= Solution
Let $q=\sigma/L$, with $L>0$, and let $Z=X^1+iX^2$. Direct differentiation gives $\ddot Z=Z''=-Z/L^2$, while the time coordinate also satisfies the <wave equation>. Using the <Minkowski metric>,
$$
\dot X\cdot X'=0,\qquad \dot X^2=-\cos^2q,\qquad X'^2=\cos^2q.
$$
Thus both <Virasoro constraints> hold, and the <induced worldsheet metric> is $g_{\mu\nu}=\cos^2q\,\operatorname{diag}(-1,1)$. The configuration solves the <Nambu–Goto action> equations in its interior.
Place the fixed end at $\sigma=0$. The first free endpoint has $Z'=e^{it/L}\cos q=0$ at $q=\pi/2$. For the usual single unfolded <rotating Nambu–Goto string>,
$$
\boxed{0\leq\sigma\leq\frac{\pi L}{2},\qquad
\ell_{\mathrm{proper}}=\int_0^{\pi L/2}|Z'|\,d\sigma=L.}
$$
This parameter interval need not equal the earlier interval $[0,\pi]$, since that interval was a coordinate convention. The <proper length> is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal <Lorentz contraction>.
All points have the same phase $t/L$, hence the segment rotates rigidly with <angular velocity> $\Omega=1/L$. At radius $r=L\sin q$, its <speed> is $v=r/L$. Therefore \b[the free tip moves at the <speed of light>]. This is consistent with its <Neumann boundary condition>: $X'=0$ and the <Virasoro constraints> imply $\dot X^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.
In this <conformal gauge>, $P_m=T\dot X_m$, so the <energy density> per $\sigma$ is $-P_0=T$. Equivalently, the <energy density> per <proper length> is $T/\sqrt{1-r^2/L^2}$. Thus
$$
E=T\int_0^{\pi L/2}d\sigma=\frac{\pi TL}{2},\qquad
E_{\mathrm{rest}}=TL,
$$
and the <rotational kinetic energy> is
$$
\boxed{E_{\mathrm{rot}}=\left(\frac\pi2-1\right)TL
=\left(1-\frac2\pi\right)E.}
$$
The <angular momentum> about the fixed point is
$$
J=\int d\sigma\,(X^1P_2-X^2P_1)
=TL\int_0^{\pi L/2}\sin^2(\sigma/L)\,d\sigma
=\frac{\pi TL^2}{4}.
$$
Therefore the <Regge trajectory> is
$$
\boxed{E=\frac{\pi TL}{2},\qquad J=\frac{E^2}{\pi T},\qquad\beta=\frac1{\pi T}.}
$$
The endpoint conditions alone also admit folded extensions with parameter length $(n+\tfrac12)\pi L$, $n\geq0$. Their <proper length>, <energy> and <angular momentum> are $(2n+1)$ times the values above, and $\beta=1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.
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