Past exam of the theoretical physics I course of the University of Cambridge / 2018 / 2 / a / Solution

Use the downward $y$ direction of the PDF diagram. The bob's Cartesian coordinates areThusThe hinge and rod are massless, so only the bob contributes kinetic energy:Because gravity points along positive $y$, its gravitational potential energy is $-mgy=-mg\ell \cos \theta$. The zero-rest-length spring has elastic potential energy $k{s}^2/2$. Put $s=\ell \eta$ and divide $T-V$ by the positive constant $m{\ell }^2$. The equivalent Lagrangian isMultiplying the action by a nonzero constant leaves its stationary paths unchanged. This derivation is a direct use of kinetic energy from Cartesian position vectors; no independent kinetic term should be assigned to the massless support.