Calculating Rotational Inertia of a Rectangular Prism

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Homework Help Overview

The discussion revolves around calculating the rotational inertia of a rectangular prism with uniform mass distribution, specifically about an axis through one corner and perpendicular to the larger faces. The original poster presents their calculations and reasoning regarding the moment of inertia, including the use of integrals and the relationship between mass, volume, and density.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants examine the contributions of differential mass elements to the moment of inertia and question the dependence of the moment of inertia on the dimension c. There are discussions about the correct interpretation of the distance from the axis of rotation and how it affects the calculations.

Discussion Status

The discussion is active, with participants providing insights and questioning assumptions made in the original calculations. Some participants suggest reconsidering the approach based on the axis of rotation and the implications of fixed density versus total mass on the moment of inertia.

Contextual Notes

There is a mention of the absence of a definitive answer in the textbook, which raises questions about the correctness of the calculations presented. The discussion also highlights the importance of assumptions regarding density and mass in determining the moment of inertia.

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Homework Statement


There is a rectangular prism of uniform mass distribution with lengths of a, b, and c (b>a>c). Calculate it's rotational inertia about an axis through one corner and perpendicular to the large faces.

Homework Equations


I = \int r^2 dm
r^2 = x^2 + y^2 + z^2
\rho = \frac{M}{V}
V = abc

The Attempt at a Solution



I am examining a cubic differential mass of dm's contribution on the moment of inertia about the axis of rotation. The radius between dm and the axis of rotation is r^2 = x^2 + y^2 + z^2. The density, \rho, is constant which is \frac{M}{V}, so dm = \rho dV.

I = \int r^2 dm = \int (x^2 + y^2 + z^2) \rho dV
I = \rho \iiint_V x^2 dV + y^2 dV + z^2 dV = \int^a_0 \int^b_0 \int^c_0 x^2 dzdydx + \int^a_0 \int^b_0 \int^c_0 y^2 dzdydx + \int^a_0 \int^b_0 \int^c_0 z^2 dzdydx [/itex]<br /> I = \frac{\rho}{3} ( a^3 bc + ab^3 c + abc^3)<br /> I = \frac{M}{3abc} ( a^3 bc + ab^3 c + abc^3)<br /> I = \frac{M}{3} (a^2 + b^2 + c^2)<br /> <br /> This problem looked cool so I did it, but it was an even one so there is no answer in the back of the book. Does this look right?
 
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Is this the situation? (red is axis of rotation)
1657047568503.png

If this is the case, the moment of inertia can not depend on c.
There is another, simpler, shape that you can use instead...
 
The classical mistake when calculating MoIs: ##r## should not be the magnitude of the position vector of the element dV, but its distance perpendicularly to the axis of rotation.
 
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drmalawi said:
Is this the situation? (red is axis of rotation)
View attachment 303793
If this is the case, the moment of inertia can not depend on c.
There is another, simpler, shape that you can use instead...
Just to add a side note: This depends on what is considered given. If the density is known and fixed, then the MoI will depend on c because larger c means more mass. If the density is instead adjusted such that the total mass ##M## is known, then indeed the MoI will be independent of c.
 
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Orodruin said:
This depends on what is considered given
That's true, the c-dependece will show up in M if a fixed density is given (also M would have a and b dependence M = ρabc)
 

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