[User]
This is a key question which will help you understand idea 7. A spherical shell with
    radius R and uniform surface charge density σ spins with angular frequency ω about a diameter.

      (a) Find the magnetic field at the sphere’s center.

      (b) Find the magnetic dipole moment of the sphere.

      (c) It can be shown that (1) the magnetic field inside the sphere is uniform, and (2) the magnetic
          field outside the sphere is exactly that of a magnetic dipole. (It requires doing some obnoxious
          integrals, as can be seen in section 5.4 of Griffiths.) Using this information, make a qualitatively
          accurate sketch of the field.

      (d) There is a closely related question in electrostatics: suppose we had two solid spheres of the
          same radius R, with volume charge densities ±ρ, and the spheres were displaced by a tiny
          distance d ≪ R. Qualitatively, what would the electric field of this setup look like, for r < R
          and r > R? How does it differ from your answer to part (c)?

[2] Problem 18 (Purcell 6.12). A ring with radius R carries a current I. Show that the magnetic
    field due to the ring, at a point in the plane of the ring, a distance r from the center, is given by

                                         µ0 I π       (R − r cos θ)R dθ
                                              Z
                                    B=                                     .
                                          2π 0 (r + R2 − 2rR cos θ)3/2

    In the r ≫ R limit, show that
                                                 µ0 m
                                                 B≈
                                                 4π r3
    where m = IA is the magnetic dipole moment of the ring, as expected from idea 6.

[3] Problem 19 (Purcell 6.14). Consider a square loop with current I and side length a centered
    at the origin, with sides parallel to the x and y axes. Show that the magnetic field at rx̂ is
    B ≈ (µ0 /4π)(m/r3 ) for r ≫ a, as expected from idea 6. Be careful with factors of 2!

Solution:
(a) First, we find the field due to a ring of counterclockwise current I with radius a in
         the z = 0 plane at a point directly above the center at some height z. Using the Biot–Savart
         law, we see that
                                                  a
                                       µ0 I 2πa √a2 +z 2      µ0 I    a2
                                  B=                     ẑ =                   ẑ.
                                        4π a2 + z 2            2 (a2 + z 2 )3/2
         Let us work in spherical coordinates with the axis of rotation being the z axis. Then, at angle
         θ, we essentially have a ring of charge of radius a = R sin θ, z = R cos θ, and

                                     dI = σR(dθ)ω(R sin θ) = σR2 ω sin θ dθ.

         Therefore,
                                   µ0 σR2 ω sin θ dθ R2 sin2 θ     ẑ
                              dB =                        3
                                                               ẑ = µ0 σωR sin3 θ dθ.
                                           2            R          2
         Integrating from 0 to π to obtain the full field,
                                                   Z π
                                        ẑ                          2
                                   B = µ0 σωR          sin3 θ dθ = µ0 σωR ẑ.
                                        2            0              3

     (b) The magnetic dipole moment of a slice is

                                         dm = ẑπ(R sin θ)2 ωσR2 sin θ dθ.

         Integrating this gives

                                                  m = πωσR4 ẑ.

      (c) The field is as shown below.

          The key feature is that the field lines of the dipole outside and the uniform field inside match
          up perfectly, so that every field line forms a closed loop; this is Gauss’s law for magnetism.

      (d) We’ve already seen this setup in E1. It’s easy to show by the shell theorem that the electric
          field is uniform for r < R, and exactly an electric dipole field for r > R. The crucial difference
          is that the uniform field inside points along the dipole moment, rather than against it. As a
          result, the electric field switches directions upon crossing r = R. This reflects the fact that
          ∇ × E = 0, so that the line integral of E along any closed path has to vanish. If we follow
          a field line, this is only possible if the field switches direction. In addition, compared to the
          magnetic case, the field lines don’t close up, because ∇ · E isn’t zero.

[2] Problem 18 (Purcell 6.12). A ring with radius R carries a current I. Show that the magnetic
    field due to the ring, at a point in the plane of the ring, a distance r from the center, is given by

                                         µ0 I π       (R − r cos θ)R dθ
                                              Z
                                    B=                                     .
                                          2π 0 (r + R2 − 2rR cos θ)3/2

    In the r ≫ R limit, show that
                                                 µ0 m
                                                   B≈
                                                 4π r3
    where m = IA is the magnetic dipole moment of the ring, as expected from idea 6.

    Solution. Let the ring be centered at the origin, and let the field point be ax̂, and say we are at
    an angle θ. Then, dl = (−R sin θ) x̂ + (R cos θ) ŷ, and

                      r = a x̂ − R(cos θ x̂ + sin θ ŷ) = (a − R cos θ) x̂ + (−R sin θ) ŷ.
                        √
    Note that r = |r| = a2 + R2 − 2aR cos θ. Therefore, by the Biot–Savart law,

                                             µ0 I  R(R − a cos θ)ẑ dθ
                                      dB =                               ,
                                             4π (a2 + R2 − 2aR cos θ)3/2

    so integrating from 0 to 2π and noting that θ and −θ contribute the same, we arrive at the desired
    result. Now, to take the r ≫ R limit cleanly and consistently, it’s best to nondimensionalize
    everything. Defining x = R/r ≪ 1, we can pull dimensionful factors out of the integral to get
                                     µ0 I rR π         x − cos θ
                                             Z
                                 B=        3
                                                                      dθ.
                                      2π r 0 (1 + x − 2x cos θ)3/2

    Now, it’s not immediately obvious to what order in x we should expand in. If we already know the
    answer is proportional to 1/r3 , then we can see the answer must be first order in x. But if we didn’t
    know that, we could expand to zeroth order, giving
                                            µ0 I R π
                                                   Z
                                       B=              (− cos θ) dθ = 0.
                                             2π r2 0
    The fact that the answer vanishes means we need to go to higher order to find the true answer. At
    first order, applying the binomial theorem, the integrand is
          (x − cos θ)(1 + x2 − 2x cos θ)−3/2 ≈ (x − cos θ)(1 + 3x cos θ) ≈ − cos θ + x(1 − 3 cos2 θ)
    where we threw away higher order terms throughout. Then
                                        µ0 I R 2 π
                                                Z
                                   B=              (1 − 3 cos2 θ) dθ.
                                        2π r3 0
    Using the fact that cos2 θ averages to 1/2 over a cycle, the integral is −π/2, giving
                                                µ0 I πR2   µ0 m
                                           B=         3
                                                         =
                                                4π r       4π r3
    as desired.
[3] Problem 19 (Purcell 6.14). Consider a square loop with current I and side length a centered
    at the origin, with sides parallel to the x and y axes. Show that the magnetic field at rx̂ is
    B ≈ (µ0 /4π)(m/r3 ) for r ≫ a, as expected from idea 6. Be careful with factors of 2!
    Solution. This calculation is a bit subtle. It is tempting to ignore the sides parallel to x̂, because
    the current is almost parallel to r, so ds × r is small; more precisely, it is suppressed by a power of
    a/r. The sides parallel to ŷ do each give much larger contributions, but they have opposite sign and
    nearly cancel out, suppressing their sum by a power of a/r. So all four sides need to be considered.
        First consider the segments parallel to x̂. We get a factor of (a/2)/r from the sin θ factor in the
    cross product. Similarly, a appears in the Biot–Savart integral in the denominator; however, its
    effect here would give higher-order terms in a/r, which we don’t want to keep since they’re much
    smaller than the final answer. So the segments each contribute equally, for a total of
                                                    a/2
                                                                !
                                           µ0 I a r       a a/2
                                                             r        µ0 I a 2
                                   B1 = −               +         = −          .
                                           4π      r2      r2         4π r3
    Next, the segments parallel to ŷ contribute a total of
                                                                  µ0 I 2a2

                                   µ0 I       a             a
                            B2 =                     −          =
                                    4π (r − a/2)2 (r + a/2)2      4π r3
    where we work to the same accuracy as for B1 . Adding the two contributions and using m = Ia2

                    01W
    gives the desired result. If you forget to count B1 , you’ll get an answer that is two times too big.


