Derivation for Moment of Inertia of Rectangle rotated through center

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Homework Help Overview

The discussion revolves around deriving the moment of inertia for a rectangle rotated about an axis through its center, specifically for a rectangle with sides a and b. The original poster mentions the known equation as (1/12)M(a²+b²) but expresses difficulty in deriving it and seeks assistance.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to set up the problem using the equation dI = ∫r²dm and expresses uncertainty about the integration process and limits. Some participants question the setup and suggest clarifying the limits for the double integral.

Discussion Status

Participants are actively engaging with the problem, with one providing clarification on the need for a double integral. The original poster acknowledges learning about double integrals and identifies the limits for integration, indicating progress in understanding the problem.

Contextual Notes

The original poster notes a lack of prior experience with double integrals, which has contributed to their initial confusion. There is an emphasis on understanding the integration limits based on the rectangle's dimensions.

matthew487
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Homework Statement



I need to know how to derive the equation for the moment of inertia of a rectangle rotated about an axis through its center. The rectangle has sides a and b. I know the equation to be (1/12)M(a2+b2), but I am having trouble deriving it. I have searched all over the internet without finding any helpful solutions.

Homework Equations



dI = ∫r2dm
dm = λdA

The Attempt at a Solution



I started by taking a point mass of the rectangle to be the distance r from the center. This point would have the mass dm.

dm would then equal λdA
λ would equal M/A or M/ab
dA would be the area of the small point. I'm guessing this would be dydx

So, for dm we have (M/ab)(dydx)

Now for r2. By pythagorean thereom, r2 = x2+y2.

Substituting these into the equation ∫r2dm, we have (M/ab)∫(x2+y2)dydx

I'm pretty sure this is incorrect, but even if it is correct I have no idea how to integrate it. I don't even know what limits to integrate about.
 
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matthew487 said:

Homework Statement



I need to know how to derive the equation for the moment of inertia of a rectangle rotated about an axis through its center. The rectangle has sides a and b. I know the equation to be (1/12)M(a2+b2), but I am having trouble deriving it. I have searched all over the internet without finding any helpful solutions.

Homework Equations



dI = ∫r2dm
dm = λdA

The Attempt at a Solution



I started by taking a point mass of the rectangle to be the distance r from the center. This point would have the mass dm.

dm would then equal λdA
λ would equal M/A or M/ab
dA would be the area of the small point. I'm guessing this would be dydx

So, for dm we have (M/ab)(dydx)

Now for r2. By pythagorean thereom, r2 = x2+y2.

Substituting these into the equation ∫r2dm, we have (M/ab)∫(x2+y2)dydx

I'm pretty sure this is incorrect, but even if it is correct I have no idea how to integrate it. I don't even know what limits to integrate about.

Hi Matthew487...

Welcome to PF!

The item in red should be (M/ab)∫∫(x2+y2)dydx

The limit of the inner integral will be values 'y' will be taking .The limit of outer integral will be values 'x' will be taking.

Considering origin to be at the center .What should be the limits of the inner integral ?What should be the limits of the outer integral ?
 
Ahhhh! Thank You! I had never done or even heard of a double integral before. That was the problem. After quickly learning what that is, I was able to solve it. For the inside dx integral the limits are from -a/2 to a/2. The outside integral for dy is from -b/2 to b/2. Thanks a ton!
 
Excellent! :thumbs:
 

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