# Preface

This section discusses a pariticular example: rocking pendulum.

# Rocking Pendulum

plus = Polygon[{{-1/2, 0}, {-1/2, 1/4}, {-4, 1/4}, {-4, 0}}]
minus = Polygon[{{1/2, 0}, {1/2, 1/4}, {4, 1/4}, {4, 0}}]
a = Show[Graphics[{Pink, minus}], Graphics[{Pink, plus}]]
line1 = Graphics[Line[{{-2, -3.5}, {0, 0.6}}], PlotRange -> {{-2, 2}, {-4, 1}}]
line2 = Graphics[Line[{{-0.85, 1.0}, {0.8, 0.25}}], PlotRange -> {{-2, 2}, {-4, 1}}]
makeArrowPlot[g_Graphics, ah_: 0.05, dx_: 1*^-6, dy_: 1*^-6] :=
Module[{pr = PlotRange /. Options[g, PlotRange], gg, lhs, rhs},
gg = g /. GraphicsComplex -> (Normal[GraphicsComplex[##]] &);
lhs := Or @@
Flatten[{Thread[Abs[#[[1, 1, 1]] - pr[[1]]] < dx], Thread[Abs[#[[1, 1, 2]] - pr[[2]]] < dy]}] &;
rhs := Or @@ Flatten[{Thread[Abs[#[[1, -1, 1]] - pr[[1]]] < dx], Thread[Abs[#[[1, -1, 2]] - pr[[2]]] < dy]}] &;
gg = gg /.
x_Line?(lhs[#] && rhs[#] &) :> {Arrowheads[{-ah, ah}], Arrow @@ x};
gg = gg /. x_Line?lhs :> {Arrowheads[{-ah, 0}], Arrow @@ x};
gg = gg /. x_Line?rhs :> {Arrowheads[{0, ah}], Arrow @@ x};
gg]
curve = Plot[{-Sqrt[9 - x^2]}, {x, -2.2, 2.2}, PlotStyle -> {Thick, Dashed}, Axes -> False] // makeArrowPlot
Show[a, line1, line2, curve]

