= Solution
With the wedge fixed, let $r$ be the signed gap between the cylinder bottom and the surface, positive outwards. The centre height is $s\sin\alpha+(R+r)\cos\alpha$. Retaining all three coordinates before imposing constraints gives
$$
L=\frac12M_c(\dot r^2+\dot s^2)+\frac12I_c\dot\theta^2
-M_cg[s\sin\alpha+(R+r)\cos\alpha].
$$
Use the <holonomic constraints> $f_1=r=0$ and $f_2=s+R\theta-s_0=0$. With the convention $d(\partial L/\partial\dot q)/dt-\partial L/\partial q=\sum_j\lambda_j\partial f_j/\partial q$, the <Lagrange multiplier> equations are
$$
M_c\ddot r+M_cg\cos\alpha=\lambda_1,\qquad
M_c\ddot s+M_cg\sin\alpha=\lambda_2,\qquad
I_c\ddot\theta=R\lambda_2.
$$
The constraint gradients identify $\lambda_1=N$, the <normal reaction>, and $\lambda_2=f$, the uphill <static friction> force. The latter supplies the rolling <torque>. Setting $\ddot r=0$ and $\ddot\theta=-\ddot s/R$ yields
$$
\boxed{N=M_cg\cos\alpha,\qquad \ddot s=-\frac23g\sin\alpha,\qquad f=\frac13M_cg\sin\alpha.}
$$
The <static friction> condition $|f|\leq\mu_sN$ therefore becomes
$$
\boxed{\mu_s\geq\frac13\tan\alpha,}
$$
for $0\leq\alpha<\pi/2$. <Static friction> is required to create rotation, but does not dissipate <energy> here because the contact point is instantaneously at rest on the fixed plane.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2019-rolling-cylinder.png]
{title=Coordinates for a rolling cylinder on a moving wedge and the normal, friction and weight forces for the fixed-wedge case}
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