= Solution
The continuous transformation $\theta\mapsto\theta+\epsilon$, $\theta_B\mapsto\theta_B+\epsilon$, $z\mapsto z+b\epsilon$, $r\mapsto r$ moves both endpoints by the same screw motion and preserves the fixed <helix>. Both $\Delta$ and $u$ are unchanged, so $\delta L=0$. <Noether theorem> therefore gives
$$
\boxed{\mathcal Q=p_\theta+p_{\theta_B}+bp_z=mr^2\dot\theta+M(a^2+b^2)\dot\theta_B+bm\dot z=\text{constant}.}
$$
This is the <Noether conserved quantity for a mechanical point symmetry> applied to <helical symmetry>. In physical Cartesian terms it equals \b[total $z$ <angular momentum> plus $b$ times total $z$ <linear momentum>]: the bead contributes $Ma^2\dot\theta_B+b(Mb\dot\theta_B)$. Neither pure axial <translation> nor pure axial <rotation> preserves the fixed coil separately. Their conserved combination is a consequence of the <helical symmetry>, not two independently conserved momenta. Substitution of the four equations of motion verifies $\dot{\mathcal Q}=-kar\sin\Delta+k(ar\sin\Delta+bu)-bku=0$.<Classical Dynamics by David Tong>[online chapter 2, §2.4.1, equations (2.76)–(2.79): continuous coordinate symmetry and its conserved momentum]{ref}
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