= Solution
{c}
Apply the general <linear momentum> equation $d\mathbf p/dt=\sum\mathbf F$ to the whole <rigid body>. Its center of mass has <acceleration> $\ddot x\mathbf e_1$, and each rail contributes its own <static friction> and <normal force>. Hence
$$
m\ddot x\mathbf e_1=(T\cos\alpha-2F)\mathbf e_1+(T\sin\alpha+2N-mg)\mathbf e_2,
$$
so
$$
\boxed{m\ddot x=T\cos\alpha-2F,\qquad N=\frac{mg-T\sin\alpha}{2}.}
$$
Here $N$ is the <normal force> at each rail, not the combined support.
Take <torques> about the center of mass, with positive <angular velocity> along $\mathbf e_3$. The thread leaves the lower side shown in the source diagram: its contact <position vector> relative to the center is
$$
\mathbf q=a\sin\alpha\,\mathbf e_1-a\cos\alpha\,\mathbf e_2.
$$
Thus $\mathbf q\times\mathbf T=aT\mathbf e_3$. Each contact point has relative <position vector> $-R\mathbf e_2$, so each <friction> <torque> is $-RF\mathbf e_3$. The <gravitational weight> acts through the center of mass and the <normal forces> have zero <torque> about it. The <angular momentum> equation becomes
$$
\boxed{\mathcal I_{33}\ddot\phi=aT-2RF.}
$$
The condition of <rolling without slipping> makes the contact point instantaneously stationary:
$$
\mathbf v_P=\dot x\mathbf e_1+\dot\phi\mathbf e_3\times(-R\mathbf e_2)=(\dot x+R\dot\phi)\mathbf e_1=0.
$$
Integrating and using $x(0)=\phi(0)=0$ gives $x=-R\phi$. Thus rightward <translation> corresponds to clockwise <rotation>. Eliminating $F$ from the two dynamical equations with $\ddot x=-R\ddot\phi$ yields
$$
(\mathcal I_{33}+mR^2)\ddot\phi=aT-RT\cos\alpha,
$$
and therefore
$$
\boxed{\ddot\phi=-\frac{T(R\cos\alpha-a)}{mR^2+\mathcal I_{33}}.}
$$
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2021/prelims/mathematics-iv-resit-cylinder-forces.png]
{title=Cylinder side view, showing the tension moment arm and the combined forces from both rails}
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