[User]
8. Consider a circle √
                      of radius R. A point charge lies at a distance a from its center and on its axis
    such that R = a 3. If electric flux passing through the circle is φ then the magnitude of the
    point charge is
         √
     (a) 3ǫ0 φ
    (b) 2ǫ0 φ
               √
     (c) 4ǫ0 φ/ 3
    (d) 4ǫ0 φ

 HB

Solution:
8.

   HB
                                                                3
                                                           γ=
                                                                2
         9.
                                                          5V
                                                       3−
                                                          V0
                                                 C = R
                                                          2V
                                                       1−
                                                          V0

INPhO 2009 Problem 8


[User]
Ekraan, kaks peeglit ning koherentse monokromaatilise kiirguse allikas (kiiratav lainepikkus $\lambda$) on paigutatud nii, nagu näidatud joonisel. Valgusallikast jõuab ekraanile vaid peegeldunud valgus. Peeglite normaalide vaheline nurk on $2 \alpha$. Peeglite puutepunkti ning valgusallika vaheline kaugus on $a$ ning valgusallika kaugus ekraanist on samuti $a$. On teada, et punkti $O$ lähedusse tekib interferentsimuster, kus järjestikuste tumedate triipude vaheline kaugus on $d$. Leida lainepikkus $\lambda .$ Eeldada, et $a \gg d$.
\begin{center}
	\includegraphics[width=0.6\linewidth]{2004-v3g-09-yl.pdf}
\end{center}

Solution:
Leiame valgusallika kujutised peeglites. Punkti $O$ ümbrusse langevat valgust võime siis vaadelda kujutistest kiiratud valguse superpositsioonina. Leiame järgnevalt $\sin \phi .$ Antud geomeetrilisest konstruktsioonist saame:
$$
x=2 a \cos \alpha \sin \alpha, \quad y=2 a \cos \alpha \cos \alpha+a
$$
$\tan \phi=\frac{x}{y}=\frac{2 a \cos \alpha \sin \alpha}{2 a \cos \alpha \cos \alpha+a}, \quad \sin \phi=\frac{1}{\sqrt{1+(\cot \phi)^{2}}}$,
$\sin \phi=\left[1+\left(\frac{2 a \cos \alpha \cos \alpha+a}{2 a \cos \alpha \sin \alpha}\right)^{2}\right]^{-\frac{1}{2}}=\left[1+\left(\frac{\cos \alpha+1 /(2 \cos \alpha)}{\sin \alpha}\right)^{2}\right]^{-\frac{1}{2}}$
$=\frac{\sin \alpha}{\sqrt{\sin ^{2} \alpha+\left(\cos \alpha+1 /(2 \cos \alpha)^{2}\right)}}=\frac{\sin \alpha}{\sqrt{2+1 /\left(4 \cos ^{2} \alpha\right)}}=\frac{\sin 2 \alpha}{\sqrt{8 \cos ^{2} \alpha+1}}$

\begin{center}
	\includegraphics[width=0.8\linewidth]{2004-v3g-09-lah.pdf}
\end{center}

Olgu punkti $O$ läheduses mingis punktis $M$ interferentsi miinimum. Kui liiguksime sellest punktist $d$ võrra ülespoole, siis alumisest kujutisest tuleva valguse teekond pikeneb $d \sin \phi$ vōrra ning ülemisest kujutisest tuleva valguse teekond lüheneb $d \sin \phi$ võrra. Seega saame punktiga $M$ võrreldes käiguvahe $2 d \sin \phi$. Sellele vastab miinimum juhul kui $\lambda=2 d \sin \phi$. Seega saamegi vastuseks
\[
\lambda=\frac{2 d \sin 2 \alpha}{\sqrt{8 \cos ^{2} \alpha+1}}
\]

Peeglid

Ekraan, kaks peeglit ning koherentse monokromaatilise kiirguse allikas (kiiratav lainepikkus $\lambda$) on paigutatud nii, nagu näidatud joonisel. Valgusallikast jõuab ekraanile vaid peegeldunud valgus. Peeglite normaalide vaheline nurk on $2 \alpha$. Peeglite puutepunkti ning valgusallika vaheline kaugus on $a$ ning valgusallika kaugus ekraanist on samuti $a$. On teada, et punkti $O$ lähedusse tekib interferentsimuster, kus järjestikuste tumedate triipude vaheline kaugus on $d$. Leida lainepikkus $\lambda .$ Eeldada, et $a \gg d$.
\begin{center}
	\includegraphics[width=0.6\linewidth]{2004-v3g-09-yl.pdf}
\end{center}


[User]
Show explicitly that the norm of the displacement four-vector is invariant under
    Lorentz transformations, i.e. that
                             (∆s)2 = ∆s · ∆s = (∆t)2 − (∆x)2 − (∆y)2 − (∆z)2
    is Lorentz invariant. Since all four-vectors transform the same way, this proves it for all of them.

        Example 3

        Find a four-vector representing the velocity of a particle with position x(t).

        Solution
        Just as multiplying an ordinary vector with a rotational invariant produces another vector,
        multiplying or dividing a four-vector with a Lorentz invariant gives another four-vector. In
        this case, the appropriate four-vector is found by dividing displacement by the proper time
        experienced by the particle,
                                               dxµ       dxµ
                                         uµ =        =γ      = (γ, γv)
                                                dτ        dt
                                                              √
        where v = dx/dt is the spatial velocity and γ = 1/ 1 − v 2 as usual. Since its spatial part
        reduces to the spatial velocity in the limit of low speeds, it is the relativistic generalization
        of the spatial velocity. Finally, we define the four-momentum as pµ = muµ = (E, p), where
        E = γm and p = γmv are the relativistic energy and momentum.

        Example 4

        Give a simple interpretation of the squared norm of a particle’s four-velocity, u · u, and its
        four-momentum, p · p.

        Solution
        The answers have to be simple, because they must be invariants that only depend on the
        intrinsic properties of the particle, i.e. only on the invariant mass m. For the four-velocity,

                                             u · u = γ 2 − γ 2v2 = 1

        which is clearly invariant. For the four-momentum we have p · p = m2 .

       Remark
       Pop-science books usually describe the result u · u = 1 by saying that “particles always move
       with the same speed through spacetime”, just like how a particle in uniform circular motion
       always has the same spatial speed. However, this is very misleading, because it makes people
       think that if dx/dτ increases in magnitude, then dt/dτ decreases. In fact it’s the opposite:
       time dilation means more time passes for each tick of a moving clock, so dt/dτ increases.
       The analogy can’t work, because inner products of four-vectors have terms with minus signs,
       while ordinary inner products of three-vectors don’t. The point, as always, is that there are
       a lot of simple things in physics which are almost impossible to explain properly with fuzzy
       math-free analogies. With math, relativity can make sense to high school students. Without
       math, it can’t really make sense to anyone.

       Example 5

       Give a simple interpretation of the inner product of two momentum four-vectors, p1 · p2 .

       Solution
       By definition, this is equal to m1 m2 u1 · u2 , and since the inner product is invariant, we can
       evaluate u1 · u2 in any frame. Suppose we work in the frame of the first particle, where

                                  µ               µ          1         v
                                 u1 = (1, 0), u2 =       √        , √         .
                                                           1 − v2    1 − v2
       Carrying out the inner product, we have the relatively simple result
                                                        m1 m2
                                             p1 · p2 = √
                                                         1 − v2
       where v is the relative speed, meaning the speed of one particle in the frame of the other.

