Proof for moment of Inertia of thin disc

AI Thread Summary
The discussion focuses on the calculation of the moment of inertia for a thin disc with a hole, where the inner radius is R1 and the outer radius is R2. The initial approach involved integrating the mass distribution, but the user realized their calculation was incorrect, arriving at I = M(R2^2 - R1^2)/4 instead of the correct I = M(R2^2 - R1^2)/2. The error stemmed from a misunderstanding of the area element, where the user initially used dA = πr.dr instead of the correct dA = 2πr.dr. This realization led to a humorous self-reflection on the mistake. The discussion highlights the importance of correctly defining area elements in physics calculations.
Redoctober
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Homework Statement



Its actually not a homework , i am just curious about this

I have disc with a hole in middle , R1 is inner radius , R2 is outer radius , Mass M is the disc's mass . Let density of disc uniform .

I did it this way . -

λ.dA=dm
λ.pie.r.dr=dm as Area=pie.r^2

Let substute in the equation I = ∫( R^2).dm

therefore I = ∫( p.pie.r^3 ).dr
remove constants outside the integral therefore I =λ.pie∫r^3.dr
integrate (0 to R )to get I = λ.pie.r^4/4
using Mass = λ*pie*r^2

therefore i get I = (MR^2)/4 for thin Disc

As i have a hole therefore its I = M(R2^2-R1^2)/4

But sadly this is incorrent :/ . It seems that I = M(R2^2-R1^2)/2 is the correct solution

Please Help ! :O why my way is wrong :(
 
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you wrote,

λ.pie.r.dr=dm

Shouldn't there be a 2 in there? dA = 2.pie.r.dr ?
 
Spinnor said:
you wrote,

λ.pie.r.dr=dm

Shouldn't there be a 2 in there? dA = 2.pie.r.dr ?

Oh my god "Slap myself in face" Lol such a silly error :S ! .
 

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