Moment of Inertia for Square Plate on X Axis

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SUMMARY

The moment of inertia for a square uniform plate about the x-axis can be calculated using the mass element dm = σdxdy, where σ represents the surface density. The integration involves calculating the integral of x²dm over the area of the plate, specifically from -h/2 to +h/2. By leveraging symmetry, one can simplify the calculation by integrating from 0 to h/2 and multiplying the result by two. This approach streamlines the process compared to calculating the moment of inertia about the z-axis.

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  • Knowledge of mass density and its application in physics
  • Basic principles of symmetry in integration
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fightnchikn
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helllo, I am having trouble finding the moment of inertia for a square uniform plate with length h on the sides...how do you go about it when the interia is about the x axis...the z axis is perpendicular to the square in this case... i understand how to do it when it is with respect to the z axis...the double intergral and all but when its on the xaxis i get confused...how do you set that up.
 
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fightnchikn said:
helllo, I am having trouble finding the moment of inertia for a square uniform plate with length h on the sides...how do you go about it when the interia is about the x axis...the z axis is perpendicular to the square in this case... i understand how to do it when it is with respect to the z axis...the double intergral and all but when its on the xaxis i get confused...how do you set that up.
It should be easier to do it for the x-axis than for the z axis. I assume the origin is in the center of the plate, with the x-axis being a line that divides the plate into two equal rectangles. Consider a mass element dm = σdxdy and integrate x^2dm over the area of the plate. The y integral is very simple since there is no function of y in the integrand. You just get h, which is constant for all x. All that is left is the integral of x^2.
 
Now when I integrate what do I integrate from 0 to h?
 
fightnchikn said:
Now when I integrate what do I integrate from 0 to h?
If the axis is as I assumed, it is from -h/2 to +h/2. By symmetry, you could do twice the integral from 0 to h/2 and get the same result.
 

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