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

AI Thread Summary
The discussion centers on calculating the kinetic energy of a falling stick without friction. It highlights that the kinetic energy should be the sum of rotational kinetic energy about the center of mass and translational kinetic energy of the center of mass. The absence of friction means there is no horizontal motion, simplifying the analysis. Participants emphasize the need for a clear problem statement to facilitate effective assistance. Overall, the kinetic energy expression involves both rotational and translational components under ideal conditions.
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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 $$\dot{\alpha}^2$$?
 
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K.E would be sum of rotational and translational kinetic energy. Gravity is a conservative force, and if there's no air resistance and assuming that no energy dissipation happens due to heat and sound etc, then K.E of falling stick = Rotational K.E about it's Center of mass + Translational K.E of the center of mass.
 
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