Solution:
Plugging in the Lorentz transformations, we have

                   (∆s′ )2 = (∆t′ )2 − (∆x′ )2 − (∆y ′ )2 − (∆z ′ )2
                          = γ 2 (∆t − v∆x)2 − γ 2 (∆x − v∆t)2 − (∆y)2 − (∆z)2
                          = γ 2 (1 − v 2 )(∆t)2 − γ 2 (1 − v 2 )(∆x)2 − (∆y)2 − (∆z)2
                          = (∆t)2 − (∆x)2 − (∆y)2 − (∆z)2

as desired.

   Example 3

   Find a four-vector representing the velocity of a particle with position x(t).

   Solution
   Just as multiplying an ordinary vector with a rotational invariant produces another vector,
   multiplying or dividing a four-vector with a Lorentz invariant gives another four-vector. In
   this case, the appropriate four-vector is found by dividing displacement by the proper time
   experienced by the particle,
                                          dxµ       dxµ
                                     uµ =       =γ      = (γ, γv)
                                           dτ        dt
                                                         √
   where v = dx/dt is the spatial velocity and γ = 1/ 1 − v 2 as usual. Since its spatial part
   reduces to the spatial velocity in the limit of low speeds, it is the relativistic generalization
   of the spatial velocity. Finally, we define the four-momentum as pµ = muµ = (E, p), where
   E = γm and p = γmv are the relativistic energy and momentum.

   Example 4

   Give a simple interpretation of the squared norm of a particle’s four-velocity, u · u, and its
   four-momentum, p · p.

   Solution
   The answers have to be simple, because they must be invariants that only depend on the
   intrinsic properties of the particle, i.e. only on the invariant mass m. For the four-velocity,

                                         u · u = γ 2 − γ 2v2 = 1

   which is clearly invariant. For the four-momentum we have p · p = m2 .

   Remark
   Pop-science books usually describe the result u · u = 1 by saying that “particles always move
   with the same speed through spacetime”, just like how a particle in uniform circular motion
   always has the same spatial speed. However, this is very misleading, because it makes people

       think that if dx/dτ increases in magnitude, then dt/dτ decreases. In fact it’s the opposite:
       time dilation means more time passes for each tick of a moving clock, so dt/dτ increases.
       The analogy can’t work, because inner products of four-vectors have terms with minus signs,
       while ordinary inner products of three-vectors don’t. The point, as always, is that there are
       a lot of simple things in physics which are almost impossible to explain properly with fuzzy
       math-free analogies. With math, relativity can make sense to high school students. Without
       math, it can’t really make sense to anyone.

       Example 5

       Give a simple interpretation of the inner product of two momentum four-vectors, p1 · p2 .

       Solution
       By definition, this is equal to m1 m2 u1 · u2 , and since the inner product is invariant, we can
       evaluate u1 · u2 in any frame. Suppose we work in the frame of the first particle, where

                                  µ               µ          1         v
                                 u1 = (1, 0), u2 =       √        , √         .
                                                           1 − v2    1 − v2
       Carrying out the inner product, we have the relatively simple result
                                                         m1 m2
                                              p1 · p2 = √
                                                          1 − v2
       where v is the relative speed, meaning the speed of one particle in the frame of the other.

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


[User]
01W
                      USAPhO 2017, problem A2. A simple but tricky problem. (For a more involved
    problem based on the same fundamental idea, see Physics Cup 2019, problem 2.)

Solution:
USAPhO 2017, problem A2. A simple but tricky problem. (For a more involved

                         01^
    problem based on the same fundamental idea, see Physics Cup 2019, problem 2.)


[User]
In this problem, we consider a simple kinetic theory model for “transport” in a
    sparse gas. Two parallel plates are at a distance L from each other. The space between the plates is
    filled with a gas of density ρ, with molecule of mass m. Assume that the gas density is low enough
    to neglect collisions between gas molecules, and that all radiation effects can be neglected.
      (a) Suppose the two plates are held at temperatures T + ∆T and T , where ∆T ≪ T . Assume
          that when gas molecules bounce from a plate, they instantly obtain the plate’s temperature.
          Estimate the heat flux per area P/A (in units of W/m2 ) between the plates. (This determines
          the thermal conductivity, to be defined in T2.)
      (b) Suppose the top plate moves with a transverse speed v∥ parallel to itself, while the bottom
          plate is still, and mv∥2 ≪ kB T . Assume that when gas molecules bounce from a plate, they

         obtain that plate’s transverse speed. Estimate the force per area F/A on each plate. (This
         determines the viscosity, as defined in M7.)
   Both the thermal conductivity and viscosity are associated with the transport of conserved quantities
   (energy and transverse momentum, respectively) through the gas. Now, above we have neglected
   collisions entirely, which is unrealistic for most gases unless they are extremely sparse. However,
   it’s possible to get some rough estimates for what happens in the presence of collisions.
     (c) Let σ be the cross-sectional area of a gas molecule. Estimate the mean free path d of a gas
         molecule.

    (d) Our previous analysis held when L ≪ d. Now suppose, more realistically, that L ≫ d.
        Estimate P/A and F/A in the steady state. (Hint: consider a set of imaginary planes that
        are spaced by a distance d, and consider the heat/momentum flow between pairs of adjacent
        planes; for such pairs, collisions can be neglected.)

     (e) Suppose a gas molecule begins at the top plate, and both plates have temperature T . Again
         assuming that L ≫ d, estimate the typical time it takes before the gas molecule hits the
         bottom plate. (Hint: model the position of the particle as a sum of random steps, and recall
         the facts about variance in P2.)

                   01h
   For a clear discussion of the ideas of this problem, see sections 7.3 and 7.4 of The Art of Insight.

Solution:
(a) The gas molecules have speed v ∼ kB T /m, and they hit the plates every time
      t ∼ L/v. Each time a plate is hit, an energy ∆E ∼ kB ∆T is deposited or taken away. So the
      average energy flow rate for one atom is
                                                              r
                                                ∆E     kB ∆T kB T
                                       Patom =      ∼                .
                                                 t       L        m
      The number of atoms per unit area of the plate is Lρ/m, which gives
                                                       r
                                       P            ρ kB T
                                          ∼ kB ∆T           .
                                       A            m     m

 (b) This is rather similar, except that instead of an energy ∆E deposited every impact, a momen-
     tum ∆p∥ = mv∥ is deposited per impact. Then by the exact same logic,
                                                       r
                                           F       ρ       kB T
                                             ∼ mv∥              .
                                           A       m        m
      Notice that the result of this problem depends explicitly on the molecular mass m. This is
      very important, because in the early days of kinetic theory, the quantity m canceled out of
      almost every calculation. This led people to believe that atoms were unobservable in principle;
      they were just a fictitious calculational tool used to approximate continuous matter. One of
      Einstein’s most famous discoveries was showing that macroscopically observable properties of
      Brownian motion did depend on m, hence giving a way to indirectly observe the reality of
      atoms; you will do this calculation in XRev.

