[User]
Q2-1
Theory

                                                                                English (Official)

Theoretical Problem 2: A ball on a turntable [10.0 points]
Preamble

Notations and conventions: The length of a vector 𝐴 ⃗ is simply denoted as 𝐴 ≡ |𝐴|.⃗ It's 𝑥, 𝑦, 𝑧 components
are denoted by 𝐴𝑥 , 𝐴𝑦 , 𝐴𝑧 , respectively. The time derivative of a quantity is denoted by the dot over the
             ̇
quantity: 𝐴 ⃗ ≡ 𝑑𝐴/𝑑𝑡,
                  ⃗    𝐴 ̇ ≡ 𝑑𝐴/𝑑𝑡. The unit vector along the direction of vector 𝐴 ⃗ is denoted as 𝐴.̂ The unit
vectors along the Cartesian coordinates are, therefore, 𝑥,̂ 𝑦 ̂ and 𝑧.̂ The definitions of scalar and vector
products are:
(𝐴 ⃗ ⋅ 𝐵)⃗ = (𝐵⃗ ⋅ 𝐴)⃗ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧 = 𝐴𝐵 cos 𝜃

(𝐴 ⃗ × 𝐵)⃗ = −(𝐵⃗ × 𝐴)⃗
= (𝐴𝑦 𝐵𝑧 − 𝐴𝑧 𝐵𝑦 )𝑥̂ + (𝐴𝑧 𝐵𝑥 − 𝐴𝑥 𝐵𝑧 )𝑦 ̂ + (𝐴𝑥 𝐵𝑥 − 𝐴𝑦 𝐵𝑥 )𝑧 ̂

|𝐴 ⃗ × 𝐵|⃗ = 𝐴𝐵 sin 𝜃,
where 𝜃 is the angle between 𝐴 ⃗ and 𝐵.⃗ You may need the following properties of vectors and their
multiplications. Triple product rules for vectors:
(𝐴 ⃗ × 𝐵)⃗ × 𝐶 ⃗ = (𝐴 ⃗ ⋅ 𝐶)⃗ 𝐵⃗ − (𝐵⃗ ⋅ 𝐶)⃗ 𝐴,⃗
(𝐴 ⃗ × 𝐵)⃗ ⋅ 𝐶 ⃗ = (𝐵⃗ × 𝐶)⃗ ⋅ 𝐴 ⃗ = (𝐶 ⃗ × 𝐴)⃗ ⋅ 𝐵.⃗
The vector products are very useful in describing many relations in physics. For example:
𝑣 ⃗ = 𝜔⃗ × 𝑟,⃗
    ⃗
𝐹𝐿𝑜𝑟𝑒𝑛𝑡𝑧   = 𝑄𝑣 ⃗ × 𝐵,⃗
and, often, saves time combining three equations for vector components into a single equation.

The statement

                                 Figure 1. Ball rolling on the turntable without slipping

A ball of mass 𝑚 and radius 𝑟 is rolling on a horizontal turntable without slipping (see Figure 1). Its mass
density has a spherical symmetry, i.e. only depends on the distance from its center. The moment of
inertia of the ball is 𝐼 . In part B and C, where the turntable can rotate freely, the moment of inertia of
the turntable is denoted as 𝐼𝑑 . The purpose of the problem is to analyze the motion and trajectory of
the ball with respect to the laboratory frame. Throughout the problem, assume the turntable is large
enough so that the ball does not fall off. The following notations are used:
Ω - the magnitude of the turntable angular velocity,

                                                                                Q2-2
Theory

                                                                                English (Official)

𝜔⃗ - the spinning angular velocity of the ball with respect to its spinning axis,
𝑅⃗ - the horizontal position of the ball center with respect to the rotation axis of the turn table,
𝑣 ⃗ - the velocity of the ball at 𝑅⃗ with respect to the laboratory frame.
Assume that the initial position 𝑅⃗ 0 ≡ ⃗𝑅(0) and velocity 𝑣0⃗ ≡ ⃗
                                                                𝑣(0) of the ball, the angular velocity of the
turn table Ω0 ≡ Ω(0) are known. For the initial vector quantities 𝑅⃗ 0 ≡ 𝑅(0)
                                                                            ⃗    and 𝑣 ⃗0 ≡ ⃗
                                                                                𝑣(0), assume that
their directions are known. In addition, whenever you need to express a vector quantity, you may use
𝑧 ̂ in your expression. Also, if asked to write your expression in terms of the known quantity you may
use any or all of 𝑚, 𝑟, 𝐼 and 𝐼𝑑 . Unless otherwise stated, keep 𝐼 as general. The following notations are
recommended:
      𝐼          𝐼𝑑
𝛼 = 𝐼+𝑚𝑟 2 , 𝛿 = 𝑚𝑟2 ,

You may write the final answers as vector expressions involving cross product (vector product), dot prod-
uct (scalar product) and unit vectors in axis directions.

Part A: Ball on turntable with constant angular velocity [1.5 points]
First we start with the simplest case wherein the turntable angular velocity with respect to vertical axis 𝑧 ̂
is constant, therefore Ω = Ω0 .

     A.1     Express the ball's velocity 𝑣 ⃗ in terms of Ω, 𝜔,⃗ 𝑟 and 𝑅⃗ from a kinematic constraint.           0.1pt

     A.2     Using Newton's equation and torque equation with respect to its center, find                       0.2pt
             the acceleration of the ball 𝑎⃗ ≡ 𝑣 ⃗̇ in terms of Ω, 𝑣,⃗ 𝑟, 𝑚 and 𝐼.

     A.3     Find the velocity 𝑣 ⃗ in terms of Ω, 𝑅,⃗ 𝑣0⃗ , 𝑅0⃗ , 𝑟, 𝑚 and 𝐼.                                   0.2pt

     A.4     Write an explicit solution for the trajectory of the ball given the initial conditions             0.5pt
             𝑣0⃗ and 𝑅⃗ 0 .

     A.5     Assume this time that the ball has a uniform mass density, i.e. 𝐼 = 2𝑚𝑟2 /5. Tra-                  0.5pt
             jectory you have found is a circle and it's radius is 𝑅𝑡 . Choose its magnitude to
             be the same as 𝑅0 . How long does it take for the ball to approach the initial spot
             on the table (the position on the turntable at 𝑡 = 0) with the closest distance?

Part B: Ball on freely rotating turntable [4.0 points]
In this part, the turntable can rotate freely without any friction around 𝑧 -axis. Therefore its free rotation
is hindered only by the ball's friction.

     B.1     Find the velocity 𝑣 ⃗ and acceleration 𝑣 ⃗̇ of the ball in terms of Ω , 𝑅,⃗ Ω0 , 𝑅⃗ 0 , Ω,̇ 𝑟, 𝑚   0.2pt
             and 𝐼.

     B.2     Find the magnitude of the angular acceleration of the turntable Ω̇ in terms of                     0.6pt
             Ω, Ω0 , 𝑅,⃗ 𝑅⃗ 0 , 𝑣 ⃗0 , 𝑟, 𝑚, 𝐼 and 𝐼𝑑 . You may use the constants 𝛼 and 𝛿 defined in the
             beginning of the problem.

                                                                                Q2-3
Theory

                                                                                English (Official)

    B.3     Find the magnitude of the angular velocity of the turntable Ω as a function of 𝑅                   0.6pt
            only. Use this constants in your expression: Ω0 , 𝑅0 , 𝑟, 𝑚, 𝐼, 𝐼𝑑 .

    B.4     From the result of B.3, for a given Ω0 , 𝑅0 , find the maximum possible Ω .                        0.1pt

    B.5     Write down the vertical component the angular momentum ̂              𝑧𝑀𝑧 of the whole             2.5pt
            system. Subtract any constant term and rename the remaining part as 𝑧𝐿                   ̂ .
            In part B.1 you found the velocity of the ball 𝑣,⃗ which can be written as the sum
            of a part that depends on the position of the ball 𝑅⃗ and a constant vector. Let us
            call this constant vector 𝑐 ⃗ . Choose the direction of 𝑥-axis along this vector and 𝑦-
            axis along 𝑧 ̂× 𝑐 ⃗ . In this frame of reference, find Ω in terms of 𝐿, 𝑅,⃗ 𝑐,⃗ 𝑧,̂ 𝑅2 , 𝑟, 𝑚, 𝐼
            and 𝐼𝑑 . Combining this with the result of B.3, write down an equation only
            containing 𝑅2 and 𝑦 variables and 𝐿, 𝑟, 𝑚, 𝐼, 𝑐 and 𝐼𝑑 . Here 𝑐 is the magnitude of
            𝑐.⃗ Substituting 𝑅2 = 𝑥2 + 𝑦2 , write down an expression containing only 𝑥 and 𝑦
            variables and describing a curve. From this, list all possible types of trajectories.

Part C: Ball on turntable in magnetic field [4.5 points]
In this part, we consider a density profile so that 𝐼 = 𝑚𝑟2 /10 . This can be realized, for example, if the
ball is filled up to half of its radius with uniform density and the remaining part has a negligible mass.
In addition, on its outer surface, the ball has a uniform charge density 𝑄/(4𝜋𝑟2 ) , where 𝑄 is the total
surface charge. The whole setup is in a uniform magnetic field 𝐵⃗ that is in 𝑧 ̂ direction. The turntable
rotates with constant Ω like in Part A.

                  Figure 2. Ball rolling on the turntable in a constant magnetic field 𝐵⃗

    C.1     Write down Newton's equation and the torque 𝜏𝑠⃗ equation for the ball. Find                        0.5pt
            expression for the torque due to the spinning of the ball around its axis in terms
            of 𝑄, 𝑟, 𝜔⃗ and 𝐵⃗ .

    C.2     Using the results of C.1, find expression for the linear acceleration of the ball                  0.5pt
            with respect to the laboratory frame in terms of 𝑄, 𝑟, 𝜔⃗ and 𝐵⃗ .

                                                                            Q2-4
Theory

                                                                                English (Official)

  C.3    We assume all quantities of unit lenght are measured by meter, all angular ve-         1.0pt
         locities have unit of 1 Hertz, and all quantities of time have the unit of 1 second.
         The equation for the linear acceleration you found in part C.2 is a second order
         differential equation for 𝑅⃗ of the following form:
         𝑑2 𝑅⃗    𝑑𝑅⃗           ⃗
          𝑑𝑡2 − 𝛾 𝑑𝑡 × 𝑧 ̂ + 𝛽 𝑅 = 0.
         Write down 𝛾 and 𝛽 constants in terms of 𝑄, 𝑟, 𝐵, 𝐼, 𝑚, Ω. Make the following
         transformation to a polar coordinates for the components of 𝑅⃗ :
         𝑥(𝑡) = 𝜌(𝑡) cos(𝜂(𝑡)),
         𝑦(𝑡) = 𝜌(𝑡) sin(𝜂(𝑡))
         so that the new equations do not have the first time derivative term. Here the
         polar angle 𝜂(𝑡) is a function of time. Find the form of this function.
         Express the coefficient 𝛽 ′ of 𝜌(𝑡) in the new equation in terms of 𝛾 and 𝛽 .
         Write down the conditions for different types of behavior of 𝜌(𝑡) with respect to
         time: harmonic, exponential etc.

  C.4    Consider the following initial conditions for the solution found in part C.3:          0.9pt
         𝑥(0) = 1 𝑚, 𝑦 = 0 𝑚, 𝑣𝑥 (0) = 𝑥|̇ 𝑡=0 = 1 𝑚/𝑠, , 𝑣𝑦 (0) = 𝑦|̇ 𝑡=0 = −1 𝑚/𝑠.
         From these conditions, find 𝛽 and 𝛾. Using them find the corresponding Ω.
         Sketch the trajectory. Is the charge of the surface negative or positive? For the
         negative write − and for the positive write + on your answer sheet.

