Solutions for Homework 4

(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 e^wt.
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,