[User]
36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

                                            Th 1       AN ILL FATED SATELLITE

       The most frequent orbital manoeuvres performed by spacecraft
consist of velocity variations along the direction of flight, namely
accelerations to reach higher orbits or brakings done to initiate re-entering in
the atmosphere. In this problem we will study the orbital variations when the
engine thrust is applied in a radial direction.

       To obtain numerical values use: Earth radius RT = 6.37 ⋅10 6 m ,
Earth surface gravity g = 9.81 m/s 2 , and take the length of the sidereal day
to be T0 = 24.0 h .

       We consider a geosynchronous1 communications satellite of mass m
placed in an equatorial circular orbit of radius r0 . These satellites have an
“apogee engine” which provides the tangential thrusts needed to reach the                                                  Image: ESA
final orbit.

          Marks are indicated at the beginning of each subquestion, in parenthesis.

Question 1

1.1       (0.3) Compute the numerical value of r0 .

1.2       (0.3+0.1) Give the analytical expression of the velocity v 0 of the satellite as a function of g, RT , and r0 , and
          calculate its numerical value.

1.3       (0.4+0.4) Obtain the expressions of its angular momentum L0 and mechanical energy E 0 , as functions of v 0 , m, g
          and RT .

                                                                                v0            m
       Once this geosynchronous circular orbit has been reached (see Figure F-1), the satellite
                                                                                ∆v
has been stabilised in the desired location, and is being readied to do its work, an error by the
                                                                                r0
ground controllers causes the apogee engine to be fired again. The thrust happens to be
directed towards the Earth and, despite the quick reaction of the ground crew to shut the
engine off, an unwanted velocity variation ∆ v is imparted on the satellite. We characterize
this boost by the parameter β = ∆ v / v 0 . The duration of the engine burn is always negligible
with respect to any other orbital times, so that it can be considered as instantaneous.
                                                                                F-1
Question 2

          Suppose β < 1 .

2.1       (0.4+0.5) Determine the parameters of the new orbit2, semi-latus-rectum l and eccentricity ε , in terms of r0 and β.

2.2       (1.0) Calculate the angle α between the major axis of the new orbit and the position vector at the accidental misfire.

2.3       (1.0+0.2) Give the analytical expressions of the perigee rmin and apogee rmax distances to the Earth centre, as
          functions of r0 and β , and calculate their numerical values for β = 1 / 4 .

2.4       (0.5+0.2) Determine the period of the new orbit, T, as a function of T0 and β, and calculate its numerical value for
           β = 1/ 4 .
1
    Its revolution period is T0 .
2
    See the “hint”.
                                                                                Th 1 Page 1 of 3

                                           36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

Question 3

3.1      (0.5) Calculate the minimum boost parameter, β esc , needed for the satellite to escape Earth gravity.

3.2                                                                                ′ , as a
         (1.0) Determine in this case the closest approach of the satellite to the Earth centre in the new trajectory, rmin
         function of r0 .

                                                                                v0
Question 4
                                                                                ∆v
         Suppose β > β esc .
                                                                                φ
                                                                                r0
4.1      (1.0) Determine the residual velocity at the infinity, v ∞ , as a function of v 0
         and β.

4.2      (1.0) Obtain the “impact parameter” b of the asymptotic escape direction in
         terms of r0 and β. (See Figure F-2).
4.3      (1.0+0.2) Determine the angle φ of the asymptotic escape direction in terms of                                               b
                                                          3                                                     F-2                   v∞
         β. Calculate its numerical value for β =           β esc .
                                                          2

                                                               HINT

                                                                                m
        Under the action of central forces obeying the inverse-square law, bodies follow
trajectories described by ellipses, parabolas or hyperbolas. In the approximation m << M                                r
the gravitating mass M is at one of the focuses. Taking the origin at this focus, the general
polar equation of these curves can be written as (see Figure F-3)                                                   θ

                                      l
                      r (θ ) =                                                                            M
                                 1 − ε cos θ

where l is a positive constant named the semi-latus-rectum and ε is the eccentricity of the
curve. In terms of constants of motion:                                                                                F-3
                                                                            1/ 2
                                 L2                       ⎛    2 E L2 ⎞
                      l=                       and    ε = ⎜1 + 2 2 3 ⎟
                           GM m2                          ⎜ G M m ⎟
                                                          ⎝           ⎠

where G is the Newton constant, L is the modulus of the angular momentum of the orbiting mass, with respect to the origin, and E is its
mechanical energy, with zero potential energy at infinity.

         We may have the following cases:

i)     If 0 ≤ ε < 1 , the curve is an ellipse (circumference for ε = 0 ).

ii)    If ε = 1 , the curve is a parabola.

iii)   If ε > 1 , the curve is a hyperbola.

                                                                                Th 1 Page 2 of 3

                          36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

           COUNTRY CODE          STUDENT CODE             PAGE NUMBER           TOTAL No OF PAGES

                                      Th 1       ANSWER SHEET

Question   Basic formulas and               Analytical results              Numerical results              Marking
               ideas used                                                                                guideline

   1.1                                                                     r0 =                            0.3

   1.2                              v0 =                                   v0 =
                                                                                0.4

                                     L0 =                                                                  0.4
   1.3

                                     E0 =                                                                  0.4

                                    l=                                                                     0.4
   2.1                                                                                0.5
                                    ε=

   2.2                                                                     α=                              1.0

                                    rmax =                                 rmax =
   2.3                                                                                1.2
                                    rmin =                                 rmin =

   2.4                              T=                                     T=                              0.7

   3.1                                                                      β esc =                        0.5

   3.2                              r ′min =                                                               1.0

   4.1                              v∞ =                                                                   1.0

   4.2                              b=                                                                     1.0

   4.3                              φ=                                     φ=                              1.2

                                                                                Th 1 Page 3 of 3

Solution:
36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

                                                           Th1        AN ILL FATED SATELLITE
                                                                             SOLUTION

1.1 and 1.2
                      MT m            v 02       ⎫
                  G             =m               ⎪   ⎧                 1/ 3
                       r02            r0         ⎪          ⎛   2 2 ⎞
                                                     ⎪ r = ⎜ g RT T0 ⎟      ⇒ r0 = 4.22 ⋅ 10 7 m
                         2π r0                   ⎪   ⎪  0   ⎜        ⎟
                                                 ⎪   ⎪      ⎝ 4π
                                                                   2
                  v0 =                                               ⎠
                                                 ⎬ ⇒ ⎨
                          T0                     ⎪   ⎪           g
                       GM T                      ⎪   ⎪ v 0 = RT             ⇒ v 0 = 3.07 ⋅ 10 3 m/s
                  g=                             ⎪   ⎪
                                                     ⎩           r0
                          RT2                    ⎪⎭

1.3

                                             g RT2                               m g RT2
                  L0 = r0 m v 0 =                     m v0    ⇒        L0 =
                                              v0
                                                 2                                 v0

                                                                             2
                          1       M m 1      g RT m 1                                                            1
                  E0 =
                              2          2
                            mv − G T = mv0 −
                                                       2      2
                                                   = mv0 − mv 0                                    ⇒    E0 = −     mv 02
                          2 0      r0 2        r0   2                                                            2

2.1
          The value of the semi-latus-rectum l is obtained taking into account that the orbital angular momentum is the same
in both orbits. That is

                                2
                             L0               m 2 g 2 RT4         1          g RT2
                  l=                    =                                =              = r0   ⇒   l = r0
                       G MT m2                       v02     g RT2 m 2           v 02

          The eccentricity value is

                                             2
                                    2 E L0
                 ε 2 = 1+
                                G 2 M T2 m 3

where E is the new satellite mechanical energy

                  E=
                         1
                         2
                               2
                                (     2   M m 1
                                                 ) 2       1    2 1   2
                           m v 0 + ∆ v − G T = m∆ v + E 0 = m∆ v − mv 0
                                           r0 2            2      2

that is

                                 ⎛ ∆ v2 ⎞ 1
                  E=
                         1
                         2
                           mv 02 ⎜ 2 − 1⎟ = mv 02 β 2 − 1
                                 ⎜ v    ⎟ 2
                                                                 (       )
                                 ⎝ 0    ⎠

          Combining both, one gets                    ε=β

          This is an elliptical trajectory because ε = β < 1 .

