Why Do We Use \(\ell + r\theta\) in Polar Coordinates for Analytical Mechanics?

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Discussion Overview

The discussion revolves around the use of the expression \(\ell + r\theta\) in polar coordinates within the context of analytical mechanics. Participants are exploring the geometric and physical reasoning behind this formulation, particularly in relation to potential and kinetic energy calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Homework-related

Main Points Raised

  • One participant questions why the length from the tangency point to the top equals \(\ell + r\theta\), seeking a geometric explanation.
  • Another participant explains that \(r\theta\) represents the arc length corresponding to the angle \(\theta\) and relates it to the initial length \(\ell\) of the rope.
  • A different participant expresses confusion about the relationship between \(\ell + r\theta\) and the tangent, asking for a simpler mathematical explanation.
  • There is a mention of kinetic energy being expressed as \(T = \frac{1}{2} m(\ell + r\theta)^2 \dot{\theta}^2\), with one participant indicating they need more time to analyze this term.
  • Another participant encourages self-reliance in problem-solving, suggesting that the diagram provided should help clarify the concepts involved.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the geometric reasoning behind the expression \(\ell + r\theta\) and its application in kinetic energy calculations. Multiple viewpoints and levels of understanding are present, indicating that the discussion remains unresolved.

Contextual Notes

Some participants express uncertainty about the geometric laws involved and seek clarification, highlighting potential gaps in understanding the relationship between the components of the expression.

TheDestroyer
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Thanks for integral, he made the potential.pdf analyse for an analytical mechanic but i still have 2 questions,

1- Why does the length up between the tengency point and top of (h) equals:

[tex]\ell + r\theta[/tex] ?

2- Why the kinetic energy here equals:

[tex]T = \frac{1}{2} m(\ell + r\theta)^2 \dot{\theta}^2[/tex] ?

I mean why we replaced R with [tex]\ell + r\theta[/tex] in the polar coordinates?

Thanks,

TheDestroyer
 

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[itex]r \theta[/itex] is the arc length corresponding to the angle [itex]\theta[/itex] Notice that if you let [itex]\theta = 2 \pi[/itex] you get the circumference of the circle.

As you unwind the rope moving the point of tangency through an angle [itex]\theta[/itex] you add the corresponding arc length ( [itex]r \theta[/itex] to the length of rope l which is initially hanging stright down.

I need some time to look at the kinetic engery term. I concentrated on the potential energy term and have not looked into the kinetic energy. I'll get back to you, if no one else contributes.
 
Last edited:
about the [tex]r\theta[/tex] i know it's the arc length (LOL I'm a second year university physics student), but the question is why does it equal to [tex]\ell + r\theta[/tex] in the tangent,

I didn't understand integral, I'm very sorry, please explain it as a math geometric laws, And try using a simple language (I don't mean you language was complicated),

And thanks very much
 
Why no body is answeringgggg?
 
You say you are a second year university student. Did you ever learn to ride a bike? Do you remember the the first time you were given a push and told to pedal.

There is nothing left in this problem which you should not be able to figure out on your own. Please study the diagram I drew for you and think about it. YOU CAN figure it out. Get 'er done
 
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