How Do You Calculate Velocity on a Circular Path Using Polar Coordinates?

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Discussion Overview

The discussion revolves around calculating velocity on a circular path using polar coordinates, specifically in the context of a homework problem. Participants explore the application of energy conservation principles and the correct interpretation of given parameters.

Discussion Character

  • Homework-related

Main Points Raised

  • One participant mentions the equation for arc length in polar coordinates, s = r * theta, and references energy conservation with the equation 1/2 mv^2 = mgh.
  • Another participant clarifies that the radius r is given as 6m, suggesting that this is distinct from the arc length.
  • A participant expresses relief at confirming the radius value and questions whether their approach using energy conservation is correct.
  • One participant asks about the origin of the factor of 1/2 in the energy conservation equation, indicating a need for clarification on the formula.
  • A later reply corrects a previous statement about the energy equation, indicating that the factor of 1/2 was mistakenly included.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the application of the energy conservation approach, and there are corrections made regarding the formulation of the equations used.

Contextual Notes

There are unresolved questions regarding the application of energy conservation and the interpretation of the parameters provided in the problem. The discussion reflects uncertainty about the correct formulation of the equations.

nysnacc
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Homework Statement


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Homework Equations


s= r* theta
1/2 mv^2 = mgh energy conservation

The Attempt at a Solution


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I think I might be having something wrong in the attempt. And how do you do using the polar coordinate approach?

14.103_03.PNG
14.103_04.PNG
 
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Hey

The question gives r as 6m, not the arch length as 6m?

Cant help with much else tho! :)
 
OHHHH yes. r = 6m, so I can save time for getting that, good catch :)
And am I doing well on the energy conservation approach?
 
I can't see your working very clearly on this tablet I'm using but where does the 1/2 come from in 1/2*m*g*h
 
CWatters said:
I can't see your working very clearly on this tablet I'm using but where does the 1/2 come from in 1/2*m*g*h
I wrote it wrong, it shouldn't be 1/2 there
 

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