Assignment 7

Problem 1:

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 2:

A string oflength 2l is suspended at points A and B locatedon a horizontal line. The distance between A and Bis 2d (d < l). A small, heavy bead can slide on thestring without friction. Find the period of the small-amplitude oscillations of the bead in the vertical plane containing the suspensionpoints. The acceleration due to gravity is g.

Solution:

Concepts:
The Lagrangian formalism, small oscillations
Reasoning:
We are asked to find the period of small oscillations.
Details of the calculation:
For any point on the trajectory of the bead the sum of the distances from point A and point is 2l. The trajectory therefore is a section of an ellipse. The semi-major axis of this ellipse is a = l and the semi-minor axis is b = (l² - d²)^½. If the origin of the coordinate system is placed at the center of the ellipse, as shown, the equation of the ellipse is x²/a² + y²/b² = 1.

The equilibrium position of the bead is at x = 0, y = -b. For small displacements from equilibrium we have y = -b(1 - x²/a²)^½ ≈ -b(1 - x²/(2a²)).
Moving the origin of the coordinate system to the equilibrium point we have y = b(x²/(2a²). To first order the trajectory is parabolic near the equilibrium point.

Since this problem involves holonomic constraints the Lagrangian method is well suited for solving this problem. The equation of constraint is y ≈ b(x²/(2a²)).
Using the Lagrangian method we have
T = ½m((dx/dt)² + (dy/dt)²) ≈ ½m((dx/dt)². U = mgy = mgb(x²/(2a²)).
L = T - U.
[If we want only first order terms in the equation of motions we must keep terms up to second order in the small quantities in the Lagrangian.]
d/dt(∂L/∂(dx/dt)) - ∂L/∂x = 0.
d²x/dt² = -gbx/a².
Solution: x = Acos(ωt + Φ) with ω = (gb/a²)^½. T = 2π/ω= 2πa/(gb)^½.

Problem 3:

Consider a pendulum in a plane (i.e. a "2D world"), consisting of a mass m attached at the end of a weightless rope of length l₀. When the pendulum is set into motion the length of the rope is shortened at a constant rate dl/dt = -α = constant. Compute the Lagrangian, write down the equation of motion, and discuss the conservation of energy for this system. Does the sign of α matter for energy conservation.

Solution:

Concepts:
Lagrangian Mechanics
Reasoning:
The Lagrangian explicitly depends on time.
Details of the calculation:
(a) L = T - U.
As long as (l₀ - αt)> 1 we have
T = ½m(l₀ - αt)²(dθ/dt)² + ½mα², U = -mg(l₀ - αt) cosθ.
L = ½m(l₀ - αt)²(dθ/dt)² + ½mα² + mg(l₀ - αt)cosθ is the Lagrangian of the system.
∂L/∂(dθ/dt) = m(l₀ - αt)²dθ/dt, d/dt(∂L/∂(dθ/dt)) = m(l₀ - αt)²d²θ/dt² - 2mα(l₀ - αt)dθ/dt.
∂L/∂θ = -mg(l₀ - αt)sinθ.
(l₀ - αt)d²θ/dt² - 2αdθ/dt = -gsinθ is the equation of motion.
The Lagrangian explicitly depends on time, so the Hamiltonian explicitly depends on time. We have one generalized coordinate, θ, which does not explicitly depend on time.
So the Hamiltonian is the energy of the system, and the energy of the system is not conserved.

Problem 4:

A simple pendulum of mass m₂ and length l is constrained to move in a single plane. The point of support is attached to a mass m₁ which can move on a horizontal line in the same plane.
(a) Find the Lagrangian of the system in terms of suitable generalized coordinates.
(b) Derive the equations of motion.
(c) Find the frequency of small oscillations of the pendulum.

