Moment of Inertia of Disc through diameter

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SUMMARY

The discussion focuses on calculating the moment of inertia (MOI) of a disc through its diameter. The user successfully derived the MOI using a strip method, expressing the differential moment of inertia as dI = (1/3)dm*h², and integrated from -a to a. The confusion arises regarding the interpretation of the mass element dM, which is described as a small square mass at a distance x from the axis. The correct approach involves recognizing that the MOI contributions from different axes can be summed, leading to the conclusion that Iz = Ix + Iy for the entire disc.

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Homework Statement



The problem is attached in the picture.


The Attempt at a Solution



I managed to solve it using a different method. I have no idea what the answer is talking about..

My method

Found dI of a strip = (1/3)*dm*h2

then i replace h by x, then integrate from -a to a to get the same answer.

But as for the answers, I have no clue.
 

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As far as I can tell, x in the answer is perpendicular to the axis against which the moment is being determined. I think you can figure out the rest.
 
voko said:
As far as I can tell, x in the answer is perpendicular to the axis against which the moment is being determined. I think you can figure out the rest.

The problem is, I can't fingure out what is the shape of their dM.

It looks as if they are taking a small square mass dM that is distance x away from axis (perpendicularly)

so I = x2 dM

but dM = 2σ √(a2-x2)

It looks as if √(a2-x2) is referring to the distance along the axis from the origin..But how does that make any sense? r in ∫ r2 dM refers to distance from the axis and not the origin.

Then are they using dM as a strip or a small square mass?
 
They are taking the entire strip parallel to the axis at distance x from it.
 
To understand the general statement, it may be easier to consider only a small mass element, dm, of the disc placed at coordinate x, y (with z being zero since its a plate). This one mass element contributes to the MOI around the x-axis as dIx = y2dm, to the MOI around the y-axis as dIy = x2dm, and to the MOI around the z axis as dIz = (x2+y2)dm = x2dm + y2dm, which, using the two previous equations, means that dIz = dIy + dIx (make a diagram if you have trouble seing this). Since this is valid for all the mass element of the planar disc it must also be valid for the sum (or rather the integral) of all the mass elements MOI, that is, Iz = Ix + Iy.
 
voko said:
They are taking the entire strip parallel to the axis at distance x from it.

Yup, I finally figured that out. Initially I worked it out by using a strip perpendicular to the axis with an MOI of dI = (1/3)x2 dM
 

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