From the Hamiltonian of a free, non-relativistic, classical mass-point, H =
**p**^{2}/(2m_{0}),

(a) find the equation of motion d**p**/dt and d**q**/dt.

(b) Compute the total differential for an arbitrary function F(**p**,
**q**
,
t) and express in terms of Poisson brackets.

Recall that the Poisson
bracket

{a, b}_{PB} = ∑_{k}(∂a/∂q_{k} ∂b/∂p_{k} - ∂a/∂p_{k}
∂b/∂q_{k}).

Solution:

- Concepts:

Hamilton's equations, Poisson brackets - Reasoning:

Hamilton's equations are dq_{i}/dt = ∂H/∂p_{i}, dp_{i}/dt = -∂H/∂q_{i}. - Details of the calculation:

(a) H = ∑_{i}p_{i}^{2}/(2m), i = {x, y, z}.

dq_{i}/dt = v_{i}= p_{i}/m, dp_{i}/dt = 0.

Momentum is conserved, the particle moves with constant velocity.

(b) dF/dt = ∑_{i}[(∂F/∂q_{i})(dq_{i/}dt) + (∂F/∂p_{i})(dp_{i/}dt)] + ∂F/∂t.

(dq_{i/}dt) = (∂H/∂p_{i}), (dp_{i/}dt) = -(∂H/∂q_{i}).

dF/dt = ∑_{i}((∂F/∂q_{i})(∂H/∂p_{i}) - (∂F/∂p_{i})(∂H/∂q_{i})) + ∂F/∂t = {F, H} + ∂F/∂t.

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
= R_{0} + R't.

At t = 0 the speed of the particle is v_{0}.

(a) Are there any constants of motion?

(b) Let E(t_{1}) be the energy of the particle at the time the
radius of the ring is 2R_{0} and E_{0} its energy at t = 0.

Find the ration E(t_{1})/E_{0}.

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.

If the generalized coordinates and momenta we choose do not depend on time and the Lagrangian depends explicitly on time, then the energy is not conserved. - 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 = R_{0 }+ R't is the equation of constraint.

The constraint is holonomic and time dependent.

x = (R_{0}+ R't)cosφ, y = (R_{0}+ R't)sinφ.

T = ½m(v_{x}^{2}+ v_{y}^{2}) = ½m((dφ/dt)^{2}(R_{0}+ 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)(R_{0}+ R't)^{2}= constant, since (d/dt)(∂L/∂(dφ/dt)) = 0.

The angular momentum M = m(dφ/dt)(R_{0}+ R't)^{2}about the vertical axis is a constant of motion.

The energy is NOT a constant of motion.

(b) E = M^{2}/(2m(R_{0}+ R't)^{2}) + ½mR'^{2}.

E(t_{1})/E_{0}= (M^{2}/(8mR_{0}^{2}) + ½mR'^{2})/(M^{2}/(2mR_{0}^{2}) + ½mR'^{2})

= (M^{2}+ 4m^{2}R'^{2}R_{0}^{2})/ (4M^{2}+ 4m^{2}R'^{2}R_{0}^{2}).

If R'^{2}<< M^{2}/(2m^{2}R_{0}^{2}), then E(t_{1})/E_{0}= ¼.

The energy of the particle decreases, the particle does positive work.

A bead is constrained to move without friction on a helix whose equation in
cylindrical polar coordinates is ρ = b, z = aΦ under the influence of the
potential V = ½k(ρ^{2} + z^{2}).

(a)
Use the Lagrange multiplier method and find the appropriate Lagrangian including
terms expressing the constraints.

(b) Apply the Euler-Lagrange
equations to obtain the equations of motion.

(c) Next, repeat parts
(a) and (b) without using the Lagrange multiplier method. Instead, build
the constraints into the general coordinate(s).

Solution:

- Concepts:

Lagrangian Mechanics, the Lagrange multiplier method - Reasoning:

We are instructed to use the Lagrange multiplier method to solve the problem. - Details of the calculation:

(a) Lagrangian not incorporating the constraints:

L = ½m[(dρ/dt)^{2}+ ρ^{2}(dΦ/dt)^{2}+(dz/dt)^{2}] - ½k(ρ^{2}+ z^{2}).

