Finding Angular Momentum Along x-Axis for t given z(0) = 0, ˙z(0)=0

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Homework Help Overview

The discussion revolves around finding the angular momentum along the x-axis given specific initial conditions for the variable z, namely z(0) = 0 and ˙z(0) = 0. The problem appears to involve dynamics and possibly the effects of gravity, as indicated by the equations presented.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are attempting to formulate the angular momentum as a function of time, while also questioning the completeness of the problem statement. There are discussions about the correct use of LaTeX for mathematical expressions and the relevance of certain variables like R and s.

Discussion Status

There is an ongoing exploration of the problem, with participants seeking clarification on the problem's context and completeness. Some have pointed out potential errors in reasoning and formatting, while others are trying to understand the implications of the angular momentum in relation to the x-axis.

Contextual Notes

Participants have noted that the problem may be part of a larger context, and there is a call for a more complete problem statement to facilitate better assistance. There are also mentions of confusion regarding the variables involved and the physical interpretation of angular momentum in this scenario.

MyoPhilosopher
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Homework Statement
Finding the angular momentum along an axis given the eqs of motion
Relevant Equations
$${\ddot{z} = \frac{g}{1+(\frac{4R}{s})^2}}$$
with s being the distance along z axis after a revolution
given z(0) = 0 as well as

˙z(0)=0​

How would one find the angular momentum along the x-axis in terms of t.
Currently, I have formulated the following:


$${\ddot{z} = \frac{g}{1+(\frac{4R}{s})^2}}$$
 
Last edited:
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Seems like this is part of a more complete problem. State that.
MyoPhilosopher said:
Currently, I have formulated the following:
$$\ddot{z} = \frac{g}{1+(\frac{2\pi R}{k})^2}$$
Currently you have ##z(t)=0##.

And a { too many in your ##\TeX## :wink:
 
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BvU said:
Seems like this is part of a more complete problem. State that.

Currently you have ##z(t)=0##.

And a { too many in your ##\TeX## :wink:
Thanks for clearing that up I was trying to understand the Latex code -
 
Currently you have z(t)=0z(t)=0. [edit] no, ##a\,t^2##

And enclose ##\LaTeX## in double $$ or (## for in-line )
 
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Thanks for the help I hope it is now readable.
My question is how do I formulate the angular momentum from the world of the x-axis as a function of time.
 
Maybe you missed this:
BvU said:
Seems like this is part of a more complete problem. State that.
It means: please provide a complete problem statement.

From the PF guidelines:
micromass said:
Reproduce the problem statement accurately.
Type the problem statement exactly as worded. If you're only asking about one part of a long problem it may not be necessary to type up the entire problem, but you need to ensure you've provided the proper context for the sub-problem. If you paraphrase or summarize, make sure you're not changing the meaning or omitting important information. It's very frustrating trying to help with a problem only to discover that critical information is missing.
No idea why you have a picture with x,y,z, formulas with z only.
No idea about R, s (k?, ##\lambda##?),

MyoPhilosopher said:
My question is how do I formulate the angular momentum from the world of the x-axis as a function of time
I don't think the world of the x-axis has angular momentum. Usually ##\vec L = \vec r \times \vec p## :wink: .
 
BvU said:
Maybe you missed this:

It means: please provide a complete problem statement.

From the PF guidelines:

No idea why you have a picture with x,y,z, formulas with z only.
No idea about R, s (k?, ##\lambda##?),I don't think the world of the x-axis has angular momentum. Usually ##\vec L = \vec r \times \vec p## :wink: .
Yep sorry I was trying to understand an issue rather than a paper problem. It was an error in my thinking I realize now. Please feel free to remove this q I don't seem able to.
 

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