For pure translation, integrate the <slender-body force density> over each rod and take its moment about $O$. The three coordinate directions give
$$
\begin{array}{c|c|c}
\mathbf U&\mathbf F/(CL U)&\mathbf G/(CL^2U)\\ \hline
U\mathbf e_x&(5,0,0)&(0,1,0)\\
U\mathbf e_y&(0,5,0)&(1,0,0)\\
U\mathbf e_z&(0,0,5)&(0,0,0).
\end{array}
$$
Thus
$$
\mathbf F=5CL\mathbf U,
\qquad
\mathbf G=CL^2(U_y,U_x,0)
$$
when $\boldsymbol\Omega=0$.

Let
$$
K=\begin{pmatrix}0&1&0\\1&0&0\\0&0&0\end{pmatrix},
\qquad
D=\begin{pmatrix}13/3&0&0\\0&13/3&0\\0&0&4/3\end{pmatrix}.
$$
The symmetry proved in part (a) determines the force generated by rotation from the translation-generated couple. Combining this with the given rotational resistance gives the complete matrix
$$
\boxed{
\begin{pmatrix}\mathbf F\\\mathbf G\end{pmatrix}
=C
\begin{pmatrix}
5L I&L^2K\\
L^2K&L^3D
\end{pmatrix}
\begin{pmatrix}\mathbf U\\\boldsymbol\Omega\end{pmatrix}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 329 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-329/2/c/solution