubble would completely dissolve, with the
shrinkage rate growing unboundedly as the radius of the bubble reaches zero. In reality, towards the
end of the collapse, the bubble would lose its spherical shape, and the compressibility of water would
become important. However, unless a particular question explicitly asserts otherwise, we will neglect
those effects here.
Another interesting question is what happens when sound wave is transmitted through water containing
bubbles. It turns out that not only the bubbles pulsate following the pressure oscillations, but also the
sound wave induces translational motions of the bubbles. These effects can be used to manipulate
bubbles with the help of acoustic waves. For example, in acoustic cavitation, high intensity ultrasound is
employed to generate cavitation or cause collapse of bubbles.
Finally, there is a sort of paradox regarding the existence of nuclei in the first place. The theory predicts
that unless water is saturated with dissolved air, diffusion of air from any nucleus into water through the
gas-water interface must induce a complete dissolution of the nucleus in a matter of seconds. However,
in reality, micron sized nuclei exist in water and it is in fact extremely difficult to get rid of them. We will
consider one of a few potential resolutions of this paradox, namely the suggestion that small crevices
in solid walls or in solid particles carried by water are responsible for acting as micro-pockets of air and
vapour.

Potentially useful information

Vapour pressure

Let us say we have a closed jar containing water and air. If the air is too dry, then its humidity will increase
due to evaporation of water. If the air is too wet, then its humidity will decrease due to condensation. It
turns out that in equilibrium, the partial pressure 𝑝𝑣 = 𝑝𝑣 (𝑇 ) of vapour in air is a function of temperature.

                                                                                Q3-3
Theory

                                                                                English (Official)

     Figure 3. (a) Closed jar containing air and water in equilibrium (b) Diffusion flux through the
     surface S is proportional to the concentration gradient across S

Now if a bubble changes its volume in an instant, then the humidity inside the bubble will lose its equi-
librium with the surrounding water, and a new equilibrium must be reached either by condensation or
evaporation. In reality this process is so rapid that we can justifiably assume that equilibrium is main-
tained at all times. Moreover, the heat lost or gained by the surrounding water during this process is
negligible, so that the temperature remains constant. To conclude, we assume that the partial pressure
of vapour contained in a bubble remains equal to 𝑝𝑣 at all times.

Henry’s law

While the concept of vapour pressure gives us a good handle on the vapour content of a bubble, Henry
law offers at least a partial handle on the air content. Thinking of a closed jar containing water and air, it
says that in equilibrium, the concentration of dissolved air in water is proportional to the partial pressure
of air above the water:
𝑢 = 𝐻𝑞
where, 𝑢 is the concentration of air in water, 𝐻 is the so called Henry's constant, and 𝑞 is the partial
pressure of air adjacent to water. As before, we will assume that equilibrium of air content in the sense
of Henry's law is maintained at least in the immediate vicinity of the bubble at all times, and that this
maintenance does not cause any temperature change.

Fick’s law

To complement Henry's law, we need to know how dissolved air in water moves from places with high
concentration to places with low concentration. This is where Fick's law enters, which states that the
diffusion flux across an area element 𝑆 is proportional to how fast the concentration changes along the
direction perpendicular to 𝑆, see Figure 3:
𝐽 = 𝜅 𝜕𝑢
      𝜕𝑥

Here 𝐽 is the diffusion flux, that is the amount of air moving across the surface per unit area per unit
time, 𝜅 is the diffusivity coefficient, and we have assumed that the 𝑥 coordinate axis is perpendicular to
𝑆. When 𝑢 is a function of 𝑥 and possibly other variables, the notation 𝜕𝑢𝜕𝑥 means that we have taken the
derivative of 𝑢 with respect to the variable 𝑥, while holding all other variables constant.

                                                                                Q3-4
Theory

                                                                                English (Official)

Diffusion equation

If you need to find a function 𝑤 = 𝑤(𝑥, 𝑡) in the first quadrant 𝑄 = {(𝑥, 𝑡) ∶ 𝑥 > 0, 𝑡 > 0} satisfying 𝜕𝑤
                                                                                𝜕𝑡 =
                       𝑤(𝑥, 0) = 𝑓(𝑥)    for 𝑥 > 0,                                                ∞
𝜕2 𝑤
         in 𝑄, and {                                then the solution is given by 𝑤(𝑥, 𝑡) = √4𝜋𝑡
                                                                                1              2
                                                                                ∫0 (𝑒−(𝑥−𝑦) /(4𝑡) −
𝜕𝑥2
                       𝑤(0, 𝑡) = 0       for 𝑡 > 0,
          2
𝑒−(𝑥+𝑦) /(4𝑡) )𝑓(𝑦)𝑑𝑦.

Gaussian type integrals
                                                       ∞            √         ∞                 √
The following integrals may come in handy. ∫0 𝑒−𝑏𝑥 𝑑𝑥 = 2√𝜋𝑏 ,               ∫0 𝑥2 𝑒−𝑏𝑥 𝑑𝑥 = 4𝑏√𝜋𝑏
                                                            2                           2
                                                                                (𝑏 > 0).

Notations and typical values of parameters.

In Table 1, we list the notations used in the statement of the problem, and the typical values of some
important constants.

                       symbol     assigned meaning                             typical value
                       𝜌          water density                                997 kg/m3
                       𝑝∞         water pressure far from a bubble             101 kPa
                       𝑝𝑣         vapour pressure                              2340 Pa
                       𝜎          surface tension                              72.8 ⋅ 10−3 N/m
                       𝑅          bubble radius
                       𝑅0         initial radius of a bubble                   10−5 m
                       𝛿          density of air                               1.29 kg/m3
                       𝑞          partial pressure of air in a bubble
                       𝑞0         initial value of 𝑞
                       𝛾          adiabatic exponent of air                    1.4
                       𝑢          concentration of dissolved air in water
                       𝜅          diffusivity coefficient for air in water     2 ⋅ 10−9 m2 /s
                       𝐻          Henry's constant for air in water            0.24 ⋅ 10−6 s2 /m2
                       𝑡          time
                       𝑓0         natural/resonant frequency

Assumptions

Unless otherwise specified, throughout this problem we assume the following.
       • Water is incompressible, inviscid, and homogeneous.
       • Water fills the entire space.
       • Pressure variation due to gravity is negligible.
       • No spatial or temporal variation in temperature.
       • There is a single bubble.

                                                                                Q3-5
Theory

                                                                                English (Official)

   • The bubble remains spherical and without translatory motion.
   • No migration of air between the bubble cavity and the surrounding water.
   • Air is an ideal gas.

Part A: Preliminary analysis [1.5 points]
These are warm-up questions to get the initial feel of the phenomenon.

    A.1     By performing a simple dimensional analysis, estimate the collapse time 𝑇 of            0.5pt
            a pure vapour bubble, in terms of bubble's initial radius 𝑅0 , water density 𝜌,
            water pressure 𝑝∞ , and the vapour pressure 𝑝𝑣 . Evaluate the formula with the
            numerical constant implicit in the formula equal to 1, when 𝑅0 = 1 mm and the
            quantities 𝜌, 𝑝∞ , and 𝑝𝑣 take their typical values from previous Notation Table .
            Assume no surface tension: 𝜎 = 0.

    A.2     Suppose that a nucleus consisting of air and vapour, with radius 𝑅0 = 10−5 m,           1.0pt
            is in equilibrium when the external pressure 𝑝∞ = 101 kPa. Find the partial
            pressure 𝑞0 of air in the bubble. Now suppose that the external pressure 𝑝∞ was
            gradually decreased, and that the air inside the bubble follows an isothermal
            process. Find the critical pressure 𝑝𝑐 , defined by the condition that if 𝑝∞ < 𝑝𝑐
            the bubble size grows without bound. The quantities 𝑝𝑣 and 𝜎 take their typical
            values from the above Notation Table.

Part B: Main dynamics [6.0 points]
Now we will study the detailed dynamics of a spherical bubble consisting of a mixture of air and vapour.
Please assume that there is no air migration through the bubble wall, and hence that the whole dynam-
ics is governed by pressure only. Note however that as we have mentioned, there will be evaporation
and condensation of water vapour at the bubble wall, that maintains the vapour pressure 𝑝𝑣 within the
bubble.

    B.1     Suppose that a single spherical bubble resides within water that fills space uni-       1.5pt
            formly, and that the bubble may evolve in size without distorting its spherical
            shape, due to changes, e.g., in the external pressure 𝑝∞ . Derive an equation
            that relates the bubble radius 𝑅(𝑡) and its time derivatives 𝑅′ (𝑡) and 𝑅″ (𝑡) , sur-
            face tension 𝜎 , water density 𝜌 , the pressure far from the bubble 𝑝∞ , and the
            pressure inside the bubble 𝑝. Then split the pressure 𝑝 into two terms, by as-
            suming that the bubble has both vapour (with partial pressure 𝑝𝑣 ) and air in it,
            and that the air follows an adiabatic process with exponent 𝛾. To give a refer-
            ence point, the partial air pressure must be 𝑞0 when the bubble size equals 𝑅0 .
            Assume that evaporation, condensation, or transfer of air between the bubble
            cavity and the surrounding water has no effect on the water volume.

    B.2     A water tank under the external pressure 𝑝∞ −
                                                          = 101 kPa, containing a nucleus of        1.0pt
            radius 𝑅0 = 10 m initially in equilibrium, was exposed to vacuum, so that the
                          −5

            system suddenly has 𝑝∞ = 0 . Estimate the terminal (asymptotic) value of the
            growth speed 𝑅′ , as well as the time it reaches this terminal value.

                                                                              Q3-6
Theory

                                                                                English (Official)

    B.3     A water tank under the external pressure 𝑝∞−
                                                          = 1.600 kPa, containing a gas bub-     1.0pt
            ble of radius 𝑅0 = 10 m initially in equilibrium, was suddenly exposed to the
                                 −5

            atmospheric pressure 𝑝∞ = 101 kPa . Estimate the minimum radius of the bub-
            ble before it rebounds.

