What is the rotational inertia of a cube when rotated about an edge?

AI Thread Summary
The discussion centers on determining the rotational inertia of a uniform density cube when rotated about an edge. Participants suggest using the parallel axis theorem and integrating the mass distribution to derive the inertia. The formula for rotational inertia is discussed, with emphasis on converting variables into Cartesian coordinates for integration. Clarifications are made regarding the relationship between the inertia of a cube and that of a square about a corner. The conversation concludes with suggestions for further resources to aid in understanding the calculations involved.
montreal1775
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Does anyone know what the rotational inertia of a cube of uniform density is when it is rotated about an edge? Any help is appreciated!
 
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montreal1775 said:
Does anyone know what the rotational inertia of a cube of uniform density is when it is rotated about an edge? Any help is appreciated!

I suggest you start with the definition of rotational inertia of a rigid body of uniform density.
 
I=integral(r^2,m)

but then what do I do?
I'd use the parallel axis theorem but I don't know how to find the rotational inertia for a cube about it's center of mass.
 
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montreal1775 said:
I'd use the parallel axis theorem but I don't know how to find the rotational inertia for a cube about it's center of mass.

You need to be able to derive the equation; you can't look it up (in your textbook, etc.)?
 
<br /> I = \int r^2 \,dm = \rho \int r^2 \,dV<br />
where \rho is the density of the cube (assumed to be uniform).

You can then convert r into a Cartesian equivalent, then split dV into dx, dy, dz and do a triple integral.
 
I think you should consider the cube from above a plane with greater mass--that is-- 1/12(m)(2L^2). Then parallel axis it.
 
Saketh said:
<br /> I = \int r^2 \,dm = \rho \int r^2 \,dV<br />
where \rho is the density of the cube (assumed to be uniform).

You can then convert r into a Cartesian equivalent, then split dV into dx, dy, dz and do a triple integral.

So would it be <br /> I = \iiint_{0}^{L} x^2+y^2+z^2 \,dxdydz<br /> where L is the length of a side?
 
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Also, Ja4Coltrane, I don't understand what you mean. Could you please elaborate?
 
r is the distance from the axis of rotation, not from the centre. If you are rotating about the z axis, then r^2 = x^2 + y^2.

Re Ja4Coltane's post, possibly he means the inertia of the cube about one edge is essentially the same as the inertia of a square about one corner. The thickness contributes to the mass, but it doesn't affect the geometric part of the formula.
 
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So is I = \rho \iiint_{0}^{L} x^2+y^2+z^2 \,dxdydz the rotational inertial of a cube rotated about a corner?
 
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What do you mean by "the rotational inertial about the origin"? I know what rotation inertia about a line is, and I know what the rotation inertia tensor about a point is (and it's got 6 independent components, not just one).

The rotation inertia about the edge defined by (x=0, y=0) is I = \iiint_{0}^{L} x^2+y^2 \,dxdydz That's very similar to the inertia of a square about one corner.
 
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montreal1775 said:
Does anyone know what the rotational inertia of a cube of uniform density is when it is rotated about an edge? Any help is appreciated!

You can use the rotational inertia of a thin plate around the axis perpendicular to the plane of the plate to derive the moment of inertia of a cube around an edge.

Edit: this link may be of some use: http://hypertextbook.com/physics/mechanics/rotational-inertia/" .
 
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