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.
- 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(vx2 + vy2) = 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).
Inserting M = mR0v0, we have
E(t1)/E0 = (v02/4 + R'2)/(v02
+ R'2).
The energy of the particle decreases, the particle does positive work.
Aside:
Why? The force of constraint always points towards the center of the
circle. The particle has a displacement component which points away from
the center of the circle.
The force of constraint does negative work on the particle, the particle
does positive work.
Problem 2:
A particle of mass m is constrained to move along the surface of a cone with
half opening angle α under the influence of gravity. The cone is oriented
with its apex pointing down, and its symmetry axis along the direction of
gravity, g.
(a) Find the Lagrangian of the particle using (r,θ) as generalized
coordinates, where r denotes the perpendicular distance from the cone axis to
the particle, and θ is the azimuthal angle (see figure).
(b) Calculate the generalized momenta pr and pθ that are
conjugate to the coordinates r and θ, respectively. Write down the
Hamiltonian H(r, θ, pr, pθ) in terms of the coordinates
and conjugate momenta.
(c) Using the Hamiltonian from part (b), find the four corresponding Hamilton's
equations of motion for dr/dt, dθ/dt, dpr/dt, and dpθ/dt.
(d) Determine which, if any, of the following quantities are conserved: pr,
pθ, and H.
Explain why the conservation follows from
the form of the Hamiltonian, and provide a physical interpretation of any
conserved quantities.
Solution:
- Concepts:
Lagrangian and Hamiltonian mechanics
- Reasoning:
We are asked to find the Lagrangian and Hamiltonian for the system.
- Details of the calculation:
(a) L = T - U, T = ½m((dx/dt)2 + (dy/dy)2 + (dz/dt)2),
U = mgz.
x = r cosθ , y = r sinθ , z = r cotα .
The cartesian velocities are
dx/dt = dr/dt cosθ − r sinθ dθ/dt ,
dy/dt = dr/dt sinθ + r cosθ dθ/dt ,
dz/dt = dr/dt cotα ,
The Lagrangian is
L = ½m((dr/dt)2 + r2(dθ/dt)2 + (dr/dt)2 cot2α)
- mgr cotα
= ½m((dr/dt)2/sin2α + r2(dθ/dt)2)
- mgr cotα.
(b) We determine the generalized conjugate momenta from the Lagrangian
pr = ∂L/(dr/dt) = m(dr/dt)/sin2α.
pθ = ∂L/(∂θ/dt) = mr2dθ/dt.
The Hamiltonian is given by
H(r, θ, pr, pθ) = pr(dr/dt) + pθ(dθ/dt) − L .
Using the expressions of pr and pθ from above we find
H(r, θ, pr, pθ) = pr2sin2α/(2m)
+ pθ2/(2mr2) + mgr cotα.
(c) Hamilton's equations of motion are
dr/dt = ∂H/∂pr = prsin2α/m,
dθ/dt = ∂H/∂pθ = pθ/(mr2),
dpr/dt = −∂H/∂r = −mg cotα + pθ2/(mr3),
dpθ/dt = −∂H/∂θ = 0 .
(d) pθ is conserved since θ is a cyclic coordinate (i.e., it does not
appear explicitly in L or H). H is also conserved since it (or the
Lagrangian) does not depend explicitly on time.
pθ is the angular momentum of the particle. Since L or H do not
explicitly on time, and the coordinates and constraints do not explicitly depend
on time, and the potential is velocity-independent, H is equal to the total
energy H = T + U.
Problem 3:
A simple pendulum of mass m and length l is free to swing in the xy-plane.
It is fixed to a support of mass M that can move freely in the horizontal
direction. Using the coordinates x, the horizontal position of the
support, and θ, the angle of the pendulum from vertical, write down the Lagrangian for this system.
Solution:
- Concepts:
Lagrangian Mechanics
- Reasoning:
We are asked to find the Lagrangian for this system.
-
Details of the calculation:
(a) Let x denote the position of M along the line and
θ denote the angle the string makes
with the vertical.
For the mass m we have x1 = x
+ lsinθ, y1 = -lcosθ.
The Lagrangian of the system is L = T - U.