    B.4     If there is no gas other than water vapour present in a bubble, the bubble com-      0.5pt
            pletely collapses in finite time. Determine the characteristic exponent 𝛼 in
            𝑅(𝑡) ∼ (𝑇 − 𝑡)𝛼 ,
            where 𝑇 is the collapse time.

    B.5     Based on the equation derived in B3, find the natural frequency of the spherical     1.0pt
            oscillation of a bubble of radius 𝑅0 = 0.1 mm.

    B.6     Suppose that the bubble described in the previous part is subjected to a stand-      1.0pt
            ing sound wave along the 𝑥-axis, whose pressure field is given by
            𝑝(𝑥, 𝑡) = 𝑝0 + 𝐴 sin ( 2𝜋𝑓
                                    𝑐 (𝑥 + 𝑎)) sin(2𝜋𝑓𝑡),
            where 𝑓 is the frequency, and 𝑐 is the speed of sound. The parameters 𝑝0 , 𝐴
            and 𝑎 are constants, whose meanings may readily be deduced from the equa-
            tion. Find the average force exerted upon the bubble. The bubble is situated
            at the origin of the 𝑥𝑦𝑧 coordinate system, and its size is much smaller than the
            wavelength of the sound.

Part C: Dissolution of nuclei through diffusion [2.5 points]
In this final section, complementary to Part B, we focus on the effect of diffusion across the bubble wall.

    C.1     Suppose that a nucleus consisting of air and vapour, with radius 𝑅0 = 10−5 m ,       2.0pt
            is placed in water-air solution, in which the dissolved air is in equilibrium with
            the atmospheric pressure above the water. The partial pressure of air in the
            bubble is 𝑞 = 1.70 ⋅ 105 Pa, and the vapour pressure can be neglected. Estimate
            the time required for the bubble to be completely resorbed into water.
            The quantities 𝑝∞ , 𝜅, 𝛿 and 𝜎 take their typical values from Table 1. Assume that
            the region surrounding the bubble in which air diffusion takes place immedi-
            ately gets much larger than the bubble itself.

                                                                          Q3-7
Theory

                                                                                English (Official)

  C.2    Consider a conical crevice in the wall of a water container, with an aperture an-   0.5pt
         gle 𝛼, see the following Figure. A small amount of air and vapour reside within
         the cone. Write down the condition of mechanical and diffusive equilibrium.
         Determine when the pocket of air stays in the crevice without disappearing.
         The contact angle of water on the surface is 𝜃.

                                         Conical Crevice

Solution:
CAVITATION: A POSSIBLE SOLUTION

                                TSOGTGEREL GANTUMUR

   Note: The chosen units might be a bit diﬀerent from the “oﬃcial” problem statement.
For instance, we might write 1 µm for 10−6 m. “Table 1” refers to the “notation table”
from the statement.
A1. By performing a simple dimensional analysis, estimate the collapse time τ of a pure
vapour bubble, in terms of bubble’s initial radius R0 , water density ρ, water pressure p∞ ,
and the vapour pressure pv . Evaluate the formula when R0 = 1 mm and the quantities ρ,
p∞ and pv take their typical values from Table 1. Assume no surface tension: σ = 0.
Solution. It is reasonable to expect that the bubble would not collapse if p∞ ≤ pv . Hence
we take p∞ > pv , and presume that the diﬀerence p∞ − pv will feature in the final formula.
The dimensions of the quantities are
            [R0 ] = m,     [ρ] = kg/m3 ,      [p∞ − pv ] = N/m2 = kg/(m · s2 ).         (1)
The only combination of these that has the dimension of time is
                                     √
                                           ρ
                              τ ∼ R0            ≈ 0.1 s.                                (2)
                                        p∞ − pv
A2. Suppose that a micro-bubble consisting of air and vapour, with radius R0 = 10 µm,
is in equilibrium when the external pressure p∞ = 100 kPa. Find the partial pressure q0
of air in the bubble. Now suppose that the external pressure p∞ was gradually decreased,
and that the air inside the bubble follows an isothermal process. Find the critical pressure
pc , defined by the condition that if p∞ < pc the bubble size grows without bound. The
quantities pv and σ take their typical values from Table 1.
Solution. The pressure equilibrium condition is
                                                    2σ
                                    pv + q = p∞ +      ,                                (3)
                                                    R
which, under R = R0 and q = q0 , yields
                                            2σ
                              q0 = p∞ − pv +    ≈ 170 kPa.                              (4)
                                            R0
On the other hand, taking into account the isothermal law
                                       q0 R03 = qR3 ,                                   (5)
we get
                                            q0 R03 2σ
                                  p∞ = pv +        −    .                         (6)
                                              R3     R
The critical pressure corresponds to the minimum of p∞ as a function of R. The radius
at the minimum is easily found to be
                                       √
                                          3q0 R0
                               Rc = R0            ≈ 60 µm,                        (7)
                                           2σ
                                              1

2                                TSOGTGEREL GANTUMUR

and the corresponding pressure is
                                              ( R0 )3
                              pc = pv − 2q0           ≈ 700 Pa.                         (8)
                                                Rc
B1. Suppose that a single spherical bubble resides within water that fills space uniformly,
and that the bubble may evolve in size without distorting its spherical shape, due to
changes, e.g., in the external pressure p∞ . Derive an equation that relates the bubble
radius R(t) and its time derivatives R′ (t) and R′′ (t), surface tension σ, water density ρ,
the pressure far from the bubble p∞ , and the pressure inside the bubble p. Then split the
pressure p into two terms, by assuming that the bubble has both vapour and air in it, and
that the air follows an adiabatic process with exponent γ. To give a reference point, the
partial air pressure must be q0 when the bubble size equals R0 . Assume that evaporation,
condensation, or transfer of air between the bubble cavity and the surrounding water has
no eﬀect on the water volume.
Solution. Let v(r, t) denote the radial velocity of the fluid element at the distance r from
the bubble center and at the time moment t. Let also u(t) = v(R, t), that is,
                                       u(t) = R′ (t).                                   (9)
Then the incompressibility condition yields
                                                 R2 u(t)
                                     v(r, t) =           ,                             (10)
                                                   r2
and hence
                             ∂v    ∂ R2 u   2R 2 R2 ′
                                =         =    u + 2u.                                 (11)
                             ∂t    ∂t r2    r2    r
The radial acceleration can now be computed as
                            ∂v ∂v        R2      2R      2R4
                        a=      +    v = 2 u′ + 2 u2 − 5 u2 .                      (12)
                             ∂t   ∂r      r      r         r
With p = p(r) denoting the pressure field (where possible time dependence is suppressed
in the notation), Newton’s law reads
                                         ρa = −p′ ,                                    (13)
or
                              R2 ′ ( 2R 2R4 ) 2              p′
                                 u  +      −       u   =  −     .                  (14)
                              r2        r2     r5            ρ
We integrate it from r = R to r = ∞, to get
                                        3      p(R) − p∞
                                 Ru′ + u2 =                .                       (15)
                                        2          ρ
Since the pressure inside the bubble satisfies
                                                  2σ
                                      p = p(R) +     ,                             (16)
                                                  R
we conclude that
                                     3         2σ      p − p∞
                             RR′′ + (R′ )2 +       =             .                 (17)
                                     2         ρR         ρ
Finally, taking into account the fact that the bubble pressure p consists of the vapour
pressure pv and the partial air pressure q, as
                                                    ( R0 )3γ
                               p = pv + q = pv + q0           ,                    (18)
                                                       R

                               CAVITATION: A POSSIBLE SOLUTION                                     3

we have
                                 3          2σ q0 R03γ
                        ρRR′′ + ρ(R′ )2 +      − 3γ = pv − p∞ .                     (19)
                                 2          R       R
B2. A water tank under the external pressure p−     ∞ = 100 kPa, containing a nucleus of
radius R0 = 10 µm initially in equilibrium, was exposed to vacuum, so that the system
suddenly has p∞ = 0. Estimate the terminal (asymptotic) value of the growth speed R′ ,
as well as the time it reaches this terminal value.
Solution. Putting R′′ = 0 and R → ∞ in (19) yields
                              √               √
                      ′         2(pv − p∞ )     2pv
                    R (∞) =                 =       ≈ 1.24 m/s.                                 (20)
                                    3ρ          3ρ
The initial acceleration can also be found from (19) as
                              q0 + pv − 2σ/R0      p−
                       R′′ (0) =                = ∞ ≈ 107 m/s2 ,                                (21)
                                     ρR0           ρR0
and hence the time for the speed to be stabilized can be estimated as
                                      R′ (∞)
                                  t = ′′      ≈ 0.1 µs.                                         (22)
                                      R (0)
B3. A water tank under the external pressure p−  ∞ = 1600 Pa, containing a gas bubble
of radius R0 = 10 µm initially in equilibrium, was suddenly exposed to the atmospheric
pressure p∞ = 100 kPa. Estimate the minimum radius of the bubble before it rebounds.
Solution. Multiply (19) by 2R2 R′ to get
                 ρ[R3 (R′ )2 ]′ + 4σRR′ − 2q0 R03γ R2−3γ R′ = 2(pv − p∞ )R2 R′ ,                (23)
or
                                       2q0 R03γ (R3−3γ )′  2
                   3   ′ 2 ′
               ρ[R (R ) ] + 2σ(R ) +  2 ′
                                                          = (pv − p∞ )(R3 )′ .       (24)
                                           3(γ − 1)        3
This can easily be integrated, from R(0) = R0 and R′ (0) = 0 to R(t) = R and R′ (t) = R′ ,
which yields
           ′ 2                      2q0 R03γ (R3−3γ − R03−3γ )  2
       3
     ρR (R ) + 2σ(R    2
                           − R0 ) +
                              2
                                                               = (pv − p∞ )(R3 − R03 ),         (25)
                                             3(γ − 1)           3
or
                                        (                                                 )
     ′ 2 2(pv − p∞ ) 2σ 2R03                σ    p∞ − pv      q0       q0 R03γ−3
 ρ(R ) =             −     + 3                 +         +         −                          . (26)
              3         R      R            R0      3      3(γ − 1) 3(γ − 1)R3γ−3
When R ≪ R0 , it shows first that
                                            R′ ∼ −R−3/2 ,                                       (27)
and moreover that the rebound radius satisfies
                     ( R )3(γ−1) (γ − 1)(p − p + 3σ/R )
                         0                   ∞    v  0
                                =                       + 1.                                    (28)
                       R                       q0
Taking into account that
                                 q0 = p−
                                       ∞ − pv + 2σ/R0 ≈ 10 Pa,                                  (29)
we conclude that the rebound radius is
                                            R ≈ 0.4 µm.                                         (30)

