(I am asking this question here instead of Physics SE because it's more involved on the mathematical side. I have made up this question to test myself.)
Assume a solid sphere with radius $R$ and mass $M$ rotating with an angular velocity $\boldsymbol \omega = \frac14\mathbf i+\frac{\sqrt3}4\mathbf j+\frac{\sqrt3}2\mathbf k$. The axis is passing through the center of mass of the body. One of the poles of the sphere is located at the origin of the coordinate system. If you want I can provide a $3$-D plot.
The primary way to do this question would be calculate the moment of inertia tensor and then use it to calculate the angular momentum using $\mathbf L=\mathbf I\boldsymbol \omega$.
But as calculating the 6 components of the MoI tensor is time consuming, I've thought of using a different approach. I'll use the rotation matrices to bring the angular velocity vector to one of the coordinate axes, and then use the standard moment of inertia of a solid sphere, $\frac25MR^2$, to calculate the angular momentum. Then we will invert the rotations to get back the required angular momentum.
We first note that the polar angle of the vector $\phi=\pi/3$ and the azimuthal angle $\theta=\pi/6$.
Assume that if the rotation is in anti-clockwise direction, then the angle of rotation will be positive.
Now, we will construct the rotation matrices.
If we rotate our vector with an angle $\phi$ about the $z$-axis, the rotation matrix will be
$$R_z(\phi)=\begin{pmatrix}\cos\phi&-\sin\phi&0\\\sin\phi&\cos\phi&0\\0&0&1\end{pmatrix}$$
For rotating with an angle $\theta$ about the $x$-axis, the rotation matrix will be
$$R_x(\theta)=\begin{pmatrix}1&0&0\\0&\cos\theta&-\sin\theta\\0&\sin\theta&\cos\theta\end{pmatrix}$$
So, the rotation matrix for first rotating with an angle $\phi$ about the $z$-axis and then rotating with an angle $\theta$ about the $x$-axis will be
$$\begin{aligned} R_{xz}(\theta,\phi)&=R_x(\theta)R_z(\phi)\\&=\begin{pmatrix}\cos\phi&-\sin\phi&0\\\cos\theta\sin\phi&\cos\theta\cos\phi&-\sin\theta\\ \sin\theta\sin\phi&\sin\theta\cos\phi&\cos\theta\end{pmatrix}\end{aligned}$$
If we make a rotation with an angle $\pi/6$ about the $z$ axis followed with a rotation with an angle $\pi/6$ about the $x$-axis, then our angle of rotation will be $\boldsymbol\omega^\prime=\mathbf k$.
The required rotation matrix will be
$$R_{xz}(\pi/6,\pi/6)=\begin{pmatrix}\sqrt3/2&-1/2&0\\\sqrt3/4&3/4&-1/2\\1/4&\sqrt3/4&\sqrt3/2\end{pmatrix}=\mathcal R$$
The inverse of this rotation matrix will be
$$ \begin{aligned}R_{xz}^{-1}(\pi/6,\pi/6)&=R_z^{-1}(\pi/6)R_x^{-1}(\pi/6)\\ &=\begin{pmatrix}\sqrt3/2&-1/2&0\\1/2&\sqrt3/2&0\\0&0&1\end{pmatrix}^{-1}\begin{pmatrix}1&0&0\\0&\sqrt3/2&-1/2&\\0&1/2&\sqrt3/2\end{pmatrix}^{-1}\\ &=\begin{pmatrix}\sqrt3/2&1/2&0\\-1/2&\sqrt3/2&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&\sqrt3/2&1/2&\\0&-1/2&\sqrt3/2\end{pmatrix}\\ R_{xz}^{-1}(\pi/6,\pi/6)&=\begin{pmatrix}\sqrt3/2&\sqrt3/4&1/4\\-1/2&3/4&\sqrt3/4\\0&-1/2&\sqrt3/2\end{pmatrix}=\mathcal R^{-1}\end{aligned}$$
Here I've used the result $(AB)^{-1}=B^{-1}A^{-1}$.
After the rotation $\mathcal R$ on $\boldsymbol \omega$, we get $\boldsymbol\omega^\prime=\mathbf k$.
In this case, the MoI tensor will be
$$\mathbf I^\prime =\begin{pmatrix}\frac25MR^2&0&0\\0&\frac25MR^2&0\\0&0&\frac25MR^2\end{pmatrix}$$
So the angular momentum will be
$$\begin{aligned}\mathbf L^\prime&=\mathbf I^\prime\boldsymbol \omega^\prime \\ &= \begin{pmatrix}\frac25MR^2&0&0\\0&\frac25MR^2&0\\0&0&\frac25MR^2\end{pmatrix}\begin{pmatrix}0\\0\\1\end{pmatrix}\\ \mathbf L^\prime&=\begin{pmatrix}0\\0\\\frac25MR^2\end{pmatrix}\end{aligned}$$
So, our required angular momentum will be
$$\begin{aligned} \mathbf L&=\mathcal R^{-1}\mathbf L^\prime \\ &= \begin{pmatrix}\sqrt3/2&\sqrt3/4&1/4\\-1/2&3/4&\sqrt3/4\\0&-1/2&\sqrt3/2\end{pmatrix}\begin{pmatrix}0\\0\\\frac25MR^2\end{pmatrix}\\ \mathbf L&= \begin{pmatrix} \frac1{10}MR^2\\\frac{\sqrt3}{10}MR^2\\\frac{\sqrt3}5MR^2\end{pmatrix}\end{aligned}$$
Is this method valid? If yes, then is my solution correct?
If no, what can I improve in my solution?
Best Answer
The moment of inertia tensor for a sphere is $$\mathbf I = \begin{pmatrix}\frac25MR^2&0&0 \\ 0&\frac25MR^2&0 \\ 0&0&\frac25MR^2\end{pmatrix}.$$ After rotation, due to the symmetry of the sphere, the MOI tensor is unchanged: $\mathbf I' = \mathbf I.$ In particular it still has three non-zero entries on the main diagonal.
As it turns out, however, the two missing non-zero components in your MOI tensor are multiplied by zero terms in $\boldsymbol\omega',$ and therefore you get the same result with the correct tensor:
$$\mathbf L' =\begin{pmatrix}0\\0\\\frac25MR^2\end{pmatrix}.$$
In general I suppose this method can be made to work, but consider the case where the object has different moments of inertia about its principal axes. The MOI tensor would then need to be transformed to the new coordinate system along with the rotation vector. I have a hunch this is not a labor-saving procedure.
On the other hand, due to the symmetry of this problem, you can simply assume there is a rotation to take $\boldsymbol\omega$ to a principal axis and not bother to calculate the rotation matrix. You also know that since this is a sphere, the angular momentum will be parallel to the axis of rotation. You find the magnitude of the angular momentum to be $\frac25MR^2.$ Now simply scale the vector $\boldsymbol\omega$ to that magnitude in order to obtain the angular momentum vector.
Again, this works because you're dealing with a sphere.