Kinetic energy of a rotating cube

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Homework Statement


I'm trying to calculate the kinetic energy of a rotating cube about one of its face diagonals, using the moment of inertia tensor for the cube rotating around one of its corners.


Homework Equations


T=[itex]\frac{1}{2}[/itex][itex]\omega[/itex][itex]\cdot[/itex]L
L=I[itex]\omega[/itex]
(I'm not sure how to signify vectors in this interface, but I realize I'm dealing with vectors...)

The Attempt at a Solution


What I'm confused about is getting the moment of inertia tensor to apply to rotating about the face diagonal. There is an example in the book where they are using the moment of inertia tensor about a corner to calculate angular momentum around the cube's main diagonal. The author introduces a unit vector, u, in direction of rotating which he defines as u=([itex]\frac{1}{\sqrt{3}}[/itex])(1, 1, 1). Then the problem follows naturally. I believe the problem I'm dealing with can be solved similarly, though I'm not understand exactly what this unit vector represents. Can anyone explain it to me?
 

Answers and Replies

  • #2
rude man
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I don't know how this necessarily relates to tensors; been quite a while since I had them - but this is just a matter of finding rotational inertia I about the specified diagonal of the cube.

Perhaps the author suggested for the diagonal to be oriented in the (1,1,1) direction from the origin of a cartesian coordinate sustem; or maybe he thought this unit vector would be a good choice for representing the perpendicular distance from each element dV to the axis of rotation. You are of course free to choose any spin axis orientation you want. Right now I don't see any particular virtue in that particular unit vector.

Irrespective of choice of spin axis orientation, you will of course be integrating
(s^2)*ρ*dx*dy*dz where s is the perpendicular distance from a volume element dV to the axis of rotation. Not a pretty thought ...

P.S. if the intent is an exercise in tensors per se, excuse this whole writeup!
 
  • #3
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you need to find a rotation matrix that takes the x direction unit vector to the direction of the diagonal of the face of the cube. Call the matrix R. Tensors transform according to the equation:
I'=RIR-1
then once you have this transformed tensor, you have

[tex] T=\frac{1}{2} I_i'_j \omega _i \omega _j [/tex]

(summed over i and j)
 
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  • #4
SteamKing
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If you have a unit cube and you put one of its corners at the origin and place the rest of the cube in the first octant, what would be the coordinates of the main diagonal of the cube? What would the components of the unit vector u along this diagonal?
 
  • #5
rude man
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If you have a unit cube and you put one of its corners at the origin and place the rest of the cube in the first octant, what would be the coordinates of the main diagonal of the cube? What would the components of the unit vector u along this diagonal?
Would that be a good idea, considering we want I about a face diagonal?
 
  • #6
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If the cube has its corner at the origin and is contained in the first octant, we want a rotation about the axis defined by the direction

[tex] ( 1 , 1 , 0 ) [/tex]

correct? Just want to make sure we are on the same page with what the problem is asking..
 
  • #7
rude man
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If the cube has its corner at the origin and is contained in the first octant, we want a rotation about the axis defined by the direction

[tex] ( 1 , 1 , 0 ) [/tex]

correct? Just want to make sure we are on the same page with what the problem is asking..
Right. (1,1,0), not (1,1,1). My gut reaction woud be to put the diagonal along the z axis with the origin at the mid-point. Then you could exploit symmetry by integrating only 1/4 of the total volume & multiplyiog by 4. The integration in fact doesn't look that intimidating after all ....
 
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  • #8
rude man
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Even better: run the z axis thru the center of the cube, parallel to the diagonal in question, then invoke the parallel-axis theorem. Now the integration can be limited to 1/8th the total volume.
 
  • #9
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Actually now that I think of it. Why don't we just use

[tex] T=\frac{1}{2}I_{ij} \omega_i \omega_j [/tex]

(summed over i and j) with the given tensor and have

[tex] \vec \omega \rightarrow \Omega (\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},0) [/tex]

then we just have a sum of four terms for the kinetic energy
 
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  • #10
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Thank you for the assistance, everyone. I ended up figuring this one out; I did exactly what AlexChandler specified in the post above this one.
 

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