  C.5    Consider the solution you have found in part C.4. If you identified it correctly       1.6pt
         your solution should have a rotating ⃗𝑅(𝑡). Find the expressions for the total and
         per rotation changes in energy for 𝑁 ≫ 1 number of rotations. Here you may
         ignore the terms small compared to 𝑁 . In this part assume the mass and the
         radius of the ball are 𝑚 = 1 𝑘𝑔 and 𝑟 = 1 𝑚 so that 𝐼 = 1/11 𝑘𝑔 ⋅ 𝑚2 .

Solution:
A ball on a turntable
                          Enkhbat Tsedenbaljir

1     Problem 1: 10 points
1.1    Preamble
Notations and conventions: The length of a vector A   ⃗ is simply denoted as A ≡
  ⃗ The time derivative of a quantity is denoted by the dot over the quantity:
|A|.
A⃗˙ ≡ d ⃗                                                              ⃗ is denoted
       A/dt, Ȧ ≡ dA/dt. The unit vector along the direction of vector A
as Â. The unit vectors along the Cartesian coordinates are, therefore, x̂, ŷ and ẑ.
The definitions of scalar and vector products are:
         ⃗ · B)
        (A   ⃗ = (B⃗ · A)
                       ⃗ = Ax Bx + Ay By + Az Bz = AB cos θ,                       (1)
      (A⃗ × B)
             ⃗ = −(B ⃗ × A)
                          ⃗                                                        (2)
                = (Ay Bz − Az By )x̂ + (Az Bx − Ax Bz )ŷ + (Ax Bx − Ay Bx )ẑ,    (3)
        ⃗ × B|
       |A    ⃗ = AB sin θ,                                                         (4)

where θ is the angle between A ⃗ and B.
                                     ⃗ You may need the following properties of
vectors and their multiplications: Scalar products of vectors vectors:
                           ⃗ · B)
                          (A   ⃗ = (B⃗ · A)
                                         ⃗ B⃗ − (B
                                                 ⃗ · C)
                                                     ⃗ A,
                                                        ⃗                          (5)
                   ⃗ × B)
                  (A   ⃗ ·C
                          ⃗ = (B⃗ × C)
                                    ⃗ ·A ⃗ = (C⃗ × A)
                                                   ⃗ · B.
                                                        ⃗                          (6)

Triple product rules for vectors:
                       ⃗ × B)
                      (A   ⃗ ×C ⃗ = (A
                                     ⃗ · C)
                                         ⃗ B⃗ − (B
                                                 ⃗ · C)
                                                     ⃗ A,
                                                        ⃗                          (7)
                   ⃗ × B)
                  (A   ⃗ ·C⃗ = (B
                                ⃗ × C)
                                    ⃗ ·A ⃗ = (C⃗ × A)
                                                   ⃗ · B.
                                                        ⃗                          (8)

The vector products are very useful in describing many relations in physics. For
example:

                                        ⃗v = ⃗ω × ⃗r,                              (9)
                              F⃗Lorentz = QB  ⃗ × ⃗v ,                            (10)

and, often, saves time combining three equations for vector components into a
single equation.

1.2    The statement
A ball of mass m and radius r is rolling on a horizontal turntable without slipping.
Its mass density has a spherical symmetry, i.e. only depends on the distance from
its center. In part B and C, where the turntable can rotate freely, the moment
of inertia of the turntable is denoted as Id . The purpose of the problem is to
analyze the motion and trajectory of the ball with respect to an observer at rest.
Throughout the problem, assume the turntable is large enough so that the ball
does not fall off. The following notations are used:
     Ω – the magnitude of the turntable angular velocity,
     ⃗ω – the spinning angular velocity of the ball with respect to its spinning axis,
     R⃗ – the horizontal position of the ball center with respect to the rotation axis
of the turn table,
                                      ⃗
    ⃗v – the velocity of the ball at R.
     Assume that the initial position R  ⃗ 0 ≡ R(0)
                                               ⃗    and velocity ⃗v0 ≡ ⃗v (0) of the ball,
the angular velocity of the turn table Ω0 ≡ Ω(0) are known. For the initial vector
quantities R ⃗ 0 ≡ R(0)
                   ⃗     and ⃗v0 ≡ ⃗v (0), assume that their directions are known. In
addition, whenever you need to express a vector quantity, you may use ẑ in your
expression. Also, if asked to write your expression in terms of the known quantity
you may use any or all of m, r, I and Id . Unless otherwise stated, keep I as
general. The following notations are recommended:
                                    I           Id
                            α=           , δ =     ,                                 (11)
                                 I + mr2       mr2

1.3    Part A: 2 points
First we start with the simplest case wherein the turntable angular velocity with
respect to vertical axis ẑ is constant, therefore Ω = Ω0 .
A1. 0.1 point
    Express the ball’s velocity ⃗v in terms of Ω, ⃗ω , r, m, I and R⃗ from a kinematic
constraint.
A.2 0.2 point
    Using Newton’s equation and torque equation with respect to its center, find
the acceleration of the ball ⃗a ≡ ⃗v˙ in terms of Ω, ⃗v , r, m and I.
A.3 0.2 points
    Find the velocity ⃗v in terms ofΩ, R, ⃗ v⃗0 , R⃗0 , r, m and I.
A.4 0.5 points
    Find the trajectory of the ball. It means, for the given initial conditions ⃗v0 and
⃗
R0 , completely specify the trajectory.
A.5 1 point

    Assume this time that the ball has a uniform mass density, i.e. I = 2mr2 /5.
Trajectory you have found has a single defining parameter Rt for its size. Choose
its magnitude to be the same as R0 . How long does it take for the ball to approach
the initial spot on the table (the position on the turntable at t = 0) with the closest
distance?

1.4     Part B
In this part, the turntable can rotate freely, without any friction, around z–axis.
Therefore its free rotation is hindered only by the ball’s friction.
    B.1 0.2 points

   Find the velocity ⃗v and acceleration ⃗v˙ of the ball in terms of Ω, R,
                                                                        ⃗ Ω0 , R
                                                                               ⃗ 0 , Ω̇,
r, m and I.
B.2 0.2 points

     Find the magnitude of the angular acceleration of the turntable Ω̇ in terms of
         ⃗ R
 Ω, Ω0 , R, ⃗ 0 , ⃗v0 , r, m, I and Id . You may use the constants α and δ defined in
 the beginning of the problem.
 B.3 0.4 points
     Find the magnitude of the angular velocity of the turntable Ω as a function of
 R only, namely, in terms of Ω0 , R, R0 , r, m, I and Id .
 B.4 0.1 points
     From the result of B.3, for a given Ω0 , R0 , find the maximum possible Ω.
 B.5 3.1 points
     Write down the vertical component the angular momentum ẑMz of the whole
 system. Subtract any constant term and rename the remaining part as ẑL.
     In part B.1 you found the velocity of the ball ⃗v , which can be written as the
 sum of a part that depends on the position of the ball R     ⃗ and a constant vector.
 Let us call this constant vector ⃗c. Choose the direction of x-axis along this vector
                                                                            ⃗ ⃗c, ẑ, R2 ,
 and y-axis along ẑ ×⃗c. In this frame of reference, find Ω in terms of L, R,
 r, m, I and Id . Combining this with the result of B.3, write down an equation only
 containing R2 and y variables and L, r, m, I, c and Id . Here c is the magnitude of
⃗c. Substituting R2 = x2 + y 2 , write down an expression containing only x and y
 variables and describing a curve. From this, list all possible types of trajectories.

1.5     Part C: 4 points
In this part, we consider a density profile so that I = mr2 /10. This can be realized,
for example, if the ball is filled up to its half radius with uniform density and the

remaining part has a negligible mass. In addition, on its outer surface, the ball
has a uniform charge density Q/(4πr2 ), where Q is the total surface charge. The
whole setup is in a uniform magnetic field B  ⃗ that is in ẑ direction. The turntable
rotates with constant Ω like in Part A.
    It is often useful to analyze the equations governing the evolution of a system
in a unitless form so that the general behavior can be studied without worrying
about a specific values or units. For this purpose, we divide the R      ⃗ and Ω by 1
meter and 1 Hertz respectively. Also we divide the time variable by 1 second.
C.1 0.3 points
    Write down Newton’s equation and the torque equation for the ball. Find
expression for the torque τ⃗s due to the spinning of the ball around its axis in terms
                 ⃗
of Q, r, ⃗ω and B.
C.2 0.2 points
    Using the results of C.1, find expression for the linear acceleration of the ball
with respect to the laboratory frame in terms of Q, r, ⃗ω and B.   ⃗
C.3 0.3 points
    The equation for the linear acceleration you found in part C.2 is a second order
differential equation for R⃗ of the following form:

                                  ⃗
                               d2 R      ⃗
                                        dR
                                    − γ             ⃗ = 0.
                                           × ẑ + β R                            (12)
                               dt2      dt
Write down γ and β constants. From now on we assume we have made the trans-
formation to the unitless forms. This in turn, has an effect on the γ and β as
factors of 1/s =Hz and 1/s2 respectively, rendering them unitless as well. Make
the following transformation to a polar coordinates for the components of ⃗
                                                                          R:

                                  x(t) = ρ(t) cos(η(t)),                         (13)
                                  y(t) = ρ(t) sin(η(t)),                         (14)

so that the new equations do not have the first time derivative term. Here the
polar angle η(t) is a function of time. Find the form of the form of this function.
Express the coefficient β ′ of ρ(t) in the new equation in terms of γ and β. Write
down the conditions for different types of trajectories: harmonic, exponential etc.
   C.4 1.5 points
   Consider the following initial conditions for the solution found in part C.3:

          x(0) = 1, y = 0, vx (0) = ẋ|t=0 = 1, , vy (0) = ẏ|t=0 = −1.          (15)

find γ and β. Using them find the corresponding Ω. Sketch the trajectory. Is the
charge of the surface negative or positive? For the negative write − and for the
positive write + on your answer sheet.

   C.5 1.5 points
   Consider the solution you have found in part C.4. If you identified it correctly
                                     ⃗
your solution should have a rotating R(t). Find the expressions for the total and
per rotation changes in energy for N ≫ 1 number of rotations. Here you may
ignore the terms small compared to N . In this part assume the mass and the
radius of the ball are m = 1 and r = 1 so that I = 1/11 (in our unitless scheme
we divide masses by 1 kg).

