Finding angular velocity using work energy theorem

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Homework Help Overview

The discussion revolves around finding angular velocity using the work-energy theorem, specifically in the context of rotational motion and potential energy. Participants are exploring the relationship between angular acceleration, potential energy, and the equations governing rotational dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss two methods for calculating angular velocity, noting that one method fails due to the non-constant nature of angular acceleration. Questions arise regarding the validity of using certain potential energy expressions, particularly the confusion surrounding the term Mθ.

Discussion Status

The discussion is ongoing, with participants expressing uncertainty about specific variables and the correctness of their approaches. Some guidance has been offered regarding the potential energy expression, but there is no consensus on the correct method or interpretation of the variables involved.

Contextual Notes

There appears to be confusion regarding the definition of variables, particularly the meaning of M in the context of the problem. Additionally, participants are grappling with the implications of gravitational effects on their calculations.

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Homework Statement


cmPVt.png


Homework Equations


Inertia about pin = ml^2/3
Work done by rotation = M*theta = 0.5(Inertia about pin) *ω2

The Attempt at a Solution


I tried two different methods and both came out to be the same answer. However they did not correspond to any of the multiple choice answers... What am I doing wrong?

zsts9.jpg


Thanks!
 
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Method 1: The angular acceleration varies with \cos(\theta). So using constant acceleration formulas doesn't work. You simply can't use that method.

Method 2: The potential energy is still U=mgh, not U=M\theta.
 
frogjg2003 said:
Method 1: The angular acceleration varies with \cos(\theta). So using constant acceleration formulas doesn't work. You simply can't use that method.

Method 2: The potential energy is still U=mgh, not U=M\theta.

Thanks, but why does Mtheta not work?
 
I'm not even sure what M is.
 
frogjg2003 said:
I'm not even sure what M is.

Moment about the pin.
 
It's still wrong. It should be \frac{1}{2}M\omega. But that by itself contains no information about the fact it's in gravity.
 

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