More Problems

Problem 1:

Small spheres of radius "r" are incident with velocity v on a stationary hard sphere of radius "R" and scatter elastically.  What is the total scattering cross section?

Solution:

• Concepts:
  Definition of the scattering cross section, geometry
• Reasoning:
  The concept of cross section, as its name suggests, is that of effective area for collision.  The cross section of a spherical target of radius R for point particles is σ = πR².  If the scattering particles have radius r, then the cross section is σ = π(r+R)². 
  

Problem 2:

Consider a uniform disc with mass m and radius a that has a massless string wrapped around it with one end attached to a fixed support and allowed to fall with the string unwinding as it falls, as shown in the figure. 

Use y and φ as the generalized coordinates to describe the system.  Find the equations of motion for y and of the falling disc and the forces of constraint using the method of Lagrange multipliers.

Solution:

• Concepts:
  Lagrange's Equations, Lagrange multipliers
  d/dt(∂L/∂(dq_k/dt)) -  ∂L/∂q_k = ∑_lλ_la_lk,    Σ_k a_lk dq_k + a_lt dt =  0.
• Reasoning:
  The problem requires us to use the method of Lagrange multipliers. 
  The coordinate y denotes the distance the CM of the disk has fallen, the coordinate φ denotes the angle through which the disk has rotated.
  Then L = m(dy/dt)²/2 + I(dφ/dt)²/2 + mgy.
  For the disk I = Ma²/2.
  The equation of constraint is y = aφ, dy = adφ, dy - adφ = 0.
  There is only one equation of constraint, so we can drop the index k.
  a_y = 1,  a_φ = -a,
• Details of the calculations:
  The equations of motion are
  md²y/dt² - mg = λ,
  Id²φ/dt²  = -aλ.
  dy - adφ = 0.
  Solving these 3 equations we obtain
  (m + I/a²)d²y/dt² = mg.
  d²y/dt² = mg/(m + 0.5m) = 2g/3.
  Q_y = λ = -mg/3 is the force of constraint acting on m in the y-direction, it is the tension in the string pulling upwards on m.  The minus sign indicates that the force of constraint is acting upwards.
  Q_φ = -aλ = mga/3 is the torque acting on the disk.  It is the generalized force of constraint associated with the coordinate φ.  The torque is pointing into the page.

Problem 3:

A uniform hoop of mass m and radius r rolls without slipping on a fixed cylinder of radius R as shown in the figure.  The only external force is that of gravity.  If the hoop starts rolling from rest on top of the big cylinder, find, by the method of Lagrange multipliers, the point at which the hoop falls off the cylinder.

Solution:

• Concepts:
  Lagrange multipliers
• Reasoning:
  The Lagrange multiplier technique yields the radial force of constraint.  The contact between the hoop and the cylinder will be lost when the radial force of constraint changes sign.  The cylinder cannot pull.
• Details of the calculation:
  To find the radial force of constraint, we let R, θ₁ and θ₂ be the generalized coordinates.
  They are connected by constraints of the form Σ_k a_lk dq_k + a_lt dt = 0.
  We have two equations of constraint.
  R = R₁, dR = 0, therefore we can choose a_1R = 1, a_1θ1 = a_1θ2 =  0.
  (We use R₁ for the radius of the fixed sphere, to distinguish from the generalized coordinate R.)
  dθ₂ = dθ₁(R + r)/r, therefore we can choose a_2R = 0, a_2θ1 = 1, a_2θ2 = -r/(R + r).
  (Note:  θ₂ ≠ θ₁(R + r)/r until R is fixed.  If we let R be a coordinate then we only have a constraint equation connecting the differentials of the coordinates.)
  
  The three equations of motion are
  d/dt(∂L/∂(dq_k/dt)) -  ∂L/∂q_k = ∑_lλ_la_lk,
  which have to be solve together with the two equations of constraint.
  We have five equations and five unknowns, R(t), θ₁(t), θ₂(t), λ₁ and λ₂. 
  
