= Solution
Keep the signs of part <11/b/i>. The <tension> has horizontal component $T\cos\theta$ but still has counterclockwise <torque> $Tr$, since it is tangent to the axle. Thus
$$
Ma=T\cos\theta+f,\qquad I\alpha=Tr+Rf,\qquad a=-R\alpha.
$$
Solving these equations gives the <critical pulling angle of a rolling spool> law
$$
\boxed{a=\frac{TR(R\cos\theta-r)}{I+MR^2},\qquad
f=-T\frac{I\cos\theta+MrR}{I+MR^2}.}
$$
The zero-acceleration angle is therefore
$$
\boxed{\theta_c=\cos^{-1}\left(\frac rR\right).}
$$
At this angle the line of action of the pull passes through the ground contact, so its <torque> about that point vanishes. A spool initially at rest stays at rest while the contact constraints hold; one already rolling has zero <acceleration>, rather than being brought to rest.
For equilibrium the vertical <normal force> is $N=Mg-T\sin\theta_c$ and $|f|=T\cos\theta_c$. The <static friction> inequality $|f|\le\mu N$ becomes
$$
T(\cos\theta_c+\mu\sin\theta_c)\le\mu Mg.
$$
Hence, for $0<r<R$,
$$
\boxed{T_{\max}=\frac{\mu Mg}{\cos\theta_c+\mu\sin\theta_c}
=\frac{\mu MgR}{r+\mu\sqrt{R^2-r^2}}.}
$$
At this limit the <normal force> is still nonnegative; for finite $\mu$ and positive $r$ it is positive. For $\theta_c<\theta\le\pi/2$, the numerator of $a$ is negative. Released from rest, the spool rolls leftward, \b[away from the pulling hand], with counterclockwise rotation, provided the assumed <rolling without slipping> and ground contact remain possible. The nonzero <acceleration> is not obtained by assuming that <friction> simply opposes the centre's motion.
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