2     Solution
2.1    Part A
A.1
   The velocity of the ball ⃗vb with respect to the turntable from the non–slipping
condition is given by:

                                ⃗vb = ⃗ω × (rẑ).                             (16)

The ball velocity with respect to the Lab frame is then
                               ⃗ + ⃗vb , ⃗v = Ωẑ × R
                    ⃗v = Ωẑ × R                    ⃗ + ⃗ω × ẑr.             (17)

A.2
   The force F⃗ and torque τ⃗ due to friction are:

                                F⃗ = m⃗v˙ ,                                   (18)
                                                    d⃗ω
                                τ⃗ = (−rẑ) × F⃗ = I .                        (19)
                                                    dt
The time derivative of equation (16) gives

                                            d⃗ω
                           ⃗v˙ = Ωẑ × ⃗v +     × (rẑ)                       (20)
                                            dt
                                                                              (21)

and substituting Eq. (18) and (19) in results in:
                                              2
                                        mr
                       ⃗v˙ = Ωẑ × ⃗v −             ẑ × ⃗v˙ × ẑ.            (22)
                                         I
Using the triple vector product rule in the last term of the above equation and
keeping in mind that both ⃗v and d⃗v /dt are orthogonal to ẑ yields
                                                          2
                                                 mr ˙
                                ⃗v˙ = Ωẑ × ⃗v −     ⃗v →                     (23)
                                                  I
                                           Ω
                                ⃗v˙ =            ẑ × ⃗v .                    (24)
                                      1 + mr2 /I

A.3

    The last equation unequivocally shows that the motion of the ball is circular
                                                           Ω
and the corresponding angular velocity of its center is 1+ma 2 /I . Now we integrate

this equation to find the radius and its center:
                               Ω             ⃗ −R   ⃗ 0 ),
                 ⃗v − ⃗v0 =            ẑ × (R                                   (25)
                          1 + mr2 /I
                                                             2
                          Ω             ⃗ −R ⃗ 0 − 1 + mr /I ẑ × ⃗v0 ) →
                 ⃗v =            ẑ × (R                                         (26)
                      1 + mr2 /I                           Ω
                          Ω             ⃗ −R ⃗ 0 ) + ⃗v0
                 ⃗v =            ẑ × (R                                         (27)
                      1 + mr2 /I

A.4
   From this we see that the circle trajectory has radius Rt and its center is located
at
                                                          2
                                   ⃗c = R
                                   R       ⃗ 0 + 1 + mr /I ẑ × ⃗v0 .            (28)
                                                        Ω
                   ⃗0 − R
                        ⃗ c| = 1 + mr 2
                                        /I              1 + mr2 /I
             Rt = |R                       |ẑ × ⃗v | =            v0            (29)
                                   Ω                        Ω

A.5
   In the case of a solid ball of uniform density, the moment of inertia is
                                       2mr2
                                   I=       ,                                    (30)
                                         5
and therefore the angular velocity of the ball’s center is
                                        2
                                   ωc = Ω.                                       (31)
                                        7
The time to return the initial point on the turntable is then
                                          14π
                                    t=         .                               (32)
                                           Ω
This solution is true for most cases. But there are special cases where this time
is shorter. Trajectory is a circle and its size is defined by its radius Rt and , as
stated, we solve for Rt = R0 . It could happen that the red spot happens to cross
path with the ball at a moment before the turntable could make a full circle. In
this case we can find the distance between the starting and the crossing positions:

                                   ωc t                2π − Ωt
                         2R0 sin           = 2Rt sin             ,             (33)
                                    2                     2
                               2π         14π
                         t=           =       .                                (34)
                             ωc + Ω       9Ω

2.2      Part B
Now we examine the case wherein the turntable rotates freely, i.e. without fric-
tion, around vertical axis. In this case the total kinetic energy and the angular
momentum are conserved.
B.1
    Integrating the torque equation for the ball one gets:
                                                 mr
                          ⃗ω × ẑ = ⃗ω0 × ẑ −      (⃗v − ⃗v0 ).            (35)
                                                 I
Substituting this into the non slipping condition we get

                                     ⃗ + ⃗ω0 × ẑr −    mr
                          ⃗v = Ωẑ × R                      (⃗v − ⃗v0 ),    (36)
                                                         I
                                        ⃗ 0 + ⃗ω (0) × ẑr,
                          ⃗v0 = Ω0 ẑ × R                                   (37)

which gives
                                I
                                              ⃗       ⃗

                       ⃗v =          ẑ × ΩR − Ω0 R0 + ⃗v0 ,                (38)
                             I + mr2
                                I
                       ⃗v˙ =         ẑ ×   Ω̇(t)R⃗ + Ω⃗v .                 (39)
                             I + mr2

B.2
      The torque equation for the turntable is:
                                             ⃗ × ⃗v˙ .
                                 Id Ω̇ẑ = −mR                              (40)

If we substitute the velocity and accelaration in the above equation and use the
triple vector product rule we get

                                ⃗         I
                                                      ⃗
                  Id Ω̇ẑ = −mR ×             ẑ × Ω̇R + Ω⃗v
                                      I + mr2
                                mI  2
                                                      ⃗

                   Id Ω̇ = −            Ω̇R + Ω ⃗v · R                      (41)
                              I + mr2
⃗v · R can be obtained using equation 44 as:

                      ⃗ =            I
                                                  ⃗ − Ω0 R
                                                         ⃗0      ⃗
                 ⃗v · R     ⃗v0 +         ẑ × ΩR              · R,         (42)
                                  I + mr2

                                     I            ⃗    ⃗
                        =   ⃗v0 −         Ω0 ẑ × R0 · R.                   (43)
                                  I + mr2

Applying this to the turntable torque equation (41), we obtain:

               mI      2             mI                I            ⃗0 · ⃗
       Id +          R Ω̇ = −              Ω ⃗v0 −          Ω0 ẑ × R    R. (44)
            I + mr2                I + mr2          I + mr2
We may rewrite the equation into a simpler form as:

                                 αΩ/r2 C  ⃗ ·R
                                             ⃗
                          Ω̇ = −                 ,                          (45)
                                   δ + αR2 /r2
where
                                          I
                               α ≡            ,                             (46)
                                      I + mr2
                                       Id
                                δ ≡        ,                                (47)
                                      mr2
                                                    ⃗ 0.
                                ⃗c ≡ ⃗v0 − αΩ0 ẑ × R                       (48)

B.3
      Observe that the velocity of the ball can be written as
                                             ⃗ + ⃗c,
                                 ⃗v = αΩẑ × R                              (49)
and, therefore, using equation (42) we see that:
                                      ⃗ ⃗
                              ⃗ = 1 d(R · R) = 1 Ṙ2 = R
                         ⃗v · R                        ⃗ · ⃗c.              (50)
                                  2    dt      2
Substituting this in equation (45) we get:
                        1 dΩ    1      1      d (αR2 /r2 )
                             =−                            .                (51)
                        Ω dt    2 δ + αR2 /r2     dt
The integration of this leads to:
                            2
                                           δ + αR02 /r2

                             Ω
                        ln         = ln                   ,                 (52)
                            Ω0             δ + αR2 /r2
                                         δ + αR02 /r2
                                Ω2 = Ω20                                    (53)
                                         δ + αR2 /r2

B.4 From this result we see that the maximum possible Ω is achieved when R2 , i.e
when the ball crosses the center of the turntable:
                                           r
                                                 αR2
                             Ωmax = Ω0 1 + 20                              (54)
                                                  δr

B.5
   Now we determine the trajectory of the ball. The total angular momentum
along ẑ is:
                                          ⃗ × ⃗v + Iωz ẑ.
                        Mz ẑ = Id Ωẑ + mR                                     (55)
Since there is no torque along ẑ acting on the ball ωz is constant. So we define the
following conserved quantity:
                                  ⃗ × ⃗v = Id Ω0 ẑ + mR
                  Lẑ = Id Ωẑ + mR                    ⃗ 0 × ⃗v0 .              (56)
The velocity of the ball ⃗v was written as the sum of a part that depends on the
                     ⃗ and a constant vector ⃗c. Then, we have:
position of the ball R

                          ⃗ × ⃗v = R
                          R          ⃗ × αΩẑ × R  ⃗ + ⃗c                   (57)
                                                  ⃗ × ⃗c.
                                     = −αΩR2 ẑ + R                             (58)
Substituting this in equation (59) one gets:

                       Lẑ = Id Ωẑ + αΩmR2 ẑ + mR ⃗ × ⃗c,                     (59)

                                        L − mẑ · R⃗ × ⃗c
                                    Ω=                                          (60)
                                           Id + αmR2
Choosing the direction of x-axis along ĉ and y-axis along ẑ × ĉ,
                                         L/mr2 + cy/r2
                                   Ω =                 ,                        (61)
                                          δ + αR2 /r2
Combining this with equation (52) we have:
                                                                        2
                                   δ + αR02 /r2           L/mr2 + cy/r2

                               Ω20        2 2
                                                     =                     ,    (62)
                                    δ + αR /r              δ + αR2 /r2
                                                                       2
            Ω20 δ + αR02 /r2       δ + αR2 /r2           L/mr2 + cy/r2 .

                                                     =                          (63)
Observe that this is the equation for conic section. Let us elaborate on this fact.
Let us introduce the following constants:
                       k ≡ Ω20 δr2 + αR02 , λ ≡ L/m.

                                                                              (64)

Expanding in Cartesian coordinates R⃗ = xx̂ + y ŷ, we obtain:

             kα δr2 + α(x2 + y 2 ) − λ2 + 2λcy + c2 y 2 = 0,

                                                                                (65)
                   kα2 x2 + (kα2 − c2 )y 2 − 2λcy = λ2 − kαδr2                  (66)

Since kα2 > 0, the trajectory is determined by the sign of kα2 − c2 :

                                  Ellipse if kα2 > c2 .                           (67)
                                  Parabola if kα2 = c2 .                          (68)
                                  Hyperbola if kα2 < c2 .                         (69)

2.3    Part C
C.1
   Here it is given that Ω = const. In addition, for the given mass distribution
where the ball is filled up to half of its radius, the momentum of inertia becomes

                                          mr2
                                     I=       .                                   (70)
                                          10
                                                     ⃗ and if the ball is charged
. In the presence of vertical uniform magnetic field B
                                       2
with uniform surface density ρ = Q/4πr , the equation of motions are changed as
follows:

                                   m⃗v˙ = F⃗f + Q⃗v × B ⃗                         (71)
                                   I ⃗ω˙ = −rẑ × F⃗f + τ⃗s ,                     (72)

where τs = Qr2 ⃗ω × ⃗ B/3 is the torque due to spinning of the charged sphere and
Ff is the friction force. Calculation of τs is essentially identical to the mechanical
moment of inertia for thin spherical shell. The torque is calculated as:
                Z
       τ⃗s   = d cos θdϕρ⃗r × (⃗ω × ⃗r) × B    ⃗                                   (73)
                Z
             = r2 d cos θdϕρ (⃗ω × ⃗r) ⃗r · B⃗                                     (74)
                      Z
                          2
             = ρ⃗ω × r d cos θdϕ⃗r ⃗r · B  ⃗                                       (75)
                          Z
             = ρBr4 ⃗ω × d cos θdϕ cos θ (sin θ cos ϕx̂ + sin θ sin ϕŷ + cos θẑ) (76)
                            Z 1               Z 2π
                    4                      2
             = ρBr ⃗ω × ẑ      d cos θ cos θ      dϕ                              (77)
                             −1                 0
                  2
                Qr      ⃗
            =      ⃗ω × B.                                                        (78)
                 3

C.2

   In addition we have the non–slipping condition from which we get:
                              ⃗v = Ωẑ × R  ⃗ + ω × ẑr,                    (79)
                              ⃗v˙ = Ωẑ × ⃗v + ⃗ω˙ × ẑr →                  (80)
                               ⃗ω˙ r = Ω⃗v − ⃗v˙ × ẑ.                      (81)
Substituting these and Ff from the Newton’s equation into the torque equation,
one gets:
                                                2
                                       ⃗ + Qr ⃗ω × B

          I ⃗ω˙ = −rẑ m⃗v˙ − Q⃗v × B                ⃗                       (82)
                                              3
                                                             2
                                                     ⃗ + Qr B ⃗v − Ωẑ × R

          I(Ω⃗v + ẑ × ⃗v˙ ) = −r2 ẑ × m⃗v˙ − Q⃗v × B                   ⃗   (83)
                                                            3
                               4Qr2 B                  Qr2 B ⃗

