I am having trouble showing that the period (T) of this system can be expressed in terms of an elliptical integral.
Given the dynamical system governed by the differential equation below:
$(1/2)I_O(\dot\theta)^2 - mgDcos(\theta) = (1/2)I_O(\dot\theta_0)^2 - mgD$
when $\dot\theta_0 > \sqrt{4mgD \over I_O}$
where $I_O$ m, g and D are constants and $\dot\theta_0$ is the angular velocity when $\theta$ = 0
Show that $T = {4 \over\dot\theta_o} * R(x)$
R(x) is an eliptical integral => $\int_0 ^{\phi\over2}{d\phi \over\sqrt{1-xsin^2(\phi)}}$ where $x = {4mgD \over I_O(\dot\theta_0)^2} < 1$
My attempt:
$\dot\theta=d \theta/ dt = f(\theta) $
$T = \int_0^{2\pi}{{d\theta \over f(\theta)}}$
Doing a bit of algebra will lead to:
$ \dot\theta = \sqrt{{2mgD \over I_O}(cos\theta-1)+(\dot\theta_0)^2} $
Using the identity $ cos\theta = 1-2sin^2{\theta \over 2}$ gives:
$ \dot\theta = \sqrt{(\dot\theta_0)^2(1-sin^2{\theta \over 2})} $
Now I have :
$T = \int_0^{2\pi}{{d\theta \over \sqrt{(\dot\theta_0)^2(1-sin^2{\theta \over 2})}}}$
Using the transformation $\phi={\theta \over 2}$ will only give me a 2 in the numerator. I also don't know the "x" comes into play
You made a small error when using the half-angle formula, because the factor $2 m g D/I_O$ has suddenly disappeared. The substitution should read \begin{equation} \sqrt{\frac{2 g m D}{I_O}(\cos \theta - 1) + \dot{\theta}_0^2} = \sqrt{\dot{\theta}_0^2 - 2 \frac{2 g m D}{I_O} \sin^2 \frac{\theta}{2}} \tag{1}. \end{equation} If you now take $\dot{\theta}_0^2$ outside the square root, you obtain from $(1)$ \begin{equation} |\dot{\theta}_0^2| \sqrt{1 - \alpha \sin^2 \frac{\theta}{2}}, \end{equation} where \begin{equation} \alpha = \frac{4 g m D}{\dot{\theta}_0^2 I_O}. \end{equation}