                                                                                Th 1 Solution Page 1 of 5

                                               36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

2.2
          The initial and final orbits cross at P, where the satellite engine fired instantaneously (see Figure 4). At this point

                                                r0                           π
                  r (θ = α ) = r0 =                           ⇒        α=
                                           1 − β cos α                       2

2.3                                                                                r
                                                                                v0           P
          From the trajectory expression one immediately obtains that
                                                                                r
the maximum and minimum values of r correspond to θ = 0 and                                                    ∆v
                                                                                r0            π
θ = π respectively (see Figure 4). Hence, they are given by                                                        α=
                                                                                2
                          l                           l                                            rmin                            rmax
                  rmax =                      rmin =
                         1− ε                        1+ ε

that is
                              r0                               r0
                  rmax =                  and        rmin =                                                             Figure 4
                             1− β                             1+ β

          For β = 1 / 4 , one gets

                  rmax = 5.63 ⋅ 10 7 m; rmin = 3.38 ⋅ 10 7 m

       The distances rmax and rmin can also be obtained from mechanical energy and angular momentum conservation,
                         r     r
taking into account that r and v are orthogonal at apogee and at perigee

                                  (           )
                                                                   2
                        1           1    gR m
                          mv β − 1 = mv − T
                            2 2        2
                  E=
                        2 0         2      r
                            mgRT2
                  L0 =            = mv r
                             v0

          What remains of them, after eliminating v, is a second-degree equation whose solutions are rmax and rmin .

2.4
          By the Third Kepler Law, the period T in the new orbit satisfies that

                  T2        T02
                        =
                  a   3
                            r03

where a, the semi-major axis of the ellipse, is given by

                        rmax + rmin     r0
                  a=                =
                             2        1− β 2

          Therefore

                             (
                  T = T0 1 − β 2      )   −3 / 2

                                                                   −3 / 2
                                                          ⎛ 15 ⎞
          For β = 1/4                              T = T0 ⎜ ⎟               = 26.4 h
                                                          ⎝ 16 ⎠

                                                                                Th 1 Solution Page 2 of 5

                                             36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

3.1
       Only if the satellite follows an open trajectory it can escape from the Earth gravity attraction. Then, the orbit
eccentricity has to be equal or larger than one. The minimum boost corresponds to a parabolic trajectory, with ε = 1

               ε=β               ⇒                        β esc = 1

       This can also be obtained by using that the total satellite energy has to be zero to reach infinity (Ep = 0) without
residual velocity (Ek = 0)

               E=
                    1
                             (
                      mv 2 β 2 − 1 = 0
                    2 0 esc
                                             )        ⇒                   β esc = 1

       This also arises from T = ∞ or from rmax = ∞ .

3.2
       Due to ε = β esc = 1 , the polar parabola equation is

                        l
               r=
                    1− cos θ

where the semi-latus-rectum continues to be l = r0 . The minimum Earth - satellite distance corresponds to θ = π , where

                    r
                ′ = 0
               rmin
                    2

         This also arises from energy conservation (for E = 0) and from the equality between the angular momenta (L0) at the
                                                         r      r
initial point P and at maximum approximation, where r and v are orthogonal.

4.1
       If the satellite escapes to infinity with residual velocity v ∞ , by energy conservation

               E=
                    1
                    2
                             (       1
                                         )
                      mv 02 β 2 − 1 = mv ∞2
                                     2
                                                             ⇒                                      Asymptote

                         (
               v∞ = v0 β 2 − 1       )
                                     1/ 2                                                                   v0
                                                                                P

4.2                                                                                r0
                                                                                ∆v
       As ε = β > β esc = 1 the satellite trajectory will be a hyperbola.                                                     φ
                                                                                θ asym
                                                                                θ asym
       The satellite angular momentum is the same at P than at the point                                                            θ asym
where its residual velocity is v ∞ (Figure 5), thus

               m v 0 r0 = m v ∞ b
                                                                                b
       So

               b = r0
                        v0
                        v∞
                                 ⇒                (
                                         b = r0 β − 1
                                                      2
                                                             )   −1 / 2                             Asymptote
                                                                                v∞
                                                                                Figure 5

                                                                                Th 1 Solution Page 3 of 5

                                     36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

4.3
      The angle between each asymptote and the hyperbola axis is that appearing in its polar equation in the limit r → ∞ .
This is the angle for which the equation denominator vanishes

                                                              ⎛1⎞
              1 − β cos θ asym = 0     ⇒      θ asym = cos −1 ⎜⎜⎟⎟
                                                              ⎝β⎠

      According to Figure 5

                   π                               π           ⎛1⎞
              φ=       + θ asym      ⇒        φ=     + cos −1 ⎜⎜ ⎟⎟
                   2                               2           ⎝β⎠

                 3        3
      For β =      β esc = , one gets        φ = 138o = 2.41 rad
                 2        2

                                                                                Th 1 Solution Page 4 of 5

                                36th International Physics Olympiad. Salamanca (España) 2005
                                                                                R.S.E.F.

                                               Th 1          ANSWER SHEET

Question       Basic formulas and               Analytical results                   Numerical results                  Marking
                    ideas used                                                                                guideline

   1.1         MT m        v2
           G             =m 0                                                      r0 = 4.22 ⋅10 7 m                    0.3
                r02        r0
                  2π r0
           v0 =
                   T0
                                                                 g
   1.2
           g=
                GM T                           v 0 = RT                            v0 = 3.07 ⋅10 3 m/s                  0.3 + 0.1
                                                                 r0
                   RT2

           r     r r                                    mgRT2
           L = m r ×v                          L0 =                                                                     0.4
                                                         v0
   1.3            1          Mm
           E=       mv 2 − G
                  2           r                             1
                                               E0 = −         mv 02                                                     0.4
                                                            2

                                               l = r0                                                                   0.4
   2.1
                                               ε=β                                                                      0.5
           Hint on the conical curves
   2.2                                                                                π
                                                                                α=                                   1.0
                                                                                2
                                                      r0
                                               rmax =                              rmax = 5.63 ⋅10 7 m
           Results of 2.1, or                        1− β                                                               1.0 + 0.2
   2.3
           conservation of E and L                    r                            rmin = 3.38 ⋅10 7 m
                                               rmin = 0
                                                     1+ β

   2.4     Third Kepler's Law                            (
                                               T = T0 1 − β            )
                                                                      2 −3 / 2     T = 26.4 h                           0.5 + 0.2

   3.1     ε = 1, E = 0, T = ∞ or
                                                                                β esc = 1                            0.5
           rmax = ∞

                                                    r
   3.2     ε = 1 and results of 2.1             ′ = 0
                                               rmin                                                                     1.0
                                                    2

   4.1     Conservation of E                   v∞ = v0 β − 1 (   2
                                                                           )1/ 2
                                                                                1.0

   4.2     Conservation of L                            (
                                               b = r0 β − 1
                                                             2
                                                                      )   −1 / 2
                                                                                1.0

                                                    π           ⎛1⎞                φ = 138o = 2.41rad
   4.3     Hint on the conical curves          φ=     + cos −1 ⎜⎜ ⎟⎟                                                    1.0 + 0.2
                                                    2           ⎝β ⎠

                                                                                Th 1 Solution Page 5 of 5

R.S.E.F.


[User]
01h
                        IPhO 2009, problem 2. This relatively straightforward problem covers the neat
    application of Doppler laser cooling, a technique for creating ultracold gases that won the 1997
    Nobel prize. (For a very similar problem, see APhO 2006, problem 1.)

Solution:
See the official solution, but note that it has some factor of 2 errors, as pointed out by

                    01h
    Stefan Ivanov here.

