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 $$\dot{\alpha}^2$$?

 

[BvU]

Hi,

What is this about ? I only see shreds of a problem ...

https://www.physicsforums.com/threads/homework-help-guidelines-for-students-and-helpers.686781/

 

[NTesla]

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.

 

[kuruman]

If you are seeking help, please help us help you and provide a statement of the problem or ask a specific question. Most of us lack even the most basic mind reading skills.