How Does Pendulum Motion Affect Work and Kinetic Energy Calculations?

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Homework Help Overview

The discussion revolves around deriving work and kinetic energy for a pendulum using the integral of force over displacement, particularly focusing on the gravitational force acting on the pendulum. Participants are exploring the implications of directionality in their calculations, especially when the angle θ is at the vertical.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants attempt to derive kinetic energy using the work-energy theorem and question the sign of the results based on the direction of displacement and force. Some express confusion about why different paths yield seemingly contradictory results regarding kinetic energy.

Discussion Status

There is an ongoing exploration of the relationship between work done and changes in kinetic energy. Some participants suggest using gravitational potential energy to simplify calculations, while others are questioning the assumptions made about force direction and displacement. Multiple interpretations of the integral and its implications are being discussed without reaching a consensus.

Contextual Notes

Participants are grappling with the implications of integrating force in different directions and how this affects the interpretation of kinetic energy changes. There are references to specific angles and the behavior of the pendulum, indicating a focus on the nuances of the problem setup.

  • #31
no nevermind, i think it works if you consider the dot product a distribution over a negative differential.
 
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  • #32
stateofdogma said:
no nevermind, i think it works if you consider the dot product a distribution over a negative differential.
Not sure what you mean by that, but to be clear, no, ds does not mean abs(ds). If the range "a to b" is such that the variable of integration, x say, decreases from 'lower' bound a to 'upper' bound b then dx is negative. E.g. ##\int_a^b f.dx = -\int_b^a f.dx##.
 
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