Quadratic-bend turning with finite body mass (${\theta }_{\infty }=2UA\frac{C}{C+M_b(B_1-M_0)}$)

Put $B_j=[m(x)(x-\bar{x}{)}^j{]}_0^L$, $M_j=\int m(x)(x-\bar{x}{)}^{j}dx$, $A=\int \alpha (t)dt$, and $C=B_0(B_2-M_1)-B_1(B_1-M_0)$. For quadratic bending $h_0=-\alpha (t)(x-\bar{x}{)}^2$, the linear recoil correction in elongated-body theory with a stable post-stroke relaxation gives the displayed angle. It follows by setting time derivatives to zero in the integrated lateral and angular momentum equations. The simpler $2UA$ is exact in the zero-body-mass limit, or when $B_1=M_0$, but is not a generic finite-body-mass identity. For $L=1$, $\bar{x}=1/2$, and $m(x)=m_{\ast }{x}^2$, it becomes $2UA/(1+2M_b/m_{\ast })$. Body deformation inertia affects derivative forcing during the stroke, not this final integral relation when the deformation and its time derivative return to zero.