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Gravitational Field of a Uniform Rod

  1. Oct 3, 2014 #1
    1. The problem statement, all variables and given/known data

    A thin uniform rod of mass M and length L is centered at the origin and lies along
    the x axis. Find the gravitational field due to the rod at all points on the x axis in the
    region
    Picture adjunct

    2. Relevant equations
    My two question are:
    Why r is'not the difference between Xp and L/2 (although that's the conclusion) but the diference between Xp and Xs?
    What's the process to integrate the function ? i cannot find it

    3. The attempt at a solution
    Solutions adjunct
     

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    Last edited: Oct 3, 2014
  2. jcsd
  3. Oct 3, 2014 #2

    BvU

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    Xp - L/2 is a fixed number. The idea is to let r vary between Xp minus L/2 and Xp plus L/2 and to add up the contributions from all the dm on the way; in other words: to integrate.

    The process to integrate is to add up all the little contributions dg from all the little chunks dm that are at distance r from p, with r varying from Xp - L/2 to Xp + L/2

    But now I've said the same thing twice.
     
  4. Oct 3, 2014 #3
    A fixed number, thank you i can live with that, but i wasn't referring to the meaning of integration but to which method i can use, i cannot get to the same result, again thank you very much (if i have to go to the mathematics section, im sorry, its my first day).
     
    Last edited: Oct 3, 2014
  5. Oct 3, 2014 #4

    BvU

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    Welcome to PF ! :)
    If you say you can't get to the same result, you'll have to post some of your own work: the textbook pictures show the situation and the steps taken to get to ##\vec g##. Where on the way does it become obscure to you ?

    And I did try to answer your question "Why r is'not the difference between Xp and L/2". But to me it is so obvious from the picture that I find it hard to clarify further...
     
  6. Oct 3, 2014 #5
    In step 5, i don't know who the author goes from the integral to the result, i have some work from the classroom but i dont know if it is correct, my integration skills are somewhat rusty. Thanks, i'm glad to find a place to solve my doubt
     

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  7. Oct 3, 2014 #6

    BvU

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    Well, my eyesight is really letting me down. Can't distinguish much on the picture.
    But I take it you can do ##\int {dy\over y^2}## and this looks pretty similar....
     
  8. Oct 3, 2014 #7
    but that should result in -(1/y) isnt it ? and that wouldn't eliminate the L which is dividing GM (step 5), i have in those notes that the integrate should be something like

    [(Xp-L/2)-(Xp+L/2)]/(Xp-L/2)^2 is this correct ? or how should it be ?
     
  9. Oct 5, 2014 #8

    BvU

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    You can check if a primitive is correct by differentiating. There is no need to write out a referendum. Part (Important part !) of physics skills is integration skills; part of integration skills (Important part !) is differentiating skills. You need them, so the investment is well worth it (and then you don't have to ask 'is this correct' at every occasion).

    You will probably find that your primitive 1/(xp-xs) was correct, but what you then did with it [(Xp-L/2)-(Xp+L/2)]/(Xp-L/2)^2 is not correct (and it also doesn't correspond with what the author gets in the denominator).
     
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