Problem 1: 

A mechanical system known as an Atwood machine consists of three weights of mass m₁, m₂, and m₃, respectively, connected by a light (massless) inextensible cords of length l - πR and l' - πR, respectively, which pass over identical pulleys with radius R, mass M, and moment of inertia I.  Find the acceleration of m₁.  Let m₁ = m, m₂ = m, m₃ = 0.25 m, M = 0.5 m and I = 0.25 mR².

Solution:

• Concepts:
  Lagrange's equations
• Reasoning:
  All forces except the forces of constraint are derivable from a potential.  The Lagrangian formalism is well suited for such a system.
• Details of the calculation:
  T = ½[m₁(dx/dt)² + M(dx/dt)² + (I/R²)(dx/dt)² + m₂(d(x - x')/dt)² + m₃(d(x + x')/dt)² + (I/R²)(dx'/dt)²]
  = ½[(m₁ + m₂ + m₃ + M + I/R²)(dx/dt)² + (I/R² + m₃ + m₂)(dx'/dt)² + 2(m₃ - m₂)(dx/dt)(dx'/dt).
  U = [-m₁x - M(l - x) - m₂(l - x + x') - m₃ (l - x + l' - x')]g = [(-m₁ + M + m₂ + m₃)x + (m₃ - m₂)x']g + constant.
  L = T - U.
  ∂L/∂v_x = (m₁ + m₂ + m₃ + M + I/R²)(dx/dt) + (m₃ - m₂)(dx'/dt).
  ∂L/∂v_x' = (I/R² + m₃ + m₂)(dx'/dt) + (m₃ - m₂)(dx/dt).
  ∂L/∂x = -(-m₁ + M + m₂ + m₃)g,  dL/dx' = -(m₃ - m₂)g.
  (m₁ + m₂ + m₃ + M + I/R²)d²x/dt² + (m₃ - m₂)d²x'/dt² + (-m₁ + M + m₂ + m₃)g = 0.
  (I/R² + m₃ + m₂)d²x'/dt² + (m₃ - m₂)d²x/dt² + (m₃ - m₂)g = 0.
  We have two equations for two unknowns, d²x/dt², and d²x'/dt².
  Plug in numbers:
  3 d²x/dt² - 0.75 d²x'/dt² + 0.75 g = 0.
  1.5 d²x'/dt² - 0.75 d²x/dt² - 0.75 g = 0.
  d²x'/dt² = 0.5 d²x/dt² + 0.5 g.
  (3 - 0.375)d²x/dt² + (0.75 - 0.375) g = 0.
  d²x/dt² = -0.143 g.
  m₁ accelerates upward with a = 0.143 g.

Problem 2:

A bead, of mass m, slides without friction on a wire that is in the shape of a cycloid with equations
x = a(2θ + sin2θ),
y = a(1 - cos2θ),
- π/2 ≤ θ ≤ π/2.

A uniform gravitational field g points in the negative y-direction.
(a)  Find the Lagrangian and the second order differential equation of motion for the coordinate θ.
(b)  The bead moves on a trajectory s with elements of arc length ds.
Integrate ds = (dx² + dy²)^½ = ((dx/dθ)² + (dy/dθ)²)^½dθ with the condition s = 0 at θ = 0 to find s as a function of θ.
(c)  Rewrite the equation of motion, switching from the coordinate θ to the coordinate s and solve it.  Describe the motion.

Solution:

• Concepts:
  Lagrangian Mechanics
• Reasoning:
  We are asked to obtain Lagrange's equation of motion, using two different generalized coordinates.
• Details of the calculation:
  (a)  L = T - U.   T = ½m(v_x² + v_y²),  v_x = dx/dt = 2a(dθ/dt)(1 + cos2θ), 
  v_y = dy/dt = 2a(dθ/dt)sin2θ.
  T = 2ma²(dθ/dt)²[(1 + cos2θ)² + sin²2θ] = 4ma²(dθ/dt)²(1 + cos2θ)
  = 8ma²(dθ/dt)²cos²θ.
  U = mgy = mga(1 - cos2θ) = 2mga sin²θ.
  L = 8ma²(dθ/dt)²cos²θ - 2mga sin²θ.
  