A rigid pendulum, such as a long rod, has a short cross-bar rigidly attached to it. As it oscillates in vertical plane, it is supported on ends of the cross-bar as they alternately contact a fixed horizontal surface:
plus = Polygon[{{-1/2, 0}, {-1/2, 1/4}, {-4, 1/4}, {-4, 0}}]
minus = Polygon[{{1/2, 0}, {1/2, 1/4}, {4, 1/4}, {4, 0}}]
a = Show[Graphics[{Pink, minus}], Graphics[{Pink, plus}]]
line1 = Graphics[Line[{{-2, -3.5}, {0, 0.6}}], PlotRange -> {{-2, 2}, {-4, 1}}]
line2 = Graphics[Line[{{-0.85, 1.0}, {0.8, 0.25}}], PlotRange -> {{-2, 2}, {-4, 1}}]
p2 = Graphics[{Dashed, Arrow[{{-0.66, -0.8}, {-0.66, -3.8}}]}]
point = Graphics[{PointSize[Large], Green, Point[{-0.66, -0.8
t1 = Graphics[ Text[Style["$Theta]", FontSize -> 14, Red], {-1.0, -2.4}]] t2 = Graphics[Text[Style["G", FontSize -> 14, Blue], {-1.0, -0.8}]] a1 = Graphics[Text[Style["a", FontSize -> 14, Blue], {0.5, 0.7}]] a2 = Graphics[Text[Style["a", FontSize -> 14, Blue], {-0.2, 1.0}]] Q = Graphics[Text[Style["Q", FontSize -> 14, Blue], {1.0, 0.45}]] b = Graphics[Text[Style["b", FontSize -> 14, Blue], {0.0, -0.2}]] mg = Graphics[Text[Style["mg", FontSize -> 14, Black], {-0.2, -3.6}]] Show[a, line1, line2, point, p2, t1, t2, a1, a2, Q, b, mg] plus = Polygon[{{-1/2, 0}, {-1/2, 1/4}, {-4, 1/4}, {-4, 0}}] minus = Polygon[{{1/2, 0}, {1/2, 1/4}, {4, 1/4}, {4, 0}}] a = Show[Graphics[{Pink, minus}], Graphics[{Pink, plus}]] line1 = Graphics[Line[{{2, -3.5}, {0, 0.6}}], PlotRange -> {{-2, 2}, {-4, 1}}] line2 = Graphics[Line[{{0.85, 1.0}, {-0.8, 0.25}}], PlotRange -> {{-2, 2}, {-4, 1}}] p2 = Graphics[{Dashed, Arrow[{{0.66, -0.8}, {0.66, -3.8}}]}] point = Graphics[{PointSize[Large], Green, Point[{0.66, -0.8}]}] t1 = Graphics[ Text[Style["\[Theta]", FontSize -> 14, Red], {1.0, -2.4}]] t2 = Graphics[Text[Style["G", FontSize -> 14, Blue], {0.3, -0.8}]] P = Graphics[Text[Style["P", FontSize -> 14, Blue], {-0.9, 0.45}]] b = Graphics[Text[Style["b", FontSize -> 14, Blue], {0.0, -0.2}]] mg = Graphics[Text[Style["mg", FontSize -> 14, Black], {0.22, -3.6}]] Show[a, line1, line2, point, p2, t1, t2, P, b, mg] plus = Polygon[{{-1/2, 0}, {-1/2, 1/4}, {-4, 1/4}, {-4, 0}}] minus = Polygon[{{1/2, 0}, {1/2, 1/4}, {4, 1/4}, {4, 0}}] a = Show[Graphics[{Pink, minus}], Graphics[{Pink, plus}]] line1 = Graphics[Line[{{0, -3.5}, {0, 0.6}}], PlotRange -> {{-2, 2}, {-4, 1}}] line2 = Graphics[Line[{{0.85, 0.27}, {-0.85, 0.27}}], PlotRange -> {{-2, 2}, {-4, 1}}] P = Graphics[Text[Style["P", FontSize -> 14, Blue], {0.9, 0.45}]] Q = Graphics[Text[Style["Q", FontSize -> 14, Blue], {-0.9, 0.45}]] mg = Graphics[Text[Style["mg", FontSize -> 14, Black], {0.4, -3.6}]] point = Graphics[{PointSize[Large], Green, Point[{0, -0.8}]}] t2 = Graphics[Text[Style["G", FontSize -> 14, Blue], {-0.3, -0.8}]] p3 = Graphics[{Dashed, Line[{{0, -0.8}, {0.85, 0.27}}]}] b = Graphics[Text[Style["b", FontSize -> 14, Blue], {-0.3, -0.2}]] a1 = Graphics[Text[Style["a", FontSize -> 14, Blue], {0.3, 0.5}]] a2 = Graphics[Text[Style["a", FontSize -> 14, Blue], {-0.3, 0.5}]] ell = ToExpression["\ell", TeXForm] t3 = Graphics[Text[Style[ell, FontSize -> 14, Black], {0.6, -0.4}]] Show[a, line1, line2, point, p3, t2, a1, a2, P, Q, b, mg, t3] Let the mass of the pendulum be m, and the length of the attached bar be 2 a. Since the construction of the rigid pendulum is symmetric, the center of gyration, which we denote by G, is along the main rod. Let k be the radius of gyration of the pendulum about G, so that the square of its radius of gyration about P and Q is $$k^2 + \ell^2 .$$ ### The first half-cycle of the motion Suppose that the pendulum is set in motion from the central position with initial conditions \[ \theta =0 \quad \mbox{and} \quad \dot{\theta} = \omega_1 \qquad\mbox{at} \quad t=0.$
Since energy is assumed to be conservative, the total energy is a constant:
$\frac{1}{2}\, m \left( k^2 + \ell^2 \right) \dot{\theta}^2 + mg \left( a\,\sin \theta -b\,\cos \theta \right) = \mbox{constant} .$
For small oscillations, powers of θ higher than the first may be neglected, yielding
$\left( k^2 + \ell^2 \right) \dot{\theta}^2 + 2ag \,\theta = \mbox{constant} \qquad\mbox{or} \qquad \dot{\theta}^2 + c^2 \theta = \mbox{constant} ,$
where $$c^2 = 2ga \left( k^2 + \ell^2 \right) .$$ The initial conditions enable this to be written as
$\dot{\theta}^2 + c^2 \theta = \omega_1^2 ,$
When the pendulum first comes to rest, let
$t= \frac{1}{2}\,\tau_1 , \quad \theta = \theta_1 , \quad \dot{\theta} =0.$
Then
$\dot{\theta}^2 + c^2 \theta = \omega_1^2 \qquad\Longrightarrow \qquad \theta_1 = \omega_1^2 / c^2 .,$
Now
\begin{align*} \frac{1}{2}\,\tau_1 &= \int_0^{\theta_1} \, \frac{{\text d} \theta}{\dot{\theta}} = \int_0^{\theta_1} \, \frac{{\text d} \theta}{\sqrt{\omega_1^2 - c^2 \theta}} = \frac{2}{c^2} \left[ \sqrt{\omega_1^2 - c^2 \theta} \right]_{0}^{\theta_1} \\ &= \frac{2\omega_1}{c^2} = \frac{2\sqrt{\theta_1}}{c} . \end{align*}
Clearly, the time taken for θ to reach zero again will also be $$\frac{1}{2}\,\tau_1 ,$$ and the angular speed will then be ω1 again. Thus, the first "half period" of oscillation will be $$\tau_1 = 4 \sqrt{\theta_1} /c .$$