Angular Momentum and Kinetic Energy of a Rigid Body
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 <PERSON> 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? 101