Moment of inertia of a thin, square plate

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    Moment of inertia
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SUMMARY

The discussion focuses on the calculation of the moment of inertia (I_x) for a thin square plate using integrals. Participants highlight the importance of correctly expressing the distance squared (r^2) in terms of x and y, specifically r^2 = x^2 + y^2, to apply the perpendicular axis theorem effectively. There are concerns regarding the accuracy of the integrals presented, including the presence of an extra dx and incorrect limits of integration for x. The conversation emphasizes the necessity of summing mass elements over the entire volume to derive the moment of inertia accurately.

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simphys
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Homework Statement
the moment of inertia about an axis through the center of and perpendicular to a uniform, thin square plate. mass M and dimension L x L.
Relevant Equations
d
I don't really understand what the 2 integrals (dx and dxdy) for I_x represent. Could I get some explanation here please? Thanks in advance.
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The second line in the derivation of ##I_x## has one ##dx## too many. Is that what bothers you?
 
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Also, the second line is missing the ##r^2## and the limits of integration for ##x## are not correct. Keep in mind that this integral represents the moment of inertia about the x-axis, and the x-axis lies in the plane of the thin plate. Think about how to express ##r^2## in terms of ##x## and/or ##y##.
 
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If you put in the correct $$r^2=x^2+ y^2$$ in the second line then you see that the x and y integrals give you the "perpendicular axis theorem" without issue. Please redo this.
 
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TSny said:
Also, the second line is missing the ##r^2## and the limits of integration for ##x## are not correct. Keep in mind that this integral represents the moment of inertia about the x-axis, and the x-axis lies in the plane of the thin plate. Think about how to express ##r^2## in terms of ##x## and/or ##y##.
Yep, that is what I did I expressed in terms of y^2 and then went ahead with that. But I just didn't understand how this solution came about basically.
 
hutchphd said:
If you put in the correct $$r^2=x^2+ y^2$$ in the second line then you see that the x and y integrals give you the "perpendicular axis theorem" without issue. Please redo this.
this is the solution. not mine :)
 
So where is your attempt? How are we to help?
 
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hutchphd said:
So where is your attempt? How are we to help?
well.. that is exactly what I don't understand..
I didn't understand why the solution has used two integrals in such a way.. I haven't used a volume integral or smtn. I did differently by using a mass element and summing over that with one definite integral.
 
One must sum each mass element (mass density times volume element) over the entire volume of the object and scaled by the square of the distance to the chosen axis . The world is three dimensional and so is the integral. Show us your work.
 

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