Past exam of the options II course of the University of Oxford / 2013 / H2 / Solution

Differentiating orthonormality gives ${\dot{e}}_i\cdot e_j=-e_i\cdot {\dot{e}}_j$. With $\omega =\frac{1}{2}{\sum }_i{e}_i\times {\dot{e}}_i$, the vector triple product yieldsThe body coordinates $x_i$ of a fixed material point are constant. Differentiating its position $R_G+{\sum }_i{x}_i{e}_i$ therefore givesThis identifies $\omega$ as the angular velocity of the rigid body.

Because $G$ is the center of mass, $\int \rho rdV=0$, so the translational contribution to the angular momentum about $G$ vanishes. The remaining vector triple product givesThis 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 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 $z=l=\sqrt{3}a$. 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 $M{R}^2/2$ about its normal and $M{R}^2/4$ about a diameter. A thin rod has moment $m{l}^2/12$ about a transverse axis through its center. The parallel axis theorem adds $m{d}^2$ for a parallel shift of distance $d$. ThereforeandThe rod contributes zero about its own longitudinal axis. Thus $\boxed{J=2M{a}^2{I}_3}$, an isotropic inertia tensor even though the shape is not spherical. This gives a nonspherical rigid body with isotropic inertia. Consequently $L=2M{a}^2\omega$. Conservation of $L$ in torque-free motion proves the angular velocity is constant for every initial orientation and spin.