4                                 TSOGTGEREL GANTUMUR

B4. If there is no gas other than water vapour present in a bubble, the bubble completely
collapses in finite time. Determine the characteristic exponent α in
                                      R(t) ∼ (T − t)α ,                             (31)
where T is the collapse time.
Solution. We have found in the previous part that
                                        R′ ∼ −R−3/2 .                               (32)
Putting R(t) ∼ (T − t) into it we get
                       α

                                 (T − t)α−1 ∼ −(T − t)−3α/2 ,                       (33)
or
                                              2
                                          α= .                                     (34)
                                              5
B5. Based on the equation derived in Part 3, find the natural frequency of the spherical
oscillation of a bubble of radius R0 = 0.1 mm.
Solution. Introducing the new variable x by R = R0 x, we write (19) as
                                 3              2σ
                   ρR02 xx′′ + ρR02 (x′ )2 +          − q0 x−3γ = pv − p∞ .         (35)
                                 2             R0 x
Now put x = 1 + y and retain the terms up to linear in y, to have
                                 2σ
                     ρR02 y ′′ +      (1 − y) − q0 (1 − 3γy) = pv − p∞ .            (36)
                                 R0
Taking into account the equilibrium condition
                                       2σ
                                          − q0 = pv − p∞ ,                          (37)
                                       R0
we infer
                                         3γq0 − 2σ/R0
                                  y ′′ +                y = 0.                      (38)
                                              ρR02
Thus the bubble is unstable if
                                3γq0 ≤ 2σ/R0 = q0 + pv − p∞ ,                       (39)
or equivalently, if
                               p∞ ≤ pv − (3γ − 1)q0 .                             (40)
On the other hand, if p∞ > pv − (3γ − 1)q0 then the bubble oscillates with the natural
frequency                      √
                               1     (3γ − 1)q0 + p∞ − pv
                      f0 =                                ≈ 33 kHz.                 (41)
                             2πR0              ρ
B6. Suppose that the bubble described in the previous part is subjected to a standing
sound wave along the x-axis, whose pressure field is given by
                                             ( 2πf        )
                        p(x, t) = p0 + A sin       (x + a) sin(2πf t),               (42)
                                                c
where f is the frequency, and c is the speed of sound. The parameters p0 , A, and a are
constants, whose meanings may readily be deduced from the equation. Find the average
force exerted upon the bubble. The bubble is situated at the origin of the xyz coordinate
system, and its size is much smaller than the wavelength of the sound.

                              CAVITATION: A POSSIBLE SOLUTION                               5

Solution. Small oscillation of the bubble is described by
                                                        ( 2πf )
                       ρR02 (y ′′ + 4π 2 f02 y) = A sin      a sin(2πf t).              (43)
                                                           c
Looking for the solution in the form
                                       y(t) = B sin(2πf t),                             (44)
we find
                                           A sin(2πf a/c)
                                    B=                           .                      (45)
                                          4π 2 ρR02 (f02 − f 2 )
Since
                                       R(t) = R0 + R0 y(t),                             (46)
the volume of the bubble is
                                   4π            4π 3
                             V (t) =   R(t)3 ≈      R [1 + 3y(t)].                      (47)
                                    3             3 0
Now recalling that the average force is
                                                   ∂p
                                       F = −⟨V        ⟩,                                (48)
                                                  ∂x
where the average is taken over time, and that the pressure gradient is
                           ∂p     2πf A       ( 2πf )
                               =          cos       a sin(2πf t),                       (49)
                           ∂x        c           c
we conclude
                                        f A2 R0          ( 4πf a )
                            F =−                    sin            .                    (50)
                                    2ρc(f0 − f )
                                           2     2           c
C1. Suppose that a nucleus consisting of air and vapour, with radius R0 = 10 µm, is
placed in water-air solution, in which the dissolved air is in equilibrium with the atmo-
spheric pressure above the water. The partial pressure of air in the bubble is q = 170 kPa,
and the vapour pressure can be neglected. Estimate the time required for the bubble to
be completely resorbed into water. The quantities p∞ , κ, δ and σ take their typical values
from Table 1. Assume that the region surrounding the bubble in which air diﬀusion takes
place immediately gets much larger than the bubble itself.
Solution. From Henry’s law, the initial concentration of dissolved air in the body of
water is
                                ui = Hp∞ ≈ 0.024 kg/m3 ,                              (51)
and the initial concentration of dissolved air in the immediate vicinity of the bubble is
                                    u = Hq ≈ 0.041 kg/m3 .                              (52)
Since u > ui , there will be diﬀusive flux directed away from the bubble, and the bubble
will start losing air. As a result, the bubble shrinks and the surface tension term 2σ/R
increases. Since the external pressure p∞ is constant, the partial air pressure q increases,
which leads to more diﬀusive flux. In the end, the bubble will get completely resorbed
into water.
   To quantify the diﬀusion of air, let us consider the region enclosed by concentric spheres
of radii r and r + ∆r with ∆r small. The rate of change of air mass in this region is
                                      dm              ∂u
                                           = 4πr2 ∆r ,                                   (53)
                                       dt             ∂t

6                               TSOGTGEREL GANTUMUR

       Figure 1. (a) Initial configuration of the system (b) The geometry used
       in the solution

where 4πr2 ∆r is the volume of the region and the air concentration u = u(r, t) is a
function of r and t. On the other hand, the same quantity can be computed as the
diﬀerence between the diﬀusive fluxes through the concentric spheres:
                    dm
                        = 4π(r + ∆r)2 J(r + ∆r) − 4πr2 J(r)
                     dt
                                        ∂u                    ∂u
                        = 4π(r + ∆r)2 κ (r + ∆r) − 4πr2 κ (r)                         (54)
                                        ∂r                    ∂r
                                     2
                                   ∂ u                ∂u
                        ≈ 4πr2 κ∆r 2 (r) + 8πrκ∆r (r),
                                   ∂r                 ∂r
where we have taken into account the directions of the fluxes, and the fact that
                         ∂u            ∂u         ∂ 2u
                            (r + ∆r) ≈    (r) + ∆r 2 (r).                             (55)
                         ∂r            ∂r         ∂r
Comparing the two equations, we infer
                                  ∂u       ∂ 2 u 2κ ∂u
                                       =κ 2 +           .                            (56)
                                   ∂t      ∂r      r ∂r
The dissolved air concentration in water at the initial time moment t = 0 is uniformly ui ,
and assuming that the radius R of the bubble remains constant, the air concentration in
the immediate vicinity of the bubble should be equal to Hq:
                               {
                                 u(r, 0) = ui    for r > R,
                                                                                (57)
                                 u(R, t) = Hq for t > 0.
Introducing a new variable v = r(u−qH), and a new time parameter τ = κt, the diﬀusion
equation (56) becomes
                                        ∂v    ∂ 2v
                                            = 2,                                 (58)
                                        ∂τ    ∂r
with                       {
                             v(r, 0) = r(ui − qH) for r > R,
                                                                                (59)
                             v(R, τ ) = 0          for τ > 0.

                           CAVITATION: A POSSIBLE SOLUTION                              7

We can go further by introducing
                      ξ =r−R        and      w(ξ, τ ) = v(R + ξ, τ ),               (60)
to write (56) as
                                      ∂w   ∂2w
                                         =      ,                                   (61)
                                      ∂τ   ∂ξ 2
with                   {
                        w(ξ, 0) = (R + ξ)(ui − qH) for ξ > 0,
                                                                                (62)
                        w(0, τ ) = 0               for τ > 0.
The solution to this problem is
                                 ∫
                        ui − qH ∞ ( −(ξ−η)2 /(4τ )                 )
                                                   − e−(ξ+η) /(4τ ) (η + R)dη.
                                                            2
             w(ξ, τ ) = √              e                                            (63)
                            4πτ 0
Since
                                      v(r, κt)         w(r − R, κt)
                      u(r, t) = qH +           = qH +                 ,             (64)
                                         r                   r
we have
                                    ∂u      1 ∂w    w
                                         =       − 2,                               (65)
                                    ∂r      r ∂ξ    r
and hence
                      ∂u         1 ∂w                  (1        1 )
                              =            = (ui − qH)      +√          .           (66)
                      ∂r r=R R ∂ξ ξ=0                   R       πκt
The second term corresponds to the width of the diﬀusion layer surrounding the bubble,
which we assume to be much larger than the bubble itself. Hence the rate of change of
the mass of the bubble can be estimated as
                         dm            ∂u
                              = 4πR2 κ         ≈ 4πκR(ui − qH).                     (67)
                          dt           ∂r r=R
On the other hand, the mass of the bubble is related to the air density, which in turn is
proportional to the pressure:
                                     4π 3       4π 3 δ0 q
                                m=       R δ=      R ·    ,                         (68)
                                      3          3     p∞
where δ0 = 1.2 kg/m3 is the air density at the atmospheric pressure p∞ = 105 Pa. Fur-
thermore, neglecting vapour pressure, we have the mechanical equilibrium condition
                                                 2σ
                                      q = p∞ +      ,                                (69)
                                                  R
leading to
                                        4π 3          ε
                                  m=       R · (δ0 + ),                              (70)
                                         3            R
with
                                  2σδ0
                              ε=        ≈ 1.73 · 10−6 kg/m2 .                        (71)
                                   p∞
After taking the derivative of (70) with respect to t, we equate it to (67), and get
                                       2σH              dR 8π dR
                  4πκR(ui − p∞ H −          ) = 4πδ0 R2    +    εR .                 (72)
                                         R              dt    3      dt
Since ui − p∞ H = 0, we have
                                                      2
                             −2σκHdt = δ0 R2 dR + εRdR,                              (73)
                                                      3

8                                  TSOGTGEREL GANTUMUR

and a direct integration yields
                                             δ0 R03 4εR02
                                   2σκHt =         +      ,                              (74)
                                               3      3
finally giving
                                   (δ0 R0 + 4ε)R02
                                t=                 ≈ 9 s.                                (75)
                                        6σκH
Thus the nucleus collapses in a matter of seconds.
C2. Consider a conical crevice in the wall of a water container, with an aperture angle α.
A small amount of air and vapour is trapped within the cone. Write down the condition
of mechanical and diﬀusive equilibrium. Determine when the pocket of air stays in the
crevice without disappearing.

