Inertia tensor of rotated object Suppose we have computed the inertia tensor of an object about its COM. Suppose the object is then rotated, is there a simple transformation that connects the new inertia tensor to the old one through the rotation matrix?
 A: Lets say $I$ is the moment of Inertia tensor before rotation (or any orthogonal transformation).
This means the angular momentum and angular velocity relates as :
$\vec L = I \vec \omega$
and the components of $L$ are given by :
$L=I\omega$         ...(1)
where $L$ and $\omega$ are $3\times 1$ matrices and $I$ is a $3\times 3$ matrix.
Now eq(1) would remain true even in the rotated frame. In rotated frame, the components of $L$ would be $R\times L$ and that of $\omega$ be $R\times \omega$.
so premultiplying eq(1) with the rotation matrix $R$ :
$R\times L = R \times I\times \omega$
$R\times L = R \times I\times (R^T\times R) \times \omega$
$(R\times L) = R \times I\times R^T\times (R \times \omega) $
$L' = R \times I\times R^T\times \omega ' $         ...(2)
So given we know the rotation matrix R, Inertia tensor in a new coordinate system can be found out using (2)
A: If a 3D rotation $R$ rotates Cartesian vectors according to
$$V_i’=R_{ik}V_k$$
then it rotates Cartesian tensors with two indices (including the inertia tensor) according to
$$I_{ij}’=R_{ik}R_{jl}I_{kl},$$
and similarly for more indices. The way to remember this is “each index gets rotated” via contraction with the rotation matrix.
The index notation here uses the Einstein convention that a repeated index is summed over all possible values (in this case, from 1 to 3).
Index notation is more flexible than the matrix notation
$$\mathbf I’=\mathbf R\,\mathbf I\,\mathbf R^\top$$
because it works for tensors with any number of indices.
