Solution (source code)

= Solution
{c}

Put $D=mR^2+\mathcal I_{33}=m(R^2+a^2/2)$. The <angular acceleration> is constant. The initial rest conditions and $x(0)=\phi(0)=0$ give
$$
\boxed{\phi(t)=-\frac{T(R\cos\alpha-a)}{2D}t^2,\qquad x(t)=\frac{TR(R\cos\alpha-a)}{2D}t^2.}
$$
For the upward rightward pull in the diagram, $0\leq\alpha\leq\pi/2$, the sign of $x(t)$ is the sign of $R\cos\alpha-a$. Consequently \b[it rolls towards the horizontal pull when $0\leq\alpha<\arccos(a/R)$]. At $\alpha=\arccos(a/R)$ it remains at rest, with balanced <forces> and <torques>. For $\arccos(a/R)<\alpha\leq\pi/2$, it rolls to the left. In particular \b[a horizontal pull, $\alpha=0$, makes it roll to the right], since $R>a$. If other pulling <angles> are allowed, the general condition for motion in the same horizontal direction as the pull is $\cos\alpha(R\cos\alpha-a)>0$.

The required <static friction> at each rail follows from the translational equation:
$$
\boxed{F_{\mathrm{req}}=\frac{T(\mathcal I_{33}\cos\alpha+maR)}{2D}.}
$$
For $0\leq\alpha\leq\pi/2$ it is positive, agreeing with the leftward <force> convention. The assumed <rolling without slipping> is physically possible only if the rails can supply this <force> while maintaining contact. With <coefficient of static friction> $\mu_s$, the necessary contact inequalities are
$$
N=\frac{mg-T\sin\alpha}{2}\geq0,\qquad |F_{\mathrm{req}}|\leq\mu_sN.
$$
For the usual pulling range, these become
$$
\boxed{T\sin\alpha<mg,\qquad T(\mathcal I_{33}\cos\alpha+maR)\leq\mu_sD(mg-T\sin\alpha).}
$$
If either fails, the proposed motion requires lift-off or more <static friction> than the rails provide. Equality $T\sin\alpha=mg$ is also incompatible here: it gives $N=0$ but a nonzero required <friction>. The question supplies no <coefficient of static friction>, so it determines the required <friction> exactly but gives no numerical slip threshold. If $F$ is regarded as a prescribed <force> rather than an available contact response, it must equal $F_{\mathrm{req}}$ for the <rolling constraint> to hold.