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

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Homework Help Overview

The discussion revolves around a physics problem involving a falling stick with no friction, focusing on the calculation of kinetic energy. The subject area includes concepts of rotational and translational motion, as well as energy conservation principles.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the components of kinetic energy, including both rotational and translational aspects. There are attempts to express kinetic energy in terms of moment of inertia and angular velocity, while questioning the validity of these expressions. Some participants seek clarification on the problem's context and details.

Discussion Status

The discussion is ongoing, with some participants providing insights into the nature of kinetic energy in the context of the problem. However, there is a lack of clear problem statement from the original poster, leading to requests for more information to facilitate better assistance.

Contextual Notes

There appears to be missing information regarding the specific setup of the problem, which is causing confusion among participants. The absence of a clear problem statement is noted as a barrier to effective discussion.

phos19
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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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