Suppose the curve $y(x)=b(x/a)^{\lambda}$ is rotated around the $y$ axis with constant frequency $\omega$, whilst a bead is moving along the curve (without friction). I want to find the frequency of small oscillations about the equilibrium point.
Given the Lagrangian I'm happy with finding the frequency of small oscillations, but I'm unsure about finding the Lagrangian.
So at any point the bead will have coordinates $r=(x,y(x),?)$. Then once I know what the $z$ component is I know to find the Lagrangian I just need to compute $T-U$ where $T$ is the kinetic energy of the bead and $U$ is its potential.
What will the $z$ coordinate be though?
If the bead is rotating around the $y$ axis, this is, so say, the "$z$" axis you are looking for. In cylindrical coordinates the lagrangian seems easy to write.
The bead has cycindrical coordinates $(\rho,\omega t,b(\rho/a)^\lambda)$ And from the line element
$\vec v=\dfrac{\mathrm d\vec r}{\mathrm dt}=(\dot \rho,\rho\omega,\lambda(b/a)(\rho/a)^{\lambda -1}\dot \rho)$
$T=(1/2)m(\dot \rho^2+\rho^2\omega^2+\lambda^2(b/a)^2(\rho/a)^{2\lambda-2}\dot \rho^2)=$
$=(1/2)m(\rho^2\omega^2+\left(1+\lambda^2(b/a)^2(\rho/a)^{2\lambda-2}\right)\dot \rho^2)$
$L=(1/2)m(\rho^2\omega^2+\left(1+\lambda^2(b/a)^2(\rho/a)^{2\lambda-2}\right)\dot \rho^2)-mgb(\rho/a)^{\lambda}$