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Question on the equation for the period of a pendulum

  1. Sep 9, 2011 #1
    [a]1. The problem statement, all variables and given/known data[/b]

    essentially my "problem statement" is to determine what the mg in the equation T=2π√(I/mgL) is being used to represent


    2. Relevant equations
    T=2π√(I/mgL)


    3. The attempt at a solution

    in a nutshell, i solved a difficult problem in a way that my professor probably didnt intend for the problem to be solved---but i know I can get full credit if i correctly justify my answer. so the equation T=2π√(I/mgL) is for the period of a pendulum.

    my question is as follows: is the mg part of the equation being used to

    (1) to ONLY to represent Fg ?
    or
    (2) to represent the total downward forces acting on the pendulum?

    for my actual hw assignment, the pendulum is hanging from a sheet of charge, so in that case could i say that the equation for T is T=2π√(I/(Fg+Fe)L) ?

    when i do this i get the correct answer but as i said i need to be very specific in my proof/explanation for my answer since im sure this is an alternative method to solve the problem rather than using gauss' law. this method makes the most sense to me which is why i want to use it.
     
  2. jcsd
  3. Sep 10, 2011 #2
    First thing's first--mg is used to represent total downward force when only gravity is there.
    What you did is correct and your sir should accept it.However to calculate Force due to electric field you need to use Gauss law.you can't escape it.
     
  4. Sep 10, 2011 #3
    I got F_e without using Gauss' Law though i just used F_e=qE
    so is there an official way to rewrite T=2π√(I/mgL) as T=2π√(I/(Fg+Fe)L) ? Im thinking that if hes seen the equation before then ill have a better shot of convincing him. (last time i tried doing something like this i had the right answer and i knew my method would work, but because i couldnt adequately explain to him as to why my answer works he told me i had to use a different "regular" method)
     
  5. Sep 10, 2011 #4
    Is it a CONDUCTING CHARGED SHEET?
     
  6. Sep 10, 2011 #5
    that information is not given but why would it matter?
     
  7. Sep 10, 2011 #6
    It matters a lot.
    When the charged sheet is conducting then charge induced on sheet varies due to bob of pendulum varies with position.
    To understand more search the net for "method of images in electrostatics".
     
  8. Sep 10, 2011 #7
    ok so say it is conducting but the pendulum is only moving very very slightly...?

    im still a bit confused, why would the varying position matter if the formula for F is just F=qE= qσ/(2ε_0 ) and therefore not dependent upon r?

    (i read a bit on method of images just now but all the info im finding is referencing to a lot of theorems that are not part of basic electrostatics :/ )
     
  9. Sep 10, 2011 #8
    Perhaps because the equation you listed for the period is only valid for "small" angles of oscillations.
     
  10. Sep 10, 2011 #9

    vela

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    No. You can't just arbitrarily replace mg by Fg+Fe. You need a valid reason such a substitution is okay to do. "It gives me the right answer" isn't a valid reason.
    The expression [itex]E=\sigma/2\varepsilon_0[/itex] is only applicable to a sheet of constant charge density. With a conducting sheet, you don't have a constant charge density because the charges in the sheet will move in the presence of the charge on the pendulum.

    Forget about the method of images, etc. You're probably supposed to assume it's a non-conducting sheet of charge so that the charge density remains constant.
     
  11. Sep 10, 2011 #10
    "You can't just arbitrarily replace mg by Fg+Fe. You need a valid reason such a substitution is okay to do."

    ok so what would that valid reason be then?
     
  12. Sep 10, 2011 #11

    vela

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    Well, that's what your job is to figure out. Frankly, I don't think there is one.
     
  13. Sep 10, 2011 #12
    Why take it for granted that sheet is non-conducting???It changes everything and method of images is very easy to follow.
     
  14. Sep 10, 2011 #13
    The valid reason is that basic Newton's Laws of Gravitation and coulomb's law are similar and inter-convertible.Both are inverse square laws.You can solve problems of gravitation if you know the result of same problem in electrostatics just by replacing 1/4(pi)(epsilon) by G and charge by mass.Though the converse is not always possible because in electrostatics we have 2 elementary particles(+ve and -ve) while in gravitation we have only one type.However in case of similar charged systems(either all charges in system are positive or all negative) we can apply formula's derived for gravitation.The same is your case.Though more precise method should be (Fg+Fe)/m=geffthen use formula for simple pendulum where only change is g changes to geff.
     
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