Lagrangian function of a sphere on a cylindrical surface

Question:
There is an open cylinder of radius $R$., which is stationary. A hollow sphere of radius $\rho$ and mass $m$ rolls on the surface without slipping. Find the Lagrangian function.

Answer:
The Lagrangian function is defined as
$$L = KE - PE$$
where $KE$ and $PE$ represent kinetic energy and potential energy, respectively.
The mechanical potential energy is the gravitational force $\times$ height, which is taken from the center. The height varies with the angle $\theta$. Thus,
$$PE = -mg(R-\rho)\cos\theta$$
The negative sign indicates the height is directed to the center.
Now the kinetic energy has two parts, translational and rotational kinetic energies.
$$KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$$
The translational velocity, $v$, is along with the trace of circle whose radius is $R-\rho$. Therefore,
$$v = (R-\rho)\omega$$
The angular velocity can be expressed by
$$\omega = \theta' = \frac{d\theta}{dt}$$
The moment of inertia of hollow sphere is
$$I = \frac{2}{3}m\rho^2$$
Plug everything into above.
$$L = \frac{1}{2}m(R-\rho)^2 \theta '^2 + \frac{1}{2}\frac{2}{3}m\rho^2 \theta '^2 + mg(R-\rho)\cos\theta$$
We can rewrite it as
$$L = \frac{m\theta'^2}{6}(3R^2 -6R\rho + 5\rho^2)+ mg(R-\rho)\cos\theta$$

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