= Solution
For a <Lagrangian> $\mathcal L=T-U$, the <Euler-Lagrange equation> is $d(\partial\mathcal L/\partial\dot q)/dt=\partial\mathcal L/\partial q$ for each unconstrained generalized coordinate. For example, $\mathcal L=m\dot x^2/2-kx^2/2$ gives $m\ddot x+kx=0$ for a spring oscillator.
For a uniform solid sphere, slicing into disks normal to the rotation axis gives
$$
I=\frac\rho2\int_{-R}^R\pi(R^2-z^2)^2\,dz
=\frac{8\pi\rho R^5}{15}=\frac25mR^2.
$$
Use total dumb-bell mass $M$ and $\ell=L-r$. Since both centres lie on its spin axis, its <moment of inertia> about that axis is $2MR^2/5$. If $\theta$ measures displacement from the bottom of the track, <rolling without slipping> imposes spin speed $|\dot\phi|=\ell|\dot\theta|/r$. The <rolling inside a fixed circular cylinder> therefore has
$$
\mathcal L=\frac12M\ell^2\left(1+\frac{2R^2}{5r^2}\right)\dot\theta^2-Mg\ell(1-\cos\theta).
$$
The equation of motion and its small-amplitude limit are
$$
\ddot\theta+\frac{g\sin\theta}{\ell(1+2R^2/(5r^2))}=0,\qquad
\boxed{\omega=\sqrt{\frac g{L-r}}\left(1+\frac{2R^2}{5r^2}\right)^{-1/2}.}
$$
With $g=9.81\,\mathrm{m\,s^{-2}}$ and the given lengths, $\boxed{T=2\pi/\omega\simeq80.5\,\mathrm s}$. If the centre rises by height $h$, <conservation of energy> at the bottom gives
$$
Mgh=\frac12M\left(1+\frac{2R^2}{5r^2}\right)v_{\max}^2,
\qquad\boxed{v_{\max}\simeq0.0349\,\mathrm{m\,s^{-1}}\quad(h=0.010\,\mathrm m).}
$$
The speed expression uses exact height-energy conversion within the ideal rolling constraint; only the period formula uses the small-angle approximation.
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