= Solution
A coaxial shell of radius $r$ and thickness $dr$ has <mass> $dm=2\pi\rho l r\,dr$. Integrating squared radius against <mass>, the <moment of inertia of a uniform solid cylinder> about its axis is
$$
I=\int_0^a r^2dm=2\pi\rho l\int_0^a r^3dr
=\frac{\pi\rho l a^4}2=\boxed{\frac12Ma^2},
\qquad M=\pi\rho a^2l.
$$
On the smooth incline, no tangential contact <force> produces spin, so the initially stationary cylinder simply slides with <acceleration> $g\sin\theta$. Therefore $t_{\rm slide}=\sqrt{2L/(g\sin\theta)}$.
For <rolling without slipping>, let $F$ be uphill <static friction> and $A$ the downward translational <acceleration>. <Newton's second law>, the rotational <torque> equation, and the contact constraint give
$$
MA=Mg\sin\theta-F,\qquad
Fa=I\alpha,\qquad A=a\alpha.
$$
With $I=Ma^2/2$, these give $F=MA/2$ and $A=2g\sin\theta/3$. The <rolling-time factor on a straight incline> is consequently
$$
\boxed{t_{\rm roll}=\sqrt{\frac{3L}{g\sin\theta}}
=\sqrt{\frac32}\,t_{\rm slide}.}
$$
The required friction is $Mg\sin\theta/3$; the stated no-slip motion presupposes sufficient <static friction>. At an instantaneously stationary contact point it does no work. Gravitational <potential energy> is shared between translational and rotational <kinetic energy>, explaining why rolling accelerates more slowly than the smooth slide.
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