                                  Figure 2. Conical crevice

Solution. The equilibrium conditions are
                                            2σ
                            pv + q = p∞ ±           and      u = Hq.                      (76)
                                            R
The plus sign corresponds to the situation where the water surface is concave, and the
surface tension tends to squeeze the air out of the crevice through diﬀusion. When the
water surface is convex, we have the opposing sign. Supposing that initially the partial
air pressure q is large, q will decrease as the region of trapped air shrinks due to diﬀusion.
At some point, a diﬀusive equilibrium will be found. Thus the pocket of air does not
disappear when the water surface is convex, meaning that
                                         π + α < 2θ.                                     (77)

APhO 2023 Theory T3


[User]
Consider a cube with a corner at the origin, and sides parallel to the x, y, and z axes.
    If a charge q is placed at (ϵ, ϵ, ϵ) for some tiny ϵ, what’s the flux through each face of the cube?

[2] Problem 3 (BAUPC). In both parts below, take the potential to be zero at infinity.

      (a) Consider a solid sphere of uniform charge density. Find the ratio of the electrostatic potential
          at the surface to that at the center.

     (b) Consider a solid cube of uniform charge density. Find the ratio of the electrostatic potential
         at a corner to that at the center. (Hint: use symmetry.)

       Idea 2
       If you follow an electric field line, the potential monotonically decreases along it.

Solution:
There are three “opposite” faces with the same flux, and three “adjacent” faces with
    the same flux. Now consider adding seven more cubes, so that the charge is now at the center of
    a 2 × 2 × 2 cube. The total flux through the outer faces of the cube is q/ϵ0 , and there are 24 unit
    faces, so the flux out of each “opposite” face is q/24ϵ0 . Now consider the original cube. By Gauss’s
    law the total flux out must be q/ϵ0 , which means the flux out of each “adjacent” face is 7q/24ϵ0 .
        (Note that if the charge were instead exactly at one of the corners, the fluxes through the opposite
    faces would still be q/24ϵ0 , while the fluxes though the adjacent faces would technically be undefined,
    since the electric field blows up on the face. But roughly speaking, the flux ought to be zero. Then
    the total flux out of the cube is only q/8ϵ0 , and that’s because the corner cuts out one “octant” of
    the point charge’s field. Also, note that the answer crucially depends on the fact that the charge’s
    coordinates are all equal to ϵ. For example, if the charge had been at (0, 0, ϵ) instead, then a similar
    argument shows that the fluxes through the near faces are zero, zero, and q/8ϵ0 .)
        Here’s a followup question, proposed by Mike Winer and first solved by Jason Youm. If a charge
    q is at the corner of a regular tetrahedron, what fraction of its flux goes through the tetrahedron’s
    far face? You can’t solve it with the same trick as the cube, but it’s possible to get the answer
    without any explicit integration by cleverly considering the flux through combinations of simpler
    surfaces, and using a little three-dimensional geometry. The answer is
                                                         √
                                             1 3 arctan 2
                                               −              ≈ 0.044.
                                             2        2π
    You can try deriving this for yourself, but it’s quite tricky; roughly 4 points by the standards of
    this problem set. In fact, it turns out that it’s possible to generalize these kinds of arguments even
    further, to solve the more general case where the charge is displaced from a vertex of a cube in an
    arbitrary direction! For a very deep dive, see this paper.

[2] Problem 3 (BAUPC). In both parts below, take the potential to be zero at infinity.
     (a) Consider a solid sphere of uniform charge density. Find the ratio of the electrostatic potential
         at the surface to that at the center.

     (b) Consider a solid cube of uniform charge density. Find the ratio of the electrostatic potential
         at a corner to that at the center. (Hint: use symmetry.)
    Solution. (a) Let U0 be the potential at the surface. If the sphere has radius R and charge
         density ρ, it has charge Q = 43 πρR3 and the shell theroem gives U0 = Q/4πϵ0 R.
         To go from the surface to the center, we need to further increase the potential by
                                           Z 0           Z R
                                                              kQr       1
                                  ∆U = −       E(r) dr =         3
                                                                   dr = U0
                                             R             0   R        2
         So the potential at the center of the sphere is U0 + ∆U = 3U0 /2, and the desired ratio is 2/3.

     (b) Let U0 be the potential at a corner. Now, imaging dividing the cube into 8 identical smaller
         cubes. The center of the cube is at the corner of all 8. For a fixed charge density, potentials
         scale like U ∝ Q/r ∝ ρr2 , so each of these cubes contributes 1/4 as much as the corner of the
         original cube. Thus, the potential at the center is 8(U0 /4) = 2U0 , and the desired ratio is 1/2.

       Idea 2
       If you follow an electric field line, the potential monotonically decreases along it.

Kevin Zhou Physics Olympiad Handouts — educational use, attribute to Kevin Zhou (knzhou.github.io). LICENSE FLAG: must reconfirm permissions before public release.


[User]
\begin{wrapfigure}{r}{0.48\textwidth}
\begin{circuitikz} \draw

(0,2) to[resistor=$R$] (2,2)
      to[Do,l=$T_1$, *-*] (2,0) -- (0,0)
      to[battery1,l=$U$] (0,2)
(2,2) to[switch] (4,2)
      to[Do,l=$T_2$] (4,0) -- (2,0)
;
\end{circuitikz}
\end{wrapfigure}
Thyristor’s (an element similar to a diode) V-I curve is shown in the graph. Two such thyristors are connected to a voltage source and a resistor, as in the drawing. The resistance of the resistor is $R = \SI{2}{\kilo\ohm}$.\\
a) At first the switch was opened. The voltage source’s voltage is increased linearly during a time period $t=\SI{42}{\second}$ from the value $U_0=\SI{0}{\volt}$ to the value $U_a = \SI{42}{\volt}$. Sketch the current strength of this circuit as a function of time $I(t)$. What is the current’s final value $I_a$?\\
b) Find the final current strengths in each thyristor if the switch was closed without changing the voltage $U_a$ applied on the circuit.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=1.41,
axes/.style=]

\draw[step=0.2, gray, dotted, very thin] (0,0) grid (4.2,4.2);
\draw[step=1, gray, very thin] (0,0) grid (4,4);

\begin{scope}[axes]
    \draw[->] (-0.2,0) -- (4.5,0) node[right, below] {$\frac{U}{\si{\volt}}$} coordinate(x axis);
    \draw[->] (0,-0.2) -- (0,4.5) node[above, left] {$\frac{I}{\si{\milli\ampere}}$} coordinate(y axis);

    \foreach \x/\xtext in {1/10, 2/20, 3/30, 4/40}
    \draw (\x,0) node[below] {$\xtext$};

    \foreach \y/\ytext in {1/5, 2/10, 3/15, 4/20}
    \draw (0,\y) node[left] {$\ytext$};
\end{scope}

\draw[thick] (0,0) .. controls (2,0.2) and (3.2,0.2) .. (3.2,0.7) .. controls (3.2,1.2) and (5/12,1.2) .. (0.5,1.9) -- (0.75,4);

\end{tikzpicture}

\end{center}
\end{figure}

Solution:
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=1.4,
axes/.style=]

%\draw[step=0.2, gray, dotted, very thin] (0,0) grid (4,4);
\draw[step=1, gray, ultra thin] (0,0) grid (4,4);

\begin{scope}[axes]
 \draw[->] (-0.2,0) -- (4.5,0) node[below] {$\frac{V_1}{\si{\volt}}$} coordinate(x axis);
 \draw[->] (0,-0.2) -- (0,4.5) node[left] {$\frac{I}{\si{\milli\ampere}}$} coordinate(y axis);

 \foreach \x/\xtext in {1/10, 2/20, 3/30, 4/40}
 \draw (\x,0) node[below] {$\xtext$};

 \foreach \y/\ytext in {1/5, 2/10, 3/15, 4/20}
 \draw (0,\y) node[left] {$\ytext$};
\end{scope}

\draw [thick, name path=func] (0,0) .. controls (2,0.2) and (3.2,0.2) .. (3.2,0.7) .. controls (3.2,1.2) and (5/12,1.2) .. (0.5,1.9) -- (0.75,4);

\path
let
\n0 = {2.7}
in
coordinate (E0) at (\n0,0)
coordinate (W0) at (0,\n0);

\draw [dashed, thin, red, name path=example] (E0) -- (W0);

