Solution
= Solution
Let $r$ be the cylindrical radius of the point of contact and $\phi$ its azimuth. Rolling without slip or twist gives $|\omega|=v/a$ and therefore
$$
L=\frac{7m}{10}(\dot r^2\csc^2\alpha+r^2\dot\phi^2)-mgr\cot\alpha,
$$
up to an irrelevant additive constant. The factor $7/5$ combines translational and rotational inertia.