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The inertial tensor of a homogeneous rectangular sheet of mass m with sides of respective length a and b and negligible thickness is:

$$I = \frac{m}{12}\begin{bmatrix}b^2 & 0 &0 \\ 0 & a^2& 0 \\ 0 & 0&a^2+b^2\end{bmatrix}$$

The coordinate system with respect to which the expression above is calculated is such that its origin coincides with the center of mass (CM) of the sheet, its x-axis is parallel to the side of length a, its axis yes parallel to the side of length b its z-axis is perpendicular to the sheet.

Now, consider that the sheet rotates with constant angular velocity $\omega$ (in the direction of one of the diagonals of the sheet) around a fixed axis of rotation that coincides with one of the vertices whose diagonal is perpendicular to $\omega$.

Said vertex does not undergo any linear displacement and I have to obtain the kinetic energy of the sheet and its angular momentum $\vec{L}$ as well as the contribution due to the CM movement.

Attempt: I first obtained the vector expression for $\omega$, in terms of the lengths $a, b$ whose direction is parallel to one of the sheet's diagonal. The unitary vector for this direction is $\frac{1}{\sqrt{a^2+b^2}}(-a,b,0)$, so that,

$$\vec{\omega} = \frac{\omega}{\sqrt{a^2+b^2}}(-a,b,0)$$

Then I calculated a new inertia tensor for the vertex I've been asked that, in correspondence with $\omega$ direction, I chose to be $\frac{1}{2}(a,b,0)$. Now, applying the Steiner theorem for each component of the inertia tensor above I get:

$$I' = \frac{m}{6}\begin{bmatrix}2b^2 & 3ab & 0 \\ 3ab & 2a^2 & 0 \\ 0&0&2(a^2+b^2)\end{bmatrix}$$

Hence, multiplying this tensor with the $\vec{\omega}$ expression above I get both expressions of the angular momentum and kinetic energy:

$$\vec{L} = \frac{m\omega ab}{6\sqrt{a^2+b^2}}(b,-a,0)$$ $$T = -\frac{m\omega^2 a^2b^2}{6(a^2+b^2)}$$

As you can see, kinetic energy is negative, so obviously I've done something wrong that I cannot identify. Could anybody help me? And what does it mean about "the contribution due to the CM movement"?

Thanks in advance!

EDITED: The inertia tensor I' above was wrong, so after correcting it, I could finally have the answers for the Angular Momentum and KE:

$$I' = \frac{m}{12}\begin{bmatrix}4b^2 & -3ab & 0 \\ -3ab & 4a^2 & 0 \\ 0&0&4(a^2+b^2)\end{bmatrix}$$ $$\vec{L} = \frac{7m\omega ab}{12\sqrt{a^2+b^2}}(-b,a,0)$$ $$T = \frac{7m\omega^2 a^2b^2}{12(a^2+b^2)}$$

To calculate the contribution of the CM to the KE, I've considered it as a point particle of mass m rotating about the previous axis with the same $\omega$ as before, so its moment of inertia is $$I = \frac{1}{4}m(a^2+b^2)$$

Hence, its contribution to the KE is

$$T_{CM} = \frac{1}{2}I\omega^2=\frac{1}{8}m(a^2+b^2)\omega^2$$

Am I correct?

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    $\begingroup$ Perhaps, but it is the KE due to the movement around the vertex of that rigid body, so I don't know another way of calculating that. $\endgroup$
    – user9867
    Commented Jan 2, 2021 at 19:00
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    $\begingroup$ Your moment of inertia tensor $\mathbf{I}'$ cannot be correct as it is not positive definite. You can't apply Steiner's theorem to the moment of inertia tensor, as it only applies to the scalar moment of inertia (see Goldstein section 5.3). You just need to recalculate the tensor. $\endgroup$
    – najkim
    Commented Jan 2, 2021 at 19:44
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    $\begingroup$ But in Taylor's Classical Mechanics there is a formula that states that at a point $\Delta$ from the CM, whose vector is $\vec{\Delta} = (\xi,\eta,\zeta)$, the components of the inertia tensor are $I_{xx} = I_{xx}^{CM} + M(\eta^2+\zeta^2)$, $I_{yz} = I_{yz}^{CM}-M \eta \zeta$, and so on, which could be the Steiner Theorem for an Inertial tensor. Am I correct? $\endgroup$
    – user9867
    Commented Jan 2, 2021 at 19:59
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    $\begingroup$ (a). That is not Steiner's theorem. Steiner's theorem refers to the case in which you are computing $I_{xx}$. What you wrote seems to be exercise 10.24 in Taylor. (b). You miscalculated $I_{xy}'$. You should get half of what you have right now; like I said earlier, your moment of inertia tensor needs to be positive definite. $\endgroup$
    – najkim
    Commented Jan 2, 2021 at 20:53
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    $\begingroup$ OK, understood. Thank you very much! About the CM contribution, the whole relative contribution to the CM cancels because there isn't any rotation around the CM, right? $\endgroup$
    – user9867
    Commented Jan 2, 2021 at 22:32

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