# Lagrangian Bead

**Classical Mechanics**Level pending

On a track that follows the contour \(r = \theta\), a bead of mass \(1 \text{ kg}\) begins at position \(\theta = \frac{9 \pi}{2}\). How long in seconds does it take for the bead to roll back around to a height of \(\frac{9 \pi}{2}\)? In this universe, \(g = 1 \frac{\text{m}}{\text{s}^2}\), and there is no friction between the bead and the track.