  Lagrange's equations:  d/dt(∂L/∂(dq_i/dt)) -  ∂L/∂q_i = 0.
  We have only one generalized coordinate.
  ∂L/∂(dθ/dt) = 16ma²(dθ/dt)cos²θ.
  d/dt(∂L/∂(dθ/dt)) = 16ma²d²θ/dt²cos²θ - 32ma²(dθ/dt)²cosθ sinθ.
  ∂L/∂θ = -16ma²(dθ/dt)²cosθ sinθ - 4mga cosθ sinθ.
  Equation of motion: 
  16ma²(d²θ/dt²)cos²θ - 16ma²(dθ/dt)²cosθ sinθ + 4mga cosθ sinθ = 0.
  d²θ/dt²cosθ - (dθ/dt)²sinθ + (g/(4a)) sinθ = 0.

  (b)  ds = 4a cosθ dθ.  s(θ) = ∫₀^θds = ∫₀^θ4a cosθ'dθ' = 4a sinθ.

  (c)  ds/dt = 4a cosθ (dθ/dt).  d²s/dt² = 4a cosθ d²θ/dt² - 4a sinθ (dθ/dt)².
  Therefore d²θ/dt² cosθ - (dθ/dt)² sinθ = (4a)^-1d²s/dt²,
  and  (g/(4a))sinθ = (g/(4a)²)s.

  The equation of motion in terms of the coordinate s is d²s/dt² = -(g/(4a))s. 
  This is Hooke's law with the solution s(t) = s₀ cos(ωt + φ).

  ω = (g/(4a))^½, s₀ and φ depend on the initial conditions.
  The bead executes simple harmonic motion in s.

Problem 3:

A particle of mass m is described by the Lagrangian function

L = ½m[(dx/dt)² + dy/dt)² + (dz/dt)²] + ½ωl₃,

where l₃ is the z-component of the angular momentum and ω is a constant angular frequency.
(a)  Find the equations of motion, write them in terms of the variables (x + iy) and z and solve them.
(b)  Construct the Hamiltonian function and find the kinematic and canonical momenta. 
Show that the particle has only kinetic energy and that the latter is conserved.

Solution:

• Concepts:
  Lagrange's Equations: d/dt(∂L/∂(dq_i/dt)) -  ∂L/∂q_i = 0;  Canonical momenta: ∂L/∂(dq_i/dt) = p_i;
  The Hamiltonian function: H(q, p, t) = ∑_i(dq_i/dt)p_i - L 
• Reasoning:
  The Lagrangian of the system is given, we are asked to solve the equations of motion and to construct the Hamiltonian function.
  L = ½m[(dx/dt)² + dy/dt)² + (dz/dt)²] + ½ωl₃,  l₃ = m(r × v)_z = m(x(dy/dt) - y(dx/dt)).
  L = ½m[(dx/dt)² + dy/dt)² + (dz/dt)²] + ½mω(x(dy/dt) - y(dx/dt)).
  The generalized coordinates q_i are the Cartesian coordinates.  The coordinate z is cyclic.
• Details of the calculation:
  (a) ∂L/∂(dz/dt) = ∂L/∂v_z = mv_z = p_z = constant, since z is cyclic.
  From Lagrange's equations we have
  ∂L/∂(dx/dt) = mdx/dt - ½mωy,  d/dt(∂L/∂(dx/dt)) = md²x/dt² - ½mωdy/dt,
  ∂L/∂x = ½mω(dy/dt),
  d²x/dt² - ωdy/dt = 0.
  ∂L/∂(dy/dt) = mdy/dt + ½mωx,  d/dt(∂L/∂(dy/dt)) = md²y/dt² + ½mωdy/dt,
  ∂L/∂y = -½mω(dx/dt),
  d²y/dt² + ωdx/dt = 0.
  Combining these two equations we can write
  d²x/dt² +  id²y/dt² + ω(idx/dt - dy/dt) = 0.
  d²x/dt² +  id²y/dt² + Iω(dx/dt + idy/dt) = 0.
  Let c = x + iy.  Then d²c/dt² + iωdc/dt = 0.
  c = Bexp(-iωt) + C.

  B and C are complex constants determined by the initial conditions.
  Let B = Aexp(-iΦ), A = real, C = 0, then c = Aexp(-i(ωt + Φ)).
  Solve for x and y as function of time by isolating the real and the imaginary parts.
  x = Acos(ωt + Φ), y = -Asin(ωt + Φ). 
  For z we have z = z(0) + p_zt/m.
  
