Past exam of the options II course of the University of Oxford / 2012 / H / H2

Let $E(t)$ be the matrix whose columns are the moving orthonormal basis vectors. Then $E{E}^T=I$. Differentiating givesso $W=\dot{E}{E}^T$ is skew-symmetric. Every real skew-symmetric $3\times 3$ matrix has the unique formwhich acts on a vector as $Wv=\mathbf{\omega }\times v$. Since $\dot{E}=WE$, each column satisfiesThe vector $\mathbf{\omega }$ is the instantaneous angular velocity. It may also be written $\mathbf{\omega }=\frac{1}{2}{\sum }_i{\mathbf{e}}_i\times {\dot{\mathbf{e}}}_i$.

For uniform density, the centroid is the volume average of position and coincides with the centre of mass:A point fixed in the body has ${\mathbf{r}}_{OQ}={\sum }_i{q}_i{\mathbf{e}}_i$ with time-independent body coordinates $q_i$. Since $O$ is fixed, part (a) givesThe centroid is fixed relative to the body as well, so ${\mathbf{v}}_G=\mathbf{\omega }\times {\mathbf{r}}_{OG}$. Using ${\mathbf{r}}_{OQ}={\mathbf{r}}_{OG}+{\mathbf{r}}_{GQ}$ therefore gives

Write $r=(x_1,x_2,x_3{)}^T$ for position relative to the centroid. The inertia tensor there isIn particular its diagonal entries integrate the squares of distances from the corresponding axes, and its off-diagonal entries are the negative products of coordinates.

Insert ${\mathbf{v}}_Q={\mathbf{v}}_G+\mathbf{\omega }\times r$ into $T=\frac{1}{2}{\int }_A\rho |{\mathbf{v}}_Q{|}^{2}dV$. The cross term vanishes because ${\int }_A\rho rdV=0$. Using $|\mathbf{\omega }\times r{|}^2=|\mathbf{\omega }{|}^2|r{|}^2-(\mathbf{\omega }\cdot r{)}^2$ givesIf $a$ is the vector from $G$ to $O$, positions relative to $O$ are $r-a$. Expanding the defining tensor integral with this substitution again removes the terms linear in $r$. Thus the tensor parallel axis theorem isSince $O$ is fixed, ${\mathbf{v}}_G=\mathbf{\omega }\times (-a)$, so $m|{\mathbf{v}}_G{|}^2=m{\mathbf{\omega }}^T(|a{|}^{2}I-a{a}^T)\mathbf{\omega }$. Substitution into the energy decomposition yields

Choose the $z$-axis along the cylinder and the origin at its midpoint, so $x^2+y^2\le b^2$ and $-a\le z\le a$. Reflection symmetry puts the centroid at the origin and makes all products of inertia zero. The mass isCylindrical integration gives ${\int }_A\rho x^{2}dV={\int }_A\rho y^{2}dV=m{b}^2/4$, while the uniform height coordinate has ${\int }_A\rho z^{2}dV=m{a}^2/3$. ConsequentlyThe two equal transverse entries reflect the rotational symmetry about the cylinder axis.