\fill [name intersections={of=func and example, by={a,b,c}}]
[red, opacity=0.5, every node/.style={above right, black, opacity=1}] (a) circle (1pt) node {\footnotesize $A$} (b) circle (1pt) node {\footnotesize $B$} (c) circle (1pt) node {\footnotesize $C$};

\path
let
\n1 = {3.96}
in
coordinate (E1) at (\n1,0)
coordinate (W1) at (0,\n1);

\draw [name path=line 1] (E1) -- (W1);

\fill [name intersections={of=func and line 1, by={a,c,b}}]
[red, opacity=0.5, every node/.style={below left, black, opacity=1}] (a) circle (1pt) node {\footnotesize $1$} (b) circle (1pt) node {\footnotesize $2$};

\path
let
\n2 = {4.2}
in
coordinate (E2) at (\n2,0)
coordinate (W2) at (0,\n2);

\draw [name path=line 2] (E2) -- (W2);

\fill [name intersections={of=func and line 2, by={a}}]
[red, opacity=0.5, every node/.style={above right, black, opacity=1}] (a) circle (1pt) node {\footnotesize $3$};

%\draw let \n1={0.815} in (0,\n1) -- (4,\n1);
%\draw let \n1={3.296} in (0,\n1) -- (4,\n1);
%\draw let \n1={3.509} in (0,\n1) -- (4,\n1);
\end{tikzpicture}
\end{center}
\end{figure}
Kirchhoffi esimese seaduse põhjal peab suvalise pinge ja voolutugevuse korral kehtima $V_1 = U - IR$, kus $V_1$ on pinge türistoril $T_1$. Füüsikaliselt reaalsed on sellised olukorrad, kus $V_1$ ja $I$ on türistori volt-amperkarakteristiku poolt lubatud. Üldjuhul on mingi pinge $U$ korral kuni kolm lahendit (punane sirge graafikul); punkt~$B$ ei ole stabiilne, kuna vastaks negatiivsele takistusele, ning punktid~$A$ ja~$C$ on lubatud, kuid vastavad erinevatele režiimidele, mis sõltuvad sellest, kuidas punktini jõutud on. Ülaltoodu põhjal on võimalik skitseerida voolutugevuse graafik. Alguses suureneb voolutugevus ligikaudu lineaarselt, sest türistor käitub sarnaselt lineaarsele takistile takistusega $r_1=\SI{20}{\kilo\ohm}$, mistõttu on voolutugevuse graafiku algne tõus $\frac{1}{R + r_1}=\frac{1}{\SI{22}{\kilo\ohm}}$. Punkti~$1$ (pinge \SI{39.6}{\volt}, voolutugevus \SI{4.1}{\milli\ampere}) läheduses hakkab voolutugevus kiiresti kasvama ning hüppab punkti~$2$ (pinge \SI{39.6}{\volt}, voolutugevus \SI{16.5}{\milli\ampere}). Edasi käitub türistor taas lineaarselt takistusega $r_2=\SI{0.24}{k\ohm}$ kuni lõpp-punktini~$3$ (pinge \SI{42}{\volt}, voolutugevus \SI{17.5}{\milli\ampere}). Voolutugevuse $I_a$ saab täpsemalt leida, kasutades Kirchhoffi seaduseid ja türistori takistust lineaarselt lähendades. Siis saame $I_a = \SI{17.54}{\milli\ampere}$.

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=1.4,
axes/.style=]

%\draw[step=0.2, gray, dotted, very thin] (0,0) grid (5,4);
\draw[step=1, gray, ultra thin] (0,0) grid (5,4);

\begin{scope}[axes]
 \draw[->] (-0.2,0) -- (5.5,0) node[below] {$\frac{t}{\si{\second}}$} coordinate(x axis);
 \draw[->] (0,-0.2) -- (0,4.5) node[left] {$\frac{I}{\si{\milli\ampere}}$} coordinate(y axis);

 \foreach \x/\xtext in {1/10, 2/20, 3/30, 4/40, 5/50}
 \draw (\x,0) node[below] {$\xtext$};

 \foreach \y/\ytext in {1/5, 2/10, 3/15, 4/20}
 \draw (0,\y) node[left] {$\ytext$};
\end{scope}

\draw let \n1={3.9615}, \n2={0.815} in [thick] (0,0) .. controls (2.2,0.2) and ($(\n1,0.3)-(0.1,0)$) .. (\n1,\n2);
\draw let \n1={3.9615}, \n2={0.815}, \n3={3.296} in [dashed] (\n1,\n2) -- (\n1,\n3);
\draw let \n1={3.9615}, \n2={0.815}, \n3={3.296}, \n4={3.509} in [thick] (\n1,\n3) -- (4.2,\n4);

\fill let \n1={3.9615}, \n2={0.815}, \n3={3.296}, \n4={3.509} in [red, opacity=0.5, thick, every node/.style={above left, black, opacity=1}] (\n1,\n2) circle (1pt) node {\footnotesize $1$} (\n1,\n3) circle (1pt) node {\footnotesize $2$} (4.2,\n4) circle (1pt) node {\footnotesize $3$};

\end{tikzpicture}
\end{center}
\end{figure}
Pärast lüliti sulgemist lahendame vooluringi lähtuvalt Kirchhoffi seadustest. $U = IR + V_1 = IR + V_2$, $I = I_1 + I_2$, kus $I_1^0=\SI{-11.5}{mA}$ on türistori volt-amperkarakteristiku teise lineaarse osa algordinaat.
Lahendamiseks eeldame, et $T_2$ on lineaarses režiimis enne hüpet (sarnaselt punktile $A$) ning $T_1$ pärast hüpet (sarnaselt punktile $C$). Saame $I_1 = I_1^0 + \frac{V_1}{r_2}$ ja $I_2 = V_2/r_1$. Lahendades saame türistoride pingeteks $V_1 = V_2$, voolutugevuseks läbi takisti $I = I_1^0 + V_1/r_1 + V_1/r_2$, pingeks vooluallikal
\[
U = I_1^0R + V_1R/r_2 + V_1R/r_1 + V_1
\]
ning pingeteks türistoridel
\[
V_1 = V_2 = \frac{U - I_1^0R}{R/r_1 + R/r_2 + 1} = \SI{6.84}{\volt}.
\]
Seega $I_1 = \SI{17.2}{\milli\ampere}$ ja $I_2 = \SI{0.34}{\milli\ampere}$, mis õigustab tehtud eelduseid.

Türistor

\begin{wrapfigure}{r}{0.48\textwidth}
\begin{circuitikz} \draw

(0,2) to[resistor=$R$] (2,2)
 to[Do,l=$T_1$, *-*] (2,0) -- (0,0)
 to[battery1,l=$U$] (0,2)
(2,2) to[switch] (4,2)
 to[Do,l=$T_2$] (4,0) -- (2,0)
;
\end{circuitikz}
\end{wrapfigure}
Türistori (dioodisarnase elemendi) volt-amper karakteristik on juuresoleval graafikul. Kaks sellist türistori on ühendatud pingeallika ja takistiga kõrvalolevasse skeemi. Takistus $R = \SI{2}{\kilo\ohm}$.\\
{\bf a}) Alguses on lüliti avatud. Pingeallika pinget suurendatakse lineaarselt $t=\SI{42}{\second}$ jooksul väärtuselt $U_0=\SI{0}{\volt}$ kuni väärtuseni $U_a = \SI{42}{\volt}$. Skitseerige ahelat läbiva voolutugevuse $I(t)$ sõltuvus ajast. Milline on voolutugevuse lõppväärtus $I_a$?\\
{\bf b}) Leidke lõppvoolutugevused mõlemas türistoris, kui lüliti suletakse ilma ahelale rakendatud pinget $U_a$ muutmata.

\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[scale=1.41,
axes/.style=]

\draw[step=0.2, gray, dotted, very thin] (0,0) grid (4.2,4.2);
\draw[step=1, gray, very thin] (0,0) grid (4,4);

\begin{scope}[axes]
 \draw[->] (-0.2,0) -- (4.5,0) node[right, below] {$\frac{U}{\si{\volt}}$} coordinate(x axis);
 \draw[->] (0,-0.2) -- (0,4.5) node[above, left] {$\frac{I}{\si{\milli\ampere}}$} coordinate(y axis);

 \foreach \x/\xtext in {1/10, 2/20, 3/30, 4/40}
 \draw (\x,0) node[below] {$\xtext$};

 \foreach \y/\ytext in {1/5, 2/10, 3/15, 4/20}
 \draw (0,\y) node[left] {$\ytext$};
\end{scope}

\draw[thick] (0,0) .. controls (2,0.2) and (3.2,0.2) .. (3.2,0.7) .. controls (3.2,1.2) and (5/12,1.2) .. (0.5,1.9) -- (0.75,4);

\end{tikzpicture}

\end{center}
\end{figure}

Kirchhoffi esimese seaduse põhjal peab suvalise pinge ja voolutugevuse korral kehtima $V = U - IR$, kus $V$ on pinge türistoril. Kandes vastava funktsiooni volt-amperkarakteristikule on näha, et mingi pinge $U$ korral on $I$ ja $V$ jaoks kuni kolm lahendit, kusjuures negatiivse tõusuga punkt ei ole stabiilne, kuna vastaks negatiivsele takistusele. Antud informatsiooniga on võimalik määrata, kuidas türistorit läbiv vool muutub pingeallika pinge tõstmisel.


[User]
6. Sonic Sleuth
                                 CS
   During her summer vacation, Dheera decides to carry out a smartphone based experiment. She
   utilizes a smartphone’s frequency sensor that can measure the frequency of the audio signal it
   receives. She takes a long cylindrical tube closed at one end. This tube has a length of L = 30.0 cm
   and an inner diameter of d = 2.45 cm. Dheera starts ﬁlling the tube with water, which is dripping
   from a tap at a constant rate Q (measured in milliliters per second (mL/s)).

