# Falling stick problem (no friction): What is the kinetic energy?

phos19
Since there is no friction : $$m \ddot{x} = 0$$ (no x motion).

For the kinetic energy , I've tried: $$K = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m v^2_cm = 1/2 I_{cm} \dot{\alpha}^2 + 1/2 m \dot{z}^2$$ . Giving me a weird expression , shouldn't the kinetic energy just be half the the moment of inertia about the contact point times $$\dot{\alpha}^2$$?