  (b)  The canonical momenta are
  p_z = ∂L/∂(dz/dt) = mdz/dt,  p_x = ∂L/∂(dx/dt) = mdx/dt - ½mωy,
  p_y = ∂L/∂(dy/dt) = mdy/dt + ½mωx.
  The kinematic momenta are
  mdz/dt = p_z,  mdx/dt = p_x + ½mωy,  mdy/dt = p_y - ½mωx.

  Note:  The kinematic momenta are not equal to the canonical momenta.
  The Hamiltonian is a function of the generalized coordinates and momenta.
  H = p_zdz/dt +  p_xdx/dt +  p_ydy/dt - L
  = ½[p_z²/m + (p_x + ½mωy)²/m + ( p_y - ½mωx)²/m] = H(p,q) = E.
  H does not explicitly depend on time.
  ∂H/∂t = 0, E is conserved.
  In terms of the kinematic momenta we have:
  E = [(p_{x_mech})² + (p_{y_mech})² + (p_{z_mech})²]/2m = T.
  
  Think about:  Which physical situation is described by this Lagrangian?

Problem 4:

For a symmetrical prism (one in which the apex angle lies at the top of an isosceles triangle), the total deviation angle ϕ of a light ray is minimized when the ray inside the prism travels parallel to the prism's base.  
Assume that a beam of light passes through a glass equilateral prism with refractive index 1.5.  The prism is in air and is mounted on a rotation stage, as shown in the figure.  When the prism is rotated, the angle by which the beam is deviated changes.  What is the minimum angle ϕ by which the beam is deflected?

Solution:

• Concepts:
  Snell's law, geometry
• Reasoning:
  sinθ = n sinθ_glass.
  At the angle of minimum deviation, the refraction at each interface is the same, so the beam inside the prism is parallel to the base of the prism.  At both interfaces, the internal angle between the beam and the normal is thus 30^o.
• Details of the calculation
  By Snell's Law, the external angle between the beam and the normal is thus θ,
  where 1.0*sinθ =1.5*sin30^o.
  θ = sin^-1(1.5*sin30^o) = sin^-1(0.75) = 48.6^o.
  From geometry:
  
  The angle of minimum deviation is ϕ = 2θ -2*30^o.
  ϕ = 2 sin^-1(0.75) - 60^o = 37.2^o.

Problem 5:

A point particle with mass m is restricted to move on the inside surface of a horizontal ring.  The radius of the ring increases steadily with time as R = R₀ + R't.
At t = 0 the speed of the particle is v₀.
(a)  Are there any constants of motion?
(b)  Let E(t₁) be the energy of the particle at the time the radius of the ring is 2R₀ and E₀ its energy at t = 0. 
Find the ration E(t₁)/E₀.

Solution:

• Concepts:
  Lagrange's equations
• Reasoning:
  There are no net forces except the forces of constraint.  To find the constants of motion we find the Lagrangian and look for cyclic coordinates. 
  If the generalized coordinates and momenta we choose do not depend on time and the Lagrangian depends explicitly on time, then the energy is not conserved.
• Details of the calculation:
  (a)  Let x and y be the Cartesian coordinates in the plane of the ring and r and φ be the polar coordinates. 
  r = R₀ + R't is the equation of constraint.
  The constraint is holonomic and time dependent.
  x  = (R₀ + R't)cosφ, y = (R₀ + R't)sinφ. 
  T = ½m(v_x² + v_y²) = ½m((dφ/dt)²(R₀ + R't)² + R'²).
  L = T,  we have only one generalized coordinate, φ, which does not explicitly depend on time.
  ∂L/∂φ = 0,  φ is cyclic.
  ∂L/∂(dφ/dt) = m(dφ/dt)(R₀ + R't)² = constant, since  (d/dt)(∂L/∂(dφ/dt)) = 0.
  The angular momentum M = m(dφ/dt)(R₀ + R't)² about the vertical axis is a constant of motion.
  The energy is NOT a constant of motion.
  (b)  E = M²/(2m(R₀ + R't)²) + ½mR'².
  E(t₁)/E₀ = (M²/(8mR₀²) + ½mR'²)/(M²/(2mR₀²) + ½mR'²)
  = (M² + 4m²R'²R₀²)/ (4M² + 4m²R'²R₀²).
  If R'² << M²/(2m²R₀²), then E(t₁)/E₀ = ¼.
  The energy of the particle decreases, the particle does positive work.