= Solution
The <moment of inertia of a uniform solid cylinder> is taken about its symmetry axis. Write its <mass density> as $\rho_c=m/(\pi a^2L)$. A coaxial shell of radius $r$ and thickness $dr$ has <mass> $dm=2\pi\rho_c Lr\,dr$. Thus
$$
\boxed{I=\int_0^a r^2\,dm=2\pi\rho_c L\int_0^a r^3\,dr
=\frac12ma^2.}
$$
A hollow cylinder would have a different <moment of inertia>; uniform solid density is the intended assumption.
Let $s$ measure the centre's displacement uphill from its initial <position>, and choose signed <angular velocity> $\omega$ so that uphill <rolling without slipping> has $\dot s=a\omega$. Let $F$ be the signed contact <force> uphill. The translation and rotation equations are
$$
m\ddot s=-mg\sin\alpha+F,\qquad I\dot\omega=-aF.
$$
The minus sign in the <torque> equation follows from the chosen rolling sense: uphill contact <force> opposes the positive spin. Under <rolling without slipping>, $\ddot s=a\dot\omega$. Eliminating $F$ gives
$$
\left(m+\frac I{a^2}\right)\ddot s=-mg\sin\alpha,
\qquad \ddot s=-\frac23g\sin\alpha,
\qquad F=\frac13mg\sin\alpha.
$$
Using $s(0)=0$ and $\dot s(0)=V$, the no-slip motion and maximum vertical rise are
$$
\boxed{s(t)=Vt-\frac13g\sin\alpha\,t^2,\qquad
h_{\max}=\frac{3V^2}{4g}.}
$$
Indeed the centre turns at $t=3V/(2g\sin\alpha)$, reaching $s_{\max}=3V^2/(4g\sin\alpha)$, whose vertical component is the boxed height. <Conservation of energy> independently gives the same answer from initial <kinetic energy> $mV^2/2+I(V/a)^2/2=3mV^2/4$.
The available <static friction> is at most $\mu mg\cos\alpha$. It can supply the required uphill <force> exactly when $mg\sin\alpha/3\leq\mu mg\cos\alpha$. Therefore
$$
\boxed{\tan\alpha>3\mu\quad\Longrightarrow\quad\text{slip starts}.}
$$
Interpret the stated friction magnitude as limiting <static friction> and as <kinetic friction> during slip, with the same coefficient. Take $0<\alpha<\pi/2$, $\mu\geq0$. In the slipping regime the contact point moves downhill relative to the slope, so <kinetic friction> acts uphill. To verify this direction rather than assume it unchecked, solve with $F=\mu mg\cos\alpha$. Define $b=g(\sin\alpha-\mu\cos\alpha)>0$. The result is
$$
\boxed{s(t)=Vt-\frac12bt^2,\qquad
v(t)=\dot s=V-bt,\qquad
\omega(t)=\frac Va-\frac{2\mu g\cos\alpha}{a}t.}
$$
The centre's <speed> is $|V-bt|$, with the sign of $v$ specifying ascent or descent. The contact point's relative <velocity> is
$$
w(t)=v(t)-a\omega(t)=-gt(\sin\alpha-3\mu\cos\alpha)<0\qquad(t>0).
$$
Thus the assumed uphill <kinetic friction> is consistent throughout both ascent and descent, including when the centre's <velocity> changes sign. There is no later return to <rolling without slipping> before reaching the bottom. This is the characteristic <slipping solid cylinder on an incline> regime.
The centre turns at $t_*=V/b$, giving
$$
\boxed{h_{\max}=\frac{V^2\sin\alpha}{2g(\sin\alpha-\mu\cos\alpha)}.}
$$
The formulas apply up to the positive return time $t_b=2V/b$. Substitution into the signed <velocity> and <angular velocity> gives
$$
\boxed{v(t_b)=-V,\qquad\text{speed at return}=V,\qquad
\omega(t_b)=\frac Va\frac{\sin\alpha-5\mu\cos\alpha}{\sin\alpha-\mu\cos\alpha}.}
$$
The return spin can have either sign and vanishes at $\tan\alpha=5\mu$. It reverses when $3\mu<\tan\alpha<5\mu$, although the contact point still slips downhill. The unchanged centre <speed> does not mean that no <energy> was dissipated: the rotational <kinetic energy> has decreased. Indeed integrating the frictional power $Fw<0$ gives
$$
K(0)-K(t_b)=\frac{2\mu mV^2\cos\alpha(\sin\alpha-3\mu\cos\alpha)}{(\sin\alpha-\mu\cos\alpha)^2}\geq0,
$$
which agrees with the difference of the initial and final rotational <kinetic energies>.
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