This question involves some simple ideas from the Lorenz equations when $\sigma=1$ Otherwise know as the chaotic waterwheel. Basically the question is to show that the moment of Inertia in the waterwheel is a constant ( i think) Show that $I(t) \to C$ as $t \to \infty$ as follow's
i) The total moment of inertia is a sum $I=I_{wheel} + I_{water}$ where $I_{wheel}$ depends only on the apparatus itself, and not on the distribution of water around the rim. express $I_{water}$ in terms of $M= \int^{2\pi}_{0} m(\theta,t)d\theta$
my book has $I \omega^{'} =$ damping torque + gravitational torque
where dampening torque $=-v\omega$ and gravitational torque $=gr\int^{2\pi}_{0} m(\theta,t)\sin(\theta)d\theta$
then it gets particularly ugly im probably looking at the wrong part but looking at the water disruption separately isn't helping me either.
ii) Show that M satisfies $M^{'}=Q_{total} -KM$ where $Q_{total} =\int^{2\pi}_{0} Q(\theta)d\theta$ and K=leakage rate from wheel
iii) In the equation $I(t) \to C$ as $t \to \infty$ find the value of C
I have stared at this for over an hour and came to the conclusion that I have no idea.
In case someone was curious i believe i have actually figured out the answer to the question.
I belvie that $I_{water} =[ \int^{2\pi}_{0} m(\theta,t)d\theta] r^{2}$
$M^{'}= \int^{2\pi}_{0}(\partial m/\partial t) d\theta$
$M^{'}= \int^{2\pi}_{0}[Q-Km-\omega((\partial m/\partial \theta) ]d\theta$
This being the part i could wrap head around is that.
$\int^{2\pi}_{0}(\partial m/\partial \theta)d\theta=m(2\pi)-m(0)=0$
thus proving that $M^{'}=Q_{total} -KM$
iii) Once above is satisfied this become trivial. (equation for exponential decay)