Force and torque freedom of the cell--helix pair give
$$
(A+A_0)U+B(\Omega+\omega)=0,
$$
$$
BU+(D+D_0)\Omega+D\omega=0.
$$
Writing
$$
\Delta=(A+A_0)(D+D_0)-B^2,
$$
solution of this linear system yields
$$
\boxed{U=-\frac{BD_0}{\Delta}\omega,
\qquad
\Omega=\frac{B^2-(A+A_0)D}{\Delta}\omega.}
$$
The body counter-rotates relative to the motor, and the swimming direction reverses with the helix handedness through $B$.


Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 355 / 2 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-355/2/c/solution