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


[User]
A rocket starts from rest in the lab frame at the origin, then accelerates in a straight
    line at constant rate a0 as measured by an accelerometer on the ship; that is, the proper acceleration
    is always a0 .
      (a) Show that the acceleration measured in the lab frame is a0 /γ 3 . (We already proved this more
          generally in problem 23, but try to do this more explicitly by working in the comoving frame,
          then going back to the lab frame.)
      (b) Find the speed of the rocket ship in the lab frame as a function of time t in the lab frame.
      (c) Find the speed of the rocket ship in the lab frame as a function of the proper time τ elapsed
          on the rocket. Can you explain the simplicity of your result using rapidity?

                    01^
      (d) To conclude, find expressions for t(τ ), x(t), and x(τ ), and comment on their limits.

Solution:
(a) Suppose the rocket has speed v in the lab frame, and now consider the momentarily
     comoving frame S ′ moving with speed v. In that frame, in time dt′ , the rocket accelerates
     from zero speed to speed a0 dt′ . Then the new speed in the lab frame is the sum of v and
     a0 dt, which is
                             v + a0 dt′
                                         = v + a0 dt′ − v 2 a0 dt′ + O(dt2 ).
                            1 + va0 dt′
     Therefore, we have
                              dv                 dt′                     a0
                                 = a0 (1 − v 2 )     = a0 (1 − v 2 )3/2 = 3
                              dt                 dt                      γ
     as desired.

 (b) Separating and integrating, we have
                                         Z v
                                                    dv           v
                                  a0 t =              2 3/2
                                                            =√
                                            0   (1 − v )       1 − v2
     by a trigonometric substitution. (Here and below, we’re using an abuse of notation mentioned
     in M1, where the integration variable is denoted by the same letter as its upper bound.) Then
                                                     a0 t
                                           v(t) = p            .
                                                   1 + (a0 t)2

 (c) Note that the increment of proper time measured by the rocket is dτ = dt′ , because dt′ is
     always defined in the frame momentarily moving with the rocket. Therefore
                                                            Z v
                                 dv                             dv
                                    = a0 (1 − v 2 ), a0 τ =        2
                                                                     .
                                 dτ                          0 1−v

     Using hyperbolic trig substitution, we have

                                            v(τ ) = tanh(a0 τ )

     which just tells us that the rapidity changes at rate a0 .

 (d) It’s easy to crank these out given the above results. First, note that
                                               dτ
                             dt = γ dτ = q                  = cosh(a0 τ ) dτ.
                                          1 − tanh2 (a0 τ )

     Integrating both sides gives
                                                       sinh(a0 τ )
                                            t(τ ) =                .
                                                           a0

          Next, we integrating the answer to part (b),
                                          Z t                 p
                                                   a0 t        1 + (a0 t)2 − 1
                                   x(t) =   p            dt =                  .
                                          0  1 + (a0 t)2            a0

          Finally, plugging in our expression for t(τ ) gives
                                          q
                                            1 + sinh2 (a0 τ ) − 1   cosh(a0 τ ) − 1
                                  x(τ ) =                         =                 .
                                                   a0                    a0
          This results make sense. At small t and τ , they just reduce to the familiar results t = τ and
          x = a0 t2 /2. At large t, x(t) increases linearly, since the speed of the rocket approaches the
          speed of light. What is perhaps most interesting is that at large τ , x(τ ) increases exponentially,
          because of how quickly the time dilation effect increases. If it’s possible to make it to another
          star in a human lifetime, it’s actually not that much harder to cross the whole galaxy! This is

                    01^
          a neat result, recently highlighted in the 2022 IPhO and the popular book Project Hail Mary.


[User]
8. Heatsink (6 points)

Consider aheatsinkin theformofacop-
perplateofaconstantthickness(muchsmallerthanthediameter
doftheplate). Anelectroniccomponentisfixedtotheplate,anda
temperaturesensorisfixedtotheplateatsomedistancefromthat
compnent. Youmayassumethat the heat flux (i.e. powerper unit
area) from the plate to the surrounding air is proportional to the
difference of the plate temperature at the given point (the coeffi-
cientofproportionalityisconstantovertheentireplate, including
the site ofthe electronic component).
i) (2 pts) The electronic component has been dissipating energy
with a constant power of P = 35W for a long time, and the av-
erage plate temperature has stabilized at the value T0 = 49◦
C.
Now, the component is switched off, and the average plate tem-
perature starts dropping; it takes τ = 10s to reach the value
T1 = 48◦
C. Determine theheatcapacity C (units J/◦
C)ofthe
plate. Thecapacitiesoftheelectroniccomponentandthetemper-
aturesensor arenegligible.
ii) (4 pts) Now, the electronic component has been switched off
for a long time; at the moment t = 0, a certain amount of heat
Q is dissipated at it during a very short time. In the Figure and
Table, the temperature is given as a function of time, as recorded
bythe sensor. Determine the dissipated heatamount Q.
t(s) 0 20 30 100 200 300
T (◦
C) 20.0 20.0 20.4 32.9 41.6 42.2
t(s) 400 600 800 1000 1200 1400
T (◦
C) 39.9 33.4 27.9 24.4 22.3 21.2

Solution:
8. Heatsink

i)WhentheaveragetemperatureisstableatT0,allthepowerdis-
sipated at the electronic component iseventually given to the air:
theairisbeingheatedwithpowerP. Astheheatfluxdependslin-
early on the temperature difference between a point on the plate
and the air, the average heat flux and therefore the net power dis-
sipatedintotheairdependslinearlyontheaveragetemperatureof
theplate. Theaveragetemperaturedeterminestheradiatedpower.
Now consider the situation after the heating has ended. The
average temperature is initially the same, so the radiated heat
power is initially still P. By the definition of heat capacity, an
infinitesimal heat amount given to the surroundings is dQ =
−C dTavg withtheminussignencoding thedirection oftheheat
flow. Thus, at the first moment, P = dQ
dt
= −CdTavg
dt
. Assum-
ingthat during τ theaveragetemperature dependsapproximately
linearly on time (because T0 − T1 = 1◦
C is much lessthan the
usual ambient temperature),
dTavg
dt
≈ T1−T0
τ
and C ≈ Pτ
T0−T1
=
350J/◦
C. ActuallythegraphofTavg(t)isslightlycurveddown-
wards(asitisanexponentialeventually stabilizing attheambient
temperature)andinitially somewhatsteeper,soC isabit smaller.
ii)Theaveragetemperatureoftheheatsinkfallsoffexponentially,
therefore, if the “tail” of the given graph turns out to be expo-
nential, we can presumethe “tail” depicts the situation where the
sensor is sensing the average temperature and the initial “bump”
in thetemperature distribution hasevened out. Extrapolating the
exponentialtot = 0wegettheinitial averagetemperatureTavg,0
(immediately after the Q has been dissipated into the sink) and,
by Q = C(Tavg,0 − Tamb), the heat Q. The ambient temper-
ature Tamb can be read off fromthe beginning of the given graph
where the sensor’s surroundings have not yet heated up. This is
furthermorea check forthe assumptionT0 − T1 ≪ Tamb made
in the first partofthe solution. Fromthe table, Tamb = 20.0◦
C.
Letusanalysethe(yethypothetical)exponentialTavg−Tamb
ought to obey, so that eventually we expect T ∼ Tavg =
Tamb +Tce− t
tc whereTc andtc are,respectively,acharacteristic
temperatureandacharacteristictime. (The“∼”means“isasymp-
toticalto”or“approaches”.) Weplotln(T −Tamb)usingthedata
from the table. Then approximate the “tail” linearly (valuing the
end of it most) to get ln[(T − Tamb)/◦
C] ∼ 4.89 − t
300 s
.
Therefore Tc ≈ e4.89 ◦
C ≈ 133◦
C. On the other hand,
plugging t = 0 into our exponential function shows that
Tavg,0 − Tamb = Tc and,finally, Q = CTc ≈ 46700J.
Actually, quite a good result can be obtained without re-
plotting anything, by just considering the last three datapoints
of the table. Denote ∆Ti ≡ Ti − Tamb. If the times t3 −
t2 = t2 − t1, then with an exponential we should observe
that ∆T3/∆T2 = ∆T2/∆T1. The last three timepoints are
good indeed, so we check ∆T1 = 4.4◦
C, ∆T2 = 2.3◦
C
and ∆T3 = 1.2◦
C. Their ratios are ∆T3/∆T2 ≈ 0.522 and
∆T2/∆T1 ≈ 0.523, a splendid match. This confirms the ex-
ponential “tail”. As in every equal time interval the ∆T is mul-
tiplied by the same number (that is the essence of exponentials),
Tc = ∆Tavg,0 = ∆T3 ×

∆T2
∆T3
 t3
t3−t2
≈ 114◦
C. From this,
Q ≈ 39900J. This is discrepant from our previous calculation,
butnottoomuch: Tc isexponentially sensitivetotheT-intercept
of the straight line fitted to the “tail” (its crossing point with the
T-axis)on the logarithmicplot. Thebumphasstillnot yetdisap-
pearedcompletelyenough.