                    2 ˙

           I + mr ⃗v =                 − IΩ ⃗v × ẑ −       ΩR.              (84)
                                 3                       3
The last equation maybe written as:
                                ⃗
                             d2 R      ⃗
                                      dR
                                  − γ             ⃗ = 0,
                                         × ẑ + β R                         (85)
                             dt2      dt
where
                                Qr2 B
                          β≡              ,                                 (86)
                             3 (I + mr2 )
                             4Qr2 B − 3IΩ     4β
                          γ≡            2
                                            =    − αΩ.                      (87)
                              3 (I + mr )     Ω

C.3
    Here we divide R ⃗ and Ω respectively by 1 meter and 1 Hz, so we will deal
                                                       ⃗ = {x, y}, we have the
with unitless quantities. Then, in terms of components R
following unitless equations:
                                  ẍ − γ ẏ + βx = 0,                       (88)
                                  ÿ + γ ẋ + βy = 0.                       (89)
Substituting the following coordinate transformation
                                x(t) = ρ(t) cos(η(t)),                      (90)
                                y(t) = ρ(t) sin(η(t)),                      (91)
in the component equation leads to
                              ρ̈ + (β − γ η̇ − η̇ 2 )ρ = 0,                 (92)
                              ρ̇(γ + 2η̇) = 0.                              (93)

The first equation comes from the requirement that the coefficients of cos η (sin η)
and the terms containing first time derivative ρ̇ and η vanish separately. It is
straightforward to see this is equivalent to both ẋ and ẏ terms vanish. From this
we find:
                                    γ
                               η = − t + ϕ,                                       (94)
                                    2
                                                             γ2
                               β ′ ≡ β − γ η̇ − η̇ 2 = β +      .                 (95)
                                                             4
It is clear that for ρ̈ + β ′ ρ = 0 one gets three distinct behavior for ρ(t):

                            β ′ > 0, for harmonic oscillation                     (96)
                            β ′ < 0, for exponential run away                     (97)
                            β′ = 0                                                (98)

We examine the case β ′ = 0 in part C.4.
C.4
                               2
   If β ′ = 0 we have β = − γ4 . Therefore, ρ̈ = 0 and we have ρ(t) = A + Dt,
where A and D are constants to be determined.
   From the initial conditions

           x(0) = 1, y = 0, vx (0) = ẋ|t=0 = 1, , vy (0) = ẏ|t=0 = −1.          (99)

we find:

                             A = 1, D = 1, γ = 2, β = −1.                        (100)

Then the solution for the coordinates are:

                  x(t) = (1 + t) cos(t), y(t) = −(1 + t) sin(t).                 (101)

                         ⃗ can be calculated:
From this, the length of R

                          R2 = x(t)2 + y(t)2 = (1 + t)2 .                        (102)

Using the definitions of β and γ, the solutions for Ω are found as:
                                           √
                               Ω = −11 ± 77.                                     (103)

                                              ⃗ is in ẑ direction), from β < 0 we
Since the both solutions for Ω < 0 and B > 0 (B
see that Q < 0.
C.6

   From the solution we see that for every t = 2π time R  ⃗ makes one revolution.
                         2             2     2
After N ≫ 1 rotations, R = (1 + t) = t or R = 1 + t and we find the change in
R per rotation to be ∆R = ∆t = 2π.
   Scalar multiplying the acceleration by velocity and integrating it we obtain:
                           ⃗v · ⃗v˙ = −β⃗v · R⃗ →                           (104)
                             2       2          2 2     2

                            v − v0 = −β R − R0 = t .                        (105)
Then the total and per rotation changes in the kinetic energy associated to the
motion of the ball’s center per rotation are:
                                ⃗˙ · R
                   ⃗v · ⃗v˙ = β R    ⃗ →                                  (106)
                                2     2
                                           2    2
                                                        2
                              v − v0        R − R0     t
                    ∆K =                =            = ,                  (107)
                                  2           2        2
                                  2       2
                                                 2
                                vN  +1 − vN      R
                    ∆KN =                   =∆        = t∆t = 4π 2 N.     (108)
                                      2           2
Now we estimate the change in the kinetic energy associated with the spinning of
the ball. From non–slipping condition we get

                                                           ⃗ .
                           ω 2 = v 2 + Ω2 R2 + 2Ω⃗v · ẑ × R                 (109)

For our initial condition ⃗v0 · ẑ × R⃗ 0 = −v0 R0 and, for large N , ⃗v and R
                                                                             ⃗ are

approximately orthogonal to a very good approximation, so ⃗v · ẑ × R   ⃗ = −vR.
Our calculated Ω < 0, so we can write this term as |Ω|vR. So the kinetic energy
for spinning and its change are
                      Iω 2     I (v 2 + Ω2 R2 + 2|Ω|vR)
                Ks =       =                             ,                      (110)
                       2                    2
                        I ((v 2 − v02 + Ω2 (R2 − R02 ) + 2|Ω|(vR − v0 R0 ))
               ∆Ks =                                                        .   (111)
                                                  2
Finally, combining all the results we have:
                       I(ω 2 − ω02 )           I (v 2 + Ω2 R2 + 2|Ω|vR) t2
            ∆E =                     + ∆K ≃                            + ,      (112)
                             2                              2           2
                        2               2

                       t (1 + |Ω|)
                  =                       +1 ,                                  (113)
                        2       11
                           2      2
                       I(ωN   − ωN   −1 )
          ∆EN =                           + ∆KN                                 (114)
                             2
                               (1 + |Ω|)2

                  =    4πN                  + 1 with:                           (115)
                                    11
                              √
             |Ω| =     |11 ± 77|.                                               (116)

  The sketch of the trajectory looks like

References
[1] Warren Weckesser , “A ball rolling on a freely spinning turntable”
    AM. J. Phys. 65 (8), 736-738(1997).

[2] Luis Rodriguez, Comment on “A ball rolling on a freely spinning turntable”
    by Warren Weckesser AM. J. Phys. 66 (10), 927 (1998).

[3] Hector A. Munera, “A ball rolling on a freely spinning turntable: Insights
    from a solution in polar coordinates” Latin American Journal of Physical
    Education Vol. 5 (1), 49 (2011).

APhO 2023 Theory T2


[User]
Solenoid and loop
A closed circular loop of radius r consists of an ideal battery of electromotive force ξ and a wire of
resistance R. A long thin air-core
                                 solenoid  is aligned with the axis of the loop (z-axis). Its length is
                                      √
l fl r, cross-sectional area is A r fl A , and the number of turns is N . The solenoid is powered
by a constant current I provided by an ideal current source. The directions of the currents in the
solenoid and in the loop are the same (clockwise in the figure).

a) Find the force F1 acting on the solenoid when its head O1 is positioned in the loop centre O.
   What is the force F2 acting on the solenoid when its tail O2 is located in the centre of the loop?

b) Suppose now, that the solenoid is moving slowly with a constant velocity v along z-axis starting
   far away from the loop, going past its centre, and proceeding further to the right in positive
   z-direction. Plot the current J flowing in the loop as a function of time. Highlight important
   features and values on the graph. The velocity v is so small that self inductance of the loop
   can be neglected.

Solution:
T1: Solenoid and loop

Part a. Solution I. According to Newton’s third law, the
force acting on the solenoid is equal in magnitude, but
opposite in direction to the force acting on the loop. The
latter can be obtained from Lorentz’s law by summation
of infinitesimal forces ⃗ F = J∆⃗ l × ⃗ B acting on the indi-
vidual loop elements ∆⃗ l, where J = E/R is the current
in the loop.
Since the solenoid is long and thin, the magnetic field
lines inside it are directed in +z direction, and escape
only from the immediate vicinity of its ends. The mag-
netic field outside the solenoid is vortex-free and source-
free. The same requirements are satisfied by the electric
field in empty space. Hence, the magnetic field outside
the solenoid can be well approximated by a field created
by two magnetic poles: a North pole residing close to
point O1 and a South pole located near to O2. The flux
Φ emerging from the North pole (and the flux entering
the South pole) is the same as the flux passing through
the solenoid’s cross-section:
Φ = BinA = μ0
NI
l
A.
When the endpoint O1 of the solenoid is placed in the
loop centre O, the magnetic field of the other end (O2)
neartheloopisnegligible. ThefieldcreatedbytheNorth
pole located at O1 is pointing radially outwards, and its
magnitude at the loop circumference is (from spherical
symmetry):
B(r) =
Φ
4πr2
=
μ0
4π
NIA
lr2
.
Note. it is also correct to assume that the flux from the B-field
(which carries a factor of 1/2 compared to the field above at the
end of the solenoid is uniformly distributed over hemisphere.
The forces acting on all elements of the loop (and also
the net force) point in the −z direction, which can be
expected also from the same current directions (i.e. the
loop and the solenoid attract each other). Hence, the re-
action force acting on the solenoid points in the +z di-
rection, and its magnitude is given by:
F1 =
E
R
· 2πr ·
μ0
4π
NIA
lr2
=
μ0NIAE
2lRr
.
When the endpoint O2 is located at the centre of the
loop, the magnetic field produced by the South pole ex-
erts a force on the loop. Since this field is directed radi-
ally inward, the force acting on the solenoid is the same
in magnitude, but opposite in direction (−z) as the force
calculated above:
⃗ F2 = −⃗ F1 .
Grading scheme: T1 part a., Solution I.
using Newton’s third law 0.5 p
idea of approximating the outer field
with magnetic poles 0.5 p, and justi-
fication 0.5 p
1.0 p
calculating the flux Φ emerging from
the magnetic pole (for a wrong fac-
tor, 0.3 p)
1.0 p
expressing the field B(r) of the mag-
netic pole
1.0 p
findingthemagnitudeofforceacting
on the loop from Lorentz’s force law
1.0 p
correct direction for ⃗ F1 0.5 p, and ⃗ F2
antiparallel to ⃗ F1 0.5 p
1.0 p
Total for part a.: 5.5 p
Solution II. In this solution the force acting on the
solenoid is calculated, as the force acting on the mag-
netic pole placed at the centre of the current-carrying
loop. For this we need to find an expression for the mag-
netic pole strength (or magnetic charge) Qm, which is de-
fined as the ratio of the force and the magnetic field.
Thetotaldipolemomentmofthesolenoidistheprod-
uct of the number of turns and the dipole moment IA of
each turn:
m = NIA.
This can be also expressed with the magnetic pole
strength and the distance of the poles: m = Qml. From
this we arrive to the expression
Qm =
NIA
l
=
Φ
μ0
where Φ is the total flux emerging from the pole (see so-
lution I).
Note. The same result can be obtained from the analogy
between electrostatic and magnetostatic fields. The Coulomb
force between two point charges ±Q can be derived from the
principle of virtual work. The force is the derivative of the in-
teraction part of the field energy with respect to the distance
between the charges. The force between two magnetic charges
±Qm can be also calculated this way. From the expressions of
electricandmagneticenergydensitieswecanconcludethefor-
mula of the magnetic interaction force:
wE = 1
2
ε0E2
←→ wB = 1
2μ0
B2
,
FE = 1
4πε0
Q2
r2 ←→ FB = μ0
4π
Q2
m
r2 .
As it can be seen, the well-known formulae known in electro-
staticscanbealsousedinmagnetostaticswiththesubstitutions
ε−1
0 ←→ μ0, E ←→ B, Q ←→ Qm. Carrying on this analogy the
magnetic pole strength can be figured out:
Q = ε0Ψ ←→ Qm =
Φ
μ0
=
N
l
IA,
where Ψ and Φ are the electric and magnetic flux for a closed
surface containing the electric and magnetic charge, respec-
tively.
When endpoint O1 of the solenoid is located at point
O, a North pole resides at the centre of the loop. Here
the magnetic field created by the loop can be expressed
from Biot–Savart-law:
B
(at center)
loop =
μ0J
2r
=
μ0E
2Rr
,
EuPhO-2020 Online Theoretical Solutions. Language: English
pointing in the +z direction. So the magnitude of the
force acting on this end of the solenoid is:
F1 = QmB
(at center)
loop =
μ0ENIA
2lRr
,
and it is directed to +z. When the endpoint O2 is located
at the center of the loop, the force acting on the South
pole should be calculated, resulting a force of same mag-
nitude, but opposite direction.
Grading scheme: T1 part a., Solution II.
idea of approximating the outer field
with magnetic poles 0.5 p, and justi-
fication 0.5 p
1.0 p
calculating the magnetic pole
strength Qm (for dimensionally
wrong answer 0 p)
1.0 p
justification for calculation 1.0 p
calculating the field of the current
loop at its centre from Biot–Savart-
law
0.5 p
findingthemagnitudeofforceacting
on the magnetic pole
1.0 p
correct direction for ⃗ F1 0.5 p, and ⃗ F2
antiparallel to ⃗ F1 0.5 p
1.0 p
Total for part a.: 5.5 p
Solution III. Some of the magnetic field lines created
by the loop enter into the near end O1 of the solenoid;
this entering flux is given by
Φin = B
(at center)
loop A =
μ0J
2r
A =
μ0EA
2Rr
.
Since the end O2 is far from the loop, the flux created by
the loop escaping there is negligibly small. This means
that almost all the flux Φin escapes from the solenoid
through its side.
Denote the radial component of the magnetic field
vector produced by the current-carrying loop at the
perimeter of the ith turn of the solenoid by Bi. Only
this component contributes to the net force acting on the
solenoid, as the axial component produces a radial force
which is cancelled due to rotational symmetry. The axial
force acting on the ith turn of the solenoid is given by
F1,i = 2
√
AπIBi ,
where 2
√
Aπ is the circumference of one turn, and the
force points in the +z direction. Summing up both sides
gives the net force:
F1 =
X
i
Fi =
X
i
2
√
AπIBi
Take out the factor I from the summation and insert 1
written in the unusual way (l/N) · (N/l):
F1 = I
N
l
X
i
2
√
Aπ
l
N
Bi .
The sum on the right hand side is the flux escaping
through the side of the solenoid, which equals Φin, so the
force:
F1 = I
N
l
Φin =
μ0EANI
2Rrl
,
which agrees with the previous solutions.
Grading scheme: T1 part a., Solution III.
B-field at the center of loop 0.5 p
expressing force on one turn 1.0 p
realizing that all the flux escapes
through the sides of the solenoid
0.5 p
relating the force to escaping flux 1.5 p
summation and correct result 1.0 p
correct direction for ⃗ F1 0.5 p, and ⃗ F2
antiparallel to ⃗ F1 0.5 p
1.0 p
Total for part a.: 5.5 p
Solution IV. The force acting on a current loop of
magnetic moment ⃗ m placed in magnetic field ⃗ B is given
by ⃗ ∇(⃗ m · ⃗ B). Divide the solenoid into short circular coils
of equal length ∆l, then the magnetic moment of each
short coil is
∆⃗ m = IA
N∆l
l
⃗ ez ,
where ⃗ ez denotes the unit vector in z-direction. This
magnetic moment is parallel to the field ⃗ Bloop created
by the large current-carrying loop, so the force acting on
each short segment of the solenoid in z-direction can be
written as:
∆F1 = ∆m
dBloop
dz
= IA
N∆l
l
dBloop
dz
The total force on the solenoid can be determined from
integrationof the force contributions along the solenoid:
F1 =
Z 0
−l
dl
∆F1
∆l
= IA
N
l
Bloop(0) − Bloop(−l)

