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    I know this is more of a Physics question, but I posted it in Maths due to the Maths involved ...

    I've calculated the moment of inertia tensor of a cone where it's tip is at the origin and the z axis runs through its symmetry axis. The edges are at an angle alpha to the xy plane. It's mass is M and height H. The answer I get (correct) is,

    \left[\begin{array}{ccc}\frac{3}{20}MH  ^2\left(4+\cot^2\alpha\right)&0&  0\\0&\frac{3}{20}MH^2\left(4+ \cot^2\alpha\right)&0\\ 0&0&\frac{3}{10}MR^2\end{array  } \right] .


    The question now asks to prove that the moment of inertia about its outer edge is  \frac{3}{20}MR^2\left(1+5 \sin^2\alpha\right).

    I really have no idea where to start, I thought about using the 3x3 rotation matrix on my previous answer, but this didn't help at all. Please can someone help?


    EDIT: Here's a diagram Name:  Cone diagram.PNG
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    Hi sam :P


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    (Original post by KeyFingot)
    ...
    What's the "outer edge"?

    Is this a straight line from the vertex running along the curved surface of the cone?
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    (Original post by ghostwalker)
    What's the "outer edge"?

    Is this a straight line from the vertex running along the curved surface of the cone?
    Sorry it's unclear, I've uploaded a diagram to try and help. It's the line running from the vertex along the curved edge as you said
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    (Original post by KeyFingot)
    Sorry it's unclear, I've uploaded a diagram to try and help. It's the line running from the vertex along the curved edge as you said
    OK, I've not studied this, but from a bit of googling, I suspect:

    In order to use a rotation on the tensor matrix, it needs to be located on the centre of mass.

    So, translate it to the centre of mass (don't know how this is done with the tensor matrix).

    Apply a unit vector in the direction of the line of slope to get the MofI about a line parallel to the line of slope.

    And then parallel axis theorem to get the MofI about the line of slope.

    Warning: This may be complete gibberish.
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    (Original post by ghostwalker)

    Apply a unit vector in the direction of the line of slope to get the MofI about a line parallel to the line of slope.
    I'm unsure why/what I have to do this/here. The rest of it seems OK, I'll give it a go
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    (Original post by KeyFingot)
    I'm unsure why/what I have to do this/here. The rest of it seems OK, I'll give it a go
    You've got Klaus Gernoth's previous sheets I see. Enjoy them!

    So with your tensor, the moment of inertia of some object about some axis is

    I= \hat{\mathbf{u}}^T \Theta \hat{\mathbf{u}} , where \Theta is your moment of inertia tensor and \hat{\mathbf{u}} is the unit vector in the direction of the axis of rotation. In this case we can choose \hat{\mathbf{u}}= \begin{pmatrix} 0 \\ \mathrm{cos}\alpha \\ \mathrm{sin} \alpha \end{pmatrix}, then it's just matrix multiplication.

    We use the transpose of it just to get the rows and columns the right way round for multiplication.
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    (Original post by dknt)
    You've got Klaus Gernoth's previous sheets I see. Enjoy them!

    So with your tensor, the moment of inertia of some object about some axis is

    I= \hat{\mathbf{u}}^T \Theta \hat{\mathbf{u}} , where \Theta is your moment of inertia tensor and \hat{\mathbf{u}} is the unit vector in the direction of the axis of rotation. In this case we can choose \hat{\mathbf{u}}= \begin{pmatrix} 0 \\ \mathrm{cos}\alpha \\ \mathrm{sin} \alpha \end{pmatrix}, then it's just matrix multiplication.

    We use the transpose of it just to get the rows and columns the right way round for multiplication.
    Haha yeah I stumbled across them and thought why not use them for AD revision.

    Thanks for your help - both dknt and ghostwalker
 
 
 
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