Problem 1:

A stationary space station can be approximated as a hollow spherical shell of mass 6 tons (6000 kg) and inner and outer radii of 5 m and 6 m.  To change its orientation, a uniform fly wheel of radius 10 cm and mass 10 kg located at the center of the station is spun quickly from rest to 1000 rpm.
(a)  How long (in minutes) will it take the station to rotate by 10^o?
(b)  What energy (in Joules) is needed for the whole operation?

(Moment of Inertia of hollow spherical shell: [2M(R_o⁵ - R_i⁵ )]/[5(R_o³ - R_i³)] where R_o is outer radius and R_i is inner radius)

Solution:

• Concepts:
  Conservation of angular momentum
• Reasoning:
  The system consists of the shell and the flywheel.  No external torques act on the system, so the total angular momentum of the system is conserved.  If a component of L is zero before the flywheel starts to rotate, then it is zero afterwards.
• Details of the calculation:
  (a)  The flywheel receives an angular impulse, so the shell receives an angular impulse of equal magnitude and opposite direction.
  L = Iω.
  For the disk I = ½mr² = 0.05 kgm².  L = (0.05 kgm²)1000*2π/60s = 5.24 kgm²/s.
  For the shell I = [2M(R_o⁵ - R_i⁵ )]/[5(R_o³ - R_i³)] = 122664 kgm². 
  ω = dθ/dt = L/I = 4.27*10^-5/s.
  t = (10*π/(180*4.25*10^-5))s = 4089 s ≈ 68 min.
  
  (b)  The energy needed is the final kinetic energy of the flywheel and the shell.
  E = ½I_wω_w² + ½I_sω_s² = 274.2 J.

Problem 2:

Consider the matrix of the operator A,

(a)  Find the matrix sin(A).
(b)  Find the matrix cos(A - (π/2)I), where I is the identity matrix.
(c)  Find the matrix exp(Aⁿ), for arbitrary n = 0, 1, 2, ... .  (Hint: even and odd n are different.)

Solution:

• Concepts:
  Functions of operators, change of basis
• Reasoning:
  Use the eigenbasis of A to evaluate a function of the operator A.
• Details of the calculation:
  (a)  The matrix of the operator A is given in some basis {|t₁>, |t₂>, |t₃>}.
  The eigenvalues of A are λ, where
   

  	-λ	α	0
  	α	-λ	0
  	0	0	β-λ

    = 0.

  
  

  
  λ²(β - λ) - α²(β - λ) = 0.  The solutions to this equation are λ₁ = -α, λ₂ = β, λ₃ = α.
  The corresponding normalized eigenvectors of A in the {|t₁>, |t₂>, |t₃>} basis are

  |1> = 2^-½  

  	-1
  	1
  	0

  ,   |2> =  

  	0
  	0
  	1

  ,   |3> = 2^-½  

  	1
  	1
  	0

  .

   
  

   

  The matrix of A in its eigenbasis, with the eigenvectors ordered as {|1>, |2>, |3>}, is

  A' =  

  	-α	0	0
  	0	β	0
  	0	0	α

  .  

  
  

  The matrix  sin(A') in the eigenbasis of A is

  sin(A') =  

  	-sinα	0	0
  	0	sinβ	0
  	0	0	sinα

  .  

  
  

  Let U be the unitary transformation from the eigenbasis of A to the the {|t₁>, |t₂>, |t₃>} basis.  The matrices of U^† and U are

  U^†  =  

  	-2-½	0	2-½
  	2-½	0	2-½
  	0	1	0

  ,         

  U  =  

  	-2-½	2-½	0
  	0	0	1
  	2-½	2-½	0

  .  

  
  Therefore the matrix sin(A) in the original  {|t₁>, |t₂>, |t₃>} basis is A= U^†A'U,

  sin(A) =  

  	0	sinα	0
  	sinα	0	0
  	0	0	sinβ

  .  

  
  

  
  (b)  The matrix A - (π/2)I has the same eigenvectors as the matrix of A.
  Its eigenvalues are λ₁ = -α - π/2, λ₂ = β - π/2, λ₃ = α - π/2.
  Since cos(x - π/2) = sinx,  we have

  cos(A - (π/2)I) =  

  	0	sinα	0
  	sinα	0	0
  	0	0	sinβ

  .  

  
  

  
  (c)  The matrix exp(A'ⁿ) in the eigenbasis of A is

  exp(A'ⁿ) =  

  	exp(-αn)	0	0
  	0	exp(βn)	0
  	0	0	exp(αn)

  .  