.
Using the Biot–Savart-law we can compute the magnetic
fieldBloop(z)ofthecurrent-carryingloopalongthez-axis
to
Bloop(z) =
μ0E
2R
r2
(z2 + r2)3/2
.
This expression for Bloop yields
F1 =
μ0EANIr2
2Rl

1
r3
−
1
(l2 + r2)3/2

l≫r
≈
μ0EANI
2lRr
,
and F1 is directed to +z. From a similar calculation we
get ⃗ F2 = −⃗ F1.
Grading scheme: T1 part a., Solution IV.
idea of dividing the solenoid into
short segments of equal lengths
1.0 p
givingthemagneticmomentofaseg-
ment
0.5 p
expressing the force on a segment by
a derivative of the magnetic field
1.0 p
calculating the field of the current-
carrying loop from Biot–Savart-law
(alternatively, arguing that only the
fieldatcenterofloopisimportant0.5
p and calculating this field 0.5 p)
1.0 p
integrating force contributions to
find total force on solenoid (max. 0.5
if final expression still contains inte-
gral)
1.0 p
correct direction for ⃗ F1 0.5 p, and ⃗ F2
antiparallel to ⃗ F1 0.5 p
1.0 p
Total for part a.: 5.5 p
EuPhO-2020 Online Theoretical Solutions. Language: English
Note: Using the idea of dividing the solenoid into
small segments other solutions are possible as well
(e.g. considering small dipole contributions). In this case
the grading scheme of Solution IV should be adapted ac-
cordingly. One recurring solution of this type is to di-
vide the solenoid into small segments of length dz, com-
pute the B-field in z-direction effected by current-loop
from Biot–Savart, use Gauss law to relate the radial B-
fluxthroughthesegmenttodB/dz. computetheforceon
the segment using Lorentz forces and integrate the force
over the length of the solenoid. In this case the marking
scheme is adapted in the second aspect, awarding 0.5 p
for the idea of using Gauss law to find radial B-field
Solution V. In this solution we relate the force acting
on the solenoid to the change in energy of the system.
Investigate the case when point O1 is located at O first.
Due to the same current directions, the magnetic force
⃗ F1 acting on the solenoid points in direction +z. While
keeping the solenoid in equilibrium with external force
−⃗ F1, let it move by a small displacement δz in the posi-
tive z direction. The work done by the external force is
equal to the change in energy of the system:
−⃗ F1 · δ⃗ z = −F1δz = δEtotal,.
We should be aware of the fact that the system is not
closed: there is also a battery and a current source in-
cluded in the circuits. Hence, δEtotal contains the change
in field energy and the change of energy of the power
sources:
δEtotal = δEfield + δEsources .
Since the force does not depend on what kind of power
supplies we have, let us replace the battery with a cur-
rent source providing constant current J = E/R.
Now we find a relation between δEfield and δEsources
The energy stored in the field can be expressed as
Efield =
1
2
L1I2
+
1
2
L2J2
+ L12IJ ,
where L1 is the inductance of the solenoid, L2 is that of
the loop and L12 is the mutual inductance of the system.
Upon small displacement δz, only the last term changes,
so
δEfield = δL12 · IJ .
The small displacement results in a change of the flux
enclosed by the loop and the solenoid. The flux created
by the solenoid on the loop is L12I, and the flux created
by the loop through the solenoid is L21J = L12J (here
we used the symmetry property of mutual inductance).
Duringtheshorttimeδtofthedisplacementδz, thee.m.f.
inducedintheloop(V
loop
ind )andthesolenoid(V solenoid
ind )can
be expressed with Faraday’s law:
V
loop
ind = −
δL12
δt
I , V solenoid
ind = −
δL12
δt
J .
In order to keep the current in the circuits constant, the
current sources need to provide an additional power, so
they give away extra energy (in addition to Joule heat).
This energy change of the sources is given by:
δEsources =

V
loop
ind J + V solenoid
ind I

δt.
Using the previous results we finally get:
δEsources = −2δL12IJ = −2δEfield ,
which means δEtotal = −δEfield, and hence F1δz =
δEfield = δL12IJ.
Note 1. Naively, one might think that we get the result
δEtotal = δEfield if we imagine superconducting wires without
power supplies. One can show with detailed calculation that
in that case the currents in the loop and the coil change, as the
total flux enclosed by a superconducting circuit must remain
constant. The correct physical justification of the appearing
negative sign is an important part of the solution.
Now we calculate the change in mutual inductance
δL12. The small displacement δz can be imagined as we
take a short segment from the tail O2 of the coil (con-
sisting of Nδz/l turns) and move it to the head O1. As
a result, the flux produced by the loop on the solenoid
increases by
δΦ12 = δL21J =
μ0J
2r |{z}
B(at center)
loop
A
N
l
δz .
From this we get:
F1 =
δL12IJ
δz
=
μ0ENAI
2lRr
.
If the tail O2 of the solenoid is located at point O, the
coefficient of mutual inductance decreases upon small
displacement, which results in ⃗ F2 = −⃗ F1.
Note 2. The field energy can be also calculated from the en-
ergy density integrated for the whole space. Instead of calcu-
lating the total field energy, it is easier to find its change using
the same idea presented above, i.e. take a segment of length
δz from the tail and move it to the head of the solenoid. As-
suming l ≫ δz ≫
√
A, the field created by the solenoid inside
that segment is μ0NI/l (because the field differs from this only
at distance ∼
√
A from the ends). At the end we get the same
result for the change in field energy using the expression:
δEfield =
1
2μ0

(B(at centre)
loop + Bsol)2
− (B(at centre)
loop )2
− B2
sol

Aδz .
Note 3. A third possibility is to calculate the potential en-
ergy change of the displaced few turns of the solenoid. The
magnetic moment of a segment of length δz is ⃗ m = ⃗ ezIANδz/l,
and its energy in external field is Epot = −⃗ m⃗ B. Important to
highlight that this potential energy already contains the factor
of −1 discussed at the beginning of the solution, so the force
acting on the solenoid can be expressed as
F1 = −
δEpot
δz
.
The external field is the superposition of the field ⃗ Bloop cre-
ated by the loop and the field ⃗ Bsol created by the coil (note that
this latter contains a factor of 1/2 compared to the field in the
middle of the solenoid). Since ⃗ Bsol is the same at the two ends
O1 and O2, the energy change is:
δEpot = −⃗ m⃗ B(at center)
loop − ⃗ m⃗ Bloop(z = l).
The second term can be neglected, and we get
δEpot = −
IANδz
l
μ0E
2Rr
,
EuPhO-2020 Online Theoretical Solutions. Language: English
which gives the same answer for F1 as the other ideas.
Grading scheme: T1 part a., Solution V.
Equating force to energy change in
system
0.5 p
Formulating energy equation (if
sources are missing 0.2 p)
0.5 p
expressing δEfield through contribu-
tions from currents and an interac-
tion term
1.0 p
deriving that δEsources = −2δEfield 1.0 p
computingthechangeinfieldenergy
and final result for force.
1.5 p
correct direction for ⃗ F1 0.5 p, and ⃗ F2
antiparallel to ⃗ F1 0.5 p
1.0 p
Total for part a.: 5.5 p
Part b. In order to plot a graph displaying the im-
portant features, it is beneficial to make some calcula-
tions. The problem text does not specify the zero point
of time, so take t = 0 in the moment when the center of
thesolenoidislocatedatpointO. Thismeansthatattime
t the head O1 of the solenoid is located at z1 = l/2 + vt,
while the tail O2 is located at z2 = −l/2 + vt.
The current flowing in the loop at an arbitrary mo-
ment of time is given by
J(t) =
E + Vind(t)
R
,
where Vind(t) is the induced electromotive force in the
loop as a function of time. Two different approaches can
befoundbelowwhichgiveananalyticalformulaforthis
induced e.m.f.
Solution I. We may again approximate the magnetic
field outside the solenoid by a field created by two mag-
netic poles at z1 and z2, respectively. The resulting mag-
netic flux through the loop can be calculated by consid-
eringthesolidangletheloopextendsasseenfromeither
of the poles. Using the total magnetic flux of the poles as
calculated in Solution I for part a. we get
Φloop =
μ0NIA
2l
(
z1
p
z2
1 + r2
+ 1 −
z2
p
z2
2 + r2
− 1
)
.
Usingthisresultand  ̇ z1 =  ̇ z2 = v theinducede.m.f.canbe
calculated with Faraday’s law Vind = −dΦloop/dt, which
gives
Vind = −v
μ0NIA
2l