Put the equations of constraint into the form

Σ_{k}a_{lk }dq_{k }+ a_{lt }dt = 0, l = 1, ..., m.

We have two equations of constraint.

(i) dρ = 0; a_{1ρ}= 1, a_{1Φ}= a_{1z}= 0.

(ii) dz - adΦ = 0 ; a_{2ρ}= 0, a_{2Φ}= -a, a_{2z}= 1.

(b) Lagrange's equation:

d/dt(∂L/∂(dq_{k}/dt)) - ∂L/∂q_{k}= ∑_{l}λ_{l}a_{lk}, Σ_{k}a_{lk }dq_{k }+ a_{lt }dt = 0.

We have:

md^{2}ρ/dt^{2}- mρ(dΦ/dt)^{2}= λ_{1}.

mρ^{2}d^{2}Φ/dt^{2}+ 2mρ(dρ/dt)(dΦ/dt) = -aλ_{2}.

md^{2}z/dt^{2}+ kz = λ_{2}.

ρ = b, dρ/dt = d^{2}ρ/dt^{2}= 0.

z = aΦ, dz/dt = adΦ/dt, d^{2}z/dt^{2}= ad^{2}Φ/dt^{2}.

We therefore have:

-mb(dΦ/dt)^{2}+ kb = λ_{1}.

mb^{2}d^{2}Φ/dt^{2}= -aλ_{2}.

ma d^{2}Φ/dt^{2}+ kaΦ = λ_{2}.

Using the last two equations to solve for λ_{2}we have λ_{2}= Φ kab^{2}/(a^{2}+ b^{2}).

Therefore the equation of motion is d^{2}Φ/dt^{2}+ (ka^{2}/(m(b^{2}+ a^{2}))Φ = 0.

λ_{1}= -(ka^{2}b/((b^{2}+ a^{2}))Φ + kb.

λ_{1}is the force of constraint in the radial direction, λ_{2}is the force of constraint in the z- direction.

(c) L = ½m[(adΦ/dt)^{2}+ (bdΦ/dt)^{2}] - ½k(b^{2}+ (aΦ)^{2}).

d/dt(∂L/∂(dq_{i}/dt)) - ∂L/∂q_{i}= 0.

m(b^{2}+ a^{2})d^{2}Φ/dt^{2}+ ka^{2}Φ = 0.

d^{2}Φ/dt^{2}+ (ka^{2}/(m(b^{2}+ a^{2}))Φ = 0.

Let C^{2}= ka^{2}/(m(b^{2}+ a^{2})).

Φ = A exp(-iCt) + exp(iCt) = Φ_{0}sin(Ct + δ).

The particle oscillates about z = Φ = 0. The amplitude and phase of the oscillation are determined by the initial conditions. The angular frequency of the oscillation is C.

An very long rod is being rotated in a
vertical plane at a constant angular velocity ω about a fixed
horizontal axis (the z-axis) passing through the origin. The angular velocity is
maintained at the value ω for all times by an external agent.
At t = 0 the rod passes through zero-inclination, (θ = 0
at t = 0) where θ is the angle the rod makes with the x-axis.
There is a mass m on the rod. The mass' coordinates and velocity components at
t = 0 are r(0) = g/(2ω^{2}), θ(0) = 0, dr/dt|_{0} = 0, dθ/dt|_{0} = ω,

where g is the acceleration due to gravity. The mass m is
free to slide along the rod. Neglect friction.

Hint: Recall that in plane polar
coordinates the unit vectors **r**/rand
**θ**/θ are
not constant.

(a) Find an expression for r(t), the radial coordinate of the mass, which holds
as long as the mass remains on the rod.

(b) Show that r(t) > r(0) for small t (t > 0).

(c) There is a component of the mass' weight acting down the inclined rod, but no force
component acting up the rod. With this in mind, explain why the mass begins moving farther
out along the rod instead of down the rod.

Solution:

- Concepts:

Lagrangian mechanics - Reasoning:

All forces except the forces of constraint are derivable from a potential. The Lagrangian formalism is well suited for such a system. - Details of the calculation:

(a) T = ½m[(dr/dt)^{2}+ r^{2}(dθ/dt)^{2}] = ½m[(dr/dt)^{2}+ r^{2}ω^{2}] = kinetic energy of the mass m.

