# 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 $I_{cm} ||\vec \omega||$?

• You first claimed that then equation gives the angular momentum about the center of mass axis. Later you ask if the same equation gives the angular momentum about the center of mass axis. Can you clarify that part? – Yashas Feb 21 '17 at 12:15
• -1. Unclear what you are asking. I agree with Yashas. The only difference between your 2nd and 3rd sentences is that one is rotation about an axis and the other is rotation about a point. – sammy gerbil Feb 21 '17 at 21:38

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.

• In fixed axis rotation does this treatment as scalar apply? – Madhuchhanda Mandal Feb 21 '17 at 13:26