           HB
   Dheera positions her smartphone near the open end of the tube to measure the frequency of the
   sound emitted as water ﬁlls the tube. An app on the phone captures a range of frequencies in the
   recorded audio at any given time. At randomly chosen values of time t, one of the frequencies at
   that time is shown in the following table.

                                    Time t (s)     Frequency f (Hz)         Time t (s)   Frequency f (Hz)

                                       5.0                  915                36.0            434

                                       7.6                  320                39.6            481

                                       16.2                 345                41.9            500

                                       16.7                 1008               42.5           1454

                                       20.9                 360                51.1           1618

                                       25.7                 1148               51.6            574

                                       28.9                 1196               56.1           1782

                                       31.5                 410                60.2            680

                                       33.3                 1290               66.3            820

Page 8     INPhO 2024                    Questions           Last four digits of Roll No.:

  Help her to analyse the experiment.
   (a) [3 marks] Derive the expression for the velocity of sound cs in terms of f, t, and constants.
   (b) [8 marks] Choose a pair of suitable variables and plot a linear graph. Specify the axis labels.
       Obtain the speed of sound cs and the rate Q from this plot.

                      **** END OF THE QUESTION PAPER ****

                                                         E
                                      CS
                 HB

Page 9   INPhO 2024        Questions

     Space for rough work — will NOT be submitted for evaluation

                                  E
                      CS
         HB

Solution:
6. Sonic Sleuth
   During her summer vacation, Dheera decides to carry out a smartphone based experiment. She
   utilizes a smartphone’s frequency sensor that can measure the frequency of the audio signal it
   receives. She takes a long cylindrical tube closed at one end. This tube has a length of L = 30.0 cm
   and an inner diameter of d = 2.45 cm. Dheera starts ﬁlling the tube with water, which is dripping
   from a tap at a constant rate Q (measured in milliliters per second (mL/s)).
   Dheera positions her smartphone near the open end of the tube to measure the frequency of the
   sound emitted as water ﬁlls the tube. An app on the phone captures a range of frequencies in the
   recorded audio at any given time. At randomly chosen values of time t, one of the frequencies at
   that time is shown in the following table.

                                    Time t (s)     Frequency f (Hz)        Time t (s)   Frequency f (Hz)

                                       5.0                 915                36.0               434

                                       7.6                 320                39.6               481

                                       16.2                345                41.9               500

                                       16.7

                                       20.9                  E
                                                          1008

                                                           360
                                                                              42.5

                                                                              51.1
                                                                                1454

                                                                                1618

                                       CS
                                       25.7

                                       28.9

                                       31.5

                                       33.3
                                                          1148

                                                          1196

                                                           410

                                                          1290
                                                                              51.6

                                                                              56.1

                                                                              60.2

                                                                              66.3
                                                                                1782
                                                                                574

                                                                                680

                                                                                820

                 HB
   Help her to analyse the experiment.
   (a) [3 marks] Derive the expression for the velocity of sound cs in terms of f, t, and constants.

          Solution: The frequency of the sound in the tube is determined by the formula:

                                                          ncs
                                                 f=                                                    (6.1)
                                                      4(h + 0.3d)

          Here, h and d represent the length of the air column and the diameter of the tube,
          respectively. The variable n is an odd integer representing the fundamental, third, ﬁfth
          harmonics, and so on. The speed of sound is denoted by cs , and the term 0.3d in the
          denominator accounts for the end correction in the tube.
          When water falls at a constant rate (Q), it creates the disturbances in the air column of
          the tube. These disturbances travel as sound waves through the air column, which we
          detect and analyze. If the length of the tube is L, the equation (6.1) can be modiﬁed as:

                                                 f 4(L − ( Q
                                                           A )t) + 0.3d)
                                         cs =                                                          (6.2)
                                                            n

          Here, A = πd2 /4 represents the area of the base, and n is an odd integer representing the
          fundamental, third, ﬁfth harmonics, and so on.

Page 21         INPhO 2024               Questions

   (b) [8 marks] Choose a pair of suitable variables and plot a linear graph. Specify the axis labels.
       Obtain the speed of sound cs and the rate Q from this plot.

          Solution: Since the height of the water level varies linearly with time, we expect the
          frequency of the sound to increase with time. By linearizing the equation (6.1), we get:

                                                        ncs

                                        1
                                        f
                                         f=

                                          =−
                                                      E Q
                                              4(L − ( A )t) + 0.3d)

                                             ncs
                                                4( Q
                                                 + A )t   4(L + 0.3d)
                                                             ncs
                                                                                (6.3)

                                                                                (6.4)

                                 CS
          Plotting the relationship between 1/f and t will yield a linear graph. The values for the
          plot are as follows.

           Fundamental

           Time(s)   Frequency(Hz)     1/f (s)
                                                  3rd Harmonic

                                                  Time(s)    Frequency(Hz)    1/f (s)

           HB
           7.6

           16.2

           20.9

           31.5
                     320

                     345

                     360

                     410
                                       0.00313

                                       0.00290

                                       0.00278

                                       0.00244
                                                  5

                                                  16.7

                                                  25.7

                                                  28.9
                                                             915

                                                             1008

                                                             1148

                                                             1196
                                                                              0.00109

                                                                              0.00099

                                                                              0.00087

                                                                              0.00084

           36        434               0.00230    33.3       1290             0.00078

           39.6      481               0.00208    42.5       1454             0.00069

           41.9      500               0.00200    51.1       1618             0.00062

           51.6      574               0.00174    56.1       1782             0.00056

           60.2      680               0.00147

           66.3      820               0.00122
          The presence of two distinct straight lines in the graph indicates the existence of two
          harmonic frequencies in the dataset. Upon examining the ratio of these frequencies, it
          becomes evident that these two lines correspond to n = 1 and n = 3. The intercept and
          slope of the graph can be used to determine the speed of sound and the rate of water
          ﬁlling, respectively.

Page 22       INPhO 2024               Questions    Last four digits of Roll No.:

                      Fundamental

                      −2.84 × 10−5
                                     3rd Harmonic

                                     −1.12 × 10−5
                                                    E
          Slope

          Intercept

          Cs

          Q
                      3.42 × 10−3

                      359 m/s

                      1.20 mL/s
                                     CS
                                     1.12 × 10−3

                                     366 m/s

                                     1.24 mL/s

                  HB  **** END OF THE QUESTION PAPER ****

Page 23   INPhO 2024       Questions

     Space for rough work — will NOT be submitted for evaluation

                                  E
                       CS
          HB

INPhO 2024 Problem 6


[User]
5. PLANCK’S CONSTANT (8 points)

In a
simplistic model, light emitting diodes can be
considered to only pass current when lit, and
then they have a constant voltage drop Vt =
E
e across them. E = hf is the energy of the
light quanta emitted and e = 1.60×10−19
C
is the elementary charge. Speed of light in
vacuum c =3.00×108
m/s.
You have a assorted light emitting diodes
numbered 1–6, each with a R =680Ω series
resistor. From the datasheets it is known
for the peak wavelengths of the diodes to
be 940nm, 620nm, 590nm, 525nm, 470nm,
and 450nm.
i) (2 points) Find out which wavelength cor-
responds to which diode.
ii) (4 points) Measure the Planck’s con-
stant h that corresponds to our simplistic
model. This does not have to correspond to
real Planck’s constant.
iii) (2 points) Estimate the uncertainty.
Equipment: voltage source (batteries)
with an unknown voltage, ammeter, assor-
ted light emitting diodes with series resistor.
Take care not to short the battery with the
ammeter. You may ignore the internal resist-
ances of the batteries and the ammeter.
Estonian-Finnish Olympiad 2014

Solution:
5. Planck's constant

1) When we connect each of the diodes to the
battery, we can observe the light of the emit-
ted light; the mapping is as follows: 940nm
invisible (infrared), 620nm  red, 590nm  or-
ange, 525nm  green; 470nm  blue; 450nm
 violet.
2) We can measure the current I through the
diode (which is also the current through the res-
istor R), so that the voltage on the diode would
be Vd = E − IR, but we don't know the battery
voltage. However, we do know that the diode's
voltage equals approximately to the photon's
energy Ep expressed in electron volts, Ep =
hc/(λe). Since we expect that IR = E − Vd,
if we plot IR versus 1/λ, we should obtain a
straight line
IR = E −
1
λ
hc
e
.
We can measure the slope of the straight line
A = hc/e, which allows us to calculate h =
eA/c.
3) The major source of the uncertainty is not
the instrument uncertainties, but the departure
of the real diode data from the simplistic model.
Therefore we can try to fit the data points with
dierent straight lines making the slope A as
steep as possible (while still keeping a reason-
able fit with the data points, and also as flat
as possible; the uncertainty of A is found as
∆A = 1
2
(Amax − Amin ), and ∆h = h∆A/A.

PLANCK’S CONSTANT


[User]
Theoretical Question 2: Creaking Door

The phenomenon of creaking is very common, and can be found in doors, closets, chalk squeaking on a blackboard,
playing a violin, new shoes, car brakes and other systems from everyday life. Here in Israel, a similar phenomenon
causes violent earthquakes with a period of several decades. These originate in the Dead Sea rift, located right above
the deepest known break in the earth's crust.

The physical mechanism for creaking is based on elasticity combined with the difference between the static and the
kinetic friction coefficients. In this question, we will study this mechanism and its application to the case of an
opening door.

Part I: General model (7.5 points)
Consider the following system (see Figure 1):

A box with mass is attached to a long ideal spring with spring constant , whose other end is driven at a constant
velocity . The static and the kinetic friction coefficients between the box and the floor are given respectively by
and , where            .

