# Is Lcm really equal to Icm x w?

Let's suppose a body is rotating about some axis passing through the centre of mass (cm) with a angular velocity $\vec \omega$.

The angular momentum about the centre of mass axis is given by $L = I_{cm} ||\vec \omega||$, with $I_{cm}$ being moment of inertia about the axis.

Now, does it, in any way, mean that angular momentum about the centre of mass (see, I stress the cm point here) is indeed $I_{cm} ||\vec \omega||$?

The angular momentum of a rigid body rotating about some point is given by $\vec L = \mathrm I_\text{cm}\, \vec \omega$, where $\mathrm I_\text{cm}$ is the moment of inertia tensor about the center of mass. It's neither a vector nor a scalar. There are some special cases where the moment of inertia can be treated as if it were a scalar. Introductory physics students are only given problems where these special cases apply.
Those special cases do not apply in the case of a precessing top or a tumbling textbook. The problem with pretending $\mathrm I_\text{cm}$ is a scalar is that it misses some interesting physics. The tensorial nature of the inertia tensor means that angular velocity and angular momentum are not necessarily parallel to one another.