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:
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) + pzt/m.
(b) The canonical momenta are
pz =
∂L/∂(dz/dt) = mdz/dt, px =
∂L/∂(dx/dt) = mdx/dt - ½mωy,
py =
∂L/∂(dy/dt) = mdy/dt + ½mωx.
The kinematic momenta are
mdz/dt = pz, mdx/dt = px + ½mωy, mdy/dt =
py - ½mωx.
Note: The kinematic momenta are not equal to the canonical momenta.
The Hamiltonian is a function of the generalized coordinates and momenta.
H = pzdz/dt + pxdx/dt + pydy/dt
- L
= ½[pz2/m + (px + ½mωy)2/m + ( py
- 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 = [(px_mech)2 + (py_mech)2 +
(pz_mech)2]/2m = T.
Think about: Which physical situation is described by this Lagrangian?
A string of length 2l is suspended at points A and B located on a horizontal line. The distance between A and B is 2d (d < l). A small, heavy bead can slide on the string without friction. Find the period of the small-amplitude oscillations of the bead in the vertical plane containing the suspension points. The acceleration due to gravity is g.
Solution:
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 .
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:
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:
(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.
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: