= Quadratic-bend turning with finite body mass
{title2=$\theta_\infty=2UA\,\frac{C}{C+M_b(B_1-M_0)}$}
Put $B_j=[m(x)(x-\bar x)^j]_0^L$, $M_j=\int m(x)(x-\bar x)^jdx$, $A=\int\alpha(t)dt$, 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 $m(x)=m_*x^2$, it becomes $2UA/(1+2M_b/m_*)$. Body deformation inertia affects <derivative> forcing during the stroke, not this final integral relation when the deformation and its time <derivative> return to zero.
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