Review

Problem 1:

A rugby player runs with the ball directly towards his opponent's goal, along the positive direction of an x axis.  He can legally pass the ball to a teammate as long as the ball's velocity relative to the field does not have a positive x component.  Suppose the player runs at speed 4.0 m/s relative to the field while he passes the ball with a speed of 6.0 m/s relative to himself.  What is the smallest angle (relative to the x axis) the ball can be passed in (as seen from the player) in order for the pass to be legal?

Solution:

• Concepts:
  Kinematics, frame transformations
• Reasoning:
  
  Let the velocity of the player be v₁ and let the relative velocity between the player and the ball be v₂. 
  Then the velocity of the ball relative to the field is v = v₁ + v₂ = i(v₁ + v₂ cosθ) + j v₂sinθ.
• Details of the calculation:
  For the smallest angle θ_min, v has no x-component, v₁ + v₂ cosθ_min = 0.
  cosθ_min = -v₁/v₂ = -4/6.  θ_min = 131.8^o.

Problem 2:

A tankless water heater heats water as it flows through the device.  The water circulates through a copper heat exchanger.  For a particular water heater the flow rate is 8.7 liter per minute.  The temperature rise of the water is 25 ^oC.   The water heater uses gas to heat the water in the copper coils and is 80% efficient.  What is the gas energy input rate in units of kW?
conversion:  1 kcal = 4186 J

Solution:

• Concepts:
  Energy conservation, specific heat c
• Reasoning:
  Water flow rate (kg/s) * c(J/(kg^oC) * temperature change (^oC) = power input (W) * efficiency.
• Details of the calculation:
  Water flow rate:  8.7 kg/minute = (8.7/60) kg/s.
  Specific heat of water: c = 4186 J/(kg ^oC).
  P = ((8.7/60) kg/s)*(4186 J/(kg^oC))*(25 ^oC)/0.8 = 18967 W ≈ 19 kW.

Problem 3:

A mass m is attached to a horizontal spring.  Initially the mass is at rest.  At t = 0, a force F = F₀cosωt starts acting on the mass.  For t > 0, find the displacement x of the mass from its equilibrium position, assuming there is no friction and damping.  Assume that ω² ≠ k/m.

Solution:

• Concepts:
  Newton's second law, forced oscillations
• Reasoning:
  Equation of motion md²x/dt² = -kx + F₀cosωt.
  We find a solution to the inhomogeneous equation d²x/dt² + (k/m)x = (F₀/m)cosωt
  and add the solution of the homogeneous equation d²x/dt² + (k/m)x = 0.
• Details of the calculation:
  d²x/dt² + ω₀x = (F₀/m)cosωt,  with ω₀² = k/m. 
  Try a solution x = Bcos(ωt), a particular solution to the inhomogeneous equation. 
  This yields B = (F₀/m)/(ω₀² - ω²). 
  To find the most general solution we add the homogeneous solution x = Ccos(ω₀t + δ).
  Most general solution:  x(t) = Ccos(ω₀t + δ) + [(F₀/m)/(ω₀² - ω²)]cos(ωt).
  Initial conditions:  x(0) = dx/dt|_t=0 = 0.
  Ccos(δ) + [(F₀/m)/(ω₀² - ω²)] = 0,  -ω₀Csin(δ) = 0, --> δ = 0, C = -[F₀/m)/(ω₀² - ω²).
  x(t) = -[(F₀/m)/(ω₀² - ω²)]cos(ω₀t) + [(F₀/m)/(ω₀² - ω²)]cos(ωt).
  The motion is a superposition of two oscillations with different frequencies.  If |ω₀ - ω| << ω, we observe "beats".

Problem 4:

The volume between two concentric spherical surfaces of radii a and b (a < b) is filled with an inhomogeneous dielectric with permittivity
ε = ε₀/(1 + κr),
where ε₀ and κ are constants and r is the radial coordinate.
Thus D(r) = εE(r).  A charge Q is placed on the inner surface, while the outer surface is grounded.  Find:
(a)  The displacement D(r) and the field E(r) in the region a < r < b.
(b)  The capacitance of the device.
(c)  The volume polarization charge density in the region a < r <  b.
(d)  The surface polarization charge density at r = a and at r = b.

Solution:

• Concepts:
  Gauss law and symmetry
• Reasoning:
  The problem has spherical symmetry. We can find the field between the surfaces in terms of the free charge Q on the surface using Gauss' law for D.
• Details of the calculation:
  (a)  Gauss law for D:  D(r) = Q/(4πr²) e_r.   E(r) = Q(1 + κr)/(4πε₀r²) e_r.
  
  (b)  C = Q/ΔV.
  ΔV = V_a = ∫_a^bE∙dr = [Q/(4πε₀)](1/a - 1/b + κ ln(b/a)).
  C = (4πε₀ab)/(b - a + abκ ln(b/a)).
  
  (c)  D = ε₀E + P  
  P = (ε - ε₀)E = (ε - ε₀) Q(1 + κr)/(4πε₀r²) e_r = -Qκ/(4πr) e_r.
  The volume polarization charge density is
  ρ_P = -∇∙P_.   ρ_P = Qκ/(4πr²).
  Total volume polarization charge = Qκ(b - a)
  
  (d)  σ_P = P∙n.
  At r = a, n = e_r.  σ_P = Qκ/(4πa).
  At r = b, n = -e_r.  σ_P = -Qκ/(4πb).
  
  The total polarization charge is zero, as required.