Solution:

Concepts:
Lagrangian Mechanics
Reasoning:
We are asked to compute the Lagrangian and find and solve Lagrange's equations.
Details of the calculation:
(a) Let x₁ denote the position of m₁ along the line and φ denote the angle the string makes with the vertical.
For the mass m₂ we have x₂ = x₁ + lsinφ, y₂ = -lcosφ.
The Lagrangian of the system is
L = ½(m₁(dx₁/dt)² + m₂(dx₂/dt)² + m₂(dy₂/dt)²) - m₂gy₂
= ½(m₁ + m₂)(dx₁/dt)² + ½m₂(l²(dφ/dt)² + 2l(dx₁/dt)(dφ/dt)cosφ) + m₂glcosφ.
(b) ∂L/∂(dx₁/dt) = (m₁ + m₂)(dx₁/dt) + m₂l(dφ/dt)cosφ.
x₁ is cyclic.
∂L/∂(dφ/dt) = m₂(l²(dφ/dt) + l(dx₁/dt)cosφ).
d/dt[∂L/∂(dφ/dt)] = m₂(l²(d²φ/dt²) + l(d²x₁/dt²)cosφ - l(dx₁/dt)(dφ/dt)sinφ).
∂L/∂φ = -m₂glsinφ - m₂l(dx₁/dt)(dφ/dt)sinφ.
Equations of motion:
(m₁ + m₂)(dx₁/dt) + m₂l(dφ/dt)cosφ = constant (first integral)
or
(m₁ + m₂)(d²x₁/dt²) + m₂l(d²φ/dt²)cosφ - m₂l(dφ/dt)²sinφ = 0.
(second-order equation)

m₂(l²(d²φ/dt²) + l(d²x₁/dt²)cosφ - l(dx₁/dt)(dφ/dt)sinφ) + m₂glsinφ + m₂l(dx₁/dt)(dφ/dt)sinφ = 0,
or
l(d²φ/dt²) + (d²x₁/dt²)cosφ - (dx₁/dt)(dφ/dt)sinφ + gsinφ + (dx₁/dt)(dφ/dt)sinφ = 0

(c) In the small angle approximation we neglect terms higher than first order in the small quantities dx/dt, d²x/dt², φ, dφ/dt, and d²φ/dt².
Then sinφ = φ and cosφ = 1.
(m₁ + m₂)(d²x₁/dt²) + m₂l(d²φ/dt²) = 0.
l(d²φ/dt²) + (d²x₁/dt²) + gφ = 0.
Combining
l(d²φ/dt²) - m₂l(d²φ/dt²)/(m₁ + m₂) + gφ = 0.
m₁l(d²φ/dt²)/(m₁ + m₂) + gφ = 0.
(d²φ/dt²)) + [(m₁ + m₂)g/(m₁l)]φ = 0.
(d²φ/dt²)) + ω²φ = 0.
The angular frequency of small oscillations is
ω = [(m₁ + m₂)g/(m₁l)]^½.

Problem 5:

Consider a system consisting of a mass m, a spring, and a rigid, massless lever arm of length L with one end fixed at the origin. The spring has unstretched length l₀ with force constant k and joins mass m with the lever arm. The entire assembly rests on a frictionless surface.

(a) Calculate the Lagrangian function in terms of the lengths L and l and the angles θ and φ and their derivatives.
(b) Now impose the additional constraints that φ = π/2. (One can do this by making m move in a track attached to L.) Obtain the equation of motion for this constrained system.
(c) Consider the case where d²θ/dt² = 0. Describe the motion of m.

