The angular velocity $\vec{\omega}$ lies along the axis of rotation. And the angular momentum $\vec{J}$ is the cross product of $\vec{r} \times \vec{p}$. Which according to me should also lie along the axis of rotation. But I read in a book that the direction of angular momentum vector and angular velocity vector are not the same. Why is it so?
[Physics] Angular momentum and angular velocity
angular momentumangular velocityvectors
Related Solutions
There are no "other" examples. The condition that $\vec \omega$ and $$ \vec L = I_{\rm tensor} \cdot \vec \omega $$ point to the same direction i.e. $$ (\vec L=) I_{\rm tensor} \cdot \vec \omega = k \vec \omega $$ where $k$ is a real number (and no longer a tensor) is a definition of an eigenvector of $I_{\rm tensor}$: both $\vec \omega$ and $\vec L$ are eigenvectors of the moment of inertia in such a case.
Eigenvectors of the moment of inertia tensor are called the principal axes. They may always be chosen to be orthogonal to each other and the tensor has the form $$ I_{\rm tensor} = {\rm diag} (I_1,I_2,I_3) $$ in the coordinate system given by these principal axes. In a generic case, the three principal axes are uniquely determined by the tensor (and the axes may be shown to be orthogonal to each other, as eigenvectors of any Hermitian operator).
The only subtlety appears if some of the entries $I_1,I_2,I_3$ coincide. If two of them are equal, any combination of the two vectors is an eigenvector and there is a freedom in choosing these two principal axes (the third axis corresponding to a different eigenvalue is still unambiguous).
If $I_1=I_2=I_3$, then the tensor is proportional to the unit matrix and any vector is an eigenvector of it with the same eigenvalue $I_1=I_2=I_3$. In that case, the choice of principal axes is totally arbitrary. If 2 or 3 eigenvalues coincide, the orthogonality of the basis isn't guaranteed but we still usually impose the extra conditions that the "principal axes" should be orthogonal to each other.
One could discuss various examples – particular shapes with particular tensors of the moment of inertia. But the important fact here is that for the purposes of the spin and angular momentum, only the tensor of the moment of inertia matters. It may be calculated from any matter distribution but everything else besides the directions of the three axes and the three eigenvalues $I_1,I_2,I_3$ is irrelevant for discussions on the angular momentum.
The essence of the question is the definition of the angular velocity and its relation to the axis of rotation.
First of all, to describe the rotation of a single particle, one requires an axis of rotation $\hat{\mathbf{n}}$, and a rate of change of ‘orientation’ (or ‘angular position’) $\frac{d \theta( \hat{\mathbf{n}}) }{d t}$ around that axis. Without these 2 pieces of information, “rotation” is meaningless. Having the axis of rotation and a rate of change of orientation, one can define the angular velocity $\vec{\omega}$ as $$ \boldsymbol{\omega} = \frac{d \theta}{d t} \, \hat{\mathbf{n}} $$ which includes both pieces of information. Notice that the angular velocity is parallel to the axis of rotation by definition, and has the same amount of ‘information’ as the particle velocity $\mathbf{v}$.
To clarify this further, consider the following figure which depicts the rotation in 3d space, and conforms to the example given in the question,
where the spherical-polar coordinates $\theta$ and $\phi$ are used as the azimuthal and polar angles, respectively and the zenith ($z$-axis) is taken to be parallel to $\hat{\mathbf{n}}$.
In the limit of infinitesimal change in time, $\Delta t \rightarrow 0$, $$ \Delta \mathbf{r} \approx r \, \sin \phi \, \Delta \theta ~, $$ and thus, $$ \left| \frac{d \mathbf{r}}{d t} \right| = r \sin \phi \frac{d \theta}{d t} ~. $$
One clearly observes that both the magnitude and the direction of $\frac{d \mathbf{r}}{d t}$ (which is perpendicular to the plane defined by the particle position $\mathbf{r}$ and the rotation axis $\hat{\mathbf{n}}$) are given correctly by the cross product
$$ \frac{d \mathbf{r}}{d t} = \hat{\mathbf{n}} \times \mathbf{r} \frac{d \theta}{d t} ~. $$
Since $\frac{d \mathbf{r}}{d t} \equiv \mathbf{v}$ and $\hat{\mathbf{n}} \frac{d \theta}{d t} \equiv \vec{\omega}$, one obtains
$$ \mathbf{v} = \frac{d \mathbf{r}}{d t} = \boldsymbol{\omega} \times \mathbf{r} ~. $$
Therefore, the instantaneous axis of rotation $\hat{\mathbf{n}}(t)$ and the instantaneous angular velocity $\vec{\omega}(t)$ are indeed parallel.
Finally, as shown by Gary Godfrey in a comment, one can obtain the correct relation for the angular velocity $\boldsymbol{\omega}$ in terms of the velocity $\mathbf{v}$ and position $\mathbf{r}$ as
\begin{align} \mathbf{r} \times \mathbf{v} &= \mathbf{r} \times ( \boldsymbol{\omega} \times \mathbf{r} ) \stackrel{\text{BAC-CAB}}{=} \boldsymbol{\omega} \, (\mathbf{r} \cdot \mathbf{r} ) - \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega}) \\ \Rightarrow \boldsymbol{\omega} &= \frac {(\mathbf{r} \times \mathbf{v}) + \mathbf{r} \, (\mathbf{r} \cdot \boldsymbol{\omega})}{r^2} ~. \end{align}
This answer is based on Kleppner, D., and R. Kolenkow. “An introduction to mechanics”, (2ed, 2014), pp. 294–295. The figure is taken from the same source.
Best Answer
The formula you have specified $\vec{L} = \vec{r} \times m\vec{v}$ is the definition of angular momentum of a point-like particle which respect to a point P. In this case of course angular momentum and angular velocity have the same direction. When dealing with rigid bodies (assemblies of many point-like particles), the correct full angular momentum is proved to be:
$$ \vec{L} = I \vec{\omega} $$ where in general $I$ is a tensor (a matrix for simplicity) depending on shape and mass distribution of the rigid body, called inertia tensor. It is possible, in general, that once you choose a reference frame (say with $\omega$ parallel to the $\hat{z}$ axis), the inertia tensor could be diagonal or not in that basis. If it's diagonal, then $\vec{L}$ is parallel to $\vec{\omega}$, otherwise it is not. As an example, you can consider a cylinder forced to rotate about an axis which is not parallel to any of its principal axes of symmetry.
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