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When the block is free, $x$ becomes a dynamical generalized coordinate and its kinetic energy $M{\dot{x}}^2/2$ must be included. Expanding to quadratic order about the downward stationary state and dropping a constant givesThe coordinate $x$ is cyclic. Its conserved canonical momentum is the total horizontal momentum,The two linear Euler-Lagrange equations areEliminating $\ddot{x}$ givesThus the pendulum with a freely sliding support has a zero-frequency translational mode and an oscillatory mode. With four independent constants $X_c,V,C,D$, the general solution isHere $V=P/(M+m)$ and $X_c+Vt$ is the horizontal centre of mass coordinate to first order, since $X_{\mathrm{CM}}=x+[ml/(M+m)]\theta$. The oscillation is faster than that of a pendulum with a fixed pivot; the limit $M\to \infty$ restores $\Omega \to \sqrt{g/l}$.

The time-independent Lagrangian conserves $E=T+V$. Substitute $\dot{x}=-m_{2}l\dot{\varphi }\cos \varphi /M$ into the kinetic energy to obtainThis is the angle-dependent effective inertia of a pendulum with a freely sliding support. Solving for the speed givesOn a monotone increasing segment, the travel time is thereforeThe printed oriented integral assumes that increasing-angle branch. For decreasing motion use $|d\varphi |$ or interchange the limits; if a turning point lies between the two events, split the trajectory into monotone segments and add their positive times. The numerator remains positive for positive masses. At an ordinary turning point the inverse-square-root singularity is integrable; reaching the unstable upright separatrix takes infinite time.