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

[MrKriss]

Homework Statement

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 )

Homework Equations

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

 

[SammyS]

Homework Statement

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 )

Homework Equations

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

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.
.

 

[MrKriss]

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.
.

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.

 

[SteamKing]

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.

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.

 

[MrKriss]

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

 

[SteamKing]

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

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

 

[MrKriss]

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

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

 

[SteamKing]

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

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."

 

[MrKriss]

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 . I am 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 ...

 

[SteamKing]

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 . I am 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 ...

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.