= Solution
Differentiating orthonormality gives $\dot e_i\cdot e_j=-e_i\cdot\dot e_j$. With $\omega=\tfrac12\sum_i e_i\times\dot e_i$, the <vector> triple product yields
$$
\omega\times e_j=\tfrac12\left(\dot e_j-\sum_i e_i(\dot e_i\cdot e_j)\right)=\dot e_j.
$$
The body coordinates $x_i$ of a fixed material point are constant. Differentiating its position $R_G+\sum_ix_ie_i$ therefore gives
$$
\boxed{v=v_G+\omega\times r.}
$$
This identifies $\omega$ as the <angular velocity> of the <rigid body>.
Because $G$ is the <center of mass>, $\int\rho r\,dV=0$, so the translational contribution to the <angular momentum> about $G$ vanishes. The remaining <vector> triple product gives
$$
L=\int\rho\{(r\cdot r)\omega-(r\cdot\omega)r\}\,dV,
\qquad
\boxed{J_{ij}=\int\rho\left(|r|^2\delta_{ij}-x_ix_j\right)dV,\quad L_i=\sum_jJ_{ij}\omega_j.}
$$
This symmetric <matrix> is the <inertia tensor> about $G$. The angular-momentum balance about the center of mass is $dL/dt=\tau_{\mathrm{ext},G}$ in an inertial frame. With no external forces, that torque is zero, so \b[the inertial <vector> $L$ is constant]. Its components in rotating body axes need not individually be constant.
For the tool, let the disc lie at $z=0$ and the rod extend to $z=l=\sqrt3a$. The total mass is $3M$ and its center is at $z_G=l/3$. Axial symmetry makes the <inertia tensor> diagonal with $I_x=I_y$.
The standard disc moments are $MR^2/2$ about its normal and $MR^2/4$ about a diameter. A thin rod has moment $ml^2/12$ about a transverse axis through its center. The <parallel axis theorem> adds $md^2$ for a parallel shift of distance $d$. Therefore
$$
I_z=\tfrac12M(2a)^2=2Ma^2,
$$
and
$$
\begin{aligned}
I_x=I_y
&=Ma^2+M(l/3)^2+\frac{2Ml^2}{12}+2M(l/6)^2\\
&=Ma^2+\frac{Ml^2}{3}=2Ma^2.
\end{aligned}
$$
The rod contributes zero about its own longitudinal axis. Thus $\boxed{J=2Ma^2I_3}$, an isotropic <inertia tensor> even though the shape is not spherical. This gives a <nonspherical rigid body with isotropic inertia>. Consequently $L=2Ma^2\omega$. Conservation of $L$ in torque-free motion proves \b[the <angular velocity> is constant for every initial orientation and spin].
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