  (c) If a molecule moves a distance x, it’ll sweep out a volume of σx. The number density is ρ/m,
      so the molecule should expect ρσx/m collisions. The mean free path d is when you would
      expect 1 collision:
                                                      m
                                                  d=     .
                                                      σρ

    (d) The results for when collisions can be neglected,
                                              r                     r
                                P           ρ kB T        F       ρ kB T
                                   ∼ kB δT                  ∼ mδv
                                A           m    m        A       m   m
        can still give us the heat/momentum flow between adjacent planes separated by distance d
        (so a total of L/d of these planes). Thus ∆T = Ld δT and v = Ld δv. This transfer will be the
        same throughout in steady state, and plugging in the previously obtained d = m/σρ yields
                                               r                  r
                                   P    kB ∆T    kB T     F    mv kB T
                                      ∼                     ∼             .
                                   A      σL      m       A    σL     m
         Remarkably, the density ρ has completely dropped out! This was one of the most surprising
         early predictions of kinetic theory, which was confirmed by Maxwell. For higher densities,
         more molecules are available to carry energy or momentum, but they can’t go as far before
         colliding.
                                                      p
     (e) The molecule will be moving at velocity v ∼ kB T /m, so it takes time t ∼ d/v to go between
         planes of separation d. However, after this point the molecule’s velocity will be randomized,
         so its next step might go in any direction. This is a description of a random walk.
        Each step of time t changes the displacement from the original plate by approximately ±d,
        and the steps are independent. Then the standard deviation of the displacement adds in
        quadrature, as covered in P2. The typical time to hit is when we get a standard deviation of
        order L,
                                               2
                                                         L2
                                                            r
                                               L     d           m
                                       thit ∼          ∼             .
                                               d     v    d     kB T


[User]
Theory
English (Official)
                                                                                T1

                                                               N
                                         R                                                                     M
                                                                                m

                                                                        M          R

                                                                             i
               fi          i   N                                                                         f0

                                              V cr
                                    fi       i = 1, . . . , N. f0   N
                                                             c

                                                                                vrms
                                                                                N fi
        i = 1, . . . , N       f0
                                                                            vrms       m

                                                      page 1 of 5

Theory
English (Official)
                                                                                T1

                                   ⟨K⟩ t −γ⟨U⟩ t

     ⟨K ⟩t                                   ⟨U⟩t
γ

                                         → →
                                   Γ = ∑ pi ⋅ r i
                                         i

                                                                       dΓ/dt
         ⟨ dΓ ⟩
           dt t             γ
                                                                   Γ

                  N mg R   v rms   mg

Rg
                                                              Rg
                                        ms
                     n

                                                     v(r)
                                                v(r)     r < Rg        r ≥ Rg

                                                                                v(r)

                                     page 2 of 5

Theory
English (Official)
                                                                                T1

                               v(r)                  r ≤ Rg            v0   r > Rg

                               mR
                 Rg            v0     Rg

                                                              r R g v0 n    ms
        r < Rg        r ≥ Rg

                                                                                mp
       n(r)                    T(r)                                                    r

                                           page 3 of 5

Theory
English (Official)
                                                      T1

                      dP/dr        m ′ (r) r   n(r)
        m′   (r)                                r

                     page 4 of 5

Theory
English (Official)
                                                             T1

                                       m ′ (r)   n(r) T(r)
                     r

                                                    T0

                                  α
                     n(r) =
                              r(β + r)2

     α      β

                                       r

                         page 5 of 5

Solution:
Solutions/                                                                           T1
 Marking Scheme

                                          Dark Matter
A. Cluster of Galaxies
   Question A.1
                                           Answer                                             Marks
     Potential energy for a system of a spherical object with mass
             4
     M (r )   r 3  and a test particle with mass dm at a distance r is given by
             3                                                                                0.2 pts
                                                    M (r )
                                         dU  G           dm
                                                     r
     Thus for a sphere of radius R

                   R   M (r )          R 4 r 3                 16          R
          U   G            dm    G          4 r 2  dr   G 2  2  r 4 dr           0.6 pts
                  0     r             0    3r                     3         0

                  16
                   G 2  2 R5
                  15
     Then using the total mass of the system

                                               4
                                            M   R3 
                                               3
                                                                                0.2 pts
     we have

                                              3 GM 2
                                       U 
                                              5 R

                                                                                Total   1.0 pts

                                           Page 1 of 14

  Solutions/                                                                       T1
Marking Scheme

 Question A.2
                                              Answer                                     Marks
   Using the Doppler Effect,
                                              1
                                   fi  f0         f 0 (1   ) ,
                                             1 

   where   v c and v  c . Thus the i‐th galaxy moving away (radial) speed
   is
                                                  fi  f0
                                        Vri             c                              0.2 pts
                                                     f0

   Alternative without approximation:
                      1
           fi  f0
                     1 

                  f      
          Vri  c 0  1
                   fi    

   All the galaxies in the galaxy cluster will be moving away together due to the
   cosmological expansion. Thus the average moving away speed of the N
   galaxies in the cluster is

                                  c N                        c N  fi      
                       Vcr               f i  f 0            1 .
                                 Nf 0 i 1                   N i 1  f 0               0.3 pts

   Alternative without approximation:

                  cf 0 N  1 1  c N  f 0 
          Vcr                1
                  N i 1  f i f 0  N i 1  f i 

                                                                                Total   0.5 pts

                                               Page 2 of 14

  Solutions/                                                                                T1
Marking Scheme

 Question A.3
                                                  Answer                                             Marks
   The galaxy moving away speed Vi , in part A.2, is only one component of the
   three component of the galaxy velocity. Thus the average square speed of each
   galaxy with respect to the center of the cluster is

   1 N                     1 N
     
   N i 1
          (V i  V c ) 2
                            
                           N i 1
                                  (Vxi  Vxc ) 2  (Vyi  Vyc ) 2  (Vzi  Vzc ) 2                   0.5 pts

   Due to isotropic assumption

   1 N                    3 N
     
   N i 1
          (V i  V c ) 2   (Vri  Vcr ) 2
                          N i 1

   And thus the root mean square of the galaxy speed with respect to the cluster
   center is

             3 N                        3 N                                    3  N 2
   vrms       
             N i 1
                    (Vri  Vrc ) 2       
                                        N i 1
                                                   2               2
                                               (Vri  2VcrVri  Vcr )            Vri   3Vcr
                                                                               N  i 1 
                                                                                2

              1 N  f       
                               2
                                  1 N  f      
                                                    2

   vrms  c 3     1       1 
                           i                  i
               N i 1  f 0  
                                   i 1  0  
                                    N       f
                                                                                              0.7 pts

                                        1 N 2                  
   
       c 3 1 N 2
                                2 
                                                    f N
              fi  2 fi f0  f0      fi   2 0  fi  f0 
                                                                2
                                                                  
        f 0  N i 1                  N i 1   N i 1        
                                        2
     c 3  N 2  N 
           N  fi     fi 
     f 0 N  i 1   i 1 

   Alternative without approximation:

                                                Page 3 of 14

  Solutions/                                                                                T1
Marking Scheme

             1 N f           1 N  f 0  
                             2                     2
             
             N
  vrms  c 3           1      1 
                        0
                                                
               i 1  f i     N i 1  f i 

                                                   1 N 1                                
                                                                       2
    c 3      1 N  1           1   1    1            2 1 1  1  1 
                                                                                N
                         2               
     f0       N i 1  f 2
                      i       f i f 0 f 0 2     N i 1 f i    N f 0 i 1 f i f 0 2 
                                                                                             

    cf 0 3  N  1    N 1 
                        2             2

             
            N      
                     f 
             i 1  i    i 1 i 
       N                          f

  The mean kinetic energy of the galaxies with respect to the center of the cluster
  is

            m 1 N                    m                                                                    0.3 pts
                
                                           2
  K ave             (V i  V c ) 2  vrms
            2 N i 1                 2

                                                                                Total   1.5 pts

                                               Page 4 of 14

  Solutions/                                                                                T1
Marking Scheme

 Question A.4
                                                                      Answer                                                                 Marks
   The time average of d  / dt vanishes

                                                                      d
                                                                           0
                                                                      dt t

   Now                                                                                0.6 pts
                                                            
                 d d                 dpi            dri
                     pi  ri               ri   pi 
                 dt dt i            i    dt          i      dt
                                                  
                     Fi  ri   mi vi  vi   Fi  ri  2 K
                            i                      i                          i

   Where K is the total kinetic energy of the system. Since the gravitational force on
   i‐th particle comes from its interaction with other particles then
                                                             
     Fi  ri   Fji  ri   Fji  ri   Fij  ri   Fji  ri   Fji  rj
    i        i , j i               i j               i j                       i j               i j
                                                       
                                        mi m j (ri  rj )          mi m j
             Fji  (ri  rj )   G   2    (ri  rj )   G    U tot
             i j                  i j | ri  rj | | ri  rj |  i j | ri  rj |

  Alternative proof:
                                                                                0.9 pts
                        .                      .                      .                  .             .                ⋯       .
                                ,

                                           .                      .                  .           ⋯                  .