Solution:

Concepts:
Lagrangian mechanics
Reasoning:
We are asked to find the Lagrangian and the equations of motion.
Details of the calculation:
Use L₀ for the length of the lever arm to not confuse it with the Lagrangian L.
(a) T = ½m(v_x² + v_y²), U = ½k(l - l₀)².
x = L₀cosθ + lcos(θ - φ), v_x = -L₀sinθ(dθ/dt) + (dl/dt)cos(θ - φ) - lsin(θ - φ)(dθ/dt - dφ/dt).
y = L₀sinθ + lsin(θ - φ), v_y = L₀cosθ(dθ/dt) + (dl/dt)sin(θ - φ) + lcos(θ - φ)(dθ/dt - dφ/dt).
v_x² + v_y² = L₀²(dθ/dt)² + l²(dθ/dt - dφ/dt)² + (dl/dt)²
+ 2L₀(dl/dt)(dθ/dt)(-sinθ cos(θ - φ) + cosθ sin(θ - φ))
+ 2L₀l(dθ/dt)(dθ/dt - dφ/dt)(sinθ sin(θ - φ) + cosθ co(θ - φ))
= (dl/dt)² + L₀²(dθ/dt)² + l²(dθ/dt - dφ/dt)² -
2L₀(dl/dt)(dθ/dt)sinφ + 2L₀l(dθ/dt)(dθ/dt - dφ/dt)cosφ.
L = ½m[dl/dt)² + L₀²(dθ/dt)² + l²(dθ/dt - dφ/dt)²
- 2L₀(dl/dt)(dθ/dt)sinφ + 2L₀l(dθ/dt)(dθ/dt - dφ/dt)cosφ]
- ½k(l - l₀)².

(b) Let φ = π/2, dφ/dt = 0. This eliminates the coordinate φ from the Lagrangian.
We have cosφ = 0, sinφ = 1.
L = ½m[(dl/dt)² + (L₀² + l²)(dθ/dt)² - 2L₀(dl/dt)(dθ/dt)] - ½k(l - l₀)².
∂L/∂(dl/dt) = m(dl/dt) - mL₀(dθ/dt), (d/dt)∂L/∂(dl/dt) = md²l/dt² - mL₀d²θ/dt².
∂L/∂l = ml(dθ/dt)² - k(l - l₀).
Equation of motion for l: d²l/dt² - L₀d²θ/dt² - l(dθ/dt)² + (k/m)(l - l₀) = 0.

∂L/∂(dθ/dt) = m(L₀² + l²)(dθ/dt) -mL₀(dl/dt). ∂L/∂dθ = 0.
m(L₀² + l²)(dθ/dt) -mL₀(dl/dt) = p_θ = constant, since θ is cyclic.
Angular momentum is conserved.
Equation of motion for θ: (L₀² + l²)d²θ/dt² + 2l(dl/dt)(dθ/dt) - L₀d²l/dt² = 0.

(c) Let d²θ/dt² = 0. Then dθ/dt = A is a constant.
It is not clear if we are to find a solution for case (b) with d²θ/dt² = 0, or if we are supposed to impose an additional constraint in the form of an external agent which holds the rotation rate constant.

(i) A solution for case (b) with d²θ/dt² = 0 requires that dθ/dt = A, where A is a constant.
Then p_θ = constant requires that dl/dt - constant, d²l/dt² = 0.
The equation of motion for θ then requires that dl/dt = 0.
The equation of motion for l then yields lA² + (k/m)(l - l₀) = 0, l = kl₀/(k - mA²).
As long as k/m > A², the mass rotates with angular velocity A and the spring is stretched.

(ii) Let d²θ/dt² = 0 be an additional constraint.
Then L = ½m[dl/dt)² + (L₀² + l²)A² - 2L₀A(dl/dt)] - ½k(l - l₀)².
The equation of motion derived from this Lagrangian is
d²l/dt² - A²l + (k/m)(l - l₀) = 0. This can be rewritten as
d²l/dt² + (k/m - A²)(l - l₀') = 0,
with l₀' = kl₀/(k - mA).
Define l' = (l - l₀'). Then d²l'/dt² + ω² l' = 0.
ω² = (k/m - A²).
The mass executes simple harmonic motion if k/m > A².

Note: While this harmonic motion satisfies the equation of motion for l in case (i), it does not satisfy the equation of motion for θ, and therefore is not a solution for case (i).