r2
(z2
1 + r2)3/2
−
r2
(z2
2 + r2)3/2

.
Solution II. The rate of change of flux produced by
the solenoid through the loop can be expressed in terms
ofthemutualinductanceL12 ofthesolenoidwithrespect
to the loop:
dΦ12
dt
=
d
dt
(L12I) = I
dL12
dt
.
Using the symmetry property of mutual inductance
(L12 = L21), instead of calculating L̇12 let us find the
time derivative of L21, i.e. the mutual inductance of the
loop with respect to the solenoid. For this, imagine that
the current in the loop is constant J0, and calculate the
changeoffluxproducedbytheloopthroughthesolenoid
during a short amount of time dt! The small displace-
ment vdt of the solenoid can be considered as moving a
shortsegment of length vdtfromthe tail to the head. The
change in flux in this segment is given by
dΦ21 = dL21J0 = vdt
NA
l
Bloop(z1) − Bloop(z2)

.
Using the formula for Bloop(z) obtained from Biot–
Savart-law (see Solution IV for part a.) we get
dL21
dt
= v
NA
l
μ0
2

r2
(z2
1 + r2)3/2
−
r2
(z2
2 + r2)3/2

.
From this the induced e.m.f. Vind = IL̇12 = IL̇21 can be
expressed:
Vind(t) = −
dΦ21
dt
= −v
μ0NIA
2l

r2
(z2
1 + r2)3/2
−
r2
(z2
2 + r2)3/2

Although the analytical result gives the correct ex-
pression for the current flowing in the loop, the task in
part b. was to plot the graph.
The graph should reflect the most important features
of the function. First, Vind(t) is an odd function, i.e.
Vind(−t) = −Vind(t). For t < 0 the flux through the loop
increases meaning that Vind < 0 and J < E/R, while
for t > 0 the flux decreases, which results Vind > 0 and
J > E/R.
In different ranges of time, Vind(t) behaves very dif-
ferently. For times t ≪ −l/(2v) and t ≫ l/(2v) the
solenoid is very far (approximately at distance vt) from
the current loop, and its field can be approximated by
dipole field (proportional to v−3
t−3
). The time derivative
of the field is proportional to the induced e.m.f. in the
loop, so in this time range Vind(t) ∝ t−4
, and has very
small value. The same result can be concluded from the
Taylor expansion of the complete analytical formula for
Vind.
When t ≈ ±l/(2v) (with the accuracy of r/v) the effect
of one pole of the solenoid can be neglected. The abso-
lute value of the induced e.m.f. is maximal here:
|Vind| = v
μ0NIA
2lr
,
so the maximal and minimal value of the current are
Jmin =
E
R
− v
μ0NIA
2Rrl
, Jmax =
E
R
+ v
μ0NIA
2Rrl
.
Around the maximum Vind is a quadratic function of
time, as it can be proved with expanding the complete
analytical formula.
When the centre of the solenoid is close to the cen-
tre of the loop, i.e. |t| ≪ l/(2v), the flux barely changes,
so Vind ≈ 0. A more careful analysis gives a very weak
linear dependence on time.
The statements above are summarized in the table
below.
EuPhO-2020 Online Theoretical Solutions. Language: English
time range Vind(t) J(t)
t ≪ −l/(2v) small, ∝ −1/t4
≈ E/R
t ≈ −l/(2v) large, ∝ − t + l
2v
2
Jmin, dip
|t| ≪ l/(2v) negligible (∝ t) ≈ E/R
t ≈ l/(2v) large, ∝ t − l
2v
2
Jmax, peak
t ≫ l/(2v) small, ∝ 1/t4
≈ E/R
− l
2v
0 l
2v
E
R − vμ0NIA
2Rrl
E
R
E
R + vμ0NIA
2Rrl
l
v
r
v
t / a.u.
J / a.u.
Figure 1: Qualitative graph of J as a function of time for
l = 5r.
Grading scheme: T1 part b.
relating J(t) to Vind or other suitable
quantity (independent of result in a.)
0.5 p
the J(t) graph is smooth everywhere
(if graph does not capture whole do-
main including asymptotics, 0 p.)
0.5 p
J(t) ≈ E/R, except if t ≈ ±l/(2v) (if
J(t) ≈ E/R only asymptotically, 0.2
p)
0.5 p
J(t) − E/R is clearly an odd function
(computing analytical expression for
J andgraphshowingoddbehavioris
considered sufficient, if reasoning is
missing, 0.2 p)
0.5 p
J(t)hasaminimumfirst,thenamax-
imum l/v time later (if reversed or
the time is incorrect, 0 p)
1.0 p
analytical formula for Jmin and Jmax
(either approximate or exact), if an-
alytical expression is given but min-
max values are not determined 0.5
p are given for deriving the formula
and realizing that min-max exist.
1.0 p
it is indicated that the duration of the
peak and dip is in the range of r/v
0.5 p
if J(z) instead of J(t) is plotted or la-
bels are missing on the axes 0.5 p are
deducted from part b. but only if 0.5
or more marks are given to the ac-
tual graph at all
(-0.5 p)
Total for part b.: 4.5 p
If no graph is drawn: max. 0.5 p for relating J(t) to Vind
and 1.0 p for Jmin and Jmax (for a general formula with-
out evaluation of min and max: 0.5 p) are given.
If B-field of the solenoid is assumed constant across loop
or area for flux is taken to be A, 0 p are awarded for min
and max.
Additional general guidelines for grading T1:
• Grading should always follow one of the solutions de-
scribed. If approaches for solutions are mixed the one
resulting in the highest marks is considered.
• Granularity for marks is 0.1 p.
• A simple numerical error resulting from a typo is pun-
ishedby0.1punlessthegradingschemeexplicitlysays
otherwise.
• Errors which cause dimensionally wrong results are
punished by at least 50 % of the marks if dimensions
can easily be checked. In more complicated cases less
marks may be deducted.
• Propagating errors are not punished repeatedly un-
less they either lead to considerable simplifications or
wrong results whose validity can easily be checked.
• No marks are given for the directions of the forces in
a. if forces are physically incorrect (e.g. resulting force
is calculated to zero) or if no forces are calculated.
• If current I in solenoid is used as current in loop in-
steadofE/Rorviceversainthecalculationoftheforce
0.5 p are substracted.
• If only z-component of B-field (either from solenoid or
loop) is determined and the force on the solenoid is de-
rived from this using Lorentz-forces, marks are only
given for ideas (max. 1.0 p.) and the use of the Biot–
Savart law (max. 1.0 p)

Solenoid and loop


[User]
In the circuit diagram given in the figure there are identical resistors with resistances $R_1 = R_2 = R_3 = R_4 = R$ and identical batteries with electromotive forces ${\cal E}_1 = {\cal E}_2
= {\cal E}$. Find the current strengths in the resistors (that means the expressions of $I_1$, $I_2$, $I_3$ and $I_4$ through the values of $R$ and ${\cal E}$).
\begin{center}
\includegraphics[width=0.35\linewidth]{2012-v3g-03-elektriline_sild}%
\end{center}

Solution:
Sümmeetria tõttu ülemises ja alumises takistis ($R_2$ ja $R_4$) vool puudub.
Tõepoolest, oletagem et ühes neist (nt $R_2$-s) läheb vool vasakult paremale. Peegeledame joonist vertikaaltelje
suhtes: $R_1$ ja $R_3$ ning ${\cal E}_1$ ja ${\cal E}_2$ vahetavad kohad,
kuid takistite ning elektromotoorjõudude võrdsuse tõttu skeem ei muutu. Peegelduse käigus
peaks nüüd vool muutma suuna vastupidiseks ning minema paremalt vasakule; esialgse ja
peegelduse tulemusel saadud skeemi ekvivalentsuse tõttu peaks aga vool olema endiselt vasakult paremale,
mis viib vastuoluni. Järelikult on vool $R_2$-s ja $R_4$-s null ning skeemilt võib vastavad juhtmed kõrvaldada.
Järele jääb $R_1$, ${\cal E}_2$, $R_4$ ja ${\cal E}_2$ jadaühendus, mistõttu on voolutugevus ahelas, sh $R_1$-s ja $R_3$-s
leitav kui $I=2{\cal E}/2R = {\cal E}/R$.

Elektriline sild

Joonisel toodud skeemis on tegemist ühesuguste takistitega takistustega $R_1 = R_2 = R_3 = R_4 = R$ ning ühesuguste
ideaalsete patareidega elektromotoorjõududega ${\cal E}_1 = {\cal E}_2
= {\cal E}$. Leidke voolutugevused takistites (st $I_1$, $I_2$, $I_3$ ja
$I_4$ avaldised suuruste $R$ ja ${\cal E}$ kaudu).

\begin{center}
\includegraphics[width=0.35\linewidth]{2012-v3g-03-elektriline_sild}%
\end{center}

Peegelsümmeetria tõttu ülemises ja alumises takistis ($R_2$ ja $R_4$) vool puudub.


[User]
01h
                         APhO 2011, problem 2. This covers “stick-slip”, which appears in many real-
    world contexts and Olympiad problems. You can see it in action on a violin string here. (The
    calculations in this problem are chosen to be relatively simple, requiring mostly intuition; for some
    other calculations, see USAPhO 2021, problem A1, which covers a similar setup.)

    4      Rotation

Solution:
APhO 2011, problem 2. This covers “stick-slip”, which appears in many real-
    world contexts and Olympiad problems. You can see it in action on a violin string here. (The
    calculations in this problem are chosen to be relatively simple, requiring mostly intuition; for some
    other calculations, see USAPhO 2021, problem A1, which covers a similar setup.)

    4      Rotation

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


[User]
A student proposes to build this set of pulleys and string, called a “fool’s tackle”.

    If the load L has mass m, and the pulleys and string are frictionless and have negligible mass, find
    the force F needed to keep the system static.

Solution:
A student proposes to build this set of pulleys and string, called a “fool’s tackle”.

    If the load L has mass m, and the pulleys and string are frictionless and have negligible mass, find
    the force F needed to keep the system static.

    Answer. This system can’t be static. It just instantly falls apart, no matter what F is.


[User]
01^   USAPhO 2021, problem B3. A cute setup with many nice lessons.

    The next two questions are about three-dimensional rotation, covered in M8.

[3] Problem 19 (BAUPC). A frictionless fixed cone stands on its tip.

     (a) A particle slides on the inside surface of the cone at height h above the tip, as shown at left
         above. Find the angular frequency of the circular motion.

     (b) Now suppose the cone has friction, and a small ring of negligible radius rolls on the surface
         without slipping at the same height. Also assume that the plane of the ring is at all times
         perpendicular to the line joining the point of contact and the tip of the cone, as shown at
         right above. Find the angular frequency of the circular motion.