                                                                          .                  .                  .           ⋯       .    ...

                                                              .                      .                      .           ⋯           .

   Collecting terms and noting that                                                 we have

                                                              Page 5 of 14

  Solutions/                                                          T1
Marking Scheme

     .                .                   .           ⋯       .

                          .           ⋯       .           ⋯       .

                                                  .

  Thus we have

                                      d
                                          U  2K
                                      dt
                                                                            0.2 pts
                                           d
  And by taking its time average we obtain     U  2 K  0 and thus
                                           dt           t

            1                    1
   K t      U t . Therefore   .
            2                    2

                                                                      Total 1.7 pts

                                      Page 6 of 14

  Solutions/                                                             T1
Marking Scheme

 Question A.5
                                       Answer                                      Marks
   Using Virial theorem, and since the dark matter has the same root mean square
   speed as the galaxy, then we have
                    1
          K t       U t                                                          0.3 pts
                    2

          M 2      1 3 GM 2
            vrms 
          2        25 R
  From which we have
                     2
                5 Rv rms                                                           0.1 pts
          M 
                  3G
   And the dark matter mass is then
                        2
                   5 Rv rms                                                        0.1 pts
          M dm              Nm g
                     3G

                                                                           Total 0.5 pts

                                      Page 7 of 14

   Solutions/                                                                       T1
 Marking Scheme

B. Dark Matter in a Galaxy
   Question B.1
                                                Answer                                       Marks
              Answer B.1: The gravitational attraction for a particle at a distance r from
              the center of the sphere comes only from particles inside a spherical
              volume of radius r. For particle inside the sphere with mass ms , assuming
              the particle is orbiting the center of mass in a circular orbit, we have       0.3 pts
                  m' (r )m s m s v02
              G             
                      r2       r

              with m' ( r ) is the total mass inside a sphere of radius r

                           4 3
              m' ( r )      r m s n
                           3

              Thus we have                                                                   0.2 pts
                                        1/ 2
                        4Gnms 
              v (r )           r
                           3   

              While for particle outside the sphere, we have
                                1/ 2                                                         0.2 pts
                 4Gnms R 3 
      v (r )              
                    3r       

                                               Page 8 of 14

  Solutions/                                                                  T1
Marking Scheme

         The sketch is given below

                                                                                0.1 pts

             Sketch of the rotation velocity vs distance from the center of galaxy

                                                                               Total 0.8 pts

 Question B.2
                                             Answer                                   Marks
         The total mass can be inferred from

             m' ( R g ) m s       ms v02
         G            2
                              
                 Rg                Rg
                                                                                0.5 pts
         Thus

                                  v02 R g
         m R  m' ( R g ) 
                                    G

                                                                               Total 0.5 pts

                                            Page 9 of 14

  Solutions/                                                                             T1
Marking Scheme

 Question B.3
                                                   Answer                                     Marks
         Base on the previous answer in B.1, if the mass of the galaxy comes only
         from the visible stars, then the galaxy rotation curve should fall
         proportional to 1        r on the outside at a distance r > Rg . But in the figure
         of problem b) the curve remain constant after r > Rg , we can infer from

                                                   m' (r )m s m s v02                         0.3 pts
                                            G                        .
                                                      r2        r

          to make v (r ) constant, then m'(r) should be proportional to r for r  R g ,
         i.e. for r > Rg , m' (r )  Ar with A is a constant.

         While for r  R g , to obtain a linear plot proportional to r , then m ' ( r )
         should be proportional to r 3 , i.e. m' (r )  Br 3 .                                0.3 pts

         Thus for r  R g we have

                     r
         m' (r )    t (r )4r ' 2 dr '  Br 3
                     0

                                                                                0.2 pts
         dm' ( r )   t ( r ) 4r 2 dr  3Br 2 dr

                                                      3B
         Thus total mass density  t (r ) 
                                                      4

                                                      v2
                Rg
                   3B                           m
         mR         4r ' 2 dr '  BR g or B  R3  0 2
                                         3

                 0
                   4                           Rg   GRg
                                                                                0.2 pts
                                                                    3v 02
         Thus the dark matter mass density  (r )                               nm s
                                                                 4GR g
                                                                            2

                                             Page 10 of 14

  Solutions/                                                           T1
Marking Scheme

      While for r  R g we have

                     Rg                        r
      m' (r )    (r ' )4r ' 2 dr '    (r ' )4r ' 2 dr '  Ar
                     0                         Rg

                                  r
      m' (r )  m R    (r ' )4r ' 2 dr '  Ar
                                  Rg

                                                                                0.2 pts

       r
        (r ' )4r ' dr '  Ar  M
                          2
                                       0
       R

                                        A
       (r )4r 2  A , or  (r )           .
                                       4r 2

      Now to find the constant A.
       r     A
       4r ' 4r ' dr '  A(r  R )  Ar  m
                              2
       R         2                         g           R

                                       v02
      Thus AR g  m R and A 
                                       G
      We can also find A from the following
                                                                                0.3 pts
           m' ( r ) m s     Arm s m s v02            v02
      G                  G              , thus A      .
               r2            r2     r                G
      Thus the dark matter mass density (which is also the total mass density
      since n  0 for r  R g .

                   v02
       (r )           for r  R g
                 4Gr 2

                                                                        Total 1.5 pts

                                             Page 11 of 14

   Solutions/                                                                   T1
 Marking Scheme

C. Interstellar Gas and Dark Matter

    Question C.1
                                               Answer                                    Marks
      Consider a very small volume of a disk with area A and thickness r, see Fig.1

                                                                                0.3 pts

                                    Figure 1. Hydrostatic equilibrium
      In hydrostatic equilibrium we have
                                  ( P(r )  P(r  r )) A   g (r ) Ar  0

             P      Gm' (r )
                 
             r       r2
                                                                                0.2 pts
             dP      Gm' ( r )                Gm' ( r )
                     2
                                 n( r ) m p           .
             dr       r                        r2
      .

                                                                                Total 0.5 pts

                                            Page 12 of 14

  Solutions/                                                                   T1
Marking Scheme

 Question C.2
                                              Answer                                  Marks
         Using the ideal gas law P = n kT where n = N/V where n is the number
         density, we have
                            dP      dn( r )           dT              GM  '
                                                                      Gm’(r)
                                kT          kn( r )      n(r )m p     2
                            dr       dr               dr                r

         Thus we have                                                                0.5 pts

                         kT  r 2 dn(r )    r 2 dT (r ) 
         m' ( r )                                  .
                        Gm p  n(r ) dr    T (r ) dr 

                                                                               Total 0.5 pts

 Question C.3
                                              Answer                                  Marks
         If we have isothermal distribution, we have dT/dr = 0 and

                        kT0  r 2 dn(r ) 
         m' ( r )                                                                0.2 pts
                        Gm p  n(r ) dr 

         From information about interstellar gas number density, we have
           1 dn(r )     3r  
                    
          n(r ) dr     r (r   )

         Thus we have                                                                0.2 pts
                     kT0 r 3r  
         m' (r ) 
                     Gm p ( r   )

                                         Page 13 of 14

  Solutions/                                                                       T1
Marking Scheme

      Mass density of the interstellar gas is

                      m p
       g (r ) 
                   r (  r )2

      Thus

                                                              kT0 r 3r  
      m' (r )    g (r ' )   dm (r ' ) 4r ' 2 dr ' 
                   r
                                                                                0.3 pts
                   0
                                                              Gm p (r   )

                    m p
                   r
                                                                  kT r 3r  
      m' (r )                      dm (r ' ) 4r ' 2 dr '  0
                0  r ' (  r ' )
                                   2
                                                                  Gm p (r   )

       m p                        2 kT0 3r 2  6r   2
                         ( r ) 4r 
        r (  r )                             (r   ) 2
                    2     dm
                                          Gm p

                         kT0 3r 2  6r   2      m p                                  0.3 pts
       dm (r )                              
                       4Gm p (r   ) r
                                       2 2
                                                r (  r )2

                                                                                Total 1.0 pts

                                            Page 14 of 14

English (Official)


[User]
A uniform stick of length L and mass M begins at rest. A massless rocket is attached
    to the end of the stick, and provides a constant force F perpendicular to the stick. Find the speed
    of the center of mass of the stick after a long time. Ignore gravity. You may find the integral
                                              Z ∞
                                                      2      √
                                                  e−x dx = π,
                                              −∞

    which we first encountered in P1, useful.
[4] Problem 16 (KK 6.41). A plank of length 2L leans nearly vertically against a wall. All surfaces
    are frictionless. The plank starts to slip downward. Find the height of the top of the plank when it
    loses contact with the wall or floor.

