Piece-wise Uniform Rod Dynamics

Building from the previous problem result, we have the following Lagrangian:

$L = \frac{5 b}{2} \Big( R^2 - \frac{b^2}{6} \Big) \dot{\theta}^2 - b g \Big(-5 R \, \sin (\theta + \beta) + \frac{3 b}{2} \sin \theta \Big)$

Equation of motion:

$\frac{d}{dt} \frac{\partial{L}}{\partial{\dot{\theta}}} = \frac{\partial{L}}{\partial{\theta}}$

Crunching this out results in:

$5b \Big( R^2 - \frac{b^2}{6} \Big) \ddot{\theta} = - b g \Big( - 5 R \cos(\theta + \beta) + \frac{3 b}{2} \cos \theta \Big)$

Numerically integrating results in a period of approximately $2.805$ . The graph of $\theta$ vs. time is included below, along with the simulation code.

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import math

g = 10.0
R = 2.0
b = 0.5

beta = math.acos(b/(2.0*R))

dt = 10.0**(-6.0)

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

theta = 0.0
thetad = 0.0
thetadd = 0.0

t = 0.0
count = 0

while t <= 10.0:

    theta = theta + thetad * dt
    thetad = thetad + thetadd * dt

    q1 = -5.0*R*math.cos(theta+beta)
    q2 = 3.0*b*math.cos(theta)/2.0
    right = -b*g*(q1+q2)

    q3 = R**2.0 - (b**2.0)/6.0
    left = 5.0*b*q3

    thetadd = right/left


    t = t + dt
    count = count + 1

    if count % 1000 == 0:
        print t,theta

Since total mechanical energy of the rod is constant, it's time derivative is zero. Simplifying this using the approximation for small angles the angle and it's sine are equal, we get the angular acceleration is negative, directly proportional to the angular displacement, with proportionality constant $\dfrac{gRsinβ}{R^2-\dfrac{b^2}{6}}=5.013$ .

Therefore the time period is $\dfrac{2π}{\sqrt {5.013}}=\boxed {2.806}$