= Solution
\b[Constraint and example.] A <holonomic constraint> can be written $f_\alpha(q_1,\ldots,q_n,t)=0$, with no irreducible dependence on <velocities>. A velocity relation is holonomic if it integrates to such an equation. For example, a particle constrained to a stationary circle satisfies $x^2+y^2=\ell^2$, reducing its two Cartesian <degrees of freedom> to one angle. A <nonholonomic constraint> cannot be integrated in this way.
\b[Kinematics and kinetic energy.] Set $d=R-r>0$. In the plane of the original diagram, choose the orbital angle $\theta$ positive counterclockwise from vertically downward. The marked radius of the inner cylinder then has absolute orientation $\alpha=\theta-\phi$ measured from the same downward direction. Its signed spin <angular velocity> is $\Omega=\dot\theta-\dot\phi$, while
$$
\mathbf X_G=d(\sin\theta,-\cos\theta),\qquad |\dot{\mathbf X}_G|=d|\dot\theta|.
$$
The <moment of inertia of a uniform solid cylinder> is $I=mr^2/2$. Applying the <kinetic energy decomposition about the center of mass> gives
$$
\boxed{T=\frac12md^2\dot\theta^2+\frac12I\Omega^2
=\frac m2(R-r)^2\dot\theta^2+\frac m4r^2(\dot\phi-\dot\theta)^2.}
$$
\b[No slip.] With tangential unit vector $\mathbf e_\theta=(\cos\theta,\sin\theta)$, the inner contact point has <velocity> $(d\dot\theta+r\Omega)\mathbf e_\theta$. The fixed wall requires this to vanish. Hence
$$
d\dot\theta+r(\dot\theta-\dot\phi)=0,\qquad R\dot\theta-r\dot\phi=0.
$$
Because $r,R$ are constant, integration gives
$$
\boxed{r\phi-R\theta=0,\qquad \Omega=-\frac{R-r}{r}\dot\theta,}
$$
where the integration constant is zero with the marked-dot reference in the diagram and continuously unwrapped angles. This is a <holonomic constraint>, even though it was first obtained from a condition on <velocities>.
\b[Oscillations.] The <gravitational potential energy> above the lowest position is $V=mgd(1-\cos\theta)$. Eliminating $\phi$ gives the reduced <Lagrangian>
$$
L=\frac34md^2\dot\theta^2-mgd(1-\cos\theta).
$$
The <Euler-Lagrange equation> becomes
$$
\frac32md^2\ddot\theta+mgd\sin\theta=0.
$$
Thus the bottom is a stable equilibrium, and for $|\theta|\ll1$ there is <simple harmonic motion>, with
$$
\boxed{\omega_0=\sqrt{\frac{2g}{3(R-r)}},\qquad
f_0=\frac1{2\pi}\sqrt{\frac{2g}{3(R-r)}}.}
$$
The extra rotational <kinetic energy> reduces the frequency compared with a particle sliding without friction along the same circular path. The <small oscillations inside a circular cylinder> formula applies with inertia ratio $\gamma=1/2$.
\b[Friction and energy.] Let $F_t$ be the signed tangential contact force in the $\mathbf e_\theta$ direction. Applying <Newton's second law> to translation and the <torque> equation to rotation gives
$$
md\ddot\theta=-mg\sin\theta+F_t,\qquad
I\dot\Omega=rF_t.
$$
Using the <rolling without slipping> condition yields
$$
\boxed{F_t=\frac13mg\sin\theta.}
$$
Except at special instants, <static friction> is needed to produce the spin acceleration. There is no sliding at the contact point, whose instantaneous <velocity> is zero; therefore this force does no total work. Its translational and rotational powers cancel: $F_td\dot\theta+F_tr\Omega=0$. The stationary <normal reaction> likewise does no work. With a rigid wall, adequate <static friction>, and no rolling resistance, <conservation of energy> gives
$$
\boxed{E=\frac34md^2\dot\theta^2+mgd(1-\cos\theta)=\text{constant}.}
$$
\b[Angular momentum.] The cylinder's spin <angular momentum> $I\Omega$ is generally not conserved because the contact friction exerts a <torque> about its <center of mass>. About the fixed outer-cylinder center $O$, the <angular momentum decomposition about the centre of mass> gives
$$
L_O=md^2\dot\theta+I\Omega=md\left(d-\frac r2\right)\dot\theta.
$$
Gravity and friction exert the external <torque>
$$
\dot L_O=-mgd\sin\theta+RF_t
=mg\left(r-\frac{2R}{3}\right)\sin\theta.
$$
Thus \b[the rolling cylinder does not generally conserve angular momentum], although it conserves mechanical <energy>. There is a special geometric cancellation at $R=3r/2$: then $L_O$ and its torque vanish identically for this planar motion. Angular momentum of a larger isolated system including the wall and Earth is a separate conservation statement.
\b[Scope of the holonomic reduction.] The integrated constraint is exact at every angle for planar <rolling without slipping> with sustained contact; it is not restricted to <small oscillations>. Completing a full turn requires tracking angle winding rather than reducing each angle separately modulo $2\pi$. This global bookkeeping does not make the local velocity relation nonintegrable.
However, arbitrary initial conditions need not sustain this ideal rolling motion. The inward <normal reaction> and the required friction are
$$
N=m(g\cos\theta+d\dot\theta^2),\qquad |F_t|=\frac13mg|\sin\theta|.
$$
Contact requires $N\geq0$, and a finite coefficient $\mu_s$ of <static friction> requires $|F_t|\leq\mu_sN$. If either fails, the cylinder detaches or slips and the one-coordinate rolling <Lagrangian> no longer applies. Slipping contact allows $\theta$ and $\phi$ to be independent; detachment removes the fixed center-distance constraint. These changes alone do not establish a <nonholonomic constraint>. Allowing general three-dimensional reorientation can introduce nonintegrable rolling constraints, but that motion is outside the diagram's two-angle planar model. Thus \b[the stated planar rolling constraint is holonomic throughout its regime of validity; it cannot be imposed on every physically possible motion].
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