Heatsink


[User]
Hinnake, milline oleks suitsu kiirus korstnast väljumisel, kui õhutakistusega (sh turbulentsest liikumisest tingitud takistusega) korstnas ning ahjulõõrides mitte arvestada.
Korstna kõrgus (mõõdetuna korstnajala juurest, kuhu siseneb ahjust tulev soe õhk) on $h=\SI{10}{m}$ ja õhu keskmine temperatuur korstnas $t=\SI{80}{\degreeCelsius}$. Lugeda, et ahju uks ja korstnajalg on samal kõrgusel. Välisõhu temperatuur on $t_0=\SI{0}{\degreeCelsius}$.

Solution:
Ahjusuusse siseneva õhu rõhk $p_1$ on võrdne õhurõhuga ahjusuu kõrgusel ning korstnast väljuva õhu rõhk $p_2$ on võrdne õhurõhuga korstnasuu kõrgusel. Seega, $p_2=p_1-\rho_0 gh$, kus $\rho_0$ tähistab välisõhu tihedust. Bernoulli seaduse kohaselt kehtib seos
\[
p_1=p_2 + \rho gh + \rho \frac{v^2}{2} = p_1 - \rho_0 gh + \rho gh+\rho \frac{v^2}{2},
\]
kus $\rho=T\rho_0/T_0$ on õhu tihedus korstnas ning $v$ on otsitav kiirus. Seega,
\[
v=\sqrt {2\left(\frac T{T_0}-1\right)gh}\approx \SI{3,5}{m/s}.
\]

\emph{Märkus:} ahjusuhu voolava õhu kiiruse võib lugeda tühiselt väikseks, sest voolava õhuga piirkonna ristlõike pindala on ilmselt hulga suurem korstna ristlõikepindalast.

Bernoulli seaduse võib tuletada ka energia jäävuse seadusest. Samuti lugeda õigeks impulsibalansi abil saadav tulemus (mis tuleb $\sqrt 2$ korda väiksem), sest nõuti vaid hindamist.

Korsten

Hinnake, milline oleks suitsu kiirus korstnast väljumisel, kui õhutakistusega (sh turbulentsest liikumisest tingitud takistusega) korstnas ning ahjulõõrides mitte arvestada.
Korstna kõrgus (mõõdetuna korstnajala juurest, kuhu siseneb ahjust tulev soe õhk) on $h=\SI{10}{m}$ ja õhu keskmine temperatuur korstnas $t=\SI{80}{\degreeCelsius}$. Lugeda, et ahju uks ja korstnajalg on samal kõrgusel. Välisõhu temperatuur on $t_0=\SI{0}{\degreeCelsius}$.

Korstna sees voolav õhk tekitab alarõhu, mis on avaldatav Bernoulli võrrandist. Rõhu langud korstna sees ja väljas peavad olema võrdsed.


[User]
white dwarfs.
                    01^   USAPhO 2024, problem A2. Rough estimates of the dynamics of stars and

       Remark
       The estimates performed in the previous problems are quite rough, basically treating the
       white dwarf as being homogeneous, with uniform density and pressure. In reality, we have
       ∇p = −ρg just like in any situation in hydrostatic equilibrium, where the degeneracy
       pressure p is determined by the local density n. It’s like the gaseous atmospheres you dealt
       with in T1, but with a different equation of state.

       If you additionally allow the white dwarf to have a net charge, then there is an additional
       contribution from the electrostatic force, and the resulting equations are called the Thomas–
       Fermi equations of structure. They can also be used to model many-electron atoms, when
       you can neglect the discreteness of the electrons.

Solution:
USAPhO 2024, problem A2. Rough estimates of the dynamics of stars and
    white dwarfs.

       Remark
       The estimates performed in the previous problems are quite rough, basically treating the
       white dwarf as being homogeneous, with uniform density and pressure. In reality, we have
       ∇p = −ρg just like in any situation in hydrostatic equilibrium, where the degeneracy
       pressure p is determined by the local density n. It’s like the gaseous atmospheres you dealt
       with in T1, but with a different equation of state.

       If you additionally allow the white dwarf to have a net charge, then there is an additional
       contribution from the electrostatic force, and the resulting equations are called the Thomas–
       Fermi equations of structure. They can also be used to model many-electron atoms, when
       you can neglect the discreteness of the electrons.


[User]
Question A1
A pair of wedges are located on a horizontal surface. The coefficient of friction (both sliding and
static) between the wedges is µ, the coefficient of friction between the bottom wedge B and the
horizontal surface is µ, and the angle of the wedge is θ. The mass of the top wedge A is m, and
the mass of the bottom wedge B is M = 2m. A horizontal force F directed to the left is applied to
the bottom wedge as shown in the figure.

                                                 A                   F
                                                          B
                                             θ

   Determine the range of values for F so that the the top wedge does not slip on the bottom
wedge. Express your answer(s) in terms of any or all of m, g, θ, and µ.

Solution:
Question             A1 A2 A3 A4 B1                B2 Total
                  Mean                11 3   5  6 12                15  53
            Standard Deviation         7 4   5  7 10                14  29
                Maximum               25 21 25 25 42                50 163
              Upper Quartile          18 4   9  8 18                26  73
                 Median               10 2   5  5 10                12  49
              Lower Quartile           4 0   0  0  3                 1  30
                Minimum                0 0   0  0  0                 0   5
   Some Trivia:

  • California had 121 test takers

  • New Jersey had 42

  • Texas had 38

  • Florida, Illinois, Massachusetts, Maryland, New York, and Virginia each had between one
    dozen and two dozen test takers

  • Alabama, Alaska, Colorado, Idaho, Iowa, Louisiana, Montana, North Dakota, South Dakota,
    and Wyoming did not have a test taker this year.

                  Copyright c 2017 American Association of Physics Teachers

USAPhO 2017 Question A1


[User]
Theoretical Question 1

                             A Swing with a Falling Weight
      A rigid cylindrical rod of radius R is held horizontal above the ground. With a string of
negligible mass and length L ( L > 2πR ), a pendulum bob of mass m is suspended from point
A at the top of the rod as shown in Figure 1a. The bob is raised until it is level with A and then
released from rest when the string is taut. Neglect any stretching of the string. Assume the
pendulum bob may be treated as a mass point and swings only in a plane perpendicular to the
axis of the rod. Accordingly, the pendulum bob is also referred to as the particle. The
                           K
acceleration of gravity is g .
                                                     x
                                 L                              A
              m                           r̂        Q
                                                            θ
                                               tˆ           O
                      K                                             R
                      g

                                      s

                                                                        Figure 1a
                             P

      Let O be the origin of the coordinate system. When the particle is at point P, the string
is tangential to the cylindrical surface at Q. The length of the line segment QP is called s.
The unit tangent vector and the unit radial vector at Q are given by tˆ and r̂ , respectively.
The angular displacement θ of the radius OQ, as measured counterclockwise from the
vertical x-axis along OA, is taken to be positive.
      When θ = 0 , the length s is equal to L and the gravitational potential energy U of the
particle is zero. As the particle moves, the instantaneous time rates of change of θ and s are
given by θ and s , respectively.
      Unless otherwise stated, all the speeds and velocities are relative to the fixed point O.

