Small Ring Oscillation

Placing the coordinate origin halfway between the two suspension points, the ring moves one the ellipse $\frac{x^2}{\ell^2} + \frac{y^2}{h^2} \; = \; 1$ with the two suspension points at the foci. If the ring is at the point with position vector $\mathbf{r} \; = \; \left(\begin{array}{c}\ell\sin\theta \\ -h\cos\theta \end{array}\right)$ then conservation of energy gives us that $\tfrac12m(\ell^2\cos^2\theta + h^2\sin^2\theta)\dot{\theta}^2 - mgh\cos\theta$ is constant. Thus $m(\ell^2\cos^2\theta + h^2\sin^2\theta)\ddot{\theta} - m(\ell^2 - h^2)\sin\theta \cos\theta \dot{\theta}^2 + mgh\sin\theta \; = \; 0$ Linearising about the equilibrium position $\theta = 0$ , we deduce that $\ell^2\ddot{\theta} + gh\theta \; \approx \; 0$ and so we have small oscillations of period $2\pi\sqrt{\frac{\ell^2}{gh}}$ and hence $\alpha = \boxed{2\pi}$ .

I derived a solution method for arbitrarily large oscillations, and then simulated some small oscillations as special cases. The sum of the distances from the ring to the two points is a constant. It therefore follows that the ring is confined to an ellipse, and the two anchor points are the ellipse foci:

$A x^2 + B y^2 = 1 \\ A = \frac{1}{\ell^2} \\ B = \frac{1}{h^2}$

Derive an acceleration constrain equation:

$A x^2 + B y^2 = 1 \\ A x \dot{x} + B y \dot{y} = 0 \\ A (x \ddot{x} + \dot{x}^2) + B (y \ddot{y} + \dot{y}^2) = 0$

Assuming the ring mass to be one, write the Newton's Second Law equations. In these equations, $N$ is the magnitude of the normal force to the curve, and $\vec{u} = (u_x,u_y)$ is the unit normal vector:

$\ddot{x} = N u_x \\ \ddot{y} = N u_y - g$

Plugging these into the acceleration constraint equation and solving yields:

$N = \frac{-A \dot{x}^2 - B \dot{y}^2 + B g y}{A x u_x + B y u_y}$

Having solved for $N$ , plug it back in to find the double-dot terms. Numerically integrate to simulate the dynamics. For small oscillations, the constant $\alpha = 2 \pi$ .

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import math

dt = 10.0**(-6.0)

h = 1.0
L = 2.0
g = 10.0

A = 1.0/(L**2.0)
B = 1.0/(h**2.0)

v0 = 0.005

#################################

t = 0.0

x = 0.0
y = -h

xd = v0
yd = 0.0

xdd = 0.0
ydd = 0.0

while xd >= 0.0:

    x = x + xd*dt
    y = y + yd*dt

    xd = xd + xdd*dt
    yd = yd + ydd*dt

    utx = 1.0
    uty = (A*x/B)/math.sqrt((1.0-A*x*x)/B)

    ut = math.hypot(utx,uty)

    utx = utx/ut
    uty = uty/ut

    ux = -uty
    uy = utx

    right = -A*(xd**2.0) - B*(yd**2.0) + B*y*g
    left = A*x*ux + B*y*uy

    N = right/left

    xdd = N*ux
    ydd = N*uy - g

    t = t + dt

#################################

T = 4.0*t

q = L*math.sqrt(1.0/(g*h))

alpha = T/q

print dt
print v0
print alpha

#>>> 
#1e-05
#0.01
#6.28319272853
#>>> ================================ RESTART ================================
#>>> 
#1e-06
#0.005
#6.28318640404
#>>>