  
  

  If n = even, then exp(-αⁿ) = exp(αⁿ).
  Therefore the matrix  exp(Aⁿ) in the original {|t₁>, |t₂>, |t₃>} basis, exp(Aⁿ) = U^†exp(A^'n)U, is

  exp(Aⁿ) =  

  	exp(αn)	0	0
  	0	exp(αn)	0
  	0	0	exp(βn)

    

  
  

  if n = even, and

  exp(Aⁿ) =  

  	cosh(αn)	sinh(αn)	0
  	sinh(αn)	cosh(αn)	0
  	0	0	exp(βn)

  .  

  
  

  if n is odd.

Problem 3:

A system consists of four point masses are located in the xy plane at the corners of a rectangle as follows:
mass m₁ = m at (0,0); mass m₂ = 2m at (a,0); mass m₃ = 3m at (a,b); and mass m₄ = 4m at (0,b).
(a)  Find the center of mass of the system.
(b)  Find the moments I_ii and products I_ij (i ≠ j) of inertia about the origin for the given choice of coordinate axes.

Solution:

• Concepts:
  The center of mass, the inertia tensor
• Reasoning:
  I_ij = Σ_km_k[Σ_l(x_k)_l²δ_ij - (x_k)_i(x_k)_j] is the inertia tensor.  The I_ii are called the moments and the I_ij (i ≠ j) are called the products of inertia.
• Details of the calculation:
  (a)  Center of mass:
  x_CM = Σm_ix_i/M,  y_CM = Σm_iy_i/M,  z_CM = Σm_iz_i/M.  M = Σm_i = 10 m.
  x_CM = ½a,  y_CM = 7b/10,  z_CM = 0.
  (b)  I_xx = Σm_i(y_i²+ z_i²) = 7mb².  I_yy = Σm_i(x_i²+ z_i²) = 5ma².
  I_zz= Σm_i(x_i²+ y_i²) = 4mb² + 2ma² + 3m(a² + b²) = 5ma² + 7mb².
  I_xy = I_yx = -Σm_ix_iy_i² = -3mab.  I_xz = I_zx = I_yz = I_zy = 0.

Problem 4:

Consider a 2 by 2 matrix M with matrix elements m₁₁, m₁₂, m₂₁, and m₂₂.
(a)  What relationships between these matrix elements must exist for the matrix M to be a Hermitian  matrix, M = H?
(b)  Find the eigenvalues of M = H and show that they are real.

Solution:

• Concepts:
  Mathematical foundations of quantum mechanics
• Reasoning:
  A matrix A is Hermitian if A = A^†.
• Details of the calculation:
  Let M = H be a Hermitian matrix.  Then H = H^†.
  

  H =  

  	m11	m12
  	m21	m22

  ,     H^† =  

  	m11*	m21*
  	m12*	m22*

  .

  
  H = H^† --> m₁₁ = real,  m₂₂ = real, m₁₂  = m₂₁^*.
  To find the eigenvalues λ of H we solve det(H - λI) = 0.
  (m₁₁ - λ) (m₂₂ - λ) - |m₁₂|² = 0.
  λ² - (m₁₁ + m₂₂)λ + m₁₁ m₂₂ - |m₁₂|² = 0.
  λ = (m₁₁ + m₂₂)/2 ± ((m₁₁ - m₂₂)²/4 + |m₁₂|²)^1/2.
  The sum inside the square root is zero or positive, so the square root yields a real number the eigenvalues λ are real.   

Problem 5:

Two identical billiard balls of radius R and mass M, rolling with CM velocities ±vi, collide elastically, head-on.  Assume that after the collision they have both reversed motion and are still rolling.
(a)  Find the impulse which the surface of the table must exert on each ball during its reversal of motion.
(b)  What impulse is exerted by one ball on the other?

Solution:

• Concepts:
  Force and torque
• Reasoning:
  Each ball reverses the direction of its momentum and angular momentum.
• Details of the calculation:
  (a) Take the left ball.  Let k point into the page and j point down.
  Before the collision: p = mvi,  L = Iωk = (Iv/R)k = (2mRv/5)k.
  After the collision: p = -mvi,  L = -(2mRv/5)k.
  (F_ball + F_table)Δt = -2mvi, (Rj × F_table)Δt  = -(4mRv/5) k.
  F_tableΔt = (4mv/5)i is the impulse the surface of the table exerts on each ball.
  The impulse the table exerts on the left ball is (4mv/5)i, and the impulse the table exerts on the right ball is -(4mv/5)i.

  (b)  F_ballΔt = -(2mv + 4mv/5)i = -(14mv/5)i is the impulse the right ball exerts on the left ball.
  The impulse the left ball exerts on the right ball is (14mv/5)i.