T = ½M(dx/dt)2 + ½m(dx1/dt)2 + (dy1/dt
)2) = ½(M + m)(dx/dt)2 + ½m[l2(dθ/dt)2
+ 2l(dx/dt)(dθ/dt)cosθ].
U = -mglcosθ.
L = ½(M + m)(dx/dt)2 + ½m[l2(dθ/dt)2 +
2l(dx/dt)(dθ/dt)cosθ] + mglcosθ.
Problem 4:
The Lagrangian for the system below can be written as
where xi denotes the displacement of mass mi from its
equilibrium position.
(a) Find the matrices T and K.
(b) Find the eigenvalues and eigenvectors for the oscillations of this
system.
(c) Write down the general solution for x1(t) and x2(t).
(d) For the initial conditions
solve for the constants in your answer for (c) and
determine for x1(t) and x2(t) for these initial
values.
Solution:
- Concepts:
Coupled oscillations, normal modes
- Reasoning:
L = ½∑ij[Tij(dqi/dt)(dqj/dt) - kijqiqj]
with Tij = Tji, kij = kji.
Solutions of the form qj = Re(Ajeiωt) can
be found.
We solve |kij - ω2Tij| = 0, for the normal
mode frequencies ω.
For a particular frequency ωα we solve ∑j[kij
- ωαTij]Ajα = 0 to find the Ajα.
The most general solution for each coordinate qj is α sum of
simple harmonic oscillations in all of the frequencies ωα.
qj = Re∑αCαAjαexp(iωαt).
- Details of the calculation:
(a) The kinetic energy is T = ½[m(dx1/dt)2 + m(dx2/dt)2],
and the potential energy is
U = ½[3k(x2 - x1)2
+ kx12 + kx22] = ½[4kx12
+ 4kx22 - 3kx1x2 -
3kx2x1].
(b) ω2 = 4ω02 ± 3ω02,
ω12 = ω02, ω22
= 7ω02, with ω02 = k/m.
(c)
with A and B complex constants is the general solution.
(d)
The normal mode with frequency ω2 by itself is consistent with
the initial conditions.
We have
Formally:
Problem 5:
A particle of mass m moves under the influence of gravity on the inner
surface of a paraboloid of revolution x2 + y2 = az which
is assumed frictionless.
(a) Introduce cylindrical coordinates ρ, φ, z. Write a Lagrangian for the
system employing ρ and φ as generalized coordinates.
(b) Find two constants of motion.
(c) Obtain the equations of motion.
(d) Show that the particle will describe a horizontal circle in the plane z
= h provided that it is given the proper angular velocity ω. What is the
magnitude of this velocity?
Solution:
- Concepts:
Lagrangian Mechanics
- Reasoning:
- Details of the calculation:
(a)
L = T - U = ½mv2 - mgz.
In cylindrical coordinates L = ½m[(dρ/dt)2 + ρ2(dφ/dt)2
+ (4ρ2/a2)(dρ/dt)2] - mgρ2/a.
(b) ∂L/∂(dφ/dt) = mρ2(dφ/dt) = M =
constant.
E = ½m[(1 + 4ρ2/a2)((dρ/dt)2] + M2/(2mρ2)
+ mgρ2/a = constant.
(c) ∂L/∂(dφ/dt) = m(1 + 4φ2/a2)(dφ/dt),
d/dt(∂L/∂(dφ/dt)) = m(1 + 4φ2/a2)d2φ/dt2
+ 8mρ(dφ/dt)2/a2.
∂L/∂φ = mρ(dφ/dt)2 + 4mρ(dφ/dt)2/a2
- 2mg/a.
Equations of motion:
ρ2d2φ/dt2 + 2ρ(dρ/dt)(dφ/dt) = 0.
(1 + 42c2ρ2)d2ρ/dt2 + 4c2ρ(dρ/dt)2
- ((dφ/dt)2- 2g/a)ρ = 0.
or
(1 + 42c2ρ2)d2ρ/dt2 + 4c2ρ(dρ/dt)2
- M2/(m2ρ3) + (2g/a)ρ = 0.
(d) d2ρ/dt2 = dρ/dt = 0 --> dφ/dt = (2g/a)½.
ω = (2g/a)½, independent of z.