Part A
      In Part A, the string is taut as the particle moves. In terms of the quantities introduced
above (i.e., s, θ, s , θ , R, L, g, tˆ and r̂ ), find:
(a) The relation between θ and s .                                                    (0.5 point)
                K
(b)The velocity vQ of the moving point Q relativeto O.                                 (0.5 point)
                            K
(c) The particle’s velocity v ′ relative to the moving point Q when it is at P         (0.7 point)
                            K
(d)The particle’s velocity v relative to O when it is at P.                            (0.7 point)

                                                        2

(e) The tˆ -component of the particle’s acceleration relative to O when it is at P.     (0.7 point)
(f) The particle’s gravitational potential energy U when it is at P.                    (0.5 point)
(g)The speed v m of the particle at the lowest point of its trajectory.                 (0.7 point)

Part B
      In Part B, the ratio L to R has the following value:
                           L 9π 2    π
                             =  + cot = 3.534 + 3.352 = 6.886
                           R   8 3   16
(h)What is the speed v s of the particle when the string segment from Q to P is both straight
    and shortest in length? (in terms of g and R)                                   (2.4 points)
(i) What is the speed v H of the particle at its highest point H when it has swung to the other
   side of the rod? (in terms of g and R)                                               (1.9 points)

Part C
       In Part C, instead of being suspended from A, the pendulum bob of mass m is
connected by a string over the top of the rod to a heavier weight of mass M, as shown in
Figure 1b. The weight can also be treated as a particle.
                                                          x
                                     L
                                                                  A
               m                                          θ
                                                      R
                                                              O

                                                                      M     Figure 1b

       Initially, the bob is held stationary at the same level as A so that, with the weight
hanging below O, the string is taut with a horizontal section of length L. The bob is then
released from rest and the weight starts falling. Assume that the bob remains in a vertical
plane and can swing past the falling weight without any interruption.
       The kinetic friction between the string and the rod surface is negligible. But the static
friction is assumed to be large enough so that the weight will remain stationary once it has
come to a stop (i.e. zero velocity).
(j) Assume that the weight indeed comes to a stop after falling a distance D and
    that (L − D ) >> R . If the particle can then swing around the rod to θ = 2π while both
   segments of the string free from the rod remain straight, the ratio α = D /L must not be
   smaller than a critical value α c . Neglecting terms of the order R /L or higher, obtain an
   estimate on α c in terms of M /m.                                              (3.4 points)

                                                 3

Answer Sheet                     Theoretical Question 1
                               A Swing with a Falling Weight

(a) The relation between θ and s is

(b) The velocity of the moving point Q relative to O is
       K
       vQ =

(c) When at P, the particle’s velocity relative to the moving point Q is
       K
       v′ =

(d) When at P, the particle’s velocity relative to O is
       K
       v =

(e) When at P, the tˆ -component of the particle’s acceleration relative to O is

(f) When at P, the particle’s gravitational potential energy is

       U=

(g) The particle’s speed when at the lowest point of its trajectory is

       vm =

(h) When line segment QP is straight with the shortest length, the particle‘s speed is
   (Give expression and value in terms of g and R )

       vs =

(i) At the highest point, the particle’s speed is (Give expression and value in terms of g and R)

       vH =

(j) In terms of the mass ratio M /m, the critical value α c of the ratio D /L is

       αc =

                                                  4

Solution:
Solution to Theoretical Question 1

                              A Swing with a Falling Weight
Part A
(a) Since the length of the string L = s + Rθ is constant, its rate of change must be zero.
    Hence we have
                                           s + Rθ = 0                               (A1)
(b) Relative to O, Q moves on a circle of radius R with angular velocity θ , so
                                        K
                                        vQ = Rθ tˆ = − s tˆ                         (A2)

(c) Refer to Fig. A1. Relative to Q, the displacement of P in a time interval ∆t
        K
    is ∆r ′ = ( s∆θ )( − rˆ) + ( ∆s )tˆ = [( sθ)( − rˆ) + s tˆ]∆t . It follows
                                               K
                                               v ′ = − sθ rˆ + s tˆ                                 (A3)

                                                                 Q

                               s+∆s                 ∆θ
                               s         K
                                   P    ∆r ′
                                                              s+∆s
                                       s∆θ
                                                                                Figure A1
                                                              − r̂
                                               tˆ

(d) The velocity of the particle relative to O is the sum of the two relative velocities given in
    Eqs. (A2) and (A3) so that
                               K K K
                              v = v ′ + vQ = ( − sθ rˆ + s tˆ ) + Rθ tˆ = − sθ rˆ       (A4)
                                                                                K
(e) Refer to Fig. A2. The ( − tˆ )-component of the velocity change ∆v is given
                 K                                                                    K    K
    by ( −tˆ) ⋅ ∆v = v∆θ = vθ ∆t . Therefore, the tˆ -component of the acceleration a = ∆v / ∆t
    is given by tˆ ⋅ aˆ = −vθ . Since the speed v of the particle is sθ according to Eq. (A4),
    we see that the tˆ -component of the particle’s acceleration at P is given by
                                   K
                                   a ⋅ tˆ = − vθ = −( sθ)θ = − sθ 2                               (A5)
                                               − r̂       − tˆ
                                                                     v ∆θ
                                               Q                                ∆v
                                                               ∆θ
                                         ∆θ              v                       K    K
                                                                                v + ∆v
                                                                K        K
                                                                v       ∆v
                                                                                Figure A2
                          P                                                 O
                                                         12

    Note that, from Fig. A2, the radial component of the acceleration may also be obtained as
    K
    a ⋅ rˆ = −dv / dt = −d ( sθ) / dt .

(f) Refer to Fig. A3. The gravitational potential energy of the particle is given by U = −mgh .
    It may be expressed in terms of s and θ as
                                 U (θ ) = −mg[ R(1 − cosθ ) + s sin θ ]                         (A6)

                                                       x
                                                                   A
                                                                           R −R cosθ
                                           Q
                                                       θ                   R cosθ
                                   h       s       R
                                                               O
                                                   s sinθ
                                                                                Figure A3
                                 P

(g) At the lowest point of its trajectory, the particle’s gravitational potential energy U must
    assume its minimum value U m . By differentiating Eq. (A6) with respect to θ and using
    Eq. (A1), the angle θ m corresponding to the minimum gravitational energy can be
    obtained.
                                 dU                     ds
                                    = − mg R sin θ +        sin θ + s cosθ
                                 dθ                     dθ
                                    = − mg [R sin θ + ( − R ) sin θ + s cosθ ]
                                       = −mgs cosθ
                    dU                        π
    At θ = θ m ,           = 0 . We have θ m = . The lowest point of the particle’s trajectory is
                    dθ θ m                    2
    shown in Fig. A4 where the length of the string segment of QP is s = L− π R /2.

                                                               x
                                                                       A

                                                       π
                                                           2
                                               Q
                                                       R            O
                                               s

                                                                              Figure A4
                                               P
  From Fig. A4 or Eq. (A6), the minimum potential energy is then
                             U m = U (π / 2 ) = −mg [ R + L − (πR / 2)]                         (A7)
  Initially, the total mechanical energy E is 0. Since E is conserved, the speed v m of the
  particle at the lowest point of its trajectory must satisfy

                                                               13

                                                     1 2
                                       E=0=           mv + U m                           (A8)
                                                     2 m
  From Eqs. (A7) and (A8), we obtain

                              v m = − 2U m / m = 2 g[ R + ( L − πR / 2)]                 (A9)

Part B
(h) From Eq. (A6), the total mechanical energy of the particle may be written as
                           1 2               1
                     E=0=    mv + U (θ ) = mv 2 − mg[ R(1 − cosθ ) + s sin θ ]           (B1)
                           2                 2
    From Eq. (A4), the speed v is equal to sθ . Therefore, Eq. (B1) implies

                                v 2 = ( sθ) 2 = 2 g[ R(1 − cosθ ) + s sin θ ]           (B2)

    Let T be the tension in the string. Then, as Fig. B1 shows, the tˆ -component of the net
    force on the particle is –T + mg sin θ . From Eq. (A5), the tangential acceleration of the
    particle is (− sθ 2 ) . Thus, by Newton’s second law, we have
                                       m(− sθ 2 ) = −T + mg sin θ                       (B3)

                                                         x
                                                             A
                                             Q
                                                      θ
                                         s       θ
                                                     R
                                                             O
                                        T
                                  P
                        mg sinθ                                         Figure B1
                                   θ mg

    According to the last two equations, the tension may be expressed as
                                                     mg
                         T = m( sθ 2 + g sin θ ) =       [2 R(1 − cosθ ) + 3s sin θ ]
                                                       s
                              2mgR        θ 3            L
                            =       [tan − (θ − )]( sin θ )                              (B4)
                                s         2 2            R
                              2mgR
                            =        ( y1 − y 2 )(sin θ )
                                s

    The functions y1 = tan(θ / 2) and y 2 = 3(θ − L / R) / 2 are plotted in Fig B2.

                                                      14

          y
                                           Figure B2
         30

         20
                                 θ                                               θ
                      y1 = tan                                        y1 = tan
         10                      2                                               2
                                          θs
          0
                                                                                θ
        -10
                                                       3     L
                                                y2 =     (θ − )
        -20                                            2     R

        -30
              0       π/2            π          3π/2       2π       5π/2             3π

From Eq. (B4) and Fig. B2, we obtain the result shown in Table B1. The angle at
which .y 2 = y 1 is called θ s ( π < θ s < 2π ) and is given by
                                           3       L       θ
                                             (θ s − ) = tan s                                 (B5)
                                           2       R        2
or, equivalently, by
                                           L        2   θ
                                             = θ s − tan s                                    (B6)
                                           R        3    2
Since the ratio L/R is known to be given by
                          L 9π 2    π      π 2     1    π
                            =  + cot = (π + ) − tan (π + )                                    (B7)
                          R   8 3   16     8 3     2    8
one can readily see from the last two equations that θ s = 9π / 8 .

                                                Table B1

                                     ( y1 − y 2 )           sin θ                tension T
              0 <θ <π                positive              positive              positive
                θ =π                     +∞                     0                positive
              π <θ <θs               negative              negative              positive
                  θ =θs                  zero              negative                  zero
           θ s < θ < 2π              positive              negative              negative

Table B1 shows that the tension T must be positive (or the string must be taut and straight)
in the angular range 0<θ < θ s . Once θ reaches θ s , the tension T becomes zero and the
part of the string not in contact with the rod will not be straight afterwards. The shortest
possible value s min for the length s of the line segment QP therefore occurs at θ = θ s and
is given by

                                                    15

                                         9π 2   π 9π   2R    π
                 s min = L − Rθ s = R(     + cot − ) =    cot = 3.352 R                                           (B8)
                                          8 3   16 8    3    16
    When θ = θ s , we have T = 0 and Eqs. (B2) and (B3) then leads to v 2s = − gs min sin θ s .
    Hence the speed v s is
                                                      2 gR    π   π                  4 gR     π
                     v s = − gs min sin θ s =              cot sin =                      cos
                                                        3     16  8                    3      16                  (B9)
                        = 1.133 gR

(i) When θ ≥ θ s , the particle moves like a projectile under gravity. As shown in Fig. B3, it is
    projected with an initial speed v s from the position P = ( x s , y s ) in a direction making
    an angle φ = (3π / 2 − θ s ) with the y-axis.
    The speed v H of the particle at the highest point of its parabolic trajectory is equal to the
    y-component of its initial velocity when projected. Thus,
                                            4 gR      π    π
                      v H = v s sin(θ s − π ) =  cos sin = 0.4334 gR                   (B10)
                                              3       16    8
    The horizontal distance H traveled by the particle from point P to the point of maximum
    height is
                              v s2 sin 2(θ s − π ) v s2     9π
                           H=                     =     sin    = 0.4535 R                                        (B11)
                                       2g           2g       4

                                                       x
                                                                                θs − π
                                     L
        m                                                                            vs
                                                  θ                        vH
                                            R                                                 P = ( xs , y s )
                                 y                         O
                                                  θs           φ                          H
                             s                                           s min
                                                               Q
                                                                                Figure B3

    The coordinates of the particle when θ = θ s are given by
                                                                   π                 π
                  x s = R cosθ s − s min sin θ s = − R cos             + s min sin        = 0.358R               (B12)
                                                                   8                 8
                                                                   π                 π
                 y s = R sin θ s + s min cosθ s = − R sin              − s min cos
                                                                         = −3.478 R         (B13)
                                                         8             8
    Evidently, we have | y s | > ( R + H ) . Therefore the particle can indeed reach its maximum
    height without striking the surface of the rod.

                                                       16

Part C
(j) Assume the weight is initially lower than O by h as shown in Fig. C1.

                                                            x
                                      L
                                                                    A
               m                                            θ
                                                        R
                                                                O

                                                                    h
                                                                          M       Figure C1

    When the weight has fallen a distance D and stopped, the law of conservation of total
    mechanical energy as applied to the particle-weight pair as a system leads to
                                 − Mgh = E ′ − Mg (h + D)                           (C1)
    where E′ is the total mechanical energy of the particle when the weight has stopped. It
    follows
                                        E ′ = MgD                                     (C2)
    Let Λ be the total length of the string. Then, its value at θ = 0 must be the same as at any
    other angular displacement θ. Thus we must have
                                      π                         π
                             Λ = L+        R + h = s + R(θ +        ) + (h + D)               (C3)
                                       2                        2
    Noting that D = α L and introducing l = L−D, we may write
                                A = L − D = (1 − α ) L                                        (C4)
    From the last two equations, we obtain
                             s = L − D − Rθ = A − Rθ                             (C5)
    After the weight has stopped, the total mechanical energy of the particle must be
    conserved. According to Eq. (C2), we now have, instead of Eq. (B1), the following
    equation:
                                           1 2
                            E ′ = MgD =      mv − mg [R(1 − cos θ ) + s sin θ ]               (C6)
                                           2
    The square of the particle’s speed is accordingly given by
                                            2 MgD                       s
                           v 2 = ( sθ) 2 =        + 2 gR (1 − cos θ ) + sin θ                (C7)
                                              m                         R
    Since Eq. (B3) stills applies, the tension T of the string is given by
                                       − T + mg sin θ = m(− sθ 2 )                           (C8)
    From the last two equations, it follows

                                                 17

                      T = m( sθ 2 + g sin θ )
                             mg 2 M
                         =          D + 2 R(1 − cos θ ) + 3s sin θ               (C9)
                              s  m
                             2mgR MD                  3 A
                         =           + (1 − cos θ ) +     − θ sin θ
                               s  mR                  2 R

where Eq. (C5) has been used to obtain the last equality.
     We now introduce the function
                                                3 A
                           f (θ ) = 1 − cos θ +       − θ sin θ                 (C10)
                                                2 R
From the fact A = ( L − D ) >> R , we may write
                                    3A
                       f (θ ) ≈ 1 +      sin θ − cosθ = 1 + A sin(θ − φ )       (C11)
                                    2R
where we have introduced
                                       3A 2                   2R
                           A = 1+ (        ) , φ = tan −1                       (C12)
                                       2R                      3A
From Eq. (C11), the minimum value of f(θ) is seen to be given by
                                                                2
                                                       3 A
                               f min = 1 − A = 1 − 1 +                          (C13)
                                                       2R
Since the tension T remains nonnegative as the particle swings around the rod, we have
from Eq. (C9) the inequality
                                                                      2
                       MD           M ( L − A)        3A
                          + f min =            +1− 1+                     ≥0    (C14)
                       mR              mR             2R

or
                                                        2
                     ML              MA      3A                     MA   3A
                        +1 ≥            + 1+                ≈          +        (C15)
                     mR              mR      2R                     mR   2R