U = mgh = mgr sinθ = mgr sin(ωt).

L = T - U, we have only one generalized coordinate, r.

∂L/∂(dr/dt) = m dr/dt, (d/dt)∂L/∂(dr/dt) = m d^{2}r/dt^{2},

∂L/∂r = mrω^{2}- mg sin(ωt).

equation of motion: d^{2}r/dt^{2}- rω^{2}= -g sin(ωt).

Let r = A sin(ωt). Then d^{2}r/dt^{2}= -ω^{2}A sin(ωt),

and -2ω^{2}A sin(ωt) = -g sin(ωt), A = g/(2ω^{2}).

We can add to this solution the solution of the homogeneous equation

d^{2}r/dt^{2}- rω^{2}= 0. r = C_{1}e^{ωt}+ C_{2}e^{-ωt}.

The most general solution of the equation of motion is

r(t) = (g/(2ω^{2})) sin(ωt) + C_{1}e^{ωt}+ C_{2}e^{-ωt}.

Initial conditions: r(0) = g/(2ω^{2}) = C_{1}+ C_{2}.

dr/dt|_{0}= 0 = (g/(2ω) + ω(C_{1}- C_{2}).

Solve for C_{1}and C_{2}: C_{1}= 0, C_{2}= g/(2ω^{2}).

r(t) = (g/(2ω^{2})) [sin(ωt) + e^{-ωt}].

(b) For small t:

r(t) = (g/(2ω^{2}))[ ωt + (ωt)^{3}/3! + ... + 1 - ωt + (ωt)^{2}/2 - (ωt)^{3}/3! + ...]

= (g/(2ω^{2}))[ 1 + (ωt)^{2}/2 ] > r(0) to third order in t.

(c) The reference frame of the rod is a rotating (non-inertial) frame.

In the inertial frame, the mass has an initial velocity of magnitude r(0)ω = g/(2ω) in the y-direction.

(a) If the Hamiltonian of a system is given by H = (1/b)p^{b},
with with b
= constant, find the corresponding Lagrangian.

(b) If a system has a Lagrangian L = ½G(q,t)(dq/dt)^{2} +
F(q,t)(dq/dt) - U(q,t),

show that the Hamiltonian can be written

H = (p - F(q,t))^{2}/(2G(q,t)) + U(q,t),

where p = G(dq/dt) + F.

Solution:

- Concepts:

Lagrange's and Hamilton's Equations - Reasoning:

L = L(q, dq/dt, t), H = H(q, p, t).

H = ∑_{i}(dq_{i}/dt)p_{i}- L, L = ∑_{i}(dq_{i}/dt)p_{i}- H.

dq_{i}/dt = ∂H/∂p_{i}, dp_{i}/dt = -∂H/∂q_{i}. - Details of the calculation:

(a) H = H = (1/b)p^{b}, dp/dt = -∂H/∂q = 0.

Therefore ∂L/∂q = dp/dt = 0, L does not depend on q.

∂H/∂p = dq/dt = p^{b-1}, p = (dq/dt)^{1/(b-1)}

L = (dq/dt)p - H = (dq/dt)^{b/(b-1)}- (1/b)(dq/dt)^{b/(b-1)}

= (dq/dt)^{b/(b-1)}(1- 1/b).

(b) L = ½G(q,t)(dq/dt)^{2}+ F(q,t)(dq/dt) - U(q,t).

∂L/∂(dq/dt) = p = G(dq/dt) + F, dq/dt = (p - F)/G.

H = p(p - F)/G - L

= p(p - F)/G - (p - F)^{2}/(2G) - F(p - F)/G + U(q,t)

= (p - F)^{2}/(2G) + U(q,t).

or

p(dq/dt) - L = G(dq/dt)^{2}- F(dq/dt) - ½G(dq/dt)^{2}- F(dq/dt) + U(q,t)

= ½G(dq/dt)^{2}+ U(q,t) = (p - F)^{2}/(2G) + U(q,t).