- #1
phos19
- 6
- 0
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 $$\dot{\alpha}^2$$?