 Critical pulling angle of a rolling spool

For a spool of outer radius $R$, axle radius $r$, mass $M$ and axial moment of inertia $I$, a lower-tangent pull of tension $T$ at upward angle $\theta$ produces horizontal acceleration $a=TR(R\cos \theta -r)/(I+M{R}^2)$ while rolling without slipping. This follows by combining $Ma=T\cos \theta +f$, $I\alpha =Tr+Rf$ and $a=-R\alpha$, with rightward translation and counterclockwise rotation positive. The acceleration reverses at ${\theta }_c$; zero acceleration means equilibrium only if initially at rest. At that angle the static friction limit gives $T_{\max }=\mu Mg/(\cos {\theta }_c+\mu \sin {\theta }_c)$.