From Eq. (C4), Eq. (C15) may be written as
                               ML            ML 3L
                                  +1 ≥         +    (1 − α )                    (C16)
                               mR            mR 2 R
Neglecting terms of the order (R/L) or higher, the last inequality leads to
                        ML       3L         2R
                            +1      −1   1−
                        mR                  3L ≈   1
                α ≥ 1−         = 2R    =                                        (C17)
                       ML 3L    ML 3L 2 M          2M
                          +        +         +1 1+
                       mR 2 R   mR 2 R   3m         3m
The critical value for the ratio D/L is therefore
                                                1
                                       αc =                                     (C18)
                                                 2M
                                             1+
                                                 3m

                                                 18

                                    Marking Scheme

                               Theoretical Question 1
                            A Swing with a Falling Weight
 Total      Sub
                                   Marking Scheme for Answers to the Problem
Scores     Scores
Part A       (a)    Relation between θ and s . ( s = − Rθ )
                    3⁄4 0.2 for θ ∝ s .
4.3 pts.    0.5     3⁄4 0.3 for proportionality constant (-R).
                                                      K
            (b)     Velocity of Q relative to O. ( v = Rθ tˆ )
                                                       Q

            0.5     3⁄4 0.2 for magnitude R θ .
                    3⁄4 0.3 for direction tˆ .
                                                              K
            (c)     Particle’s velocity at P relative to Q. ( v ′ = − sθ rˆ + s tˆ )
                    3⁄4 0.2+0.1 for magnitude and direction of r̂ -component.
            0.7
                    3⁄4 0.3+0.1 for magnitude and direction of tˆ -component.
                                                                  K K K
            (d)     Particle’s velocity at P relative to O. ( v = v ′ + vQ = − sθ rˆ )
            0.7                                      K         K
                    3⁄4 0.3 for vector addition of v ′ and vQ .
                                                                      K
                    3⁄4 0.2+0.2 for magnitude and direction of v .
            (e)     tˆ -component of particle’s acceleration at P.
                                                 K K
                    3⁄4 0.3 for relating a or a ⋅ tˆ to the velocity in a way that implies
            0.7               K
                            | a ⋅ tˆ | = v 2 / s .
                                    K
                    3⁄4 0.4 for a ⋅ tˆ = − sθ 2 (0.1 for minus sign.)
            (f)     Potential energy U.
                    3⁄4 0.2 for formula U = −mgh .
            0.5     3⁄4 0.3 for h = R(1 − cosθ ) + s sin θ or U as a function of θ, s, and R.
            (g)     Speed at lowest point vm.
                    3⁄4 0.2 for lowest point at θ = π / 2 or U equals minimum Um.
            0.7     3⁄4 0.2 for total mechanical energy E = mv m2 / 2 + U m = 0 .
                    3⁄4 0.3 for v m = − 2U m / m = 2 g[ R + ( L − πR / 2)] .
Part B      (h)     Particle’s speed vs when QP is shortest.

4.3 pts.    2.4     3⁄4 0.4 for tension T becomes zero when QP is shortest.
                    3⁄4 0.3 for equation of motion − T + mg sin θ = m(− sθ 2 ) .
                    3⁄4 0.3 for E = 0 = m( sθ) 2 / 2 − mg[ R(1 − cosθ ) + s sin θ ] .
                              3        L       θ
                    3⁄4 0.4 for    (θ s − ) = tan s .
                              2        R        2
                    3⁄4 0.5 for θ s = 9π / 8 .
                    3⁄4 0.3+0.2     for v s = 4 gR / 3 cos π / 16 = 1.133 gR

                                               19

          (i)   The speed vH of the particle at its highest point.
                3⁄4 0.4 for particle undergoes projectile motion when θ ≥ θ s .
          1.9   3⁄4 0.3 for angle of projection φ = (3π / 2 − θ s ) .
                3⁄4 0.3 for v H is the y-component of its velocity at θ = θ s .
                3⁄4 0.4 for noting particle does not strike the surface of the rod.
                3⁄4 0.3+0.2 for
                           v H = 4 gR / 3 cos(π / 16) sin(π / 8) = 0.4334 gR .
Part C    (j)   The critical value α c of the ratio D/L.
                3⁄4 0.4 for particle’s energy E ′ = MgD when the weight has stopped.
3.4 pts   3.4
                3⁄4 0.3 for s = L − D − Rθ .
                3⁄4 0.3 for E ′ = MgD = mv 2 / 2 − mg[ R(1 − cosθ ) + s sin θ ] .
                3⁄4 0.3 for − T + mg sin θ = m(− sθ 2 ) .
                3⁄4   0.3 for concluding T must not be negative.
                3⁄4   0.6 for an inequality leading to the determination of the range of D/L.
                3⁄4   0.6 for solving the inequality to give the range of α = D/L.
                3⁄4   0.6 for α c = (1 + 2 M / 3m) .

                                         20

A Swing with a Falling Weight


[User]
01W   USAPhO 2004, problem A3.

[3] Problem 25 (Kalda). In fiber optics, devices called equal ratio splitters are often used; these are
    devices where two optical fibers are brought into such a contact so that if an electromagnetic wave
    is propagating in one fiber, it splits into two equal amplitude waves traveling in each of the fibers.
    Assume that all waves propagate with the same polarization, i.e. that all electric fields are parallel.

      (a) Show that whenever a wave enters the splitter, from either fiber, one of the outgoing waves is
          advanced in phase by π/4, while the other is retarded by π/4.

      (b) From part (a) alone, it’s ambiguous which wave is advanced and which wave is retarded. Let’s
          suppose that the fibers are set up so that, when a wave enters along fiber 1, the wave that
          exits along fiber 1 is advanced. If a wave enters along fiber 2, is the wave that exits along
          fiber 1 advanced or retarded?

      (c) Now consider two sequentially positioned, identical equal ratio splitters, as shown.

          This is called a Mach–Zehnder interferometer. The optical path difference between the inter-
          splitter segments of the two fibers is 30 µm. Assuming the wavelength of the incoming
          monochromatic light varies from 610 nm to 660 nm, for what wavelengths is all the light
          energy directed into fiber 2?

       Remark
       Above, we’ve focused on cases where light can only exit a given optical element in two
       ways. But in general, you could have n “ports”, in which case you would need an
       entire n × n matrix of coefficients S to relate the n input amplitudes to the n output
       amplitudes. Generalizing problem 22 to this case shows that S is a unitary matrix, S † = S −1 .

       All of the optical elements we’ve seen so far obey optical reciprocity, i.e. the principle that
       “if I can see you, then you can see me”, related to time reversal symmetry. However, we can
       violate reciprocity by using special materials, such as permanent magnets. For example, a
        circulator is an optical element with 3 ports, so that all light entering port 1 exits from port
       2, light in port 2 exits from port 3, and light in port 3 exits from port 1. Though it’s strange,
       you can check that there’s no way to use it to violate the second law of thermodynamics.

Solution:
(a) This is a case where the heuristic argument of problem 22(c) works, because there’s
     nothing but vacuum at the splitting point. Let the ingoing electric field amplitude be Ein ,
     and let the outgoing field amplitudes be E1 and E2 . Then

                                Ein = E1 + E2 ,         |Ein |2 = |E1 |2 + |E2 |2

     from continuity of the electric field, and energy conservation. Thus, by the Pythagorean
     theorem, E1 and E2 must differ in phase by π/2. For the equal ratio splitter relevant to this
     problem, one of them is advanced in phase by π/4, while the other is delayed in phase by π/4.

