Physics Asked by Vishal M Das on July 24, 2021
Angular momentum about a point on the rotational axis is equal to sum of angular momentum perpendicular to rotational axis and angular momentum parallel to rotational axis. The perpendicular component vanishes if the body is symmetrical about the axis of rotation. So when we say $I_text{com}omega$ is angular momentum about COM isn’t it neglecting the cases where bodies are not symmetrical? The problem came when we have angular momentum about some point (say $O$) for a rolling body: the derivation leads to $L_O = I_text{cm}omega + M(rtimes V_text{cm})$. So some sources are saying $I_text{cm}omega$ is angular momentum about COM. Isn’t it wrong to call it angular momentum about COM?
This discussion is for a rigid body.
For planar motion, using the center of mass (CM) as the origin and the $hat z$ axis as the fixed axis of rotation $J_z = I_z omega$, but if the axis of rotation is not a principal axis $J_x$ and $J_y$ are not zero since the products of inertia are not zero. $vec J$ is the angular momentum about the CM, $I_z$ is the moment of inertia about the axis, and $vec omega$ is the angular velocity about the axis of rotation.
For general rotation about a point $vec J = tilde I cdot vecomega$ where $tilde I$ is the inertia tensor.
See Goldstein, Classical Mechanics or Symon, Mechanics.
Correct answer by John Darby on July 24, 2021
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