  T = ½m(r + R)²(dθ₁/dt)² + ½mr²(dθ₂/dt)² + ½m(dR/dt)².  U = mg(r + R)cosθ₁.
  L = T - U.
  ∂L/∂(dR/dt)  = mdR/dt,  d/dt(∂L/∂(dR/dt)) = md²R/dt², 
  ∂L/∂R = m(r + R)(dθ₁/dt)² - mgcosθ₁.
  d/dt(∂L/∂(dR/dt)) - ∂L/∂R = λ₁a_1R + λ₂a_2R
  (1)  md²R/dt² - m(r + R)(dθ₁/dt)² + mgcosθ₁ = λ₁,
  
  ∂L/∂(dθ₁/dt) = m(r + R)²(dθ₁/dt), 
  d/dt(∂L/∂(dθ₁/dt)) = m(r + R)²d²θ₁/dt² +  2m(r + R)(dθ₁/dt)(dR/dt),
  ∂L/∂θ₁ = mg(r + R)sinθ₁.
  d/dt(∂L/∂(dθ₁/dt)) - ∂L/∂θ₁ = λ₁a_1θ1 + λ₂a_2θ1, 
  (2)  m(r + R)²d²θ₁/dt² + 2m(r + R)(dθ₁/dt)(dR/dt) - mg(r + R)sinθ₁ = λ₂. 
  
  ∂L/∂(dθ₂/dt) = mr²(dθ₂/dt), 
  d/dt(∂L/∂(dθ₂/dt)) = mr²d²θ₂/dt²,
  ∂L/∂θ₂ = 0.
  d/dt(∂L/∂(dθ₂/dt)) - ∂L/∂θ₂ = λ₁a_1θ2 + λ₂a_2θ2, 
  (3)  mr²d²θ₂/dt² = -λ₂r/(r + R). 
  
  Constraints:
  dR/dt = 0 --> d²R/dt² = 0, r = R₁. 
  dθ₂ = dθ₁(R + r)/r --> d²θ₂/dt² = ((R₁ + r)/r)d²θ₁/dt².
  Therefore
  from (1):  -m(r + R₁)(dθ₁/dt)² + mgcosθ₁ = λ₁.
  from (2):  m(r + R₁)²d²θ₁/dt² - mg(r + R₁)sinθ₁ = λ₂. 
  from (3):  md²θ₁/dt² = -λ₂/(r + R₁)².
  
  Combining the last two equations we have
  2m(r + R₁)²d²θ₁/dt² - mg(r + R₁)sinθ₁ = 0.
                            
  Contact is lost when λ₁ = 0.  (The normal force cannot pull.)
  Then  dθ₁/dt = [gcosθ₁/(r + R₁)]^½.
  
  To find  dθ₁/dt as a function of θ₁ we need to solve the differential equation
  2m(r + R₁)²d²θ₁/dt² - mg(r + R₁)sinθ₁ = 0,
  or
  d²θ₁/dt² = gsinθ₁/(2(r + R₁))
  for dθ₁/dt.
  Denote dθ₁/dt = f(θ₁).
  df(θ₁)/dt = [df(θ₁)/dθ₁]f(θ₁) = ½df²(θ₁)/dθ₁ = gsinθ₁/(2(r + R₁)) .
  f²(θ₁) = A - (g/(r + R₁))cosθ₁.
  dθ₁/dt =  [A - (g/(r + R₁))cosθ₁]^½.
  
  Initial conditions:  dθ₁/dt = 0 for θ₁ = 0, A = g/(r + R₁).
  dθ₁/dt =  [g/(r + R₁) - g cosθ₁/(r + R₁)]^½.
  The hoop falls off when gcosθ₁/(r + R₁)  = g/(r + R₁) - gcosθ₁/(r + R₁),
  cosθ₁  = (1 - cosθ₁),
  or cosθ₁ = ½, θ₁ = 60^o.
  
  Comments:  (not part of the asked for solution)
  
  Finding λ₁:
   -m(r + R₁)(dθ₁/dt)² + mgcosθ₁ = λ₁.
   -m(r + R₁)[(g/(r + R₁) - gcosθ₁/(r + R₁)] + mgcosθ₁ = λ₁.
   mg(2cosθ₁ - 1) = λ₁.
  λ₁a_1r = F_r, F_r =  mg(2cosθ₁ - 1) is the reaction (normal) force of the cylinder on the hoop.
  
  Finding λ₂:
  md²θ₁/dt² = -λ₂/(r + R₁)².
  m[g/(2(r + R₁))sinθ₁] = -λ₂/(r + R₁).²
  mg sinθ/2 = -λ₂/(r + R₁).
  What is the significance of λ₂?
  mr²d²θ₂/dt² = -λ₂r/(r + R₁)².
  -λ₂r/(r + R₁)² = mgr sinθ₁/2 is the generalized force associated with the coordinate θ₂.  Since θ₂ is an angle, the generalized force is a torque. Rolling implies static friction.  The force of static friction exerts a torque about the center of the hoop.
  mgr sinθ₁/2 = τ_θ2.
  Note:  The λs are not unique, but the formalism yields unique generalized forces of constraint.