Two thin circular discs of mass m and 4m, having radii of a  and 2a  , respectively, are rigidly fixed by a massless, rigid rod of  length $l=\sqrt{24}\text{\hspace{0.17em}a}$ through their centers. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is $\omega$ . The angular momentum of the entire assembly about the point ‘O’ is $\overrightarrow{L}$ ( see the figure). Which of the following statement(s) is (are)true?

$\begin{array}{l}In △OAM , OM=\sqrt{l^2+a^2}=5a\\ \text{}Circumference of a circle of radius OM=2\pi \left(5a\right)\text{}\\ Circumference of a smaller disc =2\pi a\\ For completing this big circle once, the smaller disc wil have to take \frac{2\pi \left(5a\right)}{2\pi a}=5 rotations\\ \text{}Thus, the C.M. of the assembly rotates about z axis with an angular speed of \frac{\omega }{5}.\\ The angular momentum about C.M. of the system=L_C=I_C\omega =\left[\frac{1}{2}m{a}^2\right]\omega + \left[\frac{1}{2}\times 4m\times {\left(2a\right)}^2\right]\omega =\frac{17m{a}^2\omega }{2}\text{}\\ Now, v_c=\frac{m\left(\omega a\right)+4m\left(2\omega a\right)}{5m}=\frac{9\omega a}{5}\\ and r_1=\frac{m\left(l\right)+4m\left(2l\right)}{5m}=\frac{9l}{5}\text{}\\ So, L of C.M. =\left(5m\right)\left(\frac{9\omega a}{5}\right)\left(\frac{9l}{5}\right)=\frac{81\sqrt{24}m\omega a^2}{5}\\ Net angular momentum about z axis=L_Z=\frac{81\sqrt{24}m\omega a^2}{5}\cos \theta - I_C\omega \sin \theta \text{}\\ \Rightarrow L_Z=\frac{81\sqrt{24}m\omega a^2}{5}\times \frac{\sqrt{24}}{5}-\frac{17m{a}^2\omega }{10}=\frac{3803m\omega a^2}{50}\end{array}$