Problem 1:
Two-dimension polar coordinates have basis vector er and eφ.
(a) Draw a figure that shows the Cartesian basis vectors i and j, polar basis
vectors er and eφ for an arbitrary position vector in the first quadrant.
(b) Recall that the Cartesian infinitesimal line element is
dr = dx i + dy j.
Express the infinitesimal line element in terms er and eφ.
(c) Derive an expression for the velocity v in terms of polar basis vectors
er
and eφ.
Solution:
- Concepts:
Polar coordinates
- Reasoning:
We transform from Cartesian to polar coordinates.
- Details of the calculation:
(a)
(b) eφ = -sinφ i + cosφ j, er
= cosφ i + sinφ j.
i = cosφ er - sinφ eφ,
j = sinφ er + cosφ eφ.
x = r cosφ, dx = dr cosφ - r sinφ dφ, y = r sinφ, dy =
dr sinφ + r cosφ dφ,
dr = dx i + dy j
= (dr cosφ - r sinφ dφ)(cosφ er - sinφ eφ)
+ (dr sinφ + r cosφ dφ)(sinφ er + cosφ eφ).
dr = dr er + rdφ eφ.
(c) v = (dr/dt)er + r(dφ/dt) eφ.
Or: v = dr/dt = d(rer)/dt = (dr/dt)er
+ r d(er)/dt = (dr/dt)er + r(dφ/dt) eφ.
Problem 2:
Two masses, m1 and m2, lie on a frictionless,
horizontal table. They are connected by a spring of spring constant k.
(a) Write down the Lagrangian for this system in the form
where xi denotes the displacement of mass mi from its
equilibrium position.
(b) Find the equations of motion.
(c) Find the normal mode frequencies.
Solution:
- Concepts:
Lagrangian Mechanics, coupled oscillation
- Reasoning:
We are asked to find the normal modes of coupled harmonic
oscillators.
- Details of the calculation:
(a) L = T - U = ½m1(dx1/dt)2 + ½m2(dx2/dt)2
- ½k(x2 - x1)2.
(b) m1d2x1/dt2 + kx1 -
kx2 = 0. m2d2x2/dt2 + kx2
- kx2 = 0.
(c) det(K - ω2T) = 0, ω4 m1m1
- ω2k(m1 + m2) - 0.
ω = 0, or ω2 = k(m1 + m2)/m1m2.
Problem 3:
A particle of mass m moves in one dimension under the influence of a force
F(x,t) = (k/x2)exp(-t/τ), where k and τ are positive constants. Compute the Lagrangian
and the Hamiltonian, and discuss whether energy is conserved in this system.
Solution:
- Concepts:
Lagrangian Mechanics
- Reasoning:
We are asked to find the Lagrangian for this system one-dimensional system.
- Details of the calculation:
F(x) = -∂U/∂x. U(x) = (k/x)exp(-t/τ).
L = ½m(dx/dt)2 - (k/x)exp(-t/τ). L = L(x, dx/dt, t).
p = m(dx/dt), H = (dx/dt)p - L = p2/m - L = p2/(2m) + (k/x)exp(-t/τ).
L does explicitly depend on time, the coordinate x does not. So the
Hamiltonian is the energy of the system, and the energy of the system is not
conserved.
Problem 4:
A mass point glides without friction on a cycloid defined by x = a(θ - sinθ),
y = a(1 + cosθ),
with 0 ≤ θ ≤ 2π.
A uniform gravitational field g points in the negative y-direction.
The potential energy of the particle constrained to move on the given cycloid is
symmetric about
θ - π. Choose new coordinate that better reveal that symmetry.
Let 2φ = θ - π, -π/2 ≤ φ ≤ π/2. Then x = a(2φ + π + sin2φ), y = a(1 - cos2φ).
The bead moves on a trajectory s with elements of arc length ds.
ds = (dx2 + dy2)½ = ((∂x/∂φ)2
+ (∂y/∂φ)2)½dφ.
Write down the Lagrangian for this system in terms of the coordinate s and the
velocity ds/dt with the condition s = 0 at φ = 0.
Solution:
- Concepts:
Lagrangian Mechanics
- Reasoning:
We are asked to obtain and solve Lagrange's equation of motion.
- Details of the calculation:
We are given x and y as a function of φ.
∂x/∂φ = 2a(1 + cos2φ), ∂y/∂φ = 2a sin2φ.
ds = 2a(2 + 2cos2φ)½dφ = 4a cosφ dφ.
Integrate ds = 4a cosφ dφ with the condition s = 0 at φ = 0 to find s as a
function of φ.
s(φ) = ∫0φds = ∫0φ4a
cosφ'dφ' = 4a sinφ.
L = T - U.
T = ½m((dx/dt)2 + (dy/dt)2) = ½m((∂x/∂φ)2 +
(∂y/∂φ)2)(dφ/dt)2
= 8ma2 cos2φ(dφ/dt)2 = ½m(ds/dt)2.
U = mgy = 2mga sin2φ = (2mg/(16a))s2.
L = ½m(ds/dt)2 - (mg/(8a2))s2.
Problem 5:
Consider the systems below shown in its equilibrium
position below.
The springs hasspring constant k.
Consider only motion
in the plane of the figure.
For small displacements of the masses from their equilibrium positions, the
Lagrangian can be written as
L = ½∑ij[Tij(dqi/dt)(dqj/dt)
- kijqiqj]
with Tij = Tji,
kij = kji, where the qi are generalized
coordinates specifying a displacement from equilibrium.
Choose appropriate generalized coordinates qi and determine all
coefficients Tij and kij in the Lagrangian.
Solution:
- Concepts:
Lagrangian Mechanics
- Reasoning:
We keep terms up to second order in the small quantities in the Lagrangian.
- Details of the calculation:
Call the left pendulum 1 and the right pendulum 2. Let θi
denote a small angular displacement of pendulum i to the right.
For small angles cos(θ) = 1 - θ2/2, sinθ = θ.
T = ½m((dx1/dt)2 + (dy1/dt)2 + (dx2/dt)2
+ (dy2/dt)2)), U = -mgy1 - mgy2 + ½kΔ2.
Here Δ is the change in the length of the spring.
xi = Lsinθi, yi = L cosθi.
For small angles cos(θ) = 1 - θ2/2, sinθ = θ.
The length of the spring is [(y2 - y1)2 + (x2
- x1)2]½
= L[(cos(θ2) - cos(θ1))2 + (1
+ sin(θ2)
- sin(θ1))2]½
= L[1 + (θ2 - θ1)] for small displacements (keeping
only first order terms).
Δ = L(θ2 - θ1), Δ2 = L2(θ22
+ θ12 - θ2θ1 - θ2θ1)
to second order.
Define q1
= Lθ1, q2 = Lθ2
Keeping terms up to second order we have in the Lagrangian
T = ½m((dq1/dt)2 + (dq2/dt)2)),
U = (½mg/L)(q12 + q22) + ½k(q22
+ q12 - q2q1 - q2q1).
L = T - U = ½m((dq1/dt)2 + (dq2/dt)2) -
½(mg/L + k)(q12 + q22) + ½k(q1q2
+ q2q1).
L= ½∑ij[Tij(dqi/dt)(dqj/dt) - kijqiqj]
with Tij = Tji, kij = kji.
Here T11 = T22 = m, Tij(i≠j) = 0,
k11 = k22 = mg/L + k, k12 = k21 =
-k.