        (c) How general were our assumptions in part (b)? Specifically, would the described motion had
            been possible if the plane of the ring were at a different angle? Is a slightly smaller or larger
            angle to the horizontal possible? Would it be possible if the ring were exactly horizontal?

Solution:
(a) The centripetal force mω 2 h tan θ is horizontal, which equals to N cos θ. The particle
         must be vertically balanced, so N sin θ = mg, giving

                                                 ω 2 h tan θ = g cot θ

         and an answer of                                       r
                                                                    g
                                                    ω = cot θ         .
                                                                    h

     (b) Let the ring have moment of inertia βmr2 (where β = 1) and moves in a circle of radius
         R = h tan θ ≫ r. The no slip condition is ωr = ΩR. When taken about the point of contact,
         the ring has a horizontal angular momentum Lh = (1 + β)mr2 ω sin θ. Since τ = dL/dt, then
         about the point of contact (so the confusing normal and friction forces can be ignored) the
         torque is τ = mgr cos θ and |dL|/dt = ΩLh . Thus

                                         mgr cos θ = Ω(1 + β)mr2 ωΩ sin θ

                                               g           Ω2 R
                                                 = (1 + β)      tan θ
                                               r            r
                                                            r
                                                               g
                                                  Ω = cot θ
                                                              2h

     (c) There are two constraints in this problem: force balance and torque balance. As we saw in
         part (b), considering the torque of gravity about the contact point alone fixes the angular
         frequency Ω of the circular motion. That in turn gives the force balance equations (vertical
         force is zero, horizontal force is centripetal), and since the coefficient of friction is high enough
         to prevent slipping, there’s always some combination of normal and frictional forces that
         makes the problem work out. Since none of this depends very sensitively on the angle, we
         could change the angle and the problem would still work.
         There’s only one exception: you can’t have a horizontal ring. In that case, the angular
         momentum of the ring does not change at all (because it’s always spinning in a horizontal
         plane), so the torque balance equation can’t be satisfied. As a real-life example, when
         motorcyclists ride along the equator of the globe of death (mentioned in M2), they always
         tilt a bit above the horizontal.


[User]
8. BRıGHTNEss OF PLANETs (9 points)

— Topi
Löytäinen, Jaan Kalda.
In this problem, the time periods are
measured in years (y), and the distances are
measured in astronomicalunits (au); the dis‐
tance of Earth from Sun is R⊕ = 1au. The
planets and Moon are assumed to move in
the same plane, the ecliptic, around circu‐
lar orbits in the same direction. Distance of
VenusfromSunisR♀ = 0.72au,anddistance
of Mars from Sun is R♂ = 1.5au. In some
parts below, you will need to assume that
the planets are so‐called Lambertian scat‐
terers, i.e. when you can see the full disc
of a planet illuminated by sunlight, middle
parts of the disc seem exactly as bright as
the parts at the edge of the disc. Through‐
out the problem, assume that the observer
is situated in Tallinn, at the geographical co‐
ordinates59.5°Nand24.7°E.TheaxisofEarth
is tilted by 23.5°with respect to the normal of
the ecliptic plane.
i) (0.7 points) When can you see a waxing
crescent moon? Select one or more options
(A, B and/or C), and motivate your answer
with a diagram depicting Sun, Moon and
Earth. Select from the options
A: immediately after sunset;
B: during midnight;
C: immediately before sunrise.
ii) (1.2 points)What is the culmination angle
(i.e. the maximal angle above the horizon) of
afullmoon duringwinter solsticeinTallinn?
iii) (1.2 points)The apparent brightness (illu‐
minance at Earth, i.e. luminous flux dens‐
ity) of planets may vary over a wide range of
values. By how many times does the appar‐
ent brightness of Mars vary between when
it’s closest and farthest away from Earth?
iv) (1.2 points) What is the time period on
Earth between the moments when Mars is
the closest to Earth and farthest?
v) (1.2 points)From autumn to spring, Venus
can never be seen at midnight. For how long
maximally it can be seen after the sunset?
vi) (2.5 points) Express the normalised ap‐
parent brightness I/I0 as a function of R⊕,
R♀, and the distance L from Earth to Venus.
Here, the normalisation constant I0 can be
chosen arbitrarily. Hint: I/I0 should be a
polynomial of L−1
.
vii) (1 point) Find the distance L = L0
whentheapparentbrightnessofVenusisthe
highest, and the angular distance between
Venus and Sun at that moment.

Solution:
8. PLANETs

— Solution by Taavet
Kalda.
i) (0.7 points)A waxing crescent moon can be
seenimmediatelyaftersunset,illustratedbe‐
low
Grading:
• Figure with Moon between Earth and Sun
(0.3pts)
• All rotation directions consistent (0.1pts)
• Correct answer A (0.3pts)
ii) (1.2 points) Zenith in Tallinn forms an
angle φ = 59.5° with the celestial equator.
During a winter solstice, the Sun is ε =
23.5° below the celestial equator. Since full
moon occurs when it’s opposite to the Sun,
Moon must be ε above the celestial equator.
Further, the Moon is at its highest when it
crosses the plane formed by the centre of
Earth, the north pole (marked NP) and the
ray pointing towards the zenith in Tallinn. A
convenient way to imagine this is how dur‐
ing one day, stars and planets (alongside the
Moon)stayroughlythesamerelativetoeach‐
other in the sky, but the sky as a whole ro‐
tates around the axis formed by the centre
of Earth and north pole. This is depicted in
the figure below, alongside with the celes‐
tial equator. From the figure, we can now
read that the Moon is an angle φ−ε from the
zenith and hence the maximal culmination
angle is 90° − φ + ε = 54.0°.
Grading:
• Understanding of the positions of Tallinn,
Earth and Moon at the maximum point
(0.5pts)
• Correct final expression (0.6pts)
• Correct numerical value 54.0° (0.1pts)
• Expression and value to Zenith 36.0° given
instead (0.5pts)
iii) (1.2 points)The flux of an object drops in‐
verse squared to the distance from the said
object.
Mars is closest/farthest to Earth when it
alignswiththelineformedbyEarthandSun.
In both cases, Mars is at full phase. At its
closest, Mars is d− = R♂ − R⊕ from Earth,
and its farthest, d+ = R♂ + R⊕. In both
cases, the same amount of sunlight reaches
the Mars surface (because we assume Mars
to always be R♂ from the Sun). This, com‐
bined with phases being the same, gives us
the ratio of illuminances to be
Imax
Imin
=
d2
+
d2
−
=
(
R♂ + R⊕
R♂ − R⊕
)2
= 25.
Grading:
• Noting I ∝ 1/r2
(0.2pts)
• Notingminimum/maximumcorrespondto
R♂ ± R⊕ (0.2pts)
• Noting Mars is at full phase in both posi‐
tions (0.2pts)
• Correct final expression (0.5pts)
• Correct numerical value 25 (0.1pts)
iv) (1.2 points) Looking from the Sun, the
separation between Earth and Mars must
change from 0 to π between the two de‐
scribed situations. The relative angular
speed of the two planets is ∆ω = 2π/T⊕ −
2π/T♂. Mars’ period can be found from
Kepler’s III law as T♂ = T⊕
√
(R♂/R⊕)3. The
time taken is thus
t =
π
∆ω
=
T⊕
2
1
1 − (R⊕/R♂)3/2
= 1.1yrs.
Grading:
• Expressing the relative angular speed ∆ω
(0.3pts)
• Solving T♂ in terms of T⊕ and orbital radii
(0.3pts)
• Correct final expression for half a period
(0.5pts)
• Correct numerical value 1.1yrs (0.1pts)
v) (1.2 points) The maximal angular separa‐
tion between Sun and Venus, as seen from
Earth, can be found as the angle between
Earth‐Sun ray and the ray that goes through
Earth and is tangent to the orbit of Venus.
From the right angled triangle, we get the
seperation to be α = arcsin(R♀/R⊕) = 46.1°.
We therefore see Venus for a duration of
1day ·
α
2π
= 11050s = 3hrs 4min.
Grading:
• Angular separation is the angle between
the Earth‐Sun and Earth‐Venus rays
(0.2pts)
• Expressing time as a ratio to a full rotation
of the Earth (0.2pts)
• Maximal when tangent to orbit (0.4pts)
• Correct expression for α (0.3pts)
• Correctnumerical value for time 11050s =
3hrs 4min (0.1pts)
vi) (2.5 points) As discussed in part iii), the
luminosity of a planet is the product of its
phase, φ (the fraction of its area which is il‐
luminated),andtheinversesquareddistance
to Earth. In other words,
I
I0
=
φ
L2
.
In the figure below, the phase is simply
the ratio of the lengths of the “visible” to the
“dark” part.
From the geometry, we see that this is (1 +
cosβ)/2, where β is the angle between the
Sun, Venus, and Earth. From cosine law, we
get
R2
⊕ = R2
♀ + L2
− 2R♀Lcosβ
so
1 + cosβ = 1 +
R2
♀ − R2
⊕ + L2
2R♀L
.
Hence,
I
I0
=
R2
♀ − R2
⊕
2R♀L3
+
1
L2
+
1
2R♀L
= −
R2
⊕ − R2
♀
2R♀
x3
+ x2
+
1
2R♀
x,
where we set x = L−1
.
Grading:
• Realizing I/I0 is dependent on what frac‐
tion of Venus is illuminated from Earth’s
view (0.4pts)
• Dependency of the form φ/L2
(0.4pts)
• Realizing illuminated area is proportional
to 1 + cosβ (0.8pts)
• Expressing β using the cosine law (0.2pts)
• Correct final expression (0.7pts)
vii) (1 point)Our goal is to maximise I(x)/I0
in the range x ∈ [1/(R⊕ +R♀),1/(R⊕ −R♀)].
I(x) is a cubic that starts off from 0 at x = 0
andextends all thewaytonegativeinfinityat
x −→ ∞. We find the extrema by setting the
derivative to zero
d
dx
(
I
I0
)
= 0 = −3
R2
⊕ − R2
♀
2R♀
x2
+ 2x +
1
2R♀
.
This is a quadratic whose solutions are
x± =
2R♀ ±
√
3R2
⊕ + R2
♀
3(R2
⊕ − R♀)2
.
The negative solution is smaller than 0, and
hence doesn’t interest us. Now, the positive
solutionmustcorrespondtoamaximumdue
to the function tending to negative infinity
at big values. After careful rearranging, the
positive solution simplifies to
1/x+ = −2R♀ +
√
3R2
⊕ + R2
♀ = 0.436AU,
which safely falls in the range of [R⊕ −
R♀,R⊕ + R♀] and hence is the distance of
maximal intensity, L0 = 1/x+ = 0.436AU.
We find the maximal angular distance
between the Sun and Venus using cosine law
as
α0 = arccos
(
R2
⊕ + L2
0 − R2
♀
2L0R⊕
)
= 39.6°.
Grading:
• Noting maxima occurs at zero of derivative
(0.5pts)
• Correct final expression for L (0.2pts)
• Correct numerical value for L (0.1pts)
• Numerical answer within range
[0.28au,1.72au] (0.1pts)
• Correct numerical value for the angular
distance 39.6° (0.1pts)

BRıGHTNEss OF PLANETs


[User]
Maxwell’s wheel is a toy which demonstrates conservation of energy. It consists of a
    uniform disc of mass M and radius R, with a massless axle of radius r ≪ R.

    If the wheel is released from rest, it falls downward, in such a way so that the strings supporting it
    are always approximately vertical.

      (a) Find the approximate acceleration during descent.