                            Figure 1: A general model for creaking

We would like to understand why this setup supports two different forms of motion:

         1. The friction is always kinetic. This is known as a pure slip mode.
         2. Kinetic and static friction alternate. This is known as a stick-slip mode. Stick-slip motion is the source of
            the creaking sound commonly encountered.

a.    (1 pt.) Consider the case where at the initial time       , the box slides on the floor with velocity , and the
     spring’s tension exactly balances the kinetic friction. Assume                . The spring’s elongation will
     oscillate as a function of .
          a1. (0.6 points) Find the period and the amplitude of these oscillations.
          a2. (0.4 points) Sketch a qualitative graph of the spring’s elongation        for             .

b. (1.2 pts.) Now, consider the case where at          the box is at rest, while the initial spring elongation is the same
   as in part (a). Sketch a qualitative graph of the velocity      of the box with respect to the floor for              ,
   where is the (new) period of the oscillations         . Motion to the right corresponds to a positive sign of .
   Indicate approximately on your graph the horizontal line            .

c. (0.5 pts.) For the initial conditions of part (b), find the time-averaged value ̅ of the spring’s elongation after a
   sufficiently long time has passed.

                                                        Page 1 of 3

d. (2.4 pts.) For the conditions of part (b), find the period   of the oscillations     .

Generically, stick-slip motion stops at high driving velocities . We will now discuss one of the possible mechanisms
behind this effect.

e. (2.4 pts.) Suppose that during each period , a small amount of energy is dissipated into heat in the spring, via an
   additional mechanism. Let                  be the fractional amplitude loss per period due to dissipation in pure-slip
   motion. For        , find the critical driving velocity    above which periodic stick-slip becomes impossible.

The results of part (e) are not required for part II.

Part II: Application to creaking door (2.5 points)
A door hinge is a hollow, open-ended metal cylinder with radius , height and thickness . The lower end of the
cylinder lies on a metal base attached to the wall (the area of contact is a ring of radius ); see Figure 2. The static and
the kinetic friction coefficients between the cylinder and its base are and        respectively, with         . The upper
end of the cylinder is attached to the door, which can be regarded as perfectly rigid. A typical door hangs on two or
three such hinges, but its weight is concentrated on only one of them – this is the hinge that will creak. The cylinder of
that hinge presses down on its metal base with the weight of the entire door, whose mass is .

                                        Wall
             Door

  Figure 2: Schematic drawing of a door                    Figure 3: The twisted hinge cylinder
  hinge
The cylinder is not a perfectly rigid body – it can twist tangentially without changing its overall form, so that vertical
line segments become tilted with a small angle ; see Figure 3. The elastic force on a small area element         of the
base due to this deformation is given by:

                                                                    ,

where     is a material property known as the shear modulus. Use the values          ,        ,                     ,
            ,           ,               ,           ,             . You may use the approximation               .

f.   (1 pt.) We start rotating the door very slowly from equilibrium (zero torque). For small rotation angles, obtain an
     expression for the torsion coefficient          , where is the torque which must be applied to rotate the door by
     an angle .

g. (1.5 pts.) At very low angular velocity, when a transition from stick to slip occurs, a sound pulse is emitted. Find
   the angular velocity of the door for which the frequency of these pulses enters the audible range at               .

                                                        Page 2 of 3

Assume that the frequency   of pure-slip oscillations in the hinge is much higher:   . Provide an expression
and a numerical result.

                                               Page 3 of 3

Solution:
Theoretical Question 2: Creaking Door
                                                    SOLUTION

a1. The motion here is pure sliding under a constant kinetic friction. This is harmonic motion with a displaced
equilibrium point. The angular frequency is given by:

                                                            √

From here, the period is:

                                                                 √

The initial slope is given by:

                                                   (    )

Therefore, the amplitude of oscillations is:

                                                                       √

a2. The graph is sinusoidal, as shown below.

                                                       Page 1 of 5

The initial point is given by:

This is the equilibrium point of the sine function. The students are not required to find this equilibrium point, but they
are required to understand that it is positive.

b. This will be a stick-slip graph. The ”humps” are sinusoidal, with a non-continuous derivative at their intersections
with the horizontal segments. The peaks of the humps are higher than           , since must be the average velocity of
the box. In fact, they are also higher than      , but this is not required from the students.

c. Let’s pass into the reference frame of the driven end of the spring. The position of the box is then given by minus
the elongation . The motion is an oscillation around the equilibrium position . The slip phase is sinusoidal as in
part (a), while the stick phase corresponds to motion with a constant velocity – . The stick phase ends when the
elastic force balances the static friction, i.e. at            , and starts again at the symmetric point with respect to
   .

                                                       Page 2 of 5

We see that the average elongation is again the sine’s equilibrium point:

                                                              ̅

d. Again, let us pass into the reference frame of the driven end of the spring. During the stick phase, the box traverses
a distance of:

Its velocity during this phase is , so the duration of the stick phase is:

The slip phase is a sinusoidal motion around with angular frequency . The sinusoidal period is missing a phase
of , where is given by the ratio of initial position and initial velocity with respect to the equilibrium point:

                                                                                √

Then the length of the slip phase is:

                                         (            )           √   (           (           √ ))

And the total period is:

                                              √       (                   √               (          √ ))

e. Consider again stick-slip motion in the reference frame of the driven end of the spring. During the sinusoidal slip
phase, the sine’s amplitude will decrease due to the dissipation. At the beginning of the slip phase, the velocity is – ,
while the sine is at the phase , which we found in the solution to the previous part. Thus, the sine’s velocity
amplitude is           . For periodic stock-slip to occur, the sine must return to the slope – . Due to the dissipation,
this will happen at a phase larger than          . In other words, dissipation shortens the stick phase. The critical case is
when stick phase shortens to zero. This will happen if the sine reaches the slope – precisely at the equilibrium point,
i.e. at the phase . If the slope at     is less steep than    , the box will continue its damped sinusoidal motion
without ever reaching a stick phase again.

If it is to be killed by weak dissipation, the stick phase must be short to begin with. This corresponds to a large . The
slip phase then takes up almost an entire period of the sine wave. Thus, to a good approximation, the amplitude loss
during the slip phase is given by . The critical point is when the velocity amplitude drops from               to during
one period:

                                          |       |       |                   |

                                                                  Page 3 of 5

where the LHS is the change in the amplitude due to dissipation over one period. Using the results from (d) in the
limit of small , we get:

                                                                     √

Another derivation method based on the same reasoning is to use explicitly the initial amplitude       of the harmonic
motion:

                                                          ,

A third method is to consider energy losses                   in the reference frame of the spring’s driven end:

f. For small rotations, the lower edge of the cylinder will remain stuck to the base. When the cylinder is deformed by
an angle , a point on its upper edge shifts by a distance . This corresponds to a rotation angle               of the
door around the cylinder’s axis. The shear force on an area element       of the base is given by:

The corresponding torque is:

Summing over the contact area with the base, the total torque is:

Therefore, the torsion coefficient is:

The numerical result is not required from the student. Any expression which reduces to the one above in the limit
       will be accepted.

g. We neglect the duration of the slip phase. Using the results of section (d) with     instead of   and rotation instead
of linear motion, we get:

                                                       Page 4 of 5

Any expression which reduces to the one above in the limit        will be accepted. Numerical results from such
different expressions may vary significantly, since          is not really negligible. Each numerical result should
be checked against its expression.

                                                    Page 5 of 5

APhO 2011 Theory T2


[User]
An electron with kinetic energy 1 MeV travels along the z-axis and collides with a positron at rest.
The particles annihilate, producing a pair of photons with equal energies. The rest mass of an
electron is me = 511 keV/c2 .
  1. Explain why it is not possible for the collision process to produce only one photon.
  2. Numerically compute the speed of the electron.
  3. Find the angle between the momentum of the first photon and the z-axis.

Solution:
An electron with kinetic energy 1 MeV travels along the z-axis and collides with a positron at rest.
The particles annihilate, producing a pair of photons with equal energies. The rest mass of an
electron is me = 511 keV/c2 .

  1. Explain why it is not possible for the collision process to produce only one photon.

  2. Numerically compute the speed of the electron.

  3. Find the angle between the momentum of the first photon and the z-axis.

Solution. This is a modification of NBPhO 2015, problem 1, whose official solution is here. The
answers are:

  1. It is forbidden by energy and momentum conservation, as can be argued in several ways. For
     example, in the center of mass frame, there is nonzero initial energy and zero initial momentum.
     This couldn’t be true for a single photon, as E = pc.

  2. Using K = (γ − 1)me c2 , we have γ = 2.957. Solving for the velocity, we have v = 0.941 c =
     2.82 × 108 m/s.

  3. This is somewhat “troll” wording, also present in the original problem statement. How do we
     know which photon is the “first” one? It doesn’t matter, because both photons have to come
     out at the same angle to the z-axis, by momentum conservation. We hve
                                             px      γme v
                                   cos θ =      =              = 0.703
                                             E    (γ + 1)me c2

     which implies that θ = 0.791 rad = 45.3◦ .


[User]
\begin{wrapfigure}[10]{r}{0.5\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{2018-v3g-05-kullastunud-aur}
\par\end{center}
\end{wrapfigure}
The insulation of a wall consists of an inner (of thermal conductivity $k_{1}=\SI{0.07}{W/\left(m\cdot K\right)}$) and an outer layer (that has a thermal conductivity $k_{2}=\SI{0.05}{W/\left(m\cdot K\right)}$). Between those layers there is a film layer to hinder the air’s movement through the wall. What condition does the width $L_{1},$ of the inner thermal insulation layer have to satisfy to avoid the condensation of water vapor in the wall? The width of the wall $L=L_{1}+L_{2}=\SI{30}{cm}$, $L_{2}$ is the width of the outer thermal insulation layer, the room temperature is $T_{1}=\SI{20}{\degreeCelsius}$, the relative humidity in the room is $\eta_{1}=\SI{60}{\percent}$ and the outside temperature $T_{2}=\SI{-20}{\degreeCelsius}$. The relation between the saturated vapor pressure of water and the temperature is shown in the figure.\\
\emph{Note.} Assume that the temperature in the insulation layer changes linearly with distance and the speed of the change is inversely proportional to thermal conductivity.