 (b) Consider sending in waves with the same phase and equal amplitude Ein along both fibers 1
     and 2 simultaneously. If the wave that exits along fiber 1 is always advanced, then the final
     amplitudes are
                                 E0 √                                            E0 √
           E1 = (eiπ/4 + eiπ/4 ) √ = 2eiπ/4 E0 ,         E2 = (e−iπ/4 + e−iπ/4 ) √ = 2e−iπ/4 E0 .
                                  2                                               2
     On the other hand, if a wave that enters along fiber 2 exits along fiber 1 retarded instead, the
     final amplitudes are
                                          E0                                      E0
                   E1 = (eiπ/4 + e−iπ/4 ) √ = E0 ,         E2 = (e−iπ/4 + eiπ/4 ) √ = E0 .
                                           2                                       2

     Only the second option respects energy conservation, |E1 |2 + |E2 |2 = 2|E0 |2 , so that one must
     occur. In other words, we have:

                           1 → 1, 2 → 2 : advanced,         1 → 2, 2 → 1 : retarded

 (c) Let’s consider the two components of the waves that eventually exit along fiber 1.

       • Part of the incident wave stays in fiber 1 at the first splitter, getting advanced by π/4. It
         picks up some phase between the two splitters, then gets advanced by π/4 again at the
         second splitter.
       • Part of the incident wave goes into fiber 2 at the first splitter, getting retarded by π/4.
         It picks up some phase between the two splitters, then (by the result of part (b)) gets
         retarded by π/4 again at the second splitter.

     For all the light to come out along fiber 2, these two components that come out along fiber 1
     have to cancel out. That means they need opposite phases, which implies

                       π/4 + kℓ + π/4 − (−π/4 + k(ℓ + ∆ℓ) − π/4) = (2n + 1)π.

     This simplifies to k∆ℓ = 2πn, or nλ = ∆ℓ = 30 µm, from which we conclude

                           n ∈ {46, 47, 48, 49},    λ ∈ {612, 625, 638, 652} nm.

       Remark
       Above, we’ve focused on cases where light can only exit a given optical element in two
       ways. But in general, you could have n “ports”, in which case you would need an
       entire n × n matrix of coefficients S to relate the n input amplitudes to the n output
       amplitudes. Generalizing problem 22 to this case shows that S is a unitary matrix, S † = S −1 .

       All of the optical elements we’ve seen so far obey optical reciprocity, i.e. the principle that
       “if I can see you, then you can see me”, related to time reversal symmetry. However, we can
       violate reciprocity by using special materials, such as permanent magnets. For example, a
        circulator is an optical element with 3 ports, so that all light entering port 1 exits from port
       2, light in port 2 exits from port 3, and light in port 3 exits from port 1. Though it’s strange,
       you can check that there’s no way to use it to violate the second law of thermodynamics.


[User]
Physics Cup 2024, problem 3. This relatively straightforward problem reviews almost
    everything we’ve covered so far.

       Example 7: Griffiths 6.27

       How does a magnetic field line bend when it passes from one medium to another?

       Solution
       We say the field lines “bend” because of Gauss’s law for magnetism: they can’t start or end,
       so each one has to keep on going. Let’s suppose the figure above is drawn in the xz plane.
       Applying Gauss’s law for magnetism in a small pillbox spanning the interface gives

                                                  B1z = B2z .

       On the other hand, since ∇ × H is zero (assuming no additional, “free” current is around),
       considering an Amperian loop spanning the interface gives

                                                  H1x = H2x .

       Combining these results gives
                                                 tan θ2   µ2
                                                        =    .
                                                 tan θ1   µ1
       In other words, when a field line enters a medium with higher µ, it bends away from the
       normal, and when it enters a medium with lower µ, it bends towards the normal.

       This statement has two limiting cases which will be important later.

         • A magnetic field line can’t enter a superconductor (µ2 = 0) at all, so field lines approaching
           a superconductor bend away, to become tangent to them (θ1 → 90◦ ).

         • A magnetic field line entering a soft ferromagnet (µ2 → ∞) bends towards it to enter along
           the normal direction (θ1 → 0◦ ), similar to how electric field lines approach conductors.
           It’s also possible for θ1 to be nonzero if θ2 → 90◦ , but we won’t see any examples of this.

       You can see both of these behaviors in the limiting cases of problem 10. In general, we
       conclude that magnetic field lines are “attracted” to regions of higher µ, which makes sense
       because it helps minimize the energy. Soft ferromagnets tend to keep magnetic field lines
       within themselves, which is why they’re used in transformers.

[2] Problem 13 (IPhO 2012 Experiment). Water is a diamagnetic substance. A powerful cylindrical
    magnet with field B is placed below the water surface.
     (a) Which of the following shows the resulting shape of the water surface?

         The magnet is roughly 2/3 as wide as each of these sketches.

     (b) Let ρ be the density of the water. If the maximum change in height of the water surface has
         magnitude h, find an approximate expression for the magnetic susceptibility χm of water.
    For a very closely related, but more extreme problem, see EuPhO 2018, problem 2.

Solution:
See the official solutions here.

       Example 7: Griffiths 6.27

       How does a magnetic field line bend when it passes from one medium to another?

       Solution
       We say the field lines “bend” because of Gauss’s law for magnetism: they can’t start or end,
       so each one has to keep on going. Let’s suppose the figure above is drawn in the xz plane.
       Applying Gauss’s law for magnetism in a small pillbox spanning the interface gives

                                                  B1z = B2z .

       On the other hand, since ∇ × H is zero (assuming no additional, “free” current is around),
       considering an Amperian loop spanning the interface gives

                                                  H1x = H2x .

       Combining these results gives
                                                 tan θ2   µ2
                                                        =    .
                                                 tan θ1   µ1
       In other words, when a field line enters a medium with higher µ, it bends away from the
       normal, and when it enters a medium with lower µ, it bends towards the normal.

       This statement has two limiting cases which will be important later.

         • A magnetic field line can’t enter a superconductor (µ2 = 0) at all, so field lines approaching
           a superconductor bend away, to become tangent to them (θ1 → 90◦ ).

         • A magnetic field line entering a soft ferromagnet (µ2 → ∞) bends towards it to enter along
           the normal direction (θ1 → 0◦ ), similar to how electric field lines approach conductors.
           It’s also possible for θ1 to be nonzero if θ2 → 90◦ , but we won’t see any examples of this.

       You can see both of these behaviors in the limiting cases of problem 10. In general, we
       conclude that magnetic field lines are “attracted” to regions of higher µ, which makes sense
       because it helps minimize the energy. Soft ferromagnets tend to keep magnetic field lines
       within themselves, which is why they’re used in transformers.

[2] Problem 13 (IPhO 2012 Experiment). Water is a diamagnetic substance. A powerful cylindrical
    magnet with field B is placed below the water surface.
     (a) Which of the following shows the resulting shape of the water surface?

         The magnet is roughly 2/3 as wide as each of these sketches.

     (b) Let ρ be the density of the water. If the maximum change in height of the water surface has
         magnitude h, find an approximate expression for the magnetic susceptibility χm of water.
    For a very closely related, but more extreme problem, see EuPhO 2018, problem 2.
    Solution. (a) For a diamagnetic substance, µ < µ0 , so the magnetic field energy is higher when
         water is present. The water surface is an equipotential, so a higher magnetic field energy at
         some points must be compensated by a lower gravitational potential energy. Thus, the answer
         is option D.

     (b) Equating the change in gravitational potential energy with the change in field energy, both
         per volume, gives
                                        B2   B2     B2            B2
                                ρgh =      −     =      (µ0 − µ) ≈ 2 (µ0 − µ)
                                        2µ   2µ0   2µµ0           2µ0
         where the last step follows because µ ≈ µ0 . We thus have
                                                   µ − µ0    2µ0 ρgh
                                            χm =          =−         .
                                                     µ0        B2
         With a strong magnet, and a measurement of h accurate to about 0.1 mm, one can indeed
         detect this effect. Note that if you treated the dipole moment as permanent, and used a
         potential energy density −M · B, your answer here would be off by a factor of 2.