Calculating Rotational Inertia of a Rectangular Prism

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The discussion focuses on calculating the rotational inertia of a rectangular prism with uniform mass distribution about an axis through one corner. The moment of inertia is derived using the integral of the squared distance from the axis of rotation, leading to the formula I = (M/3)(a^2 + b^2 + c^2). A key point raised is that the moment of inertia's dependence on the dimension c varies based on whether density is fixed or adjusted to maintain a constant mass. If density is constant, the moment of inertia will depend on c, while if mass is fixed, it becomes independent of c. The calculations and considerations presented highlight the complexities involved in determining rotational inertia for varying conditions.
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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)
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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