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How can I derive J=pi*D^4/32 using integration?

  1. Jul 2, 2016 #1
    1. The problem statement, all variables and given/known data
    D1=6cm D2=2cm ( i have to prove that 2nd moment of area (J) of a circulal plate abouts its polax axis(zz) is equal to piD^4/32 )

    2. Relevant equations


    3. The attempt at a solution
    J=pi6^4/32 - pi2^4/32 = 125.66

    J=r1^2x2pirdr
    J=Integral 2pir^3d2 = 2pi integral(high 3 , low 1) r^3dr=2pi (r^4/4)(high3 ,low1)
    =2pi ((3^4/4)-(1^4/4) = 2pi (20.25 - 0.25) = 2pi x20 = 125.66 (answer is correct ) but i need to show how piD^4/32 is derrived using integration
     
  2. jcsd
  3. Jul 2, 2016 #2

    SammyS

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    Hello MrKriss. Welcome to PF !

    The way you have written your mathematics makes it very difficult to understand your post.

    After some experimentation on my part I see that you do literally mean
    (π D4 ) / 32 .​

    Is the disk lying in the x-y plane and centered at the origin, with the axis of rotation being the z axis ?

    By the way: The 32 does not seem to be correct.
    .
     
    Last edited: Jul 2, 2016
  4. Jul 2, 2016 #3
    Hya thanks for reply i know its is all in one line not rly sure how to add symbols here etc . The disk is laying flat with outside diameter 6cm and inner diamter 2cm.
    the answe to your question is yes(it is hollow) . the 32 part means i have to prove that 2nd moment of area "J" abouz zz axis is equal to PI*D^4/32 hence calculate the 2nd moment of area for given disk.
     

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  5. Jul 2, 2016 #4

    SteamKing

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    I think the easiest approach is to start with the definition of polar moment of inertia:

    ##J = \int_A r^2\; dA##

    for just a circular area. You can calculate J in terms of the radius of the circle, and then use the substitution r = D/2 to find J in terms of D.

    Once you have derived J for a solid circle, then you can use the additive property of integrals to remove the circle in the middle.
     
  6. Jul 2, 2016 #5
    Thank you both for reply and help i found how to derive that... which was suprisingly obvious and simple he he
    pi*(d^4)/32
    =
    pi^((2*r)^4)/32
    =
    (16*pi*r^4)/32
    =
    (pi*r^2^2)/2

    then i can simply integrate values. Thanks
     
  7. Jul 2, 2016 #6

    SteamKing

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    I'm confused by that last statement.

    Deriving (as in establishing from first principles [i.e. not taking derivatives]) means that you are supposed to use integration and the geometry of a circle to develop the formula J = π ⋅ r4 / 2
     
  8. Jul 3, 2016 #7
    Sorry about my mistake been sitting too long at this task i assume, but i think i have solved it so when i derrive (PI*d^4)/32 i get (PI*d^3)/8 then i integrate 6(high), 2(low)for (PI*d^3)/8 and i aquaire antiderrivative which i do not need which is PI*d^4/32 + C and the correct answer of Definite integral which is 40PI or 125.66

    Thank you and sorry for confusion earlier ,confused myself .Thanks
     
  9. Jul 3, 2016 #8

    SteamKing

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    No, you still haven't understood this problem.

    There is no point to taking the derivative of J and then integrating to get J back. Differentiation and integration are inverse operations of one another.

    As I mentioned before, you don't start with J = π ⋅ D4 / 32, but by setting up the proper integral over a circular area, you should obtain either the formula J = π ⋅ D4 / 32 or J = π ⋅ R4 / 2

    The proper way to prove these formulas is to start with the definition of J:

    ##J=∫_A\,r^2\;dA##

    and the region A is a circular area of radius R.

    In English, the word "derive" does not mean "to take the derivative". "Derive" means "to reason by deduction" or "to develop a series of mathematical arguments which lead to the desired result."
     
  10. Jul 3, 2016 #9
    Yes thank you for correcting me . done as u previously mentiond and started from definition of J overall got to the answer which is 20piad =62....... and got my answer exatly as needed. have it attached if any has interest . im just having hard time as i missed some classes regarding these. i do believe it is solved. just noticed the picture is made too small .....
     

    Attached Files:

  11. Jul 3, 2016 #10

    SteamKing

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    I'll take your word for it.

    Attaching little thumbnail images which are smaller than 100 K bytes is generally a waste of time. Nobody can read these small images without going blind.

    You'll have much better responses if you type out your work or make a pdf and attach it.
     
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