Mass in a tube
The Lagrangian is
L = (1/2)(1/3)M𝓁²θ˙² + (1/2)mx²θ˙² + (1/2)mẋ² + mgx sin θ + Mg(𝓁/2)sin θ. (8.82)
The Euler–Lagrange equations are then
d/dt(∂L/∂ẋ) = ∂L/∂x ⟹ mẍ = mxθ˙² + mg sin θ,
d/dt(∂L/∂θ˙) = ∂L/∂θ ⟹ d/dt[(1/3)M𝓁²θ˙ + mx²θ˙] = mgx + Mg(𝓁/2)cos θ (8.83)
⟹ [(1/3)M𝓁² + mx²]θ¨ + 2mxẋθ˙ = mgx + Mg(𝓁/2)cos θ.
In terms of η ≡ x/𝓁, these equations become
η¨ = ηθ˙² + g~ sin θ,
(1 + 3rη²)θ¨ = 3rg~η + (3g~/2)cos θ - 6rηη˙θ˙, (8.84)
where r ≡ m/M and g~ ≡ g/𝓁. Below is a Maple program that numerically finds the value of η when θ equals π/2, in the case where r = 1. As mentioned in Problem 1.2, this value of η does not depend on g or 𝓁, and hence not g~. In the program, we'll denote g~ by g, which we'll give the arbitrary value of 10. We'll use q for θ, and n for η. Also, we'll denote θ˙ by q1 and θ¨ by q2, etc. Even if you don't know Maple, this program should still be understandable. See Section 1.4 for more discussion of solving differential equations numerically.
n := 0: # initial n value
n1 := 0: # initial n speed
q := 0: # initial angle
q1 := 0: # initial angular speed
e := .0001: # small time interval
g := 10: # value of g/l
r := 1: # value of m/M
while q < 1.57079 do # do this process until
# the angle is pi/2
n2 := n*q1^2 + g*sin(q): # the first E–L equation
q2 := ((3*r*g*n + 3*g/2)*cos(q)
- 6*r*n*n1*q1)/(1 + 3*r*n^2): # the second E–L equation
n := n + e*n1: # how n changes
n1 := n1 + e*n2: # how n1 changes
q := q + e*q1: # how q changes
q1 := q1 + e*q2: # how q1 changes
end do: # stop the process
n; # print the value of n
354
Angular momentum, Part I (Constant L^)
The resulting value of η is 0.378. If you actually run this program on Maple with different values of g, you will find that the result for n doesn't depend on g, as stated above. A few results for η for various values of r are, in (r, η) notation: (0, 0.349), (1, 0.378), (2, 0.410), (10, 0.872), (20, 3.290). It turns out that r ≈ 11.25 yields η ≈ 1. That is, the mass m gets to the end of the tube right when the tube becomes vertical. For η values larger than 1, we could imagine attaching a massless tubular extension on the end of the given tube. It turns out that η → ∞ as r → ∞. In this case, the mass m drops nearly straight down, causing the tube to quickly swing down to a nearly vertical position. But m ends up slightly to one side and then takes a very long time to move over to become directly below the pivot.