     (b) Suppose each string has length L ≫ R2 /r. When the string completely unravels, the wheel
         turns around and starts climbing back up. During the turn around, what is the maximum
         magnitude of the acceleration of the wheel?

Solution:
Maxwell’s wheel is a toy which demonstrates conservation of energy. It consists of a
    uniform disc of mass M and radius R, with a massless axle of radius r ≪ R.

    If the wheel is released from rest, it falls downward, in such a way so that the strings supporting it
    are always approximately vertical.
     (a) Find the approximate acceleration during descent.
     (b) Suppose each string has length L ≫ R2 /r. When the string completely unravels, the wheel
         turns around and starts climbing back up. During the turn around, what is the maximum
         magnitude of the acceleration of the wheel?
    Answer.      (a) a ≈ −2r2 g/R2
     (b) amax ≈ 4grL/R2


[User]
In A’s frame, B has speed u, and C has speed v.

     (a) Suppose B and C have velocities in opposite directions. Find the speed of B with respect to
         C using four-vectors, by computing the inner product vB · vC in two different frames.

     (b) The answer of part (a) should look familiar, but with four-vectors we can easily go further.
         Generalize part (a) to the case where B and C have velocities an angle θ apart.

Solution:
(a) In A’s frame, the four-velocities are

                                        vB = (γu , γu u),     vC = (γv , −γv v).

         Let w be the desired answer. Then in C’s frame,

                                          vB = (γw , γw w),       vC = (1, 0).

         The inner product of vB and vC should be independent of frame, so

                                                 γu γv (1 + uv) = γw

         or equivalently
                                              1 + uv           1
                                         √        √       =√        .
                                           1 − u2 1 − v 2    1 − w2
         Solving for w gives the expected result,
                                                             u+v
                                                       w=           .
                                                             1 + uv

     (b) Taking vB to be along the x-axis for concreteness,

                                vB = (γu , γu u, 0),   vC = (γv , γv v cos θ, γv v sin θ).

         By the same logic as in part (a), we have

                                              γu γv (1 − uv cos θ) = γw

         and solving for w gives the complicated result
                                        p
                                          u2 + v 2 − 2uv cos θ − u2 v 2 sin2 θ
                                   w=                                          .
                                                    1 − uv cos θ
         This reduces to the usual velocity addition formula for θ = 0 and θ = π. If we didn’t use
         the tool of four-vectors and just applied the Lorentz transformations directly, this could have
         been quite a mess, but instead it wasn’t much harder than part (a)!


[User]
1. A parallel plate capacitor with plate area A and separation d (d ≪ lateral dimensions)
   is filled with a dielectric of dielectric constant k = eαx where α is a constant (α > 0)
   and plate is at x = 0. The capacitor is charged to a potential V .                  [Marks: 5]
   (a) Obtain the capacitance C.                                                              [2]

                                           E
         C=

   (b) Sketch C vs d.
                      CS                                                                      [1]

HB

INPhO - 2013                                     Roll Number:P 1 3 0 0 0
   (c) Obtain the charge on the plates.                                    [1]

        Charge =
                                            E
                        CS
  (d) Obtain the electric field ~
                                E(x).                                      [1]

  HB    ~
        E(x) =

                                        Page 2

INPhO - 2013                                         Roll Number:P 1 3 0 0 0

Solution:
1. (a) C =
     (b)
               C

           ε0 A α
                        1 − e−αd

                                                                           E
                                        CS
                                                 d

                     ǫ0 AαV
     (c) Charge =
                    1 − e−αd
         ~        αV e−αx
     (d) E(x) =            x̂
                  1 − e−αd

                         nh

INPhO 2013 Problem 1


[User]
Question B2
For this problem, assume the existence of a hypothetical particle known as a magnetic monopole.
Such a particle would have a “magnetic charge” qm , and in analogy to an electrically charged
particle would produce a radially directed magnetic field of magnitude
                                                     µ0 qm
                                                B=
                                                     4π r2
and be subject to a force (in the absence of electric fields)

                                                F = qm B

   A magnetic monopole of mass m and magnetic charge qm is constrained to move on a vertical,
nonmagnetic, insulating, frictionless U-shaped track. At the bottom of the track is a wire loop
whose radius b is much smaller than the width of the “U” of the track. The section of track near
the loop can thus be approximated as a long straight line. The wire that makes up the loop has
radius a  b and resistivity ρ. The monopole is released from rest a height H above the bottom of
the track.
   Ignore the self-inductance of the loop, and assume that the monopole passes through the loop
many times before coming to a rest.

  a. Suppose the monopole is a distance x from the center of the loop. What is the magnetic flux
     φB through the loop?

  b. Suppose in addition that the monopole is traveling at a velocity v. What is the emf E in the
     loop?

   c. Find the change in speed ∆v of the monopole on one trip through the loop.

  d. How many times does the monopole pass through the loop before coming to a rest?

   e. Alternate Approach: You may, instead, opt to find the above answers to within a dimen-
      sionless multiplicative constant (like 23 or π 2 ). If you only do this approach, you will be able
      to earn up to 60% of the possible score for each part of this question.

   You might want to make use of the integral
                                   Z ∞
                                            1          3π
                                              2 3
                                                  du =
                                    −∞ (1 + u )         8

or the integral                           Z π
                                                              3π
                                                sin4 θ dθ =
                                           0                   8

                   Copyright c 2012 American Association of Physics Teachers

Solution:
Question B2
For this problem, assume the existence of a hypothetical particle known as a magnetic monopole.
Such a particle would have a “magnetic charge” qm , and in analogy to an electrically charged
particle would produce a radially directed magnetic field of magnitude
                                                     µ 0 qm
                                                B=
                                                     4π r2
and be subject to a force (in the absence of electric fields)

                                                F = qm B

   A magnetic monopole of mass m and magnetic charge qm is constrained to move on a vertical,
nonmagnetic, insulating, frictionless U-shaped track. At the bottom of the track is a wire loop
whose radius b is much smaller than the width of the “U” of the track. The section of track near
the loop can thus be approximated as a long straight line. The wire that makes up the loop has
radius a  b and resistivity ρ. The monopole is released from rest a height H above the bottom of
the track.
   Ignore the self-inductance of the loop, and assume that the monopole passes through the loop
many times before coming to a rest.

  a. Suppose the monopole is a distance x from the center of the loop. What is the magnetic flux
     ΦB through the loop?

  b. Suppose in addition that the monopole is traveling at a velocity v. What is the emf E in the
     loop?

   c. Find the change in speed ∆v of the monopole on one trip through the loop.

  d. How many times does the monopole pass through the loop before coming to a rest?

   e. Alternate Approach: You may, instead, opt to find the above answers to within a dimen-
      sionless multiplicative constant (like 32 or π 2 ). If you only do this approach, you will be able
      to earn up to 60% of the possible score for each part of this question.

   You might want to make use of the integral
                                   Z ∞
                                            1          3π
                                              2 3
                                                  du =
                                    −∞ (1 + u )         8

or the integral                           Z π
                                                              3π
                                                sin4 θ dθ =
                                           0                   8

                                               Solution

                   Copyright c 2012 American Association of Physics Teachers

2012 Semifinal Exam                           Part B                                              15

  a. The most direct way is to calculate the flux through a flat, circular surface bounded by the
     loop. However, by Gauss’s law the flux will remain the same if we deform this surface, as
     long as we keep its boundary the same and don’t hit the   √ monopole. The easiest way is to
     deform the surface so it is part of a sphere of radius r = b2 + x2 centered at the monopole.
     Work in spherical coordinates, where θ is the angle to the x-axis and φ is the azimuthal angle.
     The loop itself is at θ0 = tan−1 (b/x), so
                                              Z θ0
                               µ0 qm 2π
                                     Z
                                                                 µ 0 qm
                        ΦB =       2
                                          dφ       r2 sin θ dθ =        (1 − cos(θ0 )).
                               4π r 0          0                    2
     We can easily check this by limiting cases: for small x we have µ0 qm /2, i.e. half of the total
     flux, while for large x we have zero flux.
     This form is acceptable, but we can also write it explicitly in terms of x. By drawing a right
     triangle, we have
                                                   b           x
                                        sin θ0 = , cos θ0 =
                                                   r           r
     which gives
                                            µ 0 qm          x
                                     ΦB =            1− √           .
                                               2          b2 + x2
     We will use the variable θ0 in the parts below, dropping the subscript.
  b. Since E = −dΦB /dt, differentiating both sides of the above result gives
                                     µ 0 qm v   b2         µ 0 qm v
                                E=                       =          sin3 θ
                                         2 (b2 + x2 )3/2      2b
     where we used the chain rule, with v = dx/dt. Either form is acceptable.
  c. Consider the force of the magnetic field produced by the current I in the loop on the monopole.
     We use the Biot-Savart law,
                                                        ds × r
                                                      Z
                                                 µ0 I
                                            B=
                                                 4π       r3
     where r √
             is a vector from the monopole to a point on the rim, ds integrates along the loop,
     and r = b2 + x2 as before. Since ds × r = r sin θ ds, the integral gives
                                          µ0 I 2πb sin θ   µ0 I
                                     B=             2
                                                         =      sin3 θ.
                                          4π      r         2b
     The force on the monopole is F = qm B, so the acceleration is
                                                     qm B
                                                a=        .
                                                      m
     The current I is related to E by Ohm’s law, E = IR, where we’ll compute R later. Then
                                                µ20 qm
                                                     2 v
                                           a=            sin6 θ.
                                                4b2 mR
     The change in speed in one trip is
                                   1 ∞         1 π dx          b π a
                         Z           Z          Z               Z
                   ∆v = a dt ≈          a dx =     a    dθ = −            dθ
                                   v −∞        v 0   dθ        v 0 sin2 θ

                  Copyright c 2012 American Association of Physics Teachers

2012 Semifinal Exam                               Part B                                        16

     where we approximated v as constant throughout one trip and the bottom of the track as an
     infinite straight line. Plugging in our result for a and using the second provided integral,
                                               2 Z π
                                          µ20 qm                   3π µ20 qm
                                                                           2
                                 ∆v = −              sin4 θ dθ = −           .
                                         4bmR 0                    32 bmR

     Finally, we need to find R. Since a  b, the loop is approximately a long cylindrical wire, so
                                                            2πb
                                                   R=ρ
                                                            πa2
     which gives the final result
                                                         3π µ20 qm
                                                                 2 a2
                                                ∆v = −                .
                                                         64 b2 mρ
     We haven’t kept track of the signs very carefully; all that matters is that ∆v is negative, as
     it must be by Lenz’s law.
     The problem can also be solved by energy conservation, finding the energy dissipated in the
     loop every trip by integrating P = E 2 /R over time, then relating that to the change in the
     monopole’s kinetic energy, ∆K ≈ mv∆v.
     If we had kept the integral above over dx, we would instead have to use the first provided
     integral. Incidentally, these integrals aren’t too hard to derive. The first is related to the
     second by u-substitution; for the second, note that
                                                                   4
                                                          e − e−iθ
                                    Z π              Z π  iθ
                                            4
                                          sin θ dθ =                  dθ.
                                     0                0       2i

     All of the terms in the expansion integrate to zero, by periodicity, except for the constant
     term. Then the integral is             Z π
                                         1 4              3π
                                                    dθ =
                                         16 2 0            8
     as stated.
                            √
  d. The initial speed is    2gH by energy conservation, so
                                         √           √         √
                                           2gH    64 2 b2 mρ gH
                                    N=          =                   .
                                          ∆v        3π    µ20 qm
                                                               2 a2

                  Copyright c 2012 American Association of Physics Teachers

USAPhO 2012 Question B2