### Solutions for Homework 4

1. (problem 7.8)

(a) L = T - U = ½m(x·12 + x·12) - ½kx²

(b) x1 = X + ½x + ½l   x2 = X - ½x - ½l

Differentiate, substitute into L:   L = mX·2 + ¼mx·2 - ½kx²

Hence: X equation: 2mX¨ = 0
x equation: ½mx¨ = - kx

(c) X(t) = X0 + V0t -- uniform motion (of Center of Mass)
x(t) = A cos (2k/m)½(t-t0) -- simple harmonic motion relative to each other

2. (Problem 7.16)

Since ω = v/R the K.E is T = ½mv² + ½Iω² = ½(m + I/R²)x·2, and the PE is U = -mgx sinα. From L = T - U we then get (m+I/R²)x¨ = mg sinα , or x¨ = (mg sinα)/(m+I/R²).

3. (Problem 7.18)

We'll consider the case when the "pendulum part" of the cylinder hangs down from a fixed point (ring). Let x be the length of that part, and θ its angle from the vertical. The angular velocity of the cylinder is ω = x·/R, and the square of the mass is the usual polar coordinate expression (from 1.44). The PE depends only on the height of m, U = -mgx cosθ. Therefore
 L = 1/2I( ×x /R)2 + 1/2m( ×x 2 + x2 × q 2 ) + mgx cosq
So we get the equations of motion
 x-equation:     (m + I/R2) ××x - mx ×q 2 - mg cosq = 0

 q-equation:     m(x2 ×q ) × + mgx sinq = 0 = m(2x ×x × q + x2 ×× q ) + mgx sinq.

4. (Problem 7.21)

See back of book for L. EOM works out to be r¨ – ω² r = 0. The given initial conditions imply A = B = r0/2. so for large r, r ® ½r0 ewt.
The conserved quantity (Hamiltonian) is ½mr·2 - ½mω²r²

5. (Problem 7.38)

(a) In spherical polar coordinates, the angle q is fixed at q = a, so the KE is just T = 1/2m(r·2 +r2 sin2 a f·2). Since the PE is U = mgz = mgrcosa,

 L= 1/2m( × r 2 + r2 sin2 a × f 2 ) -mgrcosa
(b) Since L does not depend on f, the f equation says simply that L/f· = mr2 sin2a f· is constant. That is, lz is conserved.
The r equation is

 m ×× r = mrsin2a × f 2 - mg cosa    or ×× r = lz  2 m2r3sin2a - g cosa
(1)
where, in the second version, I have canceled an m and replaced f· by lz/(m r2 sin2a). If lz = 0, the particle simply slides in the radial direction with the well-known acceleration g cosa. (Remember a is the angle of the incline with the vertical.) The condition that the particle can remain in a horizontal circle is that r· = 0 and hence that r = ro = [lz2/(m gsin2acosa)]1/3.
(c) If we write r = ro+ e, Eq.(1) becomes

 ×× e = lz2 m2ro3sin2a æè 1 + e ro öø -3 - g cosa   » lz2 m2ro3 sin2a æè 1 - 3 e ro öø - g cosa = -3lz2 m2ro4 sin2a e
where the first and last terms in the penultimate expression cancel because ro is the equilibrium radius. This is the equation of simple harmonic oscillations, so the circular path is stable and the particle oscillates about this path with frequency w = Ö3lz/(mro2sinα).