Solution:
Toas oleva niiske õhu levikut piirab soojustuskihtide vahel olev kile.
Selleks, et vältida kondenseerumist, ei tohi kile asukohas temperatuur
langeda alla kastepunkti.

Kastepunkti saame leida graafiku alusel, leides alguses küllastunud
veeauru osarõhu toatemperatuuril, arvutades sellest $\eta_{1}=\SI{60}{\percent}$
ja seejärel leides sellele vastava kastepunkti $T_{k}=\SI{12.0}{\degreeCelsius}.$
\begin{center}
\includegraphics[scale=0.9]{2018-v3g-05-kullastunud-aur-lah}
\par\end{center}

Järgmiseks on vaja leida avaldis temperatuuri jaoks kile asukohas
($T$). Eeldades, et temperatuur muutub soojustuskihis lineaarselt
kaugusega ja muutuse kiirus on pöördvõrdeline soojusjuhtivusega, saame
liikudes seest välja kirja panna kaks võrrandit:

\begin{equation*}
\begin{cases}
\begin{array}{c}
T=T_{1}-\frac{\alpha}{k_{1}}L_{1}\\
T_{2}=T-\frac{\alpha}{k_{2}}L_{2},
\end{array}
\end{cases}
\end{equation*}

\noindent kus $\alpha$ on tundmatu võrdetegur. Nende võrrandite lahendamine
annab

\begin{equation*}
T=\frac{k_{2}T_{2}L_{1}+k_{1}T_{1}L_{2}}{k_{2}L_{1}+k_{1}L_{2}}.
\end{equation*}

Ja lõpetuseks, tuleb leida sisemise soojustuskihi paksus $L_{1}$
piirjuhul, kui kile temperatuur võrdub kastepunktiga

\begin{equation*}
L_{1}=L\frac{k_{1}\left(T_{1}-T_{k}\right)}{k_{1}T_{1}-k_{2}T_{2}-\left(k_{1}-k_{2}\right)T_{k}}\approx\SI{7.8}{cm}
\end{equation*}

\noindent ja kondenseerumise vältimiseks peab sisemise soojustuskihi
paksus olema sellest väiksem.

Soojustus

\begin{wrapfigure}[10]{r}{0.5\textwidth}
\vspace{-30pt}
\begin{center}
\includegraphics[width=0.5\textwidth]{2018-v3g-05-kullastunud-aur}
\par\end{center}
\end{wrapfigure}

Seina soojustus koosneb sisemisest (soojusjuhtivus $k_{1}=\SI{0.07}{W/\left(m\cdot K\right)}$)
ja välimisest kihist (soojusjuhtivus $k_{2}=\SI{0.05}{W/\left(m\cdot K\right)}$).
Nende kihtide vahel on kile, et takistada õhu liikumist läbi seina.
Millist tingimust peab rahuldama sisemise soojustuskihi pakus $L_{1},$
et vältida veeauru kondenseerumist seinas? Seina paksus $L=L_{1}+L_{2}=\SI{30}{cm},$
$L_{2}$ on välimise soojustuskihi paksus, toa temperatuur $T_{1}=\SI{20}{\degreeCelsius},$
suhteline õhuniiskus toas $\eta_{1}=\SI{60}{\percent}$ ja välistemperatuur
$T_{2}=\SI{-20}{\degreeCelsius}.$ Küllastunud veeauru osarõhu sõltuvus
temperatuurist on toodud joonisel.

\emph{Märkus.} Eeldada, et temperatuur muutub soojustuskihis lineaarselt
kaugusega ja muutuse kiirus on pöördvõrdeline soojusjuhtivusega.

Selleks, soojustuskihtide vahel oleval kilel vältida kondenseerumist, ei tohi kile asukohas temperatuur
langeda alla kastepunkti. Kastepunkt on leitav graafikult, leides veeauru osarõhu toatemperatuuril ning seejärel leides sellele vastava kastepunkti.


[User]
Lugegem Maa ja Veenuse orbiidid ümber Päikese ringikujulisteks. Planeedid tiirlevad ümber Päikese samas suunas ja Veenuse maksimaalne eemaldumus (nurk Veenuse ja Päikese vahel Maalt vaadates) on 46 kraadi.\\
\osa Leidke Veenuse ja Maa orbiitide raadiuste suhe.\\
\osa Mitu päeva jääb järjestikuste maksimaalsete eemaldumuste vahele?

\emph{Vihje}: Kepleri seaduse kohaselt on taevakehade tiirlemisperioodide ruudud võrdelised vastavate orbiitide raadiuste kuupidega.

Solution:
\osa Maksimaalse eemaldumise korral moodustub Maast, Veenusest ja Päikesest täisnurkne kolmnurk, mille täisnurga tipp on Veenus. Siit saame Veenuse ja Maa orbitaalraadiuste suhte
\[
\alpha = \sin \ang{46} = \num{0,72}.
\]
\osa Veenuse tiirlemisperioodi saame Kepleri seadusest
\[
T_V = T_M \sqrt{\alpha^3}.
\]
Maa tiirlemise nurkkiirus $\omega_M = \frac{2\pi}{T_M}$ ning Veenuse tiirlemise nurkkiirus $\omega_V = \frac{2\pi}{T_V}$. Nende suhtelise liikumise nurkkiirus
\[
\Delta \omega=\omega_{V}-\omega_{M}=\omega_{M}\left(\frac{1}{\sqrt{\alpha^{3}}}-1\right)
\]
ning suhtelise liikumise periood on
\[
T_{s}=\frac{T_{M}}{\frac{1}{\sqrt{\alpha^{3}}}-1}=570 \text { päeva. }
\]
Järjestikuste eemaldumiste vahele jääb Päikeselt vaadatuna nurk $2\cdot (\ang{90} - \ang{46}) = \ang{88}$ või \ang{360} - \ang{88} = \ang{272} ehk päevades
\[
T_s \frac{88}{360} = 140\text{ päeva.}
\]
ja
\[
T_s \frac{272}{360} = 430\text{ päeva.}
\]

Veenus

Lugegem Maa ja Veenuse orbiidid ümber Päikese ringikujulisteks. Planeedid tiirlevad ümber Päikese samas suunas ja Veenuse maksimaalne eemaldumus (nurk Veenuse ja Päikese vahel Maalt vaadates) on 46 kraadi.\\
\osa Leidke Veenuse ja Maa orbiitide raadiuste suhe.\\
\osa Mitu päeva jääb järjestikuste maksimaalsete eemaldumuste vahele?

\emph{Vihje}: Kepleri seaduse kohaselt on taevakehade tiirlemisperioodide ruudud võrdelised vastavate orbiitide raadiuste kuupidega.

\osa Maksimaalse eemaldumise korral moodustub Maast, Veenusest ja Päikesest täisnurkne kolmnurk, mille täisnurga tipp on Veenus.\\
\osa Maa ja Veenuse suhtelise nurga (Päikeselt vaadatuna) muutus on avaldatav planeetide nurkkiiruste vahe kaudu.


[User] A cyclist is riding down a slope with an even inclination. If he is holding the brakes with exactly such a strength to cause the rear wheel to almost rise into air, then his speed while riding down the slope does not change. The center of mass of the system that consists of the cyclist and the bike is located exactly in the middle of two wheels, at a height $h$ from the ground. The distance between the axes of the wheels is $d$. How big is the angle $\alpha$ between the slope and the horizontal direction? How big has to be the coefficient of friction $\mu$ between the wheel and the ground so that the cyclist could brake like it was described?

Solution:
Kuna tagumine ratas on õhku tõusmas, siis sellele jõude ei rakendu. Ainsad jalgrattale mõjuvad jõud on raskusjõud ning jõud esiratta ja maapinna kontaktpunktis. Kuna jalgratas liigub ühtlase kiirusega ning ei hakka pöörlema ega ümber kukkuma, peab jalgrattast ja ratturist koosnevale süsteemile mõjuvate jõudude summa ja ka jõumoment iga punkti suhtes olema null. Raskusjõu tugevus ja suund on teada. Teine jõud peab olema sama suur ja vastassuunaline, et raskusjõudu tasakaalustada. Lisaks peavad jõud paiknema ühel sirgel, et jalgratas koos ratturiga ei hakkaks pöörlema. Järelikult peab massikese olema esiratta ja maapinna kontaktpunkti kohal ning hõõrdejõu ja toereaktsiooni summa on vertikaalne. Lihtsast geomeetriast saame $\alpha=\arctan(\frac{d}{2h})$ ja $\mu\ge\frac{d}{2h}$.
\begin{center}
\includegraphics[width=0.6\textwidth]{2014-lahg-08-ratas}
\end{center}

Jalgrattur

Jalgrattur sõidab alla ühtlase kallakuga nõlvast. Kui ta vajutab pidureid täpselt nii kõvasti, et tagumine ratas on peaaegu õhku tõusmas, siis tema kiirus mäest alla sõites ei muutu. Jalgratturist ja rattast koosneva süsteemi massikese asub täpselt kahe ratta vahel kaugusel $h$ maapinnast, rataste telgede vahekaugus on $d$. Kui suur on nõlva ja horisontaalsihi vaheline nurk $\alpha$? Kui suur peab olema ratta ja kaldpinna vaheline hõõrdetegur $\mu$, et jalgrattur saaks kirjeldatud moel pidurdada?

Kuna tagumine ratas on õhku tõusmas, siis sellele jõude ei rakendu. Ainsad jalgrattale mõjuvad jõud on raskusjõud ning jõud esiratta ja maapinna kontaktpunktis. Lisaks on teada, et peab kehtima nii jõudude kui jõumomentide tasakaal iga punkti suhtes.