 Frictionless sliding ladder

For a uniform slender ladder of length $2l$ sliding while touching a smooth wall and floor, use its angle $\theta$ from the vertical. Its centre of mass is $(l\sin \theta ,l\cos \theta )$ and its central moment of inertia is $m{l}^2/3$. Hence the Lagrangian is $L=2m{l}^2{\dot{\theta }}^2/3-mgl\cos \theta$, and its Euler-Lagrange equation is $\ddot{\theta }=3g\sin \theta /(4l)$. The contacts are physically possible only while their normal forces are nonnegative. If released from rest at ${\theta }_0$, energy conservation gives ${\dot{\theta }}^2=3g(\cos {\theta }_0-\cos \theta )/(2l)$, and the wall force becomes $3mg\sin \theta (3\cos \theta -2\cos {\theta }_0)/4$. Contact is lost at $\cos \theta =2\cos {\theta }_0/3$, after which the constrained equation no longer describes the motion.