       Example 7: EFPhO 2013

       A uniform ball and a uniform ring are both released from rest from the same height on an
       inclined plane with inclination angle θ. They arrive at the bottom of the plane in time TB
       and TR , respectively. The coefficients of friction of both objects with the plane are µk = 0.3
       and µs = 0.5. Find the ratio TB /TR as a function of the angle θ.

       Solution

       When rolling without slipping, the acceleration of an object with moment of inertia βmR2
       about its center of mass is
                                                    g sin θ
                                               a=
                                                    1+β
       as mentioned in a previous example. The tangential force from friction is thus
                                                             β
                                             f = mg sin θ
                                                            1+β
       which means rolling without slipping occurs when
                                                                  β
                                        µs mg cos θ ≥ mg sin θ
                                                                 1+β

        or equivalently
                                                               1+β
                                                  tan θ ≤ µs       .
                                                                β
        For the ball, this is when θ ≤ 60.3◦ , and for the ring θ ≤ 45◦ . Whenever either object slips,
        its acceleration is instead a = g(sin θ − µk cos θ).

        Since the motion is uniformly accelerated, TB /TR = aR /aB . For θ ≤ 45◦ , both roll without
                                                             p

        slipping, so the formula above applies, giving a ratio of
                                                 s            r
                                          TB        1 + βB        7
                                              =            =        .
                                          TR        1 + βR       10

        For θ ≥ 60.3◦ they both slip, so the ratio is unity. For the angles in between, the ring slips,
        giving a slightly more complicated expression. At the boundaries between these three regimes,
        the ratio TB /TR jumps discontinuously.

    The next two problems require careful thought, and test your understanding of the multiple ways

                       01m
    to describe rotational kinematics and dynamics. It will be useful to review idea 1.

Solution:
The uniform stick has moment of inertia M L2 /12 about its center, and has a constant
    torque of τ = F L/2 about its center. Thus if θ is the angular distance the stick has rotated, then

                                                FL  1
                                                   = M L2 θ̈
                                                 2  12
    which implies
                                           6F          3F 2
                                       θ̈ =      , θ=      t = βt2 .
                                           ML          ML
    Let the plane of motion of the stick be the complex plane, let the initial position of the center of
    mass be the origin, and let the angle between the stick and the real axis be θ + π/2 so the force
    points at an angle θ. Then in the complex plane, the unit vector of the force is just eiθ , so
                                                            Z ∞
                                               2         F           2
                                  M a = F eiβt      vf =        eiβt dt.
                                                         M 0

    This trick of using complex numbers allows us to write the two real components of Newton’s second
    law as a single equation.
       Now, if you were in a math class, you would be taught that this integral is not convergent (the
    integrand doesn’t even go to zero at infinity), so there is no answer for vf . But we’re in a physics
    class, so we’re allowed to use common sense. As time goes on, the stick will rotate faster and faster,
    so the acceleration will spin faster, so the endpoint of the velocity vector rotates in tighter and
    tighter circles. So even though the magnitude of the acceleration never gets smaller, the velocity
    does approach a limit!
                                             2                      R∞    2        √
       We now compute this limit. Since e−x is an even function, 0 e−x dx = 12 π. By u-substitution,
    we infer that                           Z ∞              √
                                                       2        π
                                                 e−(ax) dx =      .
                                             0                2a
                                                                                √
    In order to get −i = e3π/2 from a2 , we need a factor of ±e3π/4 = ±(−1 + i)/ 2. Thus
                               Z ∞           Z ∞  −1+i √ 2                 √
                       M             iβt2          − ± √    βt                 π
                          vf =      e dt =        e      2       dt = ±           p
                        F       0              0                        2(−1 + i) β/2

    from which we conclude the final speed is
                                                   √    r
                                                  F π     πF L
                                         |vf | =    √ =        .
                                                 2M β     12M

[4] Problem 16 (KK 6.41). A plank of length 2L leans nearly vertically against a wall. All surfaces
    are frictionless. The plank starts to slip downward. Find the height of the top of the plank when it
    loses contact with the wall or floor.

Solution.

Note that the normal forces at the contact points do no work, since the plank moves in the
perpendicular directions at those points. Therefore, mechanical energy is conserved.
    The center of mass P moves in a circle of radius L around O, and its speed is Lθ̇. Similarly, one
also sees that the plank rotates around its CM, P , at angular velocity θ̇ counterclockwise. Therefore,
if the plank starts at θ0 , we have by energy conservation that

                                            1           1 1                2
                     mgL(cos θ0 − cos θ) = mL2 θ̇2 +        mL2 θ̇2 = mL2 θ̇2 ,
                                            2           2 3                3
so
                                                3g
                                        θ̇2 =      (cos θ0 − cos θ).
                                                2L
Taking the time derivative, we obtain
                                            3g                  3g
                                  2θ̇θ̈ =      sin θ θ̇ =⇒ θ̈ =    sin θ.
                                            2L                  4L
Now, the plank loses contact when the normal force Nx at the high point of contact is 0. By Newton’s
second law, Nx = mẍ, so Nx = 0 when ẍ = 0. We also know that x = L sin θ, so ẋ = L cos θ θ̇, so
ẍ = L(cos θ θ̈ − sin θ θ̇2 ). Therefore, we have

                                              cos θ θ̈ = sin θ θ̇2

when contact is lost. Plugging in our earlier results, we find
                               3g               3g
                                  sin θ cos θ =    (cos θ0 − cos θ) sin θ,
                               4L               2L
or cos θ = 23 cos θ0 , so y = 23 y0 , as desired. (For completeness, we should have checked that the plank
actually loses contact with the wall before losing contact with the floor. This is fairly intuitive, but
it can be checked explicitly by using the above analysis to compute ÿ and show that Ny is positive
until Nx vanishes.)
    There is a slick alternative solution using Lagrangian mechanics, though it’s subtle enough that
I wouldn’t recommend trying it in a competition. We note that the center of mass moves on a
circle centered at the origin, and that the total kinetic energy of the ladder is proportional to θ̇2 .
In particular, we have a Lagrangian of
                                 1                                         4
                              L = meff L2 θ̇2 − mgL cos θ,           meff = m
                                 2                                         3

where the extra contribution in the first term is due to rotational kinetic energy. Multiplying the
Lagrangian by 3/4, which makes no difference to the equations of motion, we get

                                      1               3g
                                 L = mL2 θ̇2 − m           L cos θ.
                                      2                4
However, this is simply the Lagrangian for a mass m sliding on a frictionless hemisphere in gravity
3g/4. This is a classic, simple problem, and we know in that case that the normal force with the
hemisphere vanishes at height (2/3)L.
   Now, the motion of the mass in this problem is identical to the motion of the center of mass
of the ladder in the original problem, so the total external forces are the same. In particular, the
horizontal constraint force must vanish when the ladder’s center of mass is at height (2/3)L, so the
ladder loses contact with the wall at this point. On the other hand, the vertical external force must
be 3mg/4, which implies the normal force with the ground is mg/4, and hence positive; this shows
that the ladder has not lost contact with the ground.

   Example 7: EFPhO 2013

   A uniform ball and a uniform ring are both released from rest from the same height on an
   inclined plane with inclination angle θ. They arrive at the bottom of the plane in time TB
   and TR , respectively. The coefficients of friction of both objects with the plane are µk = 0.3
   and µs = 0.5. Find the ratio TB /TR as a function of the angle θ.

   Solution

   When rolling without slipping, the acceleration of an object with moment of inertia βmR2
   about its center of mass is
                                                g sin θ
                                           a=
                                                1+β
   as mentioned in a previous example. The tangential force from friction is thus
                                                        β
                                        f = mg sin θ
                                                       1+β
   which means rolling without slipping occurs when
                                                              β
                                    µs mg cos θ ≥ mg sin θ
                                                             1+β
   or equivalently
                                                      1+β
                                         tan θ ≤ µs       .
                                                       β
   For the ball, this is when θ ≤ 60.3◦ , and for the ring θ ≤ 45◦ . Whenever either object slips,
   its acceleration is instead a = g(sin θ − µk cos θ).

   Since the motion is uniformly accelerated, TB /TR = aR /aB . For θ ≤ 45◦ , both roll without
                                                        p

   slipping, so the formula above applies, giving a ratio of
                                            s            r
                                     TB        1 + βB        7
                                         =            =        .
                                     TR        1 + βR       10

       For θ ≥ 60.3◦ they both slip, so the ratio is unity. For the angles in between, the ring slips,
       giving a slightly more complicated expression. At the boundaries between these three regimes,
       the ratio TB /TR jumps discontinuously.

   The next two problems require careful thought, and test your understanding of the multiple ways

                      01m
   to describe rotational kinematics and dynamics. It will be useful to review idea 1.


[User]
Question A3
A light bulb has a solid cylindrical filament of length L and radius a, and consumes power P .
You are to design a new light bulb, using a cylindrical filament of the same material, operating at
the same voltage, and emitting the same spectrum of light, which will consume power nP . What
are the length and radius of the new filament? Assume that the temperature of the filament is
approximately uniform across its cross-section; the filament doesn’t emit light from the ends; and
energy loss due to convection is minimal.

Solution:
Question A3
A light bulb has a solid cylindrical filament of length L and radius a, and consumes power P .
You are to design a new light bulb, using a cylindrical filament of the same material, operating at
the same voltage, and emitting the same spectrum of light, which will consume power nP . What
are the length and radius of the new filament? Assume that the temperature of the filament is
approximately uniform across its cross-section; the filament doesn’t emit light from the ends; and
energy loss due to convection is minimal.

                                               Solution
Since the new bulb emits the same spectrum of light, the emitted power is simply proportional to
the surface area, where we ignore the surface area of the ends as directed,

                                            P ∝ 2πaL ∝ aL.

If the resistivity of the filament is ρ, the resistance is
                                                   ρL   ρL
                                              R=      =     .
                                                   A    πa2
Therefore, the power is also given by
                                              V2   V 2 πa2  a2
                                        P =      =         ∝ .
                                              R      ρL     L

                    Copyright ©2011 American Association of Physics Teachers

2011 Semifinal Exam                             Part A                                                7

Thus if the new power is nP , we may only satisfy both equations if the new radius and length are

                                      a0 = n2/3 a,   L0 = n1/3 L.

If the powers given by the two equations were not equal, we would have a contradiction. This
indicates that it is impossible to keep the bulb emitting the same spectrum using the same voltage.
Supposing we did keep the voltage the same, the filament would burn hotter or cooler than before,
changing the spectrum of the light.

USAPhO 2011 Question A3


[User]
7. BOUNCY BALL (5 points)

(by J. Toots)
A homogeneous elastic ball with radius R
bounces from a vertical wall in such a way
that it rebounds back exactly in the same
direction it came from. The ball’s speed be-
fore the bounce is v at an angle α to the ver-
tical; the ball also has an angular speed ω.
There was no slipping at the contact point,
but do not assume that ωR = vcosα. The
bounce was completely elastic, ie no kinetic
energy was lost; also, the horizontal compon-
ents of incident and rebound velocities have
the same magnitude. The moment of inertia
for a homogeneous ball is I = 2
5 mR2
.
i) (1 point) Find the angular speed ω2 of the
ball after the bounce.
ii) (2 points) Find the initial angular speed
ω in terms of v, α and R.
iii) (2 points) What is the minimum value
for coefficient of friction μ that such bounce
could happen?

Solution:
7. BOUNCY BALL

i) Energy is conserved: mv2
2 + Iω2
2 = mv2
2 +
Iω2
2
2 =⇒ |ω2|=|ω|. As the ball receives some
angular momentum from the wall (the net
force from the wall is not directed through the
ball’s centre), the angular velocity changes
its sign — the rotation flips its direction and
ω2 =−ω.
ii) The angular momentum is conserved with
respect to the contact point, because all the
forces are directed through the point. Let a=
Rcosα be the distance from the trajectory of
the center of mass to the contact point. Then
the aforementioned angular momentum con-
servation reads Iω−mav = Iω2 +mav. As
ω2 =−ω, this gives ω= mav
I = 5vcosα
2R .
iii) At the verge of slipping on a surface, the
angle between the surface normal and the
resultant of the reaction force and the frtic-
tion force equals arctanμ. Here this angle is
π
2 −α, thence μ≥cotα.

BOUNCY BALL


[User]
Question B3
The Mad Hatter
A frictionless hemisphere of radius R is fixed on top of a flat cylinder. One end of a spring with
zero relaxed length and spring constant k (i.e. the force from the spring when stretched to length `
is −k`) is fixed to the top of the hemisphere. Its other end is attached to a point mass of mass m.

                                                    R
                                                θ

a. The number and nature of the equilibrium points on the hemisphere depends on the value of
   the spring constant k. Consider the semicircular arc shown above as a dashed line, which is
   parameterized by angles in the range −π/2 ≤ θ ≤ π/2. Make a table indicating the number of
   equilibrium points on the arc, and the number that are stable, for each range of k values. A
   blank table for your reference is given below. (You may need more or fewer rows than shown.)

               Range of k (kmin < k < kmax )    # of Equilibria    # of Stable Equilibria
                        0<k<?

                        ? <k<∞

For the rest of the problem, suppose the value of k is such that the mass begins at stable equilibrium
on the surface of the hemisphere at angle θ0 . The mass can move on the two-dimensional surface
of the hemisphere, but a radially-inward external force prevents it from jumping off the surface.

b. At t = 0, the mass is given a speed v along a line of constant latitude θ = θ0 .

                                                     v
                                                                  θ = θ0

   i. Indicate which of the following trajectories the mass takes for a short time after t = 0 and
      briefly explain your reasoning. The differences between the paths are exaggerated.

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                  Part B                                              4

                                                       A
                                                       B         θ = θ0
                                                       C

  ii. What is the total radial force (i.e., normal to the surface of the hemisphere) on the mass at
      t = 0? Express your answer in terms of m, v, R, g, and θ0 .

c. A cylinder of radius r  Rθ0 is placed on top of the sphere. Suppose the mass is launched at an
   angle α away from the direction of the spring’s displacement with kinetic energy K, as shown.
   What is the maximum angle αmax at which the mass can be launched such that it can still hit
   the cylinder? Express your answer in terms of K, m, g, θ0 , r, and R. You may assume K is
   large enough for the mass to reach the cylinder for α = 0.

                                                          v
                                                      α

                                         (view from above)

                  Copyright ©2021 American Association of Physics Teachers

Solution:
Question B3
The Mad Hatter
A frictionless hemisphere of radius R is fixed on top of a flat cylinder. One end of a spring with
zero relaxed length and spring constant k (i.e. the force from the spring when stretched to length `
is −k`) is fixed to the top of the hemisphere. Its other end is attached to a point mass of mass m.

                                                     R
                                                 θ

a. The number and nature of the equilibrium points on the hemisphere depends on the value of
   the spring constant k. Consider the semicircular arc shown above as a dashed line, which is
   parameterized by angles in the range −π/2 ≤ θ ≤ π/2. Make a table indicating the number of
   equilibrium points on the arc, and the number that are stable, for each range of k values. A
   blank table for your reference is given below. (You may need more or fewer rows than shown.)

              Range of k (kmin < k < kmax )     # of Equilibria    # of Stable Equilibria
                       0<k<?

                        ? <k<∞

                                              Solution
     The spring force attracts the mass toward the top of the hemisphere, whereas the gravita-
     tional force tends to pull it away.
     For very small k, the spring force is negligible compared to the gravitational force for
     |θ| < π/2. For these k values, we only have one unstable equilibrium at the top of the
     hemisphere due to the gravitational force being zero there.
     For some large enough value of k, the spring force at π/2 exactly compensates for the
     gravitational force – this marks the end of the first regime. In this next regime, we still
     have an unstable equilibrium at the top of the hemisphere – because the spring force only
     compensates for the gravitational force for larger θ. Additionally, we have two equilibria at
     0 < |θ? | < π/2. These equilibria must be stable because for θ < θ? , the gravitational force
     is stronger than the spring force and forces the mass toward θ? , whereas the opposite holds
     for θ > θ? , and the mass is again forced toward θ? . Thus, we have three total equilibria
     (two stable, one unstable).
     Finally, for sufficiently large k the spring force is stronger than the gravitational force even
     for arbitrarily small θ, and so we have just one stable equilibrium point at the top of the
     hemisphere in this third regime.
     We now compute the two critical points for us to fill out the table. The first critical point

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                    Part B                                               9

     is given by the balancing of the two forces at θ = π/2, so we compute
                                                                   2mg
                               mg sin π/2 = k1 R(π/2) =⇒ k1 =          .
                                                                    πR
     The second critical point is given by the balancing of the two forces for θ  1, so we
     compute
                             mg sin θ ≈ mgθ = k2 Rθ =⇒ k2 = mg/R.
     We thus get the following table:

           Range of k values (kmin < k < kmax )     # of Equilibria   # of Stable Equilibria
                   0 < k < 2mg/(πR)                       1                     0
                2mg/(πR) < k < mg/R                       3                     2
                     mg/R < k < ∞                         1                     1

     Notice that the system is symmetric under flipping θ → −θ. Thus, one could incorrectly
     guess that the only possible equilibrium point is θ = 0 by symmetry. In fact, when k is in
     the right range, we get a pair of new equilibrium points at opposite θ, which map to each
     other under symmetry. This kind of situation, where the overall setup is still symmetric
     but the individual equilibrium points are not, is called spontaneous symmetry breaking.

For the rest of the problem, suppose the value of k is such that the mass begins at stable equilibrium
on the surface of the hemisphere at angle θ0 . The mass can move on the two-dimensional surface
of the hemisphere, but a radially-inward external force prevents it from jumping off the surface.

b. At t = 0, the mass is given a speed v along a line of constant latitude θ = θ0 .

                                                     v
                                                                 θ = θ0

   i. Indicate which of the following trajectories the mass takes for a short time after t = 0 and
      briefly explain your reasoning. The differences between the paths are exaggerated.

                                                         A
                                                         B         θ = θ0
                                                         C

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                       Part B                                             10

                                                  Solution
        The correct path is option C. Several explanations would work here. Here are two.
          • If we go to the rotating frame of reference, there is an outward centrifugal force that
            the mass experiences, pushing it down the sphere.
          • For the mass to go in a circle around the sphere, the spring force not only has to
            compensate for the gravitational force but also must provide centripetal acceleration.
            Therefore, the spring must get longer.

  ii. What is the total radial force (i.e., normal to the surface of the hemisphere) on the mass at
      t = 0? Express your answer in terms of m, v, R, g, and θ0 .

                                                  Solution
        We draw a free-body diagram. It is helpful to draw the diagram in the noninertial
        reference frame that revolves around the central axis of the hemisphere with speed v at
        the location of the mass.
                                         FN
                                                  kRθ0

                                  Fc
                             mg sin θ0
                                              mg cos θ0

                        2
        Here, Fc = Rmvsin θ0 is the centrifugal force, and Fr is the radial force from the hemisphere.
        The forces in the radial direction must balance for the mass to be constrained to the
        surface of the sphere. Thus,
                                             Fr + Fc⊥ = mg cos θ0 .
        The perpendicular part of the centrifugal force is Fc sin θ0 . so we get

                                                             mv 2
                                          Fr = mg cos θ0 −        .
                                                              R
        Incidentally, there’s a simple way to understand why the second term has to be exactly
        mv 2 /R. Consider decomposing the total force on the mass into radial and tangential
        parts. The radial part simply keeps the mass on the hemisphere; in the absence of a
        tangential force, the mass would travel in a great circle of radius R. Adding a tangential
        force deflects the mass away from this great circle trajectory, but doesn’t change the
        radial force required, so the net radial force always has to be mv 2 /R inward.
        Note: The phrase “total radial force” could also validly be interpreted as the net radial
        force. Thus, we accepted both mg cos θ0 − mv 2 /R and −mv 2 /R.

c. A cylinder of radius r  Rθ0 is placed on top of the sphere. Suppose the mass is launched at an
   angle α away from the direction of the spring’s displacement with kinetic energy K, as shown.

                   Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                  Part B                                                11

  What is the maximum angle αmax at which the mass can be launched such that it can still hit
  the cylinder? Express your answer in terms of K, m, g, θ0 , r, and R. You may assume K is
  large enough for the mass to reach the cylinder for α = 0.

                                                          v
                                                      α

                                        (view from above)

                                           Solution
    The initial energy of the system is given by
                                                1
                                    mgR cos θ0 + kR2 θ02 + K.
                                                2
    Suppose the mass is launched with speed v. Then, the speed in the θ direction is v cos α
    and the speed in the φ direction is v sin α, and therefore, the z-component of the angular
    momentum of the mass is
                                      L = mv sin α(R sin θ).
    We now compute the distance of closest approach. If the distance of closest approach is
    equal to r (as it does for αmax , then at r, the motion of the mass has no inward component,
    and the speed of the object at r is given by conservation of angular momentum:

                                                              v sin α(R sin θ)
                        mur = mv sin α(R sin θ) =⇒ u =                         .
                                                                      r
    Because r  R, at the point of closest approach, the energy of the system is roughly
                                                                                !
                  1                 1 mv 2 sin2 αR2 sin2 θ              α 2 R2 sin2 θ
           mgR + mu2 ≈ mgR +                               ≈ mgR + K                    .
                  2                 2          r2                            r2

    Equating with the initial energy gives us
                                                                                !
                                 1                                α2 R2 sin2 θ
                     mgR cos θ0 + kR2 θ02 + K = mgR + K                              .
                                 2                                    r2

    Before finishing the calculation, we now compute the required k for the object to be at
    equilibrium (since our answer cannot contain k). Setting kRθ0 = mg sin θ0 gives us k =

                 Copyright ©2021 American Association of Physics Teachers

2021 USAPhO                                Part B                                             12

    mg sin θ0
     Rθ0 . Then,
                                                                                !
                             1                                   α2 R2 sin2 θ0
                 mgR cos θ0 + mgRθ0 sin θ0 + K = mgR + K                             .
                             2                                        r2

   Solving for α gives us
                                  r
                          r            mgR(1 − cos θ0 ) − (1/2)mgRθ0 sin θ0
                    α=            1−                                        .
                       R sin θ0                         K
   Solving for α without using the small angle approximation for α also earned full credit.
   (The answer α = π, which is technically also correct, earned partial credit.)

                Copyright ©2021 American Association of Physics Teachers

USAPhO 2021 Question B3