Boundary expression for transverse filament thrust Created 2026-10-05 Updated 2026-10-06
The anisotropic-drag contribution to axial force from a small-slope transverse motion is . Using the small-slope elastohydrodynamic filament equation and integration by parts givesThus this propulsive contribution is determined by endpoint slope, bending moment, and shear. A separately imposed axial translation adds ordinary longitudinal drag. Exact second-order longitudinal motion required by inextensibility also contributes to instantaneous total drag, but its time-derivative contribution averages to zero over a deformation cycle.
Past exam of the mathematics course of the University of Cambridge 2017 iii Paper 342 1 b Solution Created 2026-10-03 Updated 2026-10-06
Use the downward-positive displacement and the filament bending modulus and load density defined above. Two applications of integration by parts give the first variationThus the elastic-plus-gravitational force density is . In resistive-force theory, transverse motion has drag per unit length . Neglecting filament and fluid inertia, instantaneous force balance gives the forced small-slope elastohydrodynamic filament equationClamping fixes position and tangent, while the free end has zero bending moment and zero transverse shear force. ThereforeThe last two conditions also follow as natural boundary conditions from the variation, since the free-end values of and are arbitrary. Distributed weight does not add a concentrated force or moment at the tip.
At steady state, . Integrating first from the free end gives and . Integrating twice more with the clamp conditions gives the uniform-load bending of a clamped filament:In particular the free-end displacement is and the total buoyancy-corrected gravitational load is . Thus the tip stiffness of a uniformly loaded cantilever isThe last expression uses as Young's modulus; if itself is flexural rigidity, the same result reads . This is Hooke's law for the specified loading pattern. It is not the stiffness for a point force at the tip, which would be ; replacing a distributed force by its resultant does not preserve its bending moment distribution.
For dimensional analysis, force/length squared, length to the fourth power, hence force times length squared. Consequently force/length, as required for a spring constant. Also force times time/length squared makes , and all forces per unit length. Neither viscosity nor density enters the stiffness, although viscous drag controls the rate of approach to the steady shape. The small-slope approximation requires , since .
Past exam of the mathematics course of the University of Cambridge 2019 iii Paper 334 2 b Solution Created 2026-10-03 Updated 2026-10-05
In the small-slope Monge representation, , , and the leading transverse velocity is . With no imposed axial prestress, the induced filament tension is second order in transverse amplitude, so its transverse contribution is higher order. The transverse component of force balance reduces to the small-slope elastohydrodynamic filament equationAn externally imposed zeroth-order tension would instead add .
For the transverse motion, the off-diagonal part of the resistive-force theory tensor gives axial propulsive force density to second order. Its integral isSubstitute the bending equation and integrate by parts:The boundary expression for transverse filament thrust is thereforeOnly endpoint slope, bending moment, and shear enter this expression. Exact inextensibility also generates second-order longitudinal material motion; if that motion is retained in instantaneous total axial drag, it adds . For a periodic deformation with fixed axial base position, that additional term has zero cycle average. Thus the displayed formula is the transverse